Symmetrization and Applications

Comparison Theorems

57

R e m a r k 3 . 2 . 1 T h e dual problem (3.2.2) is the weak formulation of the problem: -div(AtV
= g in H tp = 0 on dQ

where A1 is the transpose of the matrix A. T h e above lemma states t h a t if equality occurs in Talenti's theorem when / = 1 then it also occurs for all dual problems with t h e right-hand side radial and decreasing. • T h e next lemma is useful in obtaining identities which cannot normally be obtained by using test functions in t h e various variational formulations. L e m m a 3.2.2 Let f e L 2 (H) be non-negative. symmetric and strictly decreasing. If / f
/ Jn

Let (p G C(Cl) be radially

f*(pdx,

ijj sharing the same properties

with
f wdx = I r tftdx. Jo,

JU

P r o o f : Notice t h a t , by hypotheses, every ball centered at the origin and contained in the ball fi is a level set of tp and of ip and vice versa. Now, (cf. L e m m a 1.2.2)

/ f
JQ

I

( /

Jifmin

fdx)

\J{v»t}

dt

(3.2.3)

)

and f

/ f*ipdx JCI

f

= ipmin

rVmax

f*dx+ JO,

Jpmin

/

r

\

/ f*dx\dt. \A^»t} /

(3.2.4)

T h e first terms in the right-hand sides of (3.2.3) and (3.2.4) are equal. Since {ip > t} is a ball centered at the origin, we also have (cf. (1.3.2))

f

J{ip>t}

fdx < f

J{

t}

f*dx.

Symmetrization and Applications

58

Hence the hypothesis that the left-hand sides of (3.2.3) and (3.2.4) are equal implies that, for almost every t,

[

fdx = f

f*dx.

(3.2.5)

By our earlier observation on the level sets of ip and ip, it follows that (3.2.5) still holds if tp is replaced by ip. Hence, using the formulae corresponding to (3.2.3) and (3.2.4) with

0. Then, if u* = v, it follows that u — u* and that f = f*. Step 1. First of all, HV«1li, n < | | V u | | ! ( n < a ( W , u ) -

fnfudx

= jnf*vdx

<

= \\Vv\\la

fnf*u*dx =

||Vu*||| i n .

Thus, l | V u l ! i n = ||V«||l ( n -

a(u,u)

=

f fudx Jo.

=

[ f*u*dx.

(3.2.6)

Jo

Step 2. Let z G i/g(Q) be the solution of the dual problem with right-hand side = 1: a(x»*) =

/ Xdx for all Jo

X

e i?o( n )

and let w £ flo(fi) be the solution of the corresponding symmetrized problem —Aw — 1 in fi. By Talenti's theorem, it follows that z* < w. Again, Jnfzdx

= a(w, z) = Jn udx — f^u*dx

—

= / n /*«Kfc > fnfz'dx

f^'S7w.Vu*dx\ \ > Jnfzdx )

(3.2.7)

and so equality holds throughout. In particular, / f*wdx

Jo,

=

I

Jo

f*z*dx.

Now, since / ^ 0, the above equality holds in the ball {/* > 0} which is of positive radius. In particular, z* = w in this ball. But since both z* and w are radially decreasing (for w{x) = (R2 — |a;|2)/2iV) where R is the radius of fl), it follows that z*(0) = iu(0), i.e. the L°° norms of 2* and w are

Comparison Theorems

59

equal. But then, by Proposition 3.2.1, it follows that z* = w in Q. Thus, we are in the situation of Lemma 3.2.1 (applied to the dual problem with right-hand side = 1). Hence, we deduce that z ~ z* = w. It then follows from equality holding throughout in (3.2.7) that / fwdx = f f*wdx. (3.2.8) Jn Jn Step 3. Now both the functions iv and u* = v are strictly decreasing radial functions (cf. (3.1.12)) and so we can apply Lemma 3.2.2. Thus, we deduce from (3.2.8) that a(u,u*)

J fu*dx = f f*u*dx. (3.2.9) Ju Jn We again appeal to Lemma 3.2.1 and apply it to the dual problem. Since we already know that —Aw* = /*, u* must also solve the dual of the dual with right-hand side /*, i.e. a(u*,X)

=

/Vxdxforallxetfo1^). Jn

=

(3.2.10)

In particular, a(u*,u*)

-

/ f*u*dx Jn

(3.2.11)

and ( Vu*.Vudx Jn

= J f*udx Jn

= a{u*,u).

(3.2.12)

Now, 0 < ||V(«-u*)||!in <

a(u-u\u-u*).

Expanding this, and using (3.2.6), (3.2.9), (3.2.11) and (3.2.12), we get 0 < 2 ( / \Vu\2dx - J Vu.Vu*dx) \Jn Jn J

< ( f \Vu\2dx - f Vu.Vu*dx \Jn Jn

and so /' \Vu*\2dx Jn which yields u = u*.

=

f \Vu\2dx Jn

=

[ Vu.Vu*dx Jn

60

Symmetrization and Applications

Step 4. Finally, Since u solves the original problem (3.1.3) with right-hand side / and u* solves the same problem, but with right-hand side / * (cf. 3.2.10), we deduce from the above t h a t / = / * . • R e m a r k 3 . 2 . 2 T h e above proof just involves repeated use of t h e PolyaSzego and Hardy-Lit tie wood inequalities. Another proof found in [Kesavan (1991)] shows t h a t t h e maximum of u is attained at a unique interior point and t h a t in each ball {u > t}, the m a x i m u m occurs at the centre. Thus, all the level sets {u > t} are concentric and thus u is radial and decreasing and so u — u*. To do this, L e m m a 3.2.2 is used in a different way. [Alvino, Lions and Trombetti (1986)] also give a proof of the fact t h a t the level sets are concentric, using other arguments. Of course, one can also verify fairly easily t h a t t h e condition of Brothers and Ziemer (cf. (2.3.6)) is satisfied and so the fact t h a t ||Vu*||2,n — ||Vu||2,n will immediately imply t h a t u = u*. B u t the argument t h a t we have presented here and the ones cited above are self-contained and fairly elementary. However, later on, when studying the case of equality in nonlinear problems, we will, indeed, appeal to t h e result of Brothers and Ziemer. • One of t h e immediate consequences of t h e above theorem is t h e following characterization of radially decreasing functions. Corollary 3.2.1 Let fl = ft*, let w € H^(ft) satisfy - A w = 1 in ft. f G L2(ft) be non-negative. Then, the following are equivalent: (ii) Jn fwdx (iiijf^fipdx decreasing.

= Ja f*wdx. — f~f*(pdx

_ for any ip &" C(ft)

which is radial and

Let

strictly

P r o o f : T h a t (i) =!> (ii) is obvious. T h e fact t h a t (ii) <=*• (iii) is the statement of L e m m a 3.2.2. We now show t h a t (ii) => (i). let u,v G H Q ( ^ ) such t h a t -Au = f and —Av — f* in ft. Then, by Talenti's theorem, u* < v. By (ii) we get

Jnfwdx = jnVu.Vwdx — fn Viv.Vvdx

= jnudx = fn f*wdx

= JQu*dx < }Qvdx — /n

fwdx

and so it follows t h a t u* = v and by t h e preceding theorem, t h a t u = u* and / — / * , which proves (i). •

Comparison

61

Theorems

We now look at the matrix A associated with a problem for which the equality case occurs. Consider the following hypothesis: (H) Let A be a matrix such that there exists / > 0 in L2(Q) with the property that if u € HQ(Q) is the solution of (3.2.4), then u* satisfies -Au* = /* in H*. If (H) holds, then we just saw that Q. — Q*, u = u* and that f = f*. We also saw that if —Aw = 1, then w also satisfies a(xM

=

/xforallxetfo1^).

Proposition 3.2.3 Assume that (H) is valid- If A is symmetric, for almost every x £ Vl, we have

then,

N

2^a>ij{x)xj

= Xi.

(3.2.13)

3=1

Proof: Since w(x) — (R2 — \x\2)/2N , where R is the radius of fi, satisfies the dual problem with right-hand side = 1 as well as —Aw = 1, we get that a(w,w)

=

I \Vw\

This is the same as N

I z~-, aij{x)XjXidx

~

I y ^ x{dx.

By the ellipticity condition, we then deduce that for almost every a;6fi, N

y ^ aij(x)xjXi i,j = l

N

= ^^x2.

(3.2.14)

z=l

Now, by the ellipticity condition, A(x) is a symmetric matrix whose first eigenvalue is greater than, or equal to, 1 and the above condition then shows that the first eigenvalue is indeed unity with x as eigenvector. This completes the proof. • If A is not symmetric, we do not have a condition like (3.2.13) as the following example shows.

62

Symmetrization

and

Applications

Example 3.2.1 Let N = 2. Define a-ij(x) = 8ij-(i-j)(xl

+ xl),

l
Then if C{u) = -div(v4Vu), we have that w(x) = (R2 - \x\2)/4 satisfies C(w) = -Aw

= 1

in the ball centered at the origin and of radius R. (3.2.13) is not satisfied.

However condition

Example 3.2.2 Even if A is symmetric, the differential operator C need not even resemble the Laplacian. Again, let N = 2. Let £(x\1X2) be a non-negative, smooth and bounded function. Define MX\

=

l + ^i^i'^) ~x\X2^{xi,x2) l-x1x2£(x1,x2) l + x2£(x1,x2).

If w is as above and C the corresponding differential operator, then, in the ball centered at the origin and of radius R, we have C(w) = -Aw

= 1.

If an = 1 for 1 < i < N, then we can say more about A. Proposition 3.2.4 Assume that an = 1 for all 1 < i < N. Assume that (H) holds. Then (i) ai:i = -aji fori j= j . In particular, if A is also symmetric, then C = —A. (it) If (f is any radial function, then C(
for all £ £ MN. If { e ^ } ^ is the canonical basis for RN, we choose £ = e^+e^ and £ — e^ — e^ successively in the above inequality to get (i). By virtue of condition (H) and Lemma 3.2.1, we know that if — Aw = 1 in Q (where w is given by w(x) — (R2 — \x\2)/2N), then w also solves -div(i4*Vw) =

linfi.

Comparison

63

Theorems

Using this information together with (i), we get Yl'fofaiW**}

=

-^faT.&iifcM

= °-

(3.2.15)

We also get, again by (i), 2_\.aij{x)xix3

— 0-

(3.2.16)

i^3

Let

Now,

E w sir (««(*)&) = £** si- ( « « ( ^ ^ ) where ^ ( r ) =
Sobolev Imbeddings

In Section 2.4, we saw how we could use symmetrization to evaluate the best Sobolev constant and also saw that the classical Sobolev inequality for the case p — 1 was equivalent to the classical isoperimetric inequality. In this section, we apply Talenti's result to get estimates on the constants involved in certain other Sobolev imbeddings. We first start by estimating various norms of solutions of elliptic boundary value problems. Proposition 3.3.1 Let Q, C RN be a bounded domain. Let f e L2(S1). Let u G HQ(U) be the (weak) solution of the following problem:

-Au = f in n

| [6 6 l)

u = o on on.) Let20, we have u € L°°(ft) and \\u\Ua

< (^r20~l\nf\\f\\p>Ci;

--

l

. Then,

(3.3.2)

64

Symmetrization and Applications

(ii) if ^ > —p > 0, we have u G Lg(Q,), and 2

+

I f1

X

||u|kn < ( A ^ r | / T W * / (1-**)*<**

Ip.n

(3.3.3)

Uo when /? < 0 and ||u||,, n < (Ar«;^)- a |n|i Q f ( - l o g * ) * * ) ' l l / I U

(3.3.4)

w/ten p = 0. Proof: Without loss of generality, we can assume that / > 0 (cf. Exercise 3.1.2). Then, as we saw in Theorem 3.1.1, u*(s) < (Nv$)-* J™ £*-*(£ f#{r,)dn\dt. Case 1. Let p = oo. Then, |u#( a )| < ( J V w * r 2 l l / l l o o , n / ' % * - 1 d e from which it follows that \u#(s)\

< (iV<)"2 ( - 0

(|fi|* - , * ) 11/m.n

which yields (on setting s = 0 in the above inequality) the estimate (3.3.2) when p = oo. Case 2. If p < oo, then, by Holder's inequality, / ftirtdq Jo

< ||/ # || P l (o,|n|)€ 1 -* = 11/llp.nC1-*-

Thus, \u*(s)\ < {N4r2\\f\\p.n rt*-*-1** Js

which then yields (3.3.2) when /? > 0. Case 3. Let 0 < 0. Let tf"1 > - 0 . If /? / 0, then (3.3.5) gives |u#(s)| <

(N4)-2\\f\\p,n\&\-l{sV-\nf)

(3-3.5)

65

Comparison Theorems

and so

IMIS,n = H«#C(o,|n|) < (N4r29\P\-9\\f\\l,n

Tit0

-Vfyds

J0

which yields (3.3.3) after a change of variable t — s/\£}\ in the integral above. Case 4. Finally, if /? = 0, then (3.3.5) gives \u*(s)\

<

(JVw*)- 2 ||/|| P i n(log(|n|)-log S )

and hence

ll«li;,n < (Arw*)-a«||/n;-n jT'"' ( - log ( ^ ) y d* from which we easily deduce (3.3.4).

•

Let u e W2'p{n) n W 0 1,p (n). If we set / = - A u , then u is the solution of (3.3.1) with / e ^P(fi). Further, 11/Hp.n = ||Ati|| Pl n < HtiHa^n. The following result is, therefore, a direct consequence of the preceding proposition. T h e o r e m 3.3.1 Let £1 C RN be a bounded domain and let 2 < p < oo. Setp=%-±. Then: (i) if j3 > 0, we have

^ ( f l ) n i v 0 l , p ( f i ) c L°°(fi) and, /or all u £ W2'P{Q) n W01,p(ft), we Jiaue IMIoo.fi < C||u||2,Pjn where C

<

(NUJ")-2^-1^0;

(iij if {3 < 0 and q~l > ~{3, then

w2'p(n)nw^'p(Q)

cLg{n)

Symmetrization and Applications

66

and, for all u e W2*(Q) n W^V{U),

we have

HU.n < C||«||2>P,n where

c <

|(;v4)-2i/?rW+*l/oCi-^)'*!* >//^o

" | (tfu/*)- a |fi|i ( ^ ( - l o g i ) * * ) * 3.4

'/ 0 = 0.

The Obstacle Problem

In this section we will apply the techniques used to prove Talenti's result to study a simple example of a variational inequality. Let fi C RN be a bounded domain. Let A G M(l, /3, Q) be a matrix as in Section 3.1. Let a(.,.) be the associated bilinear form (cf. (3.1.2)). Let K = {w€H£(U)

\W>0

in fi}.

Given / € L2(Q), we consider the following variational inequality: find u € K such that, for all w G K, a(u, w — u) >

/ f(w ~ u)dx.

(3.4.1)

Jn This problem has a unique solution (cf., for instance, [Kesavan (1989)]) and, if A is symmetric, then u is the minimizer, over K, of the energy given by J(w)

— -a(w,w)

— I

fwdx.

This is a free boundary problem. The coincidence set is the set given by fio = {x e H | u(x) = 0}. One can, in M2, imagine a membrane stretched over the region Q with an obstacle along the plane and the membrane being fixed along the boundary d£l. The energy J is the strain energy and the equilibrium position of the membrane occurs when this energy is minimized over all possible configurations. It can be shown that on the set H+ = fi\fio> the function u satisfies the differential equation C(u) = / , where C is the associated second order elliptic differential operator (cf. Section 3.1).

67

Comparison Theorems

As in the case of Talenti's theorem, we can compare the solution of this problem with that of a 'symmetrized problem' and obtain useful information on the size of the coincidence set. As before, given f E L 2 (fi), we define

F(0 - / f*(v)dv. Jo

Lemma 3.4.1 Let u e K be the solution of (3.4-1)- Then, F > 0 on the interval [0, \{u > 0}|]. Farther, for all 0 < t'
H{«>*'}l

< (NLJ{[)-2

t~t'

,

^-2F(£)<e

(3.4.2)

J\{u>t}\

Proof: Let t > 0. Set w = (u - t)+. Then a(u, w) — j fwdx Jn

= I (C(u) — f)(u — t)dx — 0 Ju>t

since u satisfies C(u) = f on the set {u > t} C {u > 0}. Further, N

a(u,{u-t)+)

= /

^atj-^-^dx.

J{u>t}

ux

3

i j= 1

Udj

-<-

Thus, we get

/

yzai3^—^~dx

Hu>t} ^

=

i

dx3 dxi

f{u-t)dx.

J{u>t}

Notice that, by the ellipticity condition, the integrand on the left is nonnegative and so the integral is a decreasing function of t. Thus, we get N

0

^ -if

E ^7r-7rdx

fd* < ^(MW)

= f

where we have set // to be the distribution function of u. This immediately implies that F > 0 in the interval [0, \{u > 0}|]. The inequality (3.4,2) now follows exactly as in Step 2 of the proof of Theorem 3.1.1. • Corollary 3.4.1 u#(s)

Let u e K he the solution of (3.4-1)'• Then

< /(A^)-

\o

2

/.

1

^

0

^*-

2

^ ^ if

Se[0,|{U>0}|]

if se[|{u>o}|,|n|].

(3.4.3)

Symmetrization and Applications

68 and

— - 7 - W < {Nu>%)-*s*-'£F{s),

for s 6 ( 0 , | { u > 0 } | ) .

(3.4.4)

Proof: Let 0 < s < s' < \{u > 0}|. Then u*{s') < u*(s) since u* is decreasing and further, by definition of the rearrangement, |{u # > u#{s')}\ < s'. If e > 0, then, as u# is decreasing, we have |{t/ # > u # ( s ) - e } | > | { u # > u # ( s ) } | > s. Set t' - u#{s') and t = u#(s) - e. Then, by (3.4.2), we get i

U#(s) - £

u#{s')

r3'

2

< (Nug)-

/

f*-2F(0df

since the integrand is non-negative in the interval concerned. Letting e —> 0, we get u#(s)-u#(s')

< (Nvji)-2

f is

f*"2FK)df

(3.4.5)

If we now set s' = \{u > 0}|, then u # ( s ' ) = 0 and we get (3.4.3). The estimate (3.4.4) follows directly from (3.4.5). • Let us now consider the symmetrized problem: find v e K* C H^(Q*), such that for all w e K*, f Vv.V(w-v)dx Jo,'

>

f f*(w-v)dx Jn*

(3.4.6)

where K* = {weH^n*)

| w>0

in £T}.

This problem has a unique solution v E K* which minimizes the corresponding energy wdx over K*. Further, i; satisfies the equation —Av = f* over the set {v > 0}. By the Polya - Szego and the Hardy - Littlewood inequalities, it is clear that J*(w) > J*(w*) and so, by the uniqueness of the minimizer of J*,

Comparison Theorems

69

it follows that v = v*. Thus, the set {v > 0} is a ball with centre at the origin. It is then immediate to see that v#(s) u

=

{N4)~2 \o,

{

Theorem 3.4.1 spectively. Then

fl{V>°}] t*-'F(t)d(, s€[0,\{v> 0}|]

(3 4 7)

8€[|{i;>o}|,|n|].

K

--'

Let u and v be the solutions of (3.4-1) and (3.4-6) re-

'0

if f <0

a.e., dx

|{u>0}| < |{^>0}|

W */ Inf so

^°> / ^ °

and

the unique solution of F(s) = 0 in ( | { / > 0 } U ^ | ) 5 otherwise.

Further, u* < v. Proof: (i) If / < 0 a.e., then it is immediate to check that u — 0 and v = 0 are the solutions to problems (3.4.1) and (3.4.6) respectively. (ii) Let / n fdx > 0. Then F(0) - 0 and F(\Q\) = f^1 f*{s)ds = JQ fdx > 0. Further, since / * is decreasing, it follows that F is concave and thus F > 0 in the interval [0, |H|]. It then follows that z G HQ(Q*) such that —Az = f* is non-negative and is given by the right-hand side of (3.1.12). It is now obvious that z also solves (3.4.6) and so v = z and it follows that

{«>o}| = |n|. (iii) Let us now assume that Jafdx < 0 and that | { / > 0}| > 0. Since F is concave and F(0) — 0 while F(|fi|) < 0, it follows that there exists a unique s 0 G (|{/ > 0}|, |fi|) such that F(s0) = 0. Notice that F > 0 in (0, so) and that F < 0 in ($o, |^|)- Now, v is strictly radially decreasing in the set where {v > 0} and

^ ( S ) = -(iv4)" 2 s *- 2 F( s ) in that set. Hence it follows that \{v > 0}\ — so where the derivative vanishes for the first time. We already saw that F > 0 on the interval [0, \{u > 0}|] and so we have | { u > 0 } | < so =

\{v>0}\.

That u* < v^ is now immediate from Corollary 3.4.1. Hence u* < v.

•

70

Symmetrization and Applications

Proposition 3.4.1 respectively. Then

Let u and v be the solutions of (3.4-1) and (3.4.6)

J{u) >

J*(v).

Proof: By the ellipticity condition, we have J(u)

> \2 (

Jn

\Vu\2dx-

f

Ja

fudx.

Notice that u > 0 and so by the Polya - Szego and the Hardy - Littlewood inequalities we get J(u)

>

J*{u*).

But u* E K* and so by the variational characterization of v, we have J*(u*) > J*(v) which completes the proof.

3.5

•

Electrostatic Capacity

The isoperimetric inequality for the problem of electrostatic capacity is one of the early successes of the theory of symmetrization and is due to [Szego (1930)]. Let Qo be a bounded domain in M3 and let T C ^o be a subdomain. The region ft = £IQ\T represents an electrical condenser. We denote the boundary of UQ by To and the boundary of T by Ti- The electrostatic potential is the function u E Hl(fl) which satisfies the following: Au = 0 in Q u = Q on r 0 } u = 1 on IY

(3.5.i;

The electrostatic capacity of H is given by C{Q) = 1- / [ ^da 47T ./ r , dv where v is the unit outward normal on the boundary of tt.

(3.5.2)

Comparison Theorems

71

Multiplying the differential equation in (3.5.1) bu u and integrating by parts using Green's theorem, we easily see that C(fi) = - ^ / \Vu\2dx. 4TT Jn By the maximum principle, u attains its extreme values on dQ, and not in Q. Thus, 0 < u < 1 in Q. Further, if the boundaries r 0 and Ti are sufficiently regular, then, | j * < 0 on To and | j > 0 on T\. Proposition 3.5.1 (Dirichlet Principle) The electrostatic potential u is the minimizer of the functional J{w) =

/ \Vwf, dx JQ JQ'

over the set K = {we Hl(Q) \ w = 0 on TQ and w = 1 on Proof: Let w G K. Then, there exists ip e Then,

Ti}.

such that w = u + (p.

HQ(Q,)

/ Vu.Vt/;cfa; = - / Au.(pdx = 0. 7n Vo Thus, J \Vw\2dx JQ

= f {\Vu\2 + \Vip\2)dx

>

\Vu\2dx

f

JQ

JQ

which proves the claim.

•

The Dirichlet principle for the capacity immediately yields an isoperimetric inequality for the same. Theorem 3.5.1 (Szego) Let QQ and T* be balls centered at the origin such that | n j | - |fi 0 | and \T*\ = \T\. Set S = fi5\T*. Then, C(to) > C(fi).

(3.5.3)

Proof: Let u be the solution of (3.5.1) and denote by u its extension to T by unity. Then u > 0 and it belongs to HQ(QQ). Now, by the Polya - Szego inequality, 4nC{n)

=

f JQQ

\Vu\2dx

> [ JQ*

\Vu*\2dx

[\Vu*\2dx.

= JQ

72

Symmetrization

and

Applications

But on H, u* is non-negative; further, u* = 0 on OCIQ and u* = 1 on dT* Thus, by the Dirichlet principle applied to fi, we deduce that

L

' \Vu*\2dx n

> 47rC(n)

which completes the proof.

•

R e m a r k 3.5.1 We can formulate the problem discussed above in all dimensions N. The least capacity, given the volumes of Qo and T, will always be for the spherical annulus bounded by dT* and <9fi£. • We can explicitly calculate C(Q). Let the outer and inner radii of the annular domain H be Ro and R\ respectively. Thus 0 < R\ < RQ. AS observed in the proof of the preceding theorem, where we extended functions to the inner region by unity, the functional to be minimized will be \Vw\2dx

/

over the set of non-negative functions vanishing on OQQ and = 1 on T*. Further, the electrostatic potential, v, satisfies 0 < v < 1 in the annulus. Consequently v* will also be equal to unity in T* and will vanish on 3Qg. Since Schwarz symmetrization decreases this integral, it follows that /

\Vv\2dx

> [

\Vv*\2dx.

By the uniqueness of the electrostatic potential, it then follows that v = v*. Thus, writing (as usual, by abuse of notation) v = v(\x\), the function v(r) satisfies u " M + ^v'{r) = 0, Ri < r < Ro v(R{] = 1 ; v{Ro) = 0.

}

This gives „2-N

R2-JV

v(r) logr-log-Rp pr _ 9 logflx-logfio' JV _ Z

When N — 3, using the above formula and the definiton (3.5.2) of the

Comparison Theorems

73

capacity, we get that R0R1 Ro-Ri

C(Q)

3.6

The Saint Venant Problem

One of the important conjectures in mathematical physics solved by Polya was that of Saint Venant regarding the torsional rigidity of an elastic cable. Consider a cylindrical cable in M3 with uniform cross section Ho C K 2 . Assume that fii c Ho is a subdomain. Let \Q\\ = a. Then, the warping function is denned as the unique solution of the following problem: -Art = 2 u —0 u = c,

in Q = QoVh on <9Qo an unknown constant, on dQ\

• U %** = 2aThe torsional rigidity of the cable is defined as the quantity z

S(Sl) -

(3.6.1) / / \Vu\ dx. Jn The conjecture of Saint Venant states that of all cross sections (UQ) of given area and of all holes (fix) of given area, the torsional rigidity is maximal when QQ and Hi are concentric circles. The first proof of this conjecture is due to [Polya (1948)]. The result was extended to an arbitrary number of holes by [Polya and Weinstein (1950)] and [Payne (1962)] proved an important inequality for the torsinal rigidity. We will give below, following the treatment of [Mossino (1984)], a proof of Payne's inequality and the theorem of Polya and Weinstein. Remark 3.6.1 In the absence of any hole, i.e. if fii = 0, the result is an immediate consequence of Talenti's theorem. For, u will satisfy —Au — 2 in O and will vanish on <9H. If v satisfied the same equation on fi*, then, by Talenti's theorem, we have [ \Vu\2dx

<

f

\Vv\2dx

(cf. Proposition 3.1.1) which is precisely the conjecture of Saint Venant in this situation. •

74

Symmetrization and Applications

Let QQ C R 2 and let ft; C fto be pairwise disjoint subdomains for 1 < i < m. Let I \ = dft^, 0 < i < m. Let |ftj| = a* for 1 < i < m. Finally, set ft = f2o\ U ^ ! ft*. Let v be the unit outward normal on the boundary of ft. The warping function on ft is defined as the solution of the following problem: -Au

= 2 in ft u —0 on To u — Ci > 0, an unknown constant, on Ti, 1 < i < m Jr Wda = 2a^ l
(3.6.2)

Proposition 3.6.1 Assume that the Ti, 0 < i < m are all sufficiently smooth. Then, the problem (3.6.2) admits a unique solution. Proof: Step 1. (Uniqueness) If u\ and u-i are two possible solutions to the above problem, then, let w — u\ ~ u-i. Then, — Aw = 0. Multiplying this by w and integrating by parts over ft using Green's theorem, and using the other conditions defining the warping function in (3.6.2), we get / -Aw.wdx Ja

(du1_du1) — / \Vw\2dx ~ y^(ci,j - c2,i) / dv Jn ~r^ Jvr . \ dv

The integrals over I \ all vanish by the last condition in (3.6.2). Thus, by Poincare's inequality (w = 0 on To), it follows that w — 0 which proves the uniqueness of the solution. Step 2. Let c — (ci, ...,c m ) G R m be given. Let v = u(c) be the unique solution of the problem: -Av

= 0 in ft v — 0 on TQ v — Ci on Ti, 1 < i < m.

Let w be the unique solution of the problem - A w = 2 in ft w = 0 on 5ft. Define ft - /

^da

and * = / r . §£da. Set d = (<2i, ...,d m ) G R m . The

Comparison

75

Theorems

map A : Rm -> Rm defined by A(c) = d is thus linear. If A(c) = 0, then

0 = / -Av.vdx

Jn

= / iVvfidx-y^Ci

Ja

£f

/

-^-da.

M dv

By hypothesis, the integrals on I \ all vanish and so we get that J*n |Vv\ 2 dx = 0 and again, by Poincare's inequality, it follows that v = 0. Thus, c = 0 and so A is one-one. Therefore, it is onto as well. It is now evident that the affine linear map T : Rm -> M™ defined by T(c) = A(c) + /9 where /3 = (/?i, ...,/? m ) is also onto. Thus, if we choose di = 2a* — ft, then, if A(c) = d, we get u = v(c) + w satisfies =2 u =0 u — Cj, %** = 2a,i, -Au

in on on 1<

0 To r\, 1 < i < m i < m.

Step 3. We now show that the Ci are all strictly positive. If I \ , 1 < i < m are all sufficiently smooth, then u G H2(Q) c C(fl), by the Sobolev imbedding theorem, since Q C R 2 . Thus u will be continuous. Let xo G H be such that u attains its minimum there. If XQ G fi, then VU(XQ) = 0 and Au > 0. But Au = —2 < 0. Thus w attains its minimum on the boundary. If XQ G Ti for some 1 < i < m, then minu = c$ and so | ~ < 0 on IV But this contradicts the fact that the integral of the normal derivative on I \ is equal to 2a^ > 0. Thus XQ G TO and so u > 0 and u > 0 on fi\r0. This proves that Ci > 0 for 1 < i < m. • Remark 3.6.2 The smoothness of the boundary was used only in Step 3, to prove the positivity of the unknown constants. Otherwise the existence of u vanishing on To and constant on each r \ , 1 < i < m is alsways true. Even the smoothness required is minimal. It is enough if the boundary is Lipschitz continuous. • Remark 3.6.3 We have, in fact, proved that we can assign arbitrary values to the integrals of the normal derivatives over I \ , 1 < i < m as is evident from Step 2 of the above proof. The special values 2a\ = 2|fij| not only ensure that the c^ are all positive, but also give us a neat variational

76

Symmetrization and Applications

characterization for the problem (3.6.2). Set W

— {v E #0(^0) I v = fy, a constant, in Qi, 1 < i < m}.

Then, it is a simple exercise to check that the variational formulation of (3.6.2) is to find u e W such that, for all veW, /

Vu.Vvdx

= 2 / vdx. Jn

JQQ

If u is the warping function, we define the torsional rigidity of Q, as before, by S{Q) =

/

Jo.

\Vu\2dx.

Example 3.6.1 (The case of the circular annulus) Let m = 1. Let

n = {x e M2 1 0 < Ri < \x\ < R0}. Then, a\ —

TTR2.

Consider u(x) = (RQ ~ \x\2)/2. Then it is immediate to check that —Au — 2 in fi and that u — 0 on To = {x \ \x\ = RQ}. Further, u is a constant (being a radial function) on Y\ = {x \ \x\ — R\}. Now, if r = |a;|, du , _ = - t i (r) = r on Fi (where we have written, by abuse of notation, u — u(r)) and so /

JLfo

= 2-KR\ = 2a 2 .

Thus u is the warping function and the torsional rigidity in this case is given by S(n)

= 2TT /

r(u'(r))2dr

=

-(R40-Rt).

Let m(t) = \{u < t}\ for t > 0. Now, {u < t} = {x \ R2 - 2t < |x| 2 < R20}. Thus m{t) = 2irt

77

Comparison Theorems

Lemma 3.6.1 Let u be the warping function, i.e. the solution of (3.6.2) and let m(t) = \{u < t}\. Then, 2irt < m{t).

(3.6.3)

Equality is attained for the circular annulus. Proof: Without loss of generality, we can assume that the subdomains f2; are numbered such that 0 = c0 < c\ < c2 < ... < c m < c m + i = u m a x where umax stands for the maximum value of u. Step 1. Let 0 < t < umax.

set w - t - (u - t)~ £ H1^). ft, w =< (u,

Thus,

if it > t if u < t.

On one hand, we have / -Au.wdx in

- 2 / wdx = 2t\Q\ + 2 / (u - t)dx. Jn J{u
On the other hand, by Green's theorem, Ja -Au.wdx

= fn VuVwdx

- /an

= I{u
^wda

~ 2 E C l t a{.

Thus,

/

\Vu\2dx = 2 ]T ai°i + 2t XI ai + 2i l n l + 2 /

Hu
Ci
c

.>(

(u ~ *)da;.

J{u
Differentiating with respect to t, we get (for almost every t) — I \Vu\2dx dt •>{«<*}

= 2 V ^ + 2|Q| -2m(t). t>t

(3.6.4)

Step 2. As in the proof of Talenti's theorem, a combination of the Fleming - Rischell theorem and the Cauchy - Schwarz inequality gives

(PQ({u
= \AJu
\Vu?dx\

< m'(t)(~J

\Vu\2dxY (3.6.5)

78

Symmetrization and Applications

Step 3. Now (Pa{{u
=

PR2({u>t}UUCi>tni)

by the properties of the de Giorgi perimeter, since u > 0 in fi and vanishes on its external boundary. Hence, by the classical isoperimetric inequality, we have (Pn({u
> 4nA

where

A = \{u>t}\+^2a,i

\n\-m(t)+Ylai-

=

ci>t

a>t

Thus, in view of (3.6.4) and (3.6.5), we get

4TT

( \Q\ - m(t) + J2 a<) ^ 2m 'W I lnl ~ m ( f ) + J2 \

ci>t J

\

ai

I'

a>t J

Since the term in parantheses on both sides is strictly positive, we get m'(t) > 27r for almost every t which yields (3.6.3) on integrating between 0 and t (since m(0) = 0). • Lemma 3.6.2

Let u be the solution of (3.6.2). S(ft)

Then,

m |Q|2 < ^ + 2 ^ ^ ^

(3.6.6)

In particular, we have Payne's inequality: |Q|2

m

2u

s{n) < v~+ mBxy;oi.

(3.6.7)

1=1

Proof: We have, from (3.6.2), 2 / udx = Jn

/ -An.urfx = JQ

/ |Vu| 2 dx — / —da Jn Jan ®v

which gives .|fi|

S(fi)

2 / udx + 2 >_JaiCi = 2

u#(s)ds +

2y^ajCi

Comparison Theorems

79

where u# is the uindimensional increasing rearrangement of u (cf. Section 1.4). Now, set t = u#(s) in the preceding lemma. Thus, 2iru#(s)

< m(u#{s))

= \{u < u # ( s ) } | -

|{u # < u # ( s ) } | < s

since u and u# are equimeasurable and u# is a non-decreasing function. Therefore, m

|ftt 5(0) < — / 27r 7o

sds + 2 > a
which gives (3.6.6). Since, for all 1 < i < m, we have Q < ii m a x , the inequality (3.6.7) follows immediately. • Theorem 3.6.1 (Polya - Weinstein) Let u be the solution of (3.6.2). Let So be the torsional rigidity of the circular annulus having the same measure as H and having the hole of area Yl^Li ai- Then,

S(fi) < ^

+2 w f >
+ ^f;ai

= So.

(3.6.8)

Thus, of all multiply connected domains of fixed area with an arbitrary number of holes of fixed total area, the circular annulus has the maximum torsional rigidity. Proof: The first inequality above is Payne's inequality proved in the preceding lemma. Now, u m a x — u#(|fi|), by definition and we saw, in the proof of the preceding lemma, that 2TTU#(S)

<

s.

Thus, u#(|fi|) < |0|/27r and the second inequality in (3.6.8) follows. Finally, it remains for us to show that So is given by the quantity mentioned in (3.6.8). Indeed, if RQ and R\ are, respectively, the outer and inner radii of the annulus, then J2iLi ai = n^i anc * 1^1 = ^(^o ~~ -^l)- Thus, as seen in Example 3.6.1,

So = i(*8 - .Rf) = f ( ^ - i ? ? ) ( ^ + i??)

M^M iSli. ini)W+i om|.

80

Symmetrization

and

Applications

which completes the proof.

•

Remark 3.6.4 We saw that u#(s) < s/2ir. Now, we know that (cf. Exercise 1.4.1) u*(s) = u # (|fi| - 5). Thus, u#{s)

<

(\U\~S)/2TZ.

If v is the warping function for the annulus, then (cf. Example 3.6.1), v(x) = (R% - \x\2)/2 and so (cf. Exercise 1.1.2)

Thus, u#{s)
3.7

M

Comments

In this chapter, we have given a sampling of results to show how symmetrization could be used to produce comparison results for solutions of partial differential equations and to get estimates for these solutions. As mentioned earlier, the first results in this direction were those of [Bandle (1975)] and [Weinberger (1962)] and the result of [Talenti (1976b)] that we presented here was a refinement of these. The method of proof is quite robust and has been used in a variety of situations. The basic idea, always, is to get a differential inequality for the distribution function of the solution, which will reduce to an equality for the symmetrized problem, and then deduce the comparison result. There is an enormous amount of literature on this topic. The most general results, using Schwarz symmetrization, are given in [Alvino, Lions and Trombetti (1990)]. See also [Talenti (1985)]. Comparison results using Steiner symmetrization can be found in [Alvino, Diaz, Lions and Trombetti (1996)]. While most of these results are valid for second order elliptic problems with Dirichlet boundary conditions (because of the implicit involvement of the maximum principle), there have also been attempts to tackle other types of boundary conditions (cf. [Ferone (1986)] for results on a Neumann problem and [Brandolini, Posteraro and Volpicelh (2003)] for an elliptic problem with a mixed boundary condition) as well as fourth order operators (cf. [Ferone and Kawohl (2003)]). The above mentioned method for obtaining comparison results also works for parabolic problems. The reader is referred to the works of [Bandle (1976a)] and of [Mossino and Rakotoson (1986)].

Comparison

Theorems

81

We mentioned earlier, in Section 3.1 (cf. Remark 3.1.2) that P. L. Lions gave an alternative proof of Talenti's theorem without using the isoperimetric inequality and based on Lieb's proof of the Polya - Szego inequality for the case p — 2 (cf. Remark 2.3.2). This proof relies on the fact that the fundamental solution of the heat equation in R ^ is spherically symmetric, positive and decreasing, i.e. it is equal to its Schwarz symmetrization. This idea has been devoloped to give an alternative approach to the study of camparison results for solutions of elliptic and parabolic equations by [Alvino, Lions and Trombetti (1991)]. In Section 3.4, we presented a very elementary example of a variational inequality, viz. the obstacle problem. Schwarz symmetrization has been successfully used to obtain comparison results for various variational inequalities. We cite, in particular, the works of [Bandle and Mossino (1984)], [Maderna and Salsa (1984)], [Alvino, Matarasso and Trombetti (1992)], [Posteraro and Volpicelli (1993)] and [Posteraro (1995)] and the references contained therein. The proofs of the isoperimetric inequalities for the electrostatic capacity and the torsional rigidity, thereby settling long standing conjectures of mathematical physics, were amongst the earliest successes of the method of symmetrization. See [Mossino (1984)] for generalizations of the results on capacity, applied to condensers in series and in parallel. See also the work of [Ferone (1988)]. The equality case was independently investigated by [Alvino, Lions and Trombetti (1986)] and by [Kesavan (1988)], [Kesavan (1991)]. The bibliography cited above is, obviously, far from being exhaustive. The interested reader will find further references quoted in these works.

Chapter 4

Eigenvalue Problems

4.1

The Faber - Krahn Inequality

We now look at some examples of eigenvalue problems for elliptic partial differential operators and some isoperimetric inequalities which are associated to them. The first of these is the famous Faber-Krahn inequality which, in fact, was first conjectured by [Lord Rayleigh (1894)] in his treatise on the theory of sound in 1894, but was proved independently by [Faber (1923)] and [Krahn (1924)] towards the end of the first quarter of the twentieth century. We return to the notations of Section 3.1. Let CI C R ^ be a bounded domain. Let A E M(a,{3, fi) and let C(u) = - d i v ( ^ V u ) be the corresponding second order elliptic differential operator in divergence form. Let a(.,.) be the associated bilinear form defined on HQ(Q) X HQ(Q) (cf.(3.1.2)). Given / € L2(ft), let u e H&(tt) be the unique solution of the problem: a(u,v)

— / fvdx,

for all v e

HQ(Q).

JQ

Let us set u = Qf. Then Q can be thought of as a linear mapping of L2(U) into itself, since HQ(U) <—> L2(S7). Since, Q is bounded, the above inclusion is compact, by the Rellich - Kondrosov theorem (cf. [Kesavan (1989)]), and G is thus a compact linear operator on L2{U). Let us now assume that the matrix A is, in addition, symmetric, i.e. a ij — aji f° r a n 1 ^ i>3 ^ N. Then, for f,g G L2(Q), we have, setting 83

84

Symmetrization and Applications

u = Qf and w = Qg, (Gf,9)

= (fl»w) = a ( w , « ) = a(u,w)

= (/,w) =

(f,Gg)

where (.,.) denotes t h e inner-product in L 2 ( Q ) . Thus, Q is a self-adjoint operator on L 2 (fi) a n d hence its spectrum coinsists a sequence of eigenvalues (i£ [ 0. There exists a n orthonormal L2(fl) consisting of eigenvectors of Q. Let us now consider the eigenvalue problem for t h e operator C for (X,u) e R x (H 0 1 (fi)\{0}) such t h a t a(u,v)

= A(«,v) for all v £ H£(Q).

compact of 0 a n d basis for We look

(4.1.1)

This is equivalent t o solving u =

Q(\u).

Thus, from our preceding observations, we have a sequence \^ = (fj,^)~l of eigenvalues a n d associated eigenfunctions {
= A j J ( ^ , « ) for all v E H*(Q)

(4.1.2)

and, further, 0 < Af < \£

< \$

< ... < A^ < . . . - > o o .

(4.1.3)

T h a n k s t o t h e strong m a x i m u m principle, t h e first eigenvalue Af, also called the principal eigenvalue, is simple (hence t h e strict inequality following it in (4.1.3)) and, further, t h e eigenfunction does n o t vanish inside fi. Thus, we will henceforth assume t h a t tpf > 0 in fl. Notice, further, t h a t all t h e eigenfunctions ip£ are, in fact, in

HQ(£1).

T h e eigenvalues {A^} can be given a variational characterization using t h e Rayleigh quotient which is given by RA(u)

= ^ ^ , u^Q,

ueH^Q).

(4.1.4)

Let Vn be t h e subspace of HQ(Q) spanned by { v i S y ^ , — j ^ n } - Then, A£ = RA(
m a x RA{v) =

min

RA(V)

v±Vn~i

=

m i n m a x RA (V) dim W=n v£W

Eigenvalue Problems

85

where t h e minimum in the last relation is taken over all n - dimensional subspaces of

HQ(Q).

In particular, we have Af -

min

RAM.

(4.1.5)

Further, if w G HQ(£1) is such t h a t RA{W) = Af, then w will be an eigenfunction corresponding to Af. For a proof of all the above mentioned results, see, for instance, Kesavan [Kesavan (1989)]. T h u s , to get an upper bound for Af, it suffices to take any function v € H^ft) and compute RA(V). On the other hand, it is quite difficult to get lower bounds for \f. Towards t h e end of the nineteenth century, [Lord Rayleigh (1894)] conjectured that, when C — —A, the Laplace operator, t h e disc has the least principal eigenvalue amongst all plane domains of equal area. If we denote t h e first eigenvalue of t h e Laplace operator as Ai(fi), then, in our notation, this reads as Ai(fi) > Ai(fi*)

(4.1.6)

This was independently proved by [Faber (1923)] and [Krahn (1924)] nearly a quarter of a century later and (4.1.6) is now known as the Faber-Krahn inequality. Using the Polya- Szego theorem (cf. Theorem 2.3.1), we can prove a more general result in all space dimensions. T h e o r e m 4.1.1 Let A e M{a,(3,Q) where H C RN is a bounded domain. Let {A^(fi)} denote the sequence of eigenvalues of the operator C. Let {Xk(Q)} denote the eigenvalues of the Laplace operator. Then Af(fi)

> aAi(fi) > a A i ( £ T ) .

(4.1.7)

P r o o f : Let 0 be the first eigenfunction of C. Then, by t h e Rayleigh quotient characterization of eigenvalues, we have

1 {)

~ ~mh

-

KII2,« •

But

L\^t?dx —

. „

Jn \Vv\*dx _ >

mm

—-—=

=

Ai(iZ;,

86

Symmetrization and Applications

which proves the first inequality in (4.1.7). Now, let (pi > 0 be the first eigenfunction of the Laplace operator in ft. Since (pi > 0, we can apply the Polya - Szego theorem, and the fact that the Schwarz symmetrization preserves the L2 - norm, to get

JnlVvil 2 ^

>

Jn. IVpII'ds

Now ifl e HQ(Q*) and so, again appealing to the variational characterization of the principal eigenvalue, we easily deduce that Ai(fi) > Ai(fi*) and the proof is complete.

•

When ft is a ball, we can explicitly solve for the first eigenvalue and eigenfunction of the Laplacian in terms of Bessel functions. Let ft — H* = BR, the ball with centre at the origin and of radius R. Since symmetrization decreases the integral J*fl \Vipi\2dx and keeps the L2(ft) - norm unaltered, it follows that the Rayleigh quotient of (pi is greater than that of y?J. But Since X\(ft) is the minimum of the Rayleigh quotient, it follows that (p\ achieves this minimum, and hence that it is an eigenfunction as well. Now, the simplicity of the principal eigenvalue implies that (pi =
N - 1

u'(r) + Xu(r) = 0.

A straight forward computation shows that the solution can be written as u(r) =

cr1'^JK^i(VXr)

where c is an arbitrary constant and Jp(p) is the Bessel function (of the first kind) of order p satisfying the differential equation

The boundary condition u(R) = 0 determines A. If J P J I denotes the first positive zero of the function Jp then, clearly, we must have yf\R = JK-I I,

Eigenvalue Problems

Ai(Bfl) =

87

-£=H.

(4.1.8)

T h e eigenfunction is given by Vi(s) =

I s f - f j ^ p t ^ M ) .

(4.1.9)

We now examine the case when equality occurs in the Faber-Krahn inequality. Faber and K r a h n proved t h a t equality occurs in the case of plane domains only when H is a disc of the same area. In general, in the literature (cf., for instance, [Payne (1967)]), while explicit reference is m a d e to t h e case of plane domains, no mention is made of t h e equality case for higher dimensions. To the best of the our knowledge, t h e only discussion of this is to be found in [Kawohl (1985)] and [Kesavan (1988)]; the former uses ideas from t h e theory of Steiner symmetrization while the latter uses Schwarz symmetrization and the equality case of Talenti's theorem (cf. Section 3.2). We give this proof now. T h e o r e m 4.1.2

Let Q C E N be a bounded domain such that Ai(ft) -

A^fl*).

Then, CI is a ball P r o o f : Let (pi > 0 be t h e first normalized eigenfunction for t h e Laplace operator. T h u s , —A(fi — A i ^ i in fi tpi = 0 on dQ, where Ai = Ai(Q) solution of

Ai(fi*) and H^iH^n = 1—Aw — Xi(pl w = 0

Let

w

G # o ( ^ * ) be the

in Q* on dQ*.

Then, by Talenti's theorem (cf. Theorem 3.1.1)
Xiw.

Multiplying this inequality by w and integrating over Q*, we get /

\Vw\2dx

< Ai /

w2dx.

88

Symmetrization

and

Applications

Since Ai is also the principal eigenvalue of the Laplace operator on ft*, which is the minimum of the Rayleigh quotient, we deduce that

JQ.w2dx Since, Ai, the minimum of the Rayleigh quotient, is achieved at w, it follows, as already noted earlier, that w is an eigenfunction of the Laplacian, corresponding to Ai, i.e. —Aw = Xiw. Thus, w = tp\ and so we are in the equality case of Talenti's theorem and hence, by Proposition 3.2.2, it follows that ft is a ball. • Two domains fti and SI2 in ^tN a r e said to be isospectral if Afc(fti) = Afc(^2) for all positive integers k. A celebrated question of [Kac (1966)] was whether isospectral plane domains are isometric, i.e. congruent to each other. Of course, this question could be posed in all dimensions and also in the context of manifolds. While most cases were settled in the negative, the case of plane domains remained open for several decades and was finally settled by [Gordon, Webb and Volpert (1992)], again in the negative. Nevertheless, the spectrum of the Laplacian contains several pieces of information about the geometry of the domain. For instance, from the Weyl asymptotic formula (cf. [Protter (1987)])

it follows that two isospectral domains have the same volume. In the case of plane domains, a further refinement of this result due to Pleijel (cf. [Protter (1987)]) states that

yVAfc(n)t -

lal

L

l

£r[

47rt

4

V2TT£

where L is the perimeter of the domain. Thus, if two plane domains are isospectral, they have the same area as well as the same perimeter. If one of them is a disc, then, L 2 — 4-KA and so, from the isoperimetric theorem, the same being true for the other, it follows that the other domain is also a disc. We can now prove this in all dimensions. If two domains in *RN are isospectral, they have the same volume. If one of them is a ball, we can, without loss of generality, call the domains ft and ft*. Since they are

89

Eigenvalue Problems

isospectral, in particular, Ai(fi) = Ai(H*) and so, by Theorem 4.1.2, it follows that Q, is a ball. Exercise 4.1.1 Let Q C RN be a bounded domain and let P : O —> R be a strictly positive continuous function. Consider the eigenvalue problem: —Aw = XPw in Q w — 0 on dQ. Show that this problem admits an increasing sequence {\n,p(Q)} of positive eigenvalues which tends to infinity and that the first eigenvalue Ai ( p(fi) admits an eigenfunction of constant sign. Show that Ai lP Q) =

min

J

O#VG^(0)

"' ' • J^Pv^dx

With usual notations, if P* denotes the Schwarz symmetrization of P , show that Ai,p(fi) >

4.2

AI,P-(Q*).

The Szego - Weinberger Inequality

In the previous section, we discussed isoperimetric inequalities for the principal eigenvalue under homogeneous Dirichlet boundary conditions. We now consider Neumann boundary conditions. Let ft C R N be a bounded domain. Consider the eigenvalue problem -Au

= »u in n

fs = o

1

on m\

K

}

where v is the unit outward normal on dft. As in the previous case, we have an increasing sequence of eigenvalues 0 — Ho < Pi(^) < M2(^) ••• < A*n(^) < ... —> oo with an associated family of eigenfunctions (which now belong to H1(Q)) forming an orthonormal basis for L2(fl). The eigenfunction corresponding to fj,0 = 0 is the constant function. The principal (Neumann) eigenvalue is, therefore, pi (SI) which can be given the following variational characteriza-

Symmetrization and Applications

90

tion (via the Rayleigh quotient):

Ml

fi =

min J n f ' ' In vdx=0

•

(4.2.2)

In the case of a ball BR C M^ (JV > 2) of radius # > 0, we can explicitly write down the eigenvalues in terms of Bessel functions, by the method of separation of variables (cf., for instance, [Courant and Hilbert (1953)]). Using polar coordinates, we can write u(x) — w(\x\)v(u), where u G SN~l, the unit sphere in RN. Then, it can be seen that w satisfies the equation w"{r) H

7V_ i

w'(r)

TV - 1 — w(r) + fMl{BR)w(r)

=0,

for 0 < r < R

(4.2.3) and the conditions w(0) = w'(R) = 0. As in the case of the Dirichlet problem, we can easily verify that the solution to this equation is given by l w (r) = cr%~ Jx(^(B^r)

(4.2.4)

where c is an arbitrary constant and Jp denotes, as usual, the Bessel function of the first kind of order p. Once again, the eigenvalue is determined from the boundary condition, which, in this case, reads as w'(R) = 0. Thus, for instance, if R = 1, we get

where /xi = ni(Bi). In particular, if N = 2, /ij is the square of the first positive zero of J[. We will now express fii (BR) in terms of integrals of w and its derivative which will be useful in the sequel. We rewrite (4.2.3) as W(B«Mr)

= _ - ^ A ( r " - V ( r ) ) + ^««;(r).

Multiplying this on bothsides by w and integrating over BR, we get HI{BR)

/

w2{\x\)dx = NuN /

J B pi

- —(r*-V(r))u;(r)dr

" 0

f

N

~

1

n

(X2J

Eigenvalue Problems

91

Evaluating the integral of the first term on the right by parts and using the fact that w(Q) = w'(R) = 0, we get Nu>N f ~-^{rN~1w,(r))w(r)dr Jo dr

= NujN [ J0

I.

rN~lw'(rfdr

w'(\x\)2dx.

Thus,

^BR)

~

jBRM\*\)>d*

•

(4 2 5)

' '

Just as in the case of the Dirichlet eigenvalue problem, where the first eigenfunction could be chosen to be of constant sign, here we can choose w such that w'(r) > 0 in (0,i?). Thus, w is a non-negative and increasing function. The isoperimetric inequality for the first (Neumann) eigenvalue reads as MiW < M ^ * ) -

(4-2.6)

This was proved by [Szego (1954)] for simply connected plane domains and in full generality by [Weinberger (1956)]. In order to prove this inequality, we need two technical lemmas which we now establish. Lemma 4.2.1 LetfiC R ^ be a bounded domain, let g : [0, oo) —* [0, oo) be a continuous function. Define Pi(x) = g{\x\)^\x\

for l
(4.2.7)

Then, we can choose the origin in such a way that, for all 1 < % < N, we have

f P,{x )dx = 0.

(4.2.8)

Ju

Proof: Choose a ball B centred at the origin such that fi C B. Let y G B. Define F : B - • RN by

F^/rflx-,,,)^ dx.

92

Symmetrization and Applications

Taking the inner-product of F(y) with y, we get {x

F(y)-y = f

9(\x-y\)

Jn

f~lypdx.

\x-y\

If \y\ — R, the radius of B, then, for all X E U , x-y-\y\2

< \x\.\y\-\y\2

< o

and so F(y).y < 0 for all y on the boundary of B. It then follows from the Brouwer fixed point theorem that (cf. [Kesavan (2004)] or [Lions (1969)]) there exits y0 e B such that F(y0) — 0. We shift the origin to y0 and (4.2.8) follows. • The next lemma is the result of a simple application of the Schwarz symmetrization. Lemma 4.2.2 Let Q C M.N be a bounded domain. Let g : R —> R be a non-increasing function. Then f g(\x\)dx < f g(\x\)dx. Jn Jn* Proof: Let g(x) = g(\x\) and let h = ^|n- Then / g(\x\)dx

Jn

— I hdx — /

Jn

Jn*

(4.2.9)

h*dx.

The proof will be complete if we show that, for all x 6 SI*, we have h*(x) < g{\x\). Indeed, h*(x) and g(\x\) are both radial functions which are radially decreasing. Hence it is sufficient to show that their distribution functions satisfy the same inequality. Now, given any t G R,

\{h'>t}\

= \{h>t}\

= | { x e n | 5 ( | x | ) > t } | < \{X e RN | g(\x\) > t}\

Since g is non-increasing, the set appearing in the last term of the above chain of relations is a ball with centre at the origin. Since {h* > t} is also a ball centred at the origin and is contained in S~T, we get \{h*>t}\

< |$rn{xeR"|9(|:r|)>t}| =

which establishes the claim.

\{xen*\g(\x\)>t}\ •

Corollary 4.2.1 In the above lemma, if g is non-decreasing, then the inequality in (4-2.9) is reversed. Theorem 4.2.1

Let Cl c R ^ be a bounded domain. Then (4-2.6) holds.

Eigenvalue Problems

93

Proof: Step 1. Define w(r) for 0 < r < R w(R) for r>R where R is the radius of ft* and w is the function satisfying (4.2.3) - (4.2.5). Then g is a non-decreasing and non-negative function. Setting Pi(x) = 9 ( M ) g ,

l
we can assume that (by changing the origin, if necessary) JQ P{(x)dx — 0 for all 1 < i < N (cf. Lemma 4.2.1). Then, by virtue of the variational characterization of the principal eigenvalue (cf. (4.2.2)), we have Mi (ft) / Pfdx

< J \VPi\2dx,

l
Summing over alii, for 1 < i < N, we get

Mn)<(/ o £lv^)/(/ n E Ptdx A straightforward computation yields , , (JZWPiAw = 9'{rf +—N 3-1— g(r)

2

def

=

D,

v

B(r)

where r = |x| and

i=\

)

Thus, Mi (ft) <

fQB(\x\)dx Jng(\x\)2dx

Step 2. Now, for 0 < r < R, B'(r)

=

-2 Hi{BR)g{r)g'{r) + — - ( r f f ' ( r ) - p ( r ) ) :

by virtue of (4.2.3). Notice that BR is none other than ft*. Since 5 is non-decreasing it follows that B'(r) < 0 for 0 < r < R and hence that B is

Symmetrization and Applications

94

non-increasing for all r. Now, by Lemma 4.2.2 and its corollary, we get / B(\x\)dx < f

B(|x|)dx

/ g(\x\)2dx> f g(\x\)2dx. Thus, it follows that

"

l(n)

J n . B{\x\)dx

* fn.9(\*\ydx

= "l(n }

by virtue of (4.2.5). This completes the proof.

•

N

Corollary 4.2.2 Let ft C R he a bounded domain and let Ai(fi) and fii (Q) be, respectively, the principal Dirichlet and Neumann eigenvalues of the Laplacian. Then /ii(H) < Ai(fi).

(4.2.10)

Proof: In view of (4.2.6) and the Faber - Krahn inequality (4.1.6), it suffices to prove (4.2.10) when U = Q*. Let R be the radius of Q*. Now, the first (Dirichlet) eigenfunction ip\ associated to Ai(fi*) satisfies the differential equation
+ \i(n*)
where, by abuse of notation, we have set fi(x) this relation and setting v = (p[, we get v"{r) +

TV - 1 V(r)

= 0

~ y>i(|x|). Differentiating

TV — 1 — v(r) + A i ( f i > ( r ) = 0.

(4.2.11)

Now, v(Q) = 0. Thus, starting from (4.2.11) and proceeding exactly as in the derivation of (4.2.5) from (4.2.3), and taking into account that v(R)v'(R) < 0, v(0) = 0, we get t

JBR {AW2 + ^v(\x\?) dx

,n.^

Al(fi

>

>

fBRv(\x\)>dx

•

On the other hand, since v =
Eigenvalue Problems

95

the preceding inequality is greater than or equal to /xi(fi*). This completes the proof. •

4.3

Chiti's T h e o r e m

In this section, we present a comparison theorem (in the spirit of Talenti's theorem) for the first eigenfunction of the Laplacian (with homogeneous Dirichlet boundary conditions), due to [Chiti (1982a)]. Let O C R N (N > 2) be a bounded domain and let Ai(fi) be the principal eigenvalue of the (Dirichlet) Laplacian. Let 5 be the ball with centre at the origin such that Ai(5) = Ai(ft). By the Faber - Krahn inequality (4.1.6), it then follows that Ai (Q*) < Ai (S), with strict inequality if H is not a ball. Thus, by the variational characterization of the first eigenvalue, it follows that 5 c f l * . In fact, the radius of S, denoted 7 - 1 , is given by

-• - m where B\ is the unit ball in RN, in view of the formula (4.1.8) for the first eigenvalue in a ball. The eigenfunction associated to this eigenvalue is given by z(x) = c | i | 1 - * J f l _ 1 ( v ' M n ) | 2 ; | ) ,

(4.3.2)

where c is a normalizing constant. Lemma 4.3.1

Choose c in (4-3.2) such that 2(0) = ¥>*(0) = max.
where ip\ is the eigenfunction associated to Ai(fi) in Q. Then z < ip\ in S. Proof: The function z satisfies - A z = XiZ in 5 1 z = Q on OS J where, we have set Ai = Ai (Q). We can now proceed exactly as in the proof of Talenti's theorem to obtain

- £ ( * # W ) < \I(NLO%)~2S%~2 J* z#{v)dri

(4.3.3)

Symmetrization and Applications

96

for 0 < s < \S\ (start with the relation corresponding to (3.1.9) and proceed as in the proof of Corollary 3.4.1). We also note that z — z*. Similarly, < Xi(Nu,*)-2a*-*

-^(vfto)

j\t{v)dn

(4.3.4)

for 0 < s < |fl| = |n*|. Since | 5 | < |fi*|, we have tpf(\S\) > 0 while z # ( | S | ) = 0 . Now, by choice, z#(0) = z*{0) = v?I(°) = ¥>* (<>)• If 2 = 2* £ y>! in St we can find a f c > l such that, for s G [0, \S\], we have

k(ff(s) > z#(s). Let us choose the smallest such k such that the above inequality holds. Now, there exists SQ G (0, \S\) such that k
z#(s0).

Define v^ by #/• \

(ktpfis), f k(pf(s), for 0 < s < s 0 for so < s < | 5 |

Then, u* is monotonically decreasing and u*(|5|) — 0. Further, by virtue of (4.3.3) and (4.3.4), we easily see that —^v#(s) «s

< A i ( A ^ ^ ) - 2 s ^ - 2 / v#(rj)dn. Jo

If we set v(x) = v#{uN\x\N),

then, v G H&(S). Then,

/ 5 \Vv\2dx =,/ 0 7 " 1 I v ' M l ^ w j v r ^ - ^ r = (JVw Af ) 3 / 0 7 " 1 r 2JV - 2 |v#'(u;jvr JV )| 2 r JV - 1 dr

= (^<) 2 /d 5| |^ # '( 5 )l 2s2 "^^
Eigenvalue Problems

97

Thus

IsWvfte Is v2dx

<

Al

and, as Ai is also the minimum of the Rayleigh quotient on 5 , it follows that this minimum is achieved for v and so v is an eigenfunction associated to Ai on S. Then v has to be a multiple of z and so, from the definition of t;, it follows that k z#(s) for all 0 < s < \S\ which proves the lemma. • We now prove a result due to [Chiti (1982a)] which will be useful in the sequel. Theorem 4.3.1

Let c be chosen in (4-3.2) such that / tp\dx = /
/ z2dx.

(4.3.5)

JS

Then, there exists r\ G (0,7 _ 1 ) such that
{

}

Proof: Suppose z(0) < <£*(0). Then, for some fe > 1, we have kz(0) — (^J(O). By Lemma 4.3.1, it then follows that kz#(s) < y?;(0). If 2(0) = 0. Now, z = z* and so z& is continuous and (pf is, in fact, absolutely continuous on any interval (a, |H|), for a > 0 (cf. Remark 2.3.4). So, for a > 0, sufficiently small, z*{a) > (pf(0) > (pf{a). Consequently, there exists Si € (0, \S\) such that z*(si) = z#{s). We now claim that this holds for all s 6 (si, \S\]. If not, there exists «2 £ (s\, \S\) such that (pffa) = z#{s2) and ipf (s) > z#(s) for all s G (si,S2)* Again, these follow from the continuity of the

98

Symmetrization and Applications

unidimensional rearrangements. We now define W#(S) -

/ * # ( S ) ' f°r SG[°'Sl]U[S2,|5|] \
Then, w# is monotonically decreasing and w(x) = W&(LJN\X\N) is such that w € HQ(S). AS in the proof of the preceding lemma, we can show that

ds

Jo

Again, from this it will follow that the Rayleigh quotient (over S) of w is equal to Ai and hence that w is an eigenfunction for X\, Consequently, w# _ 2 # a n ( j s o (pY(s) = z#(s) in [si, s2] contradicting the maximality of si. This completes the proof. •

4.4

The Payne - Polya - Weinberger Conjecture

Let H C Mw be a bounded domain and consider the eigenvalue problem for the Laplacian with homogeneous Dirichlet boundary conditions: —Au — Xu in n 1 u = 0 on da J

(

'

In 1955-56, Payne, Polya and Weinberger considered bounds for the eigenvalues and showed that

W <, when N — 2 and conjectured that that the right-hand side could be replaced by Ai(fi*) This result was extended to all dimensions by Thompson (1969) who showed that A2(fi) < Ai(Q) ~

1 +

4 N

Eigenvalue Problems

99

and, again, it was conjectured that the right-hand side could be replaced by

A2(n*) Ai(fi*)

=

i Jj,i 2

2 \

\U

where, as usual, jPtk denotes the fc-th positive zero of the Bessel function Jp (of the first kind) of order p. This problem, for TV = 2, was studied by Brands who obtained the value 2.686 for the ratio of the first two eigenvalues, by de Vries who obtained 2.658 and finally by Chiti who got the estimate 2.586. The conjecture was finally settled positively by [Ashbaugh and Benguria (1992a)] in 1992 and we sketch their proof below. Their paper and the survey article of (Ashbaugh (1999)1 gi y e complete references to the previous works cited above. As in the preceding sections, we will denote the eigenfunction (which is positive in ft) associated to Ai(fi) by tp\. Lemma 4.4.1 Let P ^ 0 be a sufficiently smooth function such that Pipi G #o(fi). Assume further that fQ Pip\dx = 0. Then

Proof: Since Py?i € #o(fi) and is assumed to be orthogonal to
-

fnP'tfdx

•

Now, as V ( i V i ) — ViVP + PVffiii we have | V ( / V i ) | 2 - vflVPI 2 + P 2 | V ^ i | 2 + 22|VP|2 + P2\V
= - / n nP2^idx

- / n P a |Vy>i\ 2 dx

= \i{n)fat*
= [ \VP\2tpldx + \i(Q) [ P2(p2dx. Jn Jii

(4 4 3)

-'

100

Symmetrization and Applications

The inequality (4.4.2) now follows on substituting this in (4.4.3).

•

The idea is to use the above inequality for N trial functions of the form Pi(x) = S ( M ) g , l
(4.4.4)

for a suitable function g. This is reminiscent of the method used in the proof of the Szego - Weinberger inequality Indeed, the first necessity is then to ensure that the functions P% [0, oo) be a sufficiently smooth function. Then, for a suitable choice of the origin, we have JQ Piipfdx = 0 for all 1 < i < N, where Pi is of the form in (4.4-4)Proof: Let B b e a ball of radius R, with centre at the origin, such that f l c B . For y E B, define F(V) = [ JQ

\X

g(\x-y\)^_Mx-y)2dx. - y\

Then, it is immediate to verify that, for \y\ = R, we have F(y).y < 0 and so, by Brouwer's theorem, there exists yo 6 B such that F(yo) = 0. We shift the origin to yo- This proves the lemma. • We now set <x = 3%-\,i

= \Ai(^i)

where B\ is the unit ball in RN. Notice that 7 is precisely the quantity defined in the previous section (cf. (4.3.1)) and is the reciprocal of the radius of the ball, S, whose first (Dirichlet) eigenvalue is also Ai(H). We also define W(t) 1 ;

= J J f(WA'f-!(«*)> for o < t < i , [lim^!-w(s),

for t > 1.

Then u> is a C1 function. Finally, we set g(r) = w{~fr).

(4.4.5)

Eigenvalue Problems Lemma 4.4.3

101

(Gap Inequality) If g is defined as in (4-4-5), we have

J ng (N)V l( ^ J

^ ' ^ < - ^

•

(4A6)

Proof; If we define P e , for 1 < i < N as in (4.4.4), then, thanks to Lemma 4.4.2, we may assume that the functions Pnp\ are orthogonal to (p\ in L2(U). Thus, by Lemma 4.4.1, it follows that (A 2 (n) - A:(H)) / P?y\dx Jn Summing over i, for 1 < i < JV, we get

f \VPi\2ip\dx. Jn

<

L(ZliP?)tfdx But by definition of the Pi, we get (£P2)(Z)

= g(\x\)2

i=\

and

|a;|

i=i

as already observed in the proof of Theorem 4.2.1. Thus, we get (4.4.6). • Notice that from (4.4.5), we get (SV)) 2 + ^ S ( r )

2

=

l2B(ir)

where

B(t) = («,'(*))2 + ^ - i ^ ( t ) a . The choice of the function w has been made so that, as in the case of the proof of the Szego - Weinberger inequality, the gap inequality (4.4.6) above becomes an equality for the unit ball. We now prove this fact. Lemma 4.4.4

Let B\ be the unit ball in RN. L B(T)JN

J0

Then, Aar)2rdr

w(r)jijN__1{ar)'irar

102

Symmetrization and Applications

Proof: Step 1. In the case of the unit ball, we have already seen that Xi(Bi) = a2 and that \x\l-^J^_x{a\x\).

tpi{x) =

The eigenfunction corresponding to A2(#i) = 01 is no longer radial; by separation of variables, it has a radial part v(|x|) given by v(r) = r 1

2

(fir)

JN

which satisfies the differential equation v"(r) + — — v'ir) ~ v(r) + (32v(r) = 0 r TZ and the conditions v(0) = v(l) = 0. Proceeding exactly as in the proof of relation (4.2.5), we can again show that 2

0

= \2(Bi)

2 S* lV\(v'{\x\)) + U IJ

^v{\x\)J 2]dx " • 2~T

JBl

=

U

(4.4.8)

Step 2. In the case when fi = B\, we have 5 = i?i as well and so 7 = 1. With our preceding notations, we have, for 0 < r < 1, g(r) = w{r) =

•J$(/fr)

v(r)

^-i(ar)

^i(r)

where, by abuse of notation, we have set (p\(x) = <^?i(|x|). Step 3. We saw, in the proof of Lemma 4.4.1, that for suitable functions P , j \V{P^)\2dx Jo,

= [ \VP\2tpldx + Ai(fi) / P2
Applying this to Q = B\ and to each of the functions Pi{x) = #(|x|) A- and then summing over i for 1 < i < n, we get r

/

r

N 2

£|V(PiVi)|
/

N

£(|VPi|Vds + aa/

Now, as seen in the proof of the preceding lemma, N

(]T>P«|a)(*) = B(\x\). 1=1

r

92vldx.

(4.4.9)

Eigenvalue

Problems

103

In exactly the same way, since (Pi
it follows that

N

(£MPapi)\2)(x)

= (v>(\x\))2 +

^v(\x\)2.

1=1

Thus, as (g
v{\x\)2dx

=

JBI

/

B^ip^xfdx

+ a2 f

JBI

v(\x\)2dx.

JBX

The relation (4.4.7) now follows easily from this on passing to polar coordinates. • Lemma 4.4.5 With the preceding definitions of w and B, we have that w is an increasing function and that B is a decreasing function. The proof of this lemma is the crux of the argument of [Ashbaugh and Benguria (1992a)] and [Ashbaugh and Benguria (1992b)]); it is rather long and technical and is proved using the properties of Bessel functions. The interested reader is referred to the papers cited above. Let us now go back to the notations of the preceding section and set 5 to be the ball centred at the origin and with radius 7 _ 1 so that its first (Dirichlet) eigenvalue is exactly Ai(fi). Let z be the corresponding eigenfunction as in Theorem 4.3.1. We then have the following result. Lemma 4.4.6

Let f : R —• R be an increasing function. f {{\x\)v\(x)2dx Ja*

> I f{\x\)z{xfdx. Js

(4.4.10)

The inequality is reversed if f is decreasing. Proof: Using polar coordinates, we get J f(\x\)z{x)2dx Js

- f }{\x\)<el{xfdx Jn*

NLON I'* f{r)((z(r)2 Jo

n~x +NuN

-ipXrftr"-^

Jr\

I

=

f{r)(z{r)2-ip\(r)2)rN-xdr

/

/(r)tf(t

.2„7V-1

Then

104

Symmetrization

and

Applications

where r\ is the number occuring in Chiti's theorem (cf. Theorem 4.3.1) and R is the radius of ft*. (Here, we have, by abuse of notation, written z(x) — z(\x\) and 7 - 1 , we have z(r) = 0. Since / is increasing, f(r) < f(ri) for 0 < r < n and f(r) > f(ri) for n
f f(\x\)
<

NtvNf(r1)^\(z{r)2-
(z(r)2-
+ / Jrl

dr

= }{n)\jsz2dx- J y2dx

= 0

by the choice of the normalization of z in Theorem 4.3.1.

•

We are now in a position to give the proof of the Payne - Polya Weinberger conjecture as done by [Ashbaugh and Benguria (1992a)]. Theorem 4-4.1

Let Q C MN be a bounded domain.

xm

<

Ai(fi) "

w)

£

Ai(n*)

a2'

Then, l

;

Proof: By the gap inequality (4.4.6), we get

Now, B(7r) is a decreasing function as observed in Lemma 4.4.5. Then, as seen in the proof of Lemma 4.2.2, we have

(B(7W)|n)'(.) < B(7(.))

Eigenvalue

Problems

105

infi*. Thus, [ B(i\x\)
< I B(i\x\)
< [

B(-y\x\)z(x)2dx.

JS

The last inequality above follows from Lemma 4.4.6, since, again, B(-yr) is decreasing. Similarly w(-yr) is increasing. If w* is the increasing rearrangement of w on H* (cf. Section 1.4) we have (cf. Corollary 1.4.1)

/.(7i*i)V(*)^> / «tea*. Jn Jn* Again, as w(-yr) is increasing, we have that wm > w in H* and, by yet another application of Lemma 4.4.6, we get / wt / ^(7|x|) 2 ^i(x) 2 dx > Jn* Jn*

/ Js

w^xDzix^dx.

Using these estimates, we get A 2 (fi)-Ai(fl)

2 2JsB(y\x\)z(x) dx 2
_ Ai(Q) Jo ^ ( r ) J j ^ - i ( t t r ) 2 r d r a

J0

w(r)2JN_1(ar)2rdr

on using polar coordinates. But, by Lemma 4.4.4, we know that the ratio of the integrals in the last term is just (32 — a2 and the result now follows immediately. •

4.5

Rayleigh's Conjecture for Clamped Plates

The Dirichlet eigenvalues for the Laplacian in the plane give the frequencies of vibration of a membrane occupying the region of a domain and which is fixed along its boundary. The Neumann eigenvalues give the frequencies of vibration of the free membrane. We now look at the biharmonic operator A 2 . Let fi C M2 be a bounded domain. If this represents the middle surface of a thin plate which is clamped along the boundary, then the frequencies of vibration of the plate are given by the eigenvalues of the problem: A2u = Au

in Q

1

u = ^ = o on a n / '

, . (4 5 X)

-'

106

Symmetrization

and

Applications

The same problem can, of course, be posed in all space dimensions. Henceforth, we will assume that 0 C MN is a bounded domain and consider the problem (4.5.1). As in the case of the Laplacian, we can show that there exists an increasing sequence {An(Q)} of positive eigenvalues, tending to infinity, and an associated orthonormal basis of eigenfunctions (which now belong to H,g(fi)). The principal difference from the case of the Laplacian is that it is no longer true that the principal frequency Aj(fi) is a simple eigenvalue. The corresponding eigenfunction need not be of constant sign either. However, in the case of the ball, the principal eigenvalue is simple and the eigenfunction is of constant sign and, in addition, is radial. If we set

and if Jp and Ip are, respectively, the Bessel and modified Bessel functions, then the principal eigenvalue of a ball of radius R is given by

*•<*>-(*)'-« (i3&)* where kv is the first positive zero of the equation Mt)Iv+i(t)

+ Jv+i(t)Iv{t)

= 0.

(4.5.2)

For N = 2, [Lord Rayleigh (1894)] conjectured that of all plates of equal area, the circular plate has the least principal frequency (or fundamental tone). Of course, this conjecture can be stated in all dimensions and can be written as Aj(ft) > Ai(JT).

(4.5.3)

To this day, this conjecture has not been fully settled. Szego proved it under the assumption that the first eigenfunction could be taken to be positive in H. But, as already mentioned, this is not true in general. In fact, there is no simple criterion to determine whether a given domain admits a first eigenfunction which is positive. The result is false even for simple domains like the square. The next important attempt at its solution was due to [Talenti (1981)] who proved the weaker inequality Ai(fi) >cjvAi(fi*)

(4.5.4)

where 0 < c^ < 1 is a computable constant depending only on the dimension N.

Eigenvalue Problems

107

The inequality (4.5.3) was finally proved for all plane domains by [Nadirashvili (1995)] by modifying Talenti's approach. This was further refined by [Ashbaugh and Benguria (1995)] and they also established the inequality (4.5.3) for domains in IR3. The conjecture remains open in dimensions N > 4. Pushing the argument of Ashbaugh and Benguria further, [Ashbaugh and Laugesen (1996)] proved an inequality of type (4.5.4) but with an improved constant CN which, in fact, tends to unity as N —• oo. We will now sketch the proofs of these results below. The starting point of all investigations is the Rayleigh quotient characterization of the first eigenvalue given by the following: Ax H =

Jn

min

;

'

.

4.5.5

Theorem 4.5.1 (Szego) Let the first eigenjunction u\ be positive in Q. Then, (4.5.3) holds. Proof: Set / = — Au\. Consider the symmetrized problem: - A t ; = /* in Q* v = 0 on dQ\ Since m G Ho(fi), an elementary application of Green's formula yields

/ fdx = 0.

'n Jn Then, again by Green's formula, /

^-da

= [

Jan' ov

f*dx

= 0

yn„

and, as v is radial, | ^ is a constant on dft* and thus vanishes. Consequently, v e H$(n*). Now, since u\ > 0 in Q, by Talenti's theorem (cf. Proposition 3.1.1), we have / u\dx Jn

<

/ v2dx. Jn*

Thus, by virtue of (4.5.5), we get A m\

-

hf2dx

> in- f*2dx

> A

m M

108

Symmetrization

and

Applications

This completes the proof.

•

Since we cannot assume, in general, that the first eigenfunction u\ is positive, we will henceforth set H+ = {uf > 0} and fi_ = {uj~ > 0}. By classical regularity results, we have |fi| — |fi+| + |fi_|. Let us denote fi+ -

B a and ST_ = Bfcl

the balls centered at the origin and of radii a and 6 respectively. If R is the radius of fi*, we thus have aN+bN

= _RW.

(4.5.6)

We now set / — Au and define, for 0 < s < |S~21, G( 8 ) = ( / + ) # ( « ) - ( / - ) # ( | n | - s ) , \ F(s) = -G(\n\-s). j

. ^a-'>

Both i*1 and G are decreasing functions. Notice that (/~)#(|S7| — s) = (/")#(«) (cf. Exercise 1.4.1). Now, it is immediate to see that f\n\ / F{s)ds Jo

=

,|nj / G(s)ds Jo

0

for, /•M / G(s)
f\n\ / ((/+)#-(/")#)ds Jo

r

Jn

(f+-r)dx

= 0

since f+ — f~ = f has zero mean. We now define

«#(*) = (No;*)-2 JIB'1 £*-2 j * F f t ) * ^ 1 [ 2

w#(s) = ( i v 4 ) ~ j f ' l ^~

2

(4.5.8)

/o G(i,)«iude J

and set v(x) = ^ ( ( ^ v l ^ l ^ ) and WJ(X) = ^ ( u ^ a ; ^ ) . (Since v and w are radial functions, we will, by abuse of notation, write v(x) = v(\x\) and w(x) — w(\x\) at times, to simplify the exposition.) The following lemma lists the important properties of v and w. L e m m a 4.5.1

Let v and w be defined by (4-5.8). Then v > 0 in Ba and w > 0 in Bf>,

(4.5.9)

Eigenvalue Problems

109

0 on dBa and w ~ 0 on dBb,

> - w<°> - > - > - »•

(4.5.10)

<"•»>

aN~^(a) = bN~^(b) dr dr

(4.5.12)

= F* in Ba and ~ Aw = G* in Bh

(4.5.13)

and -Av

where F*{x) = F(LJN\X\N) and G*{x) = G(coN\x\N). Proof: Since F and G are decreasing, it follows that the indefinite integrals Jo ^(ri)^,ri a n a - Jo ^C7?)^7? a r e c o n c a v e functions and, since they vanish for s — 0 and s — |H|, it follows that they are both non-negative. Consequently v# and «;* are also non-negative and so (4.5.9) follows. By definition, it is clear that (4.5.10) is true. Now

_ ( r ) = -{Nu;N)-1rl-N

J

F(V)dV

and a similar relation holds for w with F replaced by G. The relations (4.5.11) follow easily from this. The relations (4.5.13) are also valid since we have already seen that the explicit solutions of these differential equations are indeed given by v and w (cf. Step 4 of the proof of Theorem 3.1.1). Finally, we prove (4.5.12). In view of (4.5.6), it is easy to see that / Avdx JBa Indeed, since Jl

= - J F*dx = - f G*dx = [ Awdx. Bb h JB JBa JB JEb Gds = 0, r r\Ba\ r\n\ / F*dx = - / Fds = Gds JBa JO J\EBb\ = -

Gds = - / G*dx. By Jo JB_ Thus, by Green's formula, we have f ^V ri — f ^W A JdBa dv ° JdBb dv

110

Symmetrization

and

Applications

which implies (4.5.12) since v and w are radial and so their normal derivatives on the boundaries of the respective balls are constants. • Proposition 4.5.1

[Talenti (1981)] With the above notations, we have (uj)* (ui)*


=

J (Av)2dx+ Jm

(45 14)

lit]

-

Further, J {Auifdx JQ

(Aw)2dx

(4.5.15)

= [ (Awfdx. Jn*

(4.5.16)

[ JQ-

and f {Amfdx Ju

-

f {Avfdx Jn*

Proof: Let t > 0. Retracing the proof of Theorem 3.1.1, we get (JVw*) a /i(*) 2 -* < -At)

I

(-f)dx

J{ui>t]

where (j,(t) = \{ui > t}\. Now,

hu^-f)dx

= ku^-ndx+j{ui>t} <-C)U+)#(s)ds

f-dx f^t\f-)#(s)ds

+

= - /oM(t)(/+)#(|fi| - *)ds + C\r)*(s)ds = r ( t ) F(S)ds. Thus, we get

1 < (A^rVW*-2/ Jo

F(s)ds(-n'(t))

which yields 1 t < (Nu£)-2

f^V Jn(t)

2 ^~2

ft F{ri)d^ Jo

Eigenvalue

Problems

111

from which we conclude that (cf. the proof of Theorem 3.1.1) (w + ) # (s) < v#(s) since (i(0) = \Ba\. In an identical manner, we can show that N \2

(Nu$yv(ty-"

•v\t)

<

\ fdx J{u<-t}

< -v'(t)

I J0

G(s)ds

where u(t) = \{u < —t}\. This will then give (u )*(s) < w(s). This proves (4.5.14). Now, / (Aufdx Jn

/ (/+) 2
=

=

/

•|n|

•|0|

(f+)*2ds+

JO

/

(D*2ds.

JO

(4.5.17) On the other hand, since ( / + ) * ( / ~ ) # = 0 (cf. Exercise 1.4.3), we have jf* 1 G{sfds

= f™{{f+)#(s))2ds

+

+ / d B f c l ( ( / - ) # ( | n i " *))2ds

2

(4.5.18)

2

jr\(f )#(s)) ds+fZ((nns)) ds \Q\ \Ba\

Similarly / ••IPSaall

/

f\Da\

f\a\ 2

F(s) ds

+

= /

2

((f )*(s)) ds

((r)#(s))2ds.

+

y|Bb|

JO

(4.5.19)

JO

The relation (4.5.15) follows from relations (4.5.17) - (4.5.19). Finally, since ,|n| /

,|n| ((f-)*(\n\~s))2ds

JO

= /

((/-)#(*))*&,

JO

it is easy to see that •mi /•mi / G(sfds Jo '0

=

/-mi / ((/+)#(S)2 + ( / - ) # ( s ) 2 ) ^ Jo

/• 2 / f dx. Jn

This proves one part of (4.5.16). The other part follows in the same way. This completes the proof. • Let 0 < t < 1. We set Kt =

2

lvzH (Bi

v is radial v(x) = 0 for \x\ = t, when t ^ 0 Vv(x) = 0 for |x| = 1.

Symmetrization and Applications

112

We now define, for 0 < t < 1, p(t LN ) =

max o?v€Kt

JBi(Av)2dx'

T h e o r e m 4.5.2 [Talenti (1981)] Let QcUN the above notations, we have Ai(Q*) A , ' Ai(fi)

<

be a bounded domain.

1 -TTT m a x ( p ( t ) + p ( l - t ) ) . p ( l ) 0 r\ u

\

With

(4.5.20) )

P r o o f : Let, as usual, u\ be an eigenfunction associated to Ai(fi). Then, in view of (4.5.14) and (4.5.16), we have A i ( n ) - 1 = ([(Aurfdx)

(u+)*2dx+

(f

v2dx < . ,\ ,o, fBR(Av)*dx fB

JB +

f

(ui)*2dx)

w2dx fBR(Aw)*dx-

By t h e definition of p, we then have, by a simple rescaling, Ai(H)-1 <

R4

Pl

aN\

+p

w) [w

fbN

Further, since it is known t h a t t h e principal eigenfunction in a ball is radial, Ai(fi*) = R~4/p(l) (cf. (4.5.5)). T h e result now follows by observing t h a t a and b satisfy (4.5.6). • R e m a r k 4 . 5 . 1 Using the properties of the Bessel and t h e modified Bessel functions, [Talenti (1981)] showed t h a t p is an increasing function and so we get the inequality Ai(n) >

^Ai(n*)

in all space dimensions. However, a more careful study of the maximum occuring in the right-hand side of (4.5.20) can be carried out and in fact we get the existence of a constant c^ such t h a t Ai(fi)

>

CJVAI(Q*).

[Talenti (1981)] showed t h a t c 2 ra 0.97768, c 3 « 0.73910, c 4 % 0.65242, c 5 % 0.60925 and c 6 » 0.58394. We see t h a t cN decreases

Eigenvalue Problems

113

to 0.5 as N increases. However, [Ashbaugh and Laugesen (1996)] have improved this to a constant which tends to unity as n —> oo. • In order to improve on the results of Talenti and, in fact, to prove Rayleigh's conjecture in dimensions N = 2 and TV = 3, we present another approach based on the relation (4.5.15). This leads to a new variational pinciple, first proposed by [Nadirashvili (1995)] and later refined by [Ashbaugh and Benguria (1995)]. From (4.5.14), it follows t h a t

ju\dx< f V*dx+[

^

and so, in view of (4.5.15), we get fQ. (Av)2dx

4- j o .

{Aw)2dx

We use (4.5.21) to formulate a very useful variational principle. Let a, b and R satisfy (4.5.6). Define

'<"*> =

JByd*

+

JBb^

•

(4 5 22)

--

and set J{a,b)

= inf €{(p,?p)

where the infimum is taken over all pairs of radial functions ((p,ip) ^ (0,0) with (p e H2{Ba) n H&{Ba) and ip e H2{Bb) D H&(Bb) such t h a t dl

d

uN-\ ^ .N •l Pf~\ = b"-^(b). a"-^(a) ar ar

Proposition 4.5.2 J(a,b). Further,

(4.5.23)

There exists a pair (tpo,ipo) such that S((po,ipo) —

A2ip0 2,0 A2tp

- (i<po in B, ,. in ,„ „l} = iiipQ Bb

(4-5.24)

together with the condition (in addition to (4-5.23), (po ~ 0 on dBa and ipQ = Q on dBb, built into the set of admissible functions for the minimiza-

114 tion

Symmetrization and Applications problem) Ay? 0 (a) + AipQ(b)

Finally,

= 0.

(4.5.25)

we also have fi =

J(a,b).

P r o o f : We can look for (ip, ip) satisfying the required conditions as well as /
= 1

when looking for a minimizer of £. Since, J ( a , b) < oo, if {((pn, ipn)} is such a minimizing sequence, it follows from t h e fact t h a t z •—> || A U / | | 2 , B is a norm on H2(B) n HQ(B) for any bounded domain B equivalent to the Sobolev norm ||2||2,2,B, t h a t {• o " JBb

/ ( A V o -ti<po)
+ I JdBa

A
J

du

JdBb

vipo)ipdx

AiPo^-da = 0

(4.5.26)

dv

for all admissible pairs ((p,ip). Here fi is a Lagrange multiplier. Taking, successively,

(Ba) and ip = 0 and then ip = 0 and ip G T>(Bf,), we get (4.5.24). Using this in (4.5.26) and t h e fact t h a t t h e pair ((p}ip) satisfies (4.5.23) and t h e fact t h a t all these functions are radial, we get (4.5.25). By applying Green's formula to the numerator of £(<po,ipo) using the relations (4.5.24) and the various boundary conditions, we easily deduce t h a t /J, = J(a,b). This completes the proof. • It now follows from (4.5.21) t h a t Ai(fi) > J(a,b). If we show t h a t J(a,b) > J(0,R) for all 0 < a < b < R satisfying (4.5.6), then we have (using the fact t h a t the first eigenfunction in a ball is radial)

Al(n)

> m i > : ( A t 2 ' X = Ai(n-) w

which will prove the conjecture.

Jn* w

dx

Eigenvalue

115

Problems

At this point, the symmetrization arguments cease and it now becomes an independent minimization problem treated, using different approaches, by [Nadirashvili (1995)] for N = 2 and by [Ashbaugh and Benguria (1995)] for N = 2,3. We sketch the latter method. The proofs are involved and technical, and are based on a difficult and intricate analysis involving the Bessel functions. From (4.5.24), it follows that we can write
= Mka) + fv(kb)

= 0

(4.5.27)

where

f„(x) = x

Jv+i{x) JJx)

+

I„+i{x) IJx)

Let us assume, without loss of generality, that aN + bN = 1. Let ku{a) denote the first positive zero of hu and let kv denote the first positive zero of fu. In fact, kv = Ai(i?i), which can be proved via the identity Jv{x)lv+\(x)

+ Jv+\(x)Iu(x)

= xl~N

Ju{x)Iy(x).

Thus, in order to show that J (a, b) > J(0,1), we need to show that ^ ( o ) > &i,(0) = ku. This is the main body of the work done by [Ashbaugh and Benguria (1995)]. In particular, a necessary condition for the above to hold turns out to be

where, as usual, j V t \ denotes the first positive zero of Jv. This inequality holds only for N = 2 and N — 3. Hence their proof works only for these dimensions. [Ashbaugh and Laugesen (1996)] show that, for all space dimensions iV, kv{a)

> ku(2N)

= 2&ju>i

Symmetrization and Applications

116 for 0 < a < 2 ^ . Thus,

Ai(0) > 2*j}tl

= CivAi(fi*)

where

which tends to unity as N —» oo, thus improving on Talenti's inequality. 4.6

The Buckling Problem

Let fl c I 2 be a bounded domain in the horizontal plane occupied by the middle surface of a thin plate. Let us assume that the plate is clamped along the boundary. If this plate is subjected to a uniform compressive load acting laterally on the boundary, then, upto some point, the plate will just be compressed in the plane itself. Beyond a certain critical load, the plate will then buckle out of the plane. In the language of bifurcation theory (cf., for instance, [Kesavan (2004)]), if the strength of the force is parametrized by a parameter C? then for all values of £, we have the trivial solution, viz. the zero vertical displacement. When £ crosses a critical value, nontrivial solutions appear and we are at a bifurcation point. This situation is modelled by the von Karman equations, which is a nonlinear problem, and the critical values when bifurcation takes place are the eigenvalues of the linearized problem. This eigenvalue problem is given by A2u = -CAu in n du = o |H

on on.r

(4 6 i:

"-

Once again, the above problem admits an increasing sequence {Cu(^)} of positive eigenvalues tending to infinity and an associated orthonormal basis of eigenfunctions (which belong to HQ(U)). The first eigenvalue fi(fi) is called the buckling load. Inspired by Rayleigh's conjecture for vibrations of plates, Polya and Szego conjectured that the circular plate has the minimal buckling load amongst all plates of equal area. The problem (4.6.1) and the conjecture of Polya and Szego can be readily extended to all dimensions. Thus, we assume, henceforth, that Q, C RN is a bounded domain and the conjecture can now be written as

Ci(n) > Ci(fl').

(4.6.2)

Eigenvalue

Problems

117

Szego proved this conjecture when the first eigenfunction is of constant sign in Q. Again, this is an unreasonable assumption since, even for fairly simple domains, the first eigenfunction changes sign and the first eigenvalue is not simple. Nevertheless, these properties are indeed true for a ball. This conjecture remains open even now. Unlike the case of plate vibrations, it has not been proved even for low dimensions. The best result, as of now, is that of [Ashbaugh and Laugesen (1996)] who show that there exists a constant d^ > 0 which tends to unity as N —> oo, such that C i W > dN<;(n*).

(4.6.3)

We present these arguments below. As usual, the eigenvalues admit a Rayleigh quotient characterization. In particular, =

Cl(n)

min

k^L.

(4.6.4)

Exercise 4.6.1 Show that Ci(fi) > Ai(n)* and that Ai(n) > Ai(fi)Ci(fi) where Ai(O) is the first eigenvalue of the vibrating plate problem and Ai(fi) is the first (Dirichlet) eigenvalue of the Laplacian. The proof of Szego's result follows the same lines as the one in the case of vibrating plates (cf. Theorem 4.5.1). Theorem 4.6.1 (Szego) Let us assume that the first eigenvalue Ci(^) admits an eigenfunction which is positive in H. Then (4-6.2) holds. Proof: Let w > 0 be such an eigenfunction. Set / — — Aw. Consider the symmetrized problem -Av = f* in IT v = 0 on dfl. Since w G HQ(U), we have that JQ fdx = 0 and so it follows easily that | g = 0 on dQ* (cf. the proof of Theorem 4.5.1). Thus, v e H§(Q*). Now,

118

Symmetrization and Applications

by Talenti's theorem (cf. Proposition 3.1.1), since w > 0 in f2, we have / \Vw\2dx

Jn

< [

\Vv\2dx.

Jn*

Thus,

This completes t h e proof.

•

As already mentioned, in general, the hypothesis of the above theorem is not verified for most domains and, in fact, no general criterion is known to determine whether this is true for a given domain. We now present, following [Ashbaugh and Laugesen (1996)], a proof of the weaker inequality (4.6.3). We begin with Payne's inequality connecting fi(fi) and t h e second (Dirichlet) eigenvalue of the Laplacian. T h e o r e m 4.6.2 [Payne (1955)] Let A2(H) denote the second eigenvalue of the Laplacian (with homogeneous Dirichlet boundary conditions). Then, Ci(fi) > A 2 (fi).

(4.6.5)

Proof: Let ipi G # o ( ^ ) be the eigenfunction associated to Ai(ft), the first eigenvalue of the Laplacian and let w\ G HQ(Q) be an eigenfunction associated to Ci(^)- Then, M

n )

=

m i n kp^L v±
( 4.6.6)

where the minimum is taken over all non-zero functions in HQ(Q) orthogonal (in L2(Q)) to v?i. For 1 < i < N, define ipi = aiWi -f ——. dxi T h e n ipi G HQ(£1). We choose a^ such t h a t ipi is orthogonal (in L2(Q)) to ipi. It is clear t h a t ipi =£ 0 since w\ G HQ(Q) while its first derivatives are only in HQ(Q). Now, using the boundary conditions on w\ and integration by parts, it is immediate to verify t h a t f \ViPi\2dx

Jn

= a2 J \Vwx\2dx+

Jn

[

Jn

dxi)

dx

Eigenvalue Problems

119

and that

Using ipi as a test function in (4.6.6) for each 1 < i < N, we deduce that

A simple application of Green's formula shows that, since w\ G # Q ( ^ ) >

SJC

' ( S ) * - J£ dx =

(Awi)2dx.

Also, /

\Vwi\2dx

J (-Awi)widx Jn

< ( J (Awi)2dx) \JQ

J

( f w\ dx Via

by the Cauchy - Schwarz inequality. Finally, since w\ is the an eigenfunction associated to Ci(^)> w e have

f(AWl)2dx = diU) f \Vwtf

'dx

and so

/ i v ^ i 2 < Ci(n) f w\ dx. Using these observations, we get &?=!<$) fa\Vvi\2
A2(fi) <

(Zli°Z)JnV2d*

fn(*v>i)2dx

+ +

fn\Vwi\2dx

which completes the proof. Lemma 4.6.1 tions,

< Ci(fi) •

Let ft C MN be a bounded domain. With the usual nota-

A 2 (n) > where v = y — 1.

2*jltl

\n\

(4.6.7)

120

Symmetrization and Applications

Proof: Let (p2 G Hl{£l) be an eigenfunction associated to A2(f2). Since if 2 has to change sign in Q, we have two non-empty open sets Q+ = { 0} and Q- = {(fi2 < 0}. In each component of these open sets, ^2 will be of constant sign, will vanish on its boundary and will verify — Ay>2 — ^2(^)^2- Thus, in each such component, we have that \i($£) is the first eigenvalue of the (Dirichlet) Laplacian. If C C fi± is one such component, then since the first eigenvalue is monotonically decreasing with the domain (as is easy to see from the variational characterization), we have A2(fi) = Ai(C) > Ai(H±). Thus, 2

A2(fi) > Ai(Ot) > A i ( n i ) -

jl%l

UN

|0±l

(We cannot say that A2(fi) = Ai(H±) since these subdomains may not be connected.) Thus, 2A2(fi)

>jlilU

UJW

.m+l*

in. I*

* ^ 2 , r I ffl

since |n+| + |fi_| < |fi| and the function t •—• £~"tf is monotonic increasing for t > 0 and is convex. In case equality holds, then retracing the proof, we see that both Q+ and fi_ will be balls (by the case of equality in the Faber - Krahn inequality) and both of equal volume (equal to |fi|/2). In this case, Q will no longer be connected. Thus, for bounded domains, we do have a strict inequality and this completes the proof. • Combining (4.6.5) and (4.6.7), we immediately get the following result. Corollary 4.6.1 (Bramble - Payne Inequality) Let N = 2 and let fi C R 2 be a bounded domain. Then Ci(n) > 27rj» 1 |fi|- 1 . Lemma 4.6.2

Let v = y — 1. Let n C M.N be a bounded domain. Then 2_

Ci(n*) = J2+1.1 ( j g ) " •

(4-6.8)

Eigenvalue

Problems

121

Proof: Without loss of generality, we can assume that |Q| = UJ^, i.e. 0* is the unit ball in WLN. Define w{x) = JvUv+iMW1*

-Jv(jv+i,i)-

Then w is smooth and vanishes on dQ* = {\x\ — 1}. Further, from the properties of Bessel functions, we also have that d fJu(t)\ SJv(t dt \ tu

J„+i(t) t"

Thus f^f = o on dn* as well. Thus, w € flg(fi*). Now, since w is radial, we will set, by abuse of notaion, w(x) = w(\x\). Then, _ d2w dr2

N ~ 1 dw r dr

After a straight forward computation, it also follows, from Bessel's equation, that &w + $+IA(W

+ Jvtiv+i,i))

= °-

Hence, AWjJ+i.iAti; = 0 and so j2+i,i > Ci(n*)On the other hand, by Payne's inequality (4.6.5)

Ci(n') > A2(n*) = # + u . Thus, 0(fi*) =

JH-I,I

f° r t n e

unit

ball

an

d (4.6.8) is proven.

Combining (4.6.5), (4.6.7) and (4.6.8), we get (4.6.3) with dN = 2 *

72

^

We then have dN

=

1

„4-_log4+0(iv_§)

which tends to unity as N —» oo.

122

4.7

Symmetrization

and

Applications

Comments

In this section, we survey some related results and also indicate several fascinating open problems relating to the eigenvalues of the Laplacian and the biharmonic operators. The Faber - Krahn inequality for the principal eigenvalue of the Laplacian opens up an entire line of investigation of isoperimetric inequalities for eigenvalues. In particular, one can pose the question for all the eigenvalues. For each positive integer k, is there an optimal domain which minimizes Afc, the k-th eigenvalue of the Dirichlet Laplacian, amongst all domains of given volume? To start with, the second eigenvalue is minimized, amongst all open sets of given volume, by the union of two disjoint balls each with half the given volume (cf. Lemma 4.6.1). This, however, is a set which is not connected. Unfortunately, the minimum is not attained over connected sets. One can consider two disjoint balls connected by a thin tube and thin out the tube so that, in the limit, we obtain two disjoint balls. Thus we can produce a sequence of connected open sets so that their second eigenvalue tends to the minimum value, which, however, is attained for a disconnected set. One can try asking the question for convex domains. The existence of an optimal convex domain which minimizes the second eigenvalue has been established by [Cox and Ross (1995)]. It was widely believed that the optimal convex domain, which minimizes the second eigenvalue amongst all convex domains of given area in the plane, is the 'stadium' i.e. the closed covex hull of two equal circles which touch each other. However, [Henrot and Oudet (2001)] have recently shown that this is not the case. It has been shown by [Bucur and Henrot (2000)] that there exists an optimal set (connected, in case of dimensions 2 and 3) which minimizes the third eigenvalue. Though one does not know yet what this domain looks like, it is conjectured that in the case of dimensions 2 and 3, it is the ball and that it is the union of three equal disjoint balls for higher dimensions. The problem remains open for all higher eigenvalues as well as for all eigenvalues with other boundary conditions (like the Neumann problem). Another interesting problem is the optimization of the principal eigenvalue in domains with an obstacle. Let fi C RN be a bounded domain and let w C fi be a subdomain. We consider the eigenvalue problem (with Dirichlet boundary conditions) for the Laplacian in the domain U\uJ. The conjecture is that the first eigenvalue is minimal when the obstacle u> touches the outer boundary Oil (i.e. the infimum of the first eigenvalue, amongst all possible positions of the obstacle, is reached as the obstacle

Eigenvalue

Problems

123

approaches the boundary) and that the maximum occurs when the obstacle is close to the 'centre' of ft. For results in this direction, see [Harrell, Kroger and Kurata (2001)], though much remains to be done. In the special case, when ft and u are both balls, [Kesvan (2003)] has shown that the maximum indeed occurs when the balls are concentric (see also Ramm and Shivakumar at www.math.ksu.edu/ramm/r.html, publication 383). There are several results and open questions regarding optimization of various eigenvalues of the Laplacian under diverse constraints. For a nice survey of these, see the article of [Henrot (2003)]. Coming to the ratios of eigenvalues, [Payne, Polya and Weinberger (1956)] made several conjectures in their original paper. Of these, only two remain open. They are to show that A2(ft) + A3(ft) < A2(ft*) + A3(ft*) Ai(fi) " Ax(ft*) for D e l 2 (and the corresponding result for (A2(ft) + ... +A;v+i(ft))/Ai(ft) when ft C M N ) and to show that A m +i(ft) Am (ft)

<

-

A m +i(fi*) Am(ft*)

w

for ft C R for all m > 1. Of course, we presented the proof of Ashbaugh and Benguria for the case m — 1 in Section 4.4. [Ashbaugh and Benguria (1993)] have also shown that A4(ft) A2(ft) ~

A2(ft*) Ai(ft*)

and the cases m = 2 and 3 follow from this stronger inequality. All higher cases remain open as yet. As far as isoperimetric inequalities for eigenvalues of the biharmonic operator (A 2 ) are concerned, almost all the corresponding problems are open. One of the most recent results concerning the buckling problem (cf. Section 4.6) is that of [Ashbaugh and Bucur (2003)] where they prove the existence of an optimal domain for the buckling load. Willms and Weinberger have indicated how to prove the conjecture for the buckling load (cf. (4.6.2)) if there exists an optimal domain which is regular (C2). The above result is a positive step in that direction. The proof involves variational and domain differentiation techniques. Finally, there are several results and open problems regarding universal inequalities for eigenvalues, i.e. inequalities between various eigenvalues

124

Symmetrization

and

Applications

that are valid for all domains. The interested reader is referred to the survey article of [Ashbaugh (1999)].

Chapter 5

Nonlinear Problems

5.1

Payne - Rayner Type Inequalities

In this chapter, we will look at some isoperimetric inequalities associated to solutions of some nonlinear problems and apply them to get symmetry results for solutions in a ball. Let n C M^ be a bounded domain. Then, as a consequence of the Holder inequality, we have the continuous inclusion

Thus, for all u G L2(Q,), we have ||U||I,Q < C||u|l2,n- If u = yi > 0, the first eigenfunction of the Dirichlet Laplacian (cf. Section 4.1), and if N = 2, [Payne and Rayner (1972)] showed that

This inequality is known as the Payne-Rayner inequality or as a reverse Holder inequality. This inequality has been extended in several directions. It was generalized to higher dimensions by [Payne and Rayner (1973)] themselves but their inequality involves the first eigenvalue of an auxiliary problem and, for this reason, their generalization is not entirely satisfactory. It was shown by [Kohler-Jobin (1977)] that

Jd


" !N_2

where, as in the previous chapter, we have set v = y — 1. A generalization of the Payne-Rayner inequality for the first eigenfunction of second order elliptic differential operators was also obtained by [Chiti (1982b)]. 125

Symmetrization and Applications

126

A more interesting generalization of this result is for solutions of semilinear equations of the form -Au

= f(u) u>0 u = 0

in O \ in 0 > on dCl J

(5.1.2)

where / : K —» M is a positive continuous function. In this case, [Payne, Sperb and Stakgold (1977)] showed t h a t , when N = 2, 8TT

/ F(u)rfx Jo.

<

( / /(u)da: ) \Jn )

(5.1.3)

where F is the primitive of / such t h a t F(Q) = 0. Notice t h a t , if we set /(£) = Ai(Q)t, we recover the original Payne - Rayner inequality (5.1.1). In this section, we will prove a further generalization of this inequality to all dimensions and for the p-Laplacian. This result is due to [Mossino (1983)]. We will also discuss the equality case and use it to derive a symmetry result for t h e solution of a class of nonlinear equations following the work of [Kesavan and Pacella (1994)]. Let H C WLN be a bounded domain. Let / : M. —> R be a continuous function. Let 1 < p < oo. Consider the following boundary value problem:

u = 0

on m.

} }

(5 1 4)

' -

T h e weak formulation of this problem reads as follows: Find u G WQ,P(Q) such t h a t , for all v G W 0 1,p (ft), / |Vu]p~2Vu.V^:r =

f f(u)vdx.

(5.1.5)

We assume, henceforth, t h a t the following hypotheses are satisfied by the nonlinearity / : (HI) There exist constants A > 0 and B > 0 such t h a t f(t) < A\t\s + B, where s e

N —v

and s > 0 otherwise. (H2) For t > 0, we have f(t)

> 0.

if p<

N

Nonlinear Problems

127

T h e hypothesis (HI) guarantees that a weak solution of (5.1.4) in W 0 (fi) is, in fact, in Cha{Q) for a suitable a G (0,1) (cf. [Guedda, and Veron (1989)]). 1,p

We can now state the generalised Payne - Rayner type inequality for positive solutions of (5.1.4). In all t h a t follows, we assume t h a t 1 < p < co and t h a t q is its conjugate exponent, i.e. v~x + q~l

=

i.

It will be immediate to see t h a t if p = q = 2 and if N = 2, the inequality (5.1.6) below is precisely t h a t of Payne, Sperb and Stakgold (5.1.3). T h e o r e m 5.1.1 [Mossmo (1983)1 Assume that the hypotheses (Hi) and (H2) are satisfied. Let F be the primitive of f such that F(0) — 0. Ifu > 0 is a solution of (5.1-4), then i2\i

i \ Jr)(N^N)q

/*,f3'

i

,i i x S9("^)F(u#(s))ds

<

[ r / f{u)
P r o o f : Let t > 0. We set //(t) = \{u > t}\. Since (u\{u>ty)# we have

*(*) =f /

/(«)<& = / "

J{u>t}

= ti#|{u#>(},

f(u*)ds.

JO

Step 1. Since u e W 0 1 , p (fi), if , for some t such t h a t 0 < t < M where M is t h e m a x i m u m of u on U (recall t h a t u is, in fact, Holder continuous, thanks to ( H i ) ) , the set {u — t) has positive measure, then, by Stampacchia's theorem (cf., for instance, [Kesavan (1989)]), Vu = 0 almost everywhere on this set, and hence, on a set of positive measure as well. Thus, f(u) = 0 on this set as well, contradicting (H2). Thus, \{u = t}\ = 0 for all 0 < t < M and so ji(t) is continuous on (0, M ) . Consequently, we have u * ( / i ( t ) ) — t on this interval and so, *'(*) = / ( « # ( / i ( t ) ) / i ' ( t ) = f(t)n'(t).

(5.1.7)

Step 2. We now use t h e variational formulation of (5.1.4), viz. (5.1.5). We set v = (u — t ) + , as in the proof of Theorem 3.1.1, for 0 < t < M to get ~

dt

f \Vu\vdx J{u>t}

=

f f(u)dx J{u>t}

=

*(t).

128

Symmetrization

and

Applications

Again, an application of Holder's inequality yields (as in Theorem 3.1.1), ~ [ \Vu\dx) at J{u>t} J

< -/i'(t)*(t)5

Then, by a combination of the co-area formula and the isoperimetric theorem (cf. Corollary 2.2.3), we get ( J V a ; | ) V W ' - * < (Px»({u>t}))q

< -//(«)*(*)*•

(5.1.8)

Multiplying this inequality throughout by / ( t ) , which is strictly positive, and using (5.1.7), we get ( M 4 ) W * / W

< -*(i)2*'(i).

(5.1.9)

Step 3. We now integrate (5.1.9) on both sides with respect to t between 0 and M. The right-hand side gives

f0M-Ht)iv(t)dt =

-±f0M£mr)dt

< i*(o)9 since 3>(M) = 0 (the inequality appears in the above computations since we only know that $ is monotonically increasing). On the other hand, we have / 0 M / ( « ) / * ( * ) ' - * * = /o M f(t) (Q ~ # ) / 0 " W = l(l-jf)

J0M f(t) C

J-*-1** s"-^X{u#>t)(S)dsdt

= i (i - jr) C /W C s«i-^X{u*>t}(s)dsdt

=«(i4)/rl/;,l"/«''('"i)^ Combining these two results, we get (5.1.6) and the proof is complete. • We now treat the equality case.

Nonlinear Problems

129

Theorem 5.1.2 [Kesavan and Pacella (1994)1 Equality is achieved in (5.1.6) if, and only if, Q is a ball and (up to a translation of the origin) u = u*. Proof: Step 1. Retracing the proof of the preceding theorem, we see that, if equality is achieved in (5.1.6), then it also the case with (5.1.9) for almost every t such that 0 < t < M. Since $'(t) = /(£)/*'(t) and since f(t) > 0, we also have equality throughout in (5.1.8). Thus, for almost every t, the level sets {u > t} obey equality in the classical isoperimetric inequality and hence are balls. In particular, as in the proof of Proposition 3.2.2, we can deduce that all level sets {u > t} and S7 are balls. So, by translating the origin, if necessary, we may assume, henceforth, that n = fi*. Step 2. Let 0 < t < M. Then, the set of all points where Vu(x) = 0 and u(x) = t is empty. To see this, we first observe that if >£ > 0 in a bounded domain G, and if the set C = {xeG\

*{x) = 0}

has empty interior, then, by a result analogous to the Hopf boundary lemma (cf. [Guedda, and Veron (1989)]), §^ < 0 at all points of dG, where w G Cl{G) is a positive solution of the problem: -div(|Vw| p - 2 Vw) = * in G w — 0 on dG.

(5.1.10)

Now, setting G = fit — {u > t}, and w — u — t, we see that w satisfies (5.1.10) for *(x) = f(u{x)) = f(w(x) + t) > 0. So, if G = fit, the set C is empty and thus, | ^ < 0 on <9fif, or, in other words, Vu / 0 on dft,t — {u = t}. (In fact, if there exists x G {u = t} but not on 5fit., then, for every e > 0, we have x G fit-e. By the convexity of level sets (all level sets are balls), the entire line segment joining x to any point of dCtt would then have the same property and so u would be constant on that line segment (= t). This would contradict the fact that u G C1,a(Q) and that ^ < 0 on d£lf This proves our claim. In particular, the measure of the set of all points x G fi such that Vu(x) ~ 0 and 0 < u(x) < M is zero. Step 3. As a result of Step 2, it follows that (cf. Remark 2.3.6) ji(t) is an absolutely continuous function. Now, from the fact that we have equality

130

Symmetrization

and

Applications

in (5.1.9), we get

i =

(5.1.11)

-(^*)-VWMW*"**W

We can integrate this to get / Jo

/(u#( 5 ))
dg,

(5.1.12)

using the change of variable £ — //(£). Step 4. Consider the problem: -div(|V^| p - 2 V^) - f(u*) v= 0

in n - f2* on W .

(5.1.13)

As in Step 4 of the proof of Theorem 3.1.1, it is easy to show that this problem has a radial, and radially decreasing solution given by v(x) = (Nu^)~q

/

Ju>N\x\N

£*-#

/

f{u#(s))ds

JO

#

and its distribution function also satisfies (5.1.12). Since u* and v are both radially decreasing and equimeasurable, it follows that v = u*. Now, the function t *-> f(t)t is a Borel function and so f \Vu\pdx

= f f{u)udx

JQ

JQ

= f JQ*

f(u*)u*dx

=

/

\Vu*\pdx

JQ*

since u* solves (5.1.13). Since u* = v is now given by the formula above, a straight forward computation reveals that, thanks to the hypothesis (H2), Vu*(x) ^ 0 when 0 < u*(x) < M. Thus, by the theorem of Brothers and Ziemer (cf. Theorem 2.3.3) we have that u = u*. • We conclude this section with an application of the above result. Theorem 5.1.3 [Kesavan and Pacella (1994)] Assume that (HI) and (H2) are satisfied and that p — N. Then, every positive solution of (5.1-4) when Q — BR, the ball of radius R, is radial and radially decreasing. Proof: For a solution u G WQ'P(Q) of (5.1.4), we have the following Pohozaev type identity (cf. [Guedda, and Veron (1989)]) [ F(u)dx+(l JQ

\

) / }{u)udx P J JQ

du

= - J (XM) dv Q JdQ

da

(5.1.14)

Nonlinear Problems

131

where v is the unit outward normal on the boundary. If H = BR, v = x/R. By Green's formula, we get [ f(u)dx Jn

=

f \Vur2^da av Jdn

=

du

f Jdsi

then

da.

Applying Holder's inequality to the last term, we get )dx f f(u)dx <<

du

(NuNRN-l)i(f

(NLJN

Thus, Pohozaev's identity (5.1.14) now gives N f F(u)dx+(l- —) f f{u)udx JBR \ P J J BR

1
R

> q(NujNRN-l)p

)dx JBR

(5.1.15) since x.v — R on t h e boundary of BR. Now setting p — N, we get q — p/(p - 1) = N/(N - 1) so t h a t (5.1.15) becomes

/

N2

f(u)dx

N

JBR

-1

(NuN)"~ w1-1 /

F(u)dx.

(5.1.16)

JBR

On the other hand, (5.1.6) yields,

[ /(«)dx

N JV-1

N2

-(NUJN)^ J F(u)dx. (5.1.17) 1 N JBR J BR Thus, from the above two inequalities, we see t h a t equality is attained in (5.1.6) and so, by the preceding theorem, we have t h a t u — u*t which completes the proof. •

>

R e m a r k 5 . 1 . 1 W h e n p = 2, the p - Laplacian reduces to the Laplace operator. T h e result of Theorem 5.1.3 was first given by [Lions (1981b)] in this case and, for quite some time, several unsuccesful a t t e m p t s were made to produce such a proof for all dimensions for semilinear equations involving t h e Laplacian. T h e real generalization of Lions' result is as above, for the iV-Laplacian in TV dimensions. • R e m a r k 5 . 1 . 2 It was shown by [Gidas, Ni and Nirenberg (1979)] that, in t h e case of the Laplacian, the symmetry result is true for all nonlinearities (albeit with some smoothness assumptions). However, we need the hypothesis (H2) t h a t f(t) > 0 for t > 0. This is not unreasonable considering the

132

Symmetrization and Applications

fact t h a t [Kichenassamy and Smoller (1990)] have constructed an example of a function / with zero mean (hence of changing sign) such t h a t the problem (5.1.4) admits a non-negative and non-radial solution in t h e ball for p ^ 2. This difference in the behaviour of the p-Laplacian when p / 2 is because in those cases, in general, we do not have t h e strong maximum principle due to the degenerate n a t u r e of the operator. For more comments on symmetry of solutions, see Section 5.3. •

5.2

A S y s t e m of S e m i l i n e a r E q u a t i o n s

In this section, we will establish an isoperimetric inequality of Payne Rayner type for a system of semilinear equations. We will also show t h a t equality occurs only in the spherically symmetric case. As in the previous section, with t h e help of a Pohozaev type identity, we will show t h a t the solutions in a ball are spherically symmetric. However, in this case, we will need to restrict our attention to the case when t h e space dimension N = 2. Let fi C MN be a bounded domain. Let fi : R —> K, 1 < i < m be a family of continuous functions which are positive and monotonically increasing. Let A — {a%j) be a real symmetric matrix of order m with non-negative constant coefficients. Let u = (ui) € (HQ(Q.) r\C(Q))m be a (weak) solution vector to the problem: -Aui

= fi(Yl™=i aijuj) U{ = 0

in

on

^ \ dQ

\
(5.2.1)

Since the fi are all positive, it follows from the maximum principle t h a t all t h e Ui are positive. To exclude the case where an equation decouples itself from the rest, we assume t h a t in each row of the matrix A, there is at least one off-diagonal coefficient which is non-zero. We now define w G ( ^ ( f i ) ) ™ by m Wi = y j a j j U j , 1 < i < m.

(5.2.2)

j=i

Since the coefficients of A are all non-negative, it follows t h a t t h e Wi are positive as well. Let Fi denote the primitive of fi such t h a t Fi (0) — 0 for 1 < i < m. We then define mi(r)

=

f

fi(w*)dx

and M{(r)

-

f

F{w*)dx

(5.2.3)

Nonlinear

Problems

133

for 0 < r < R and 1 < i < m, where i? is the radius of fi*. Lemma 5.2.1 have

For all I < i < m and for almost every 0 < r < R, we m N x

-NuNr - w*\r)

< ^Oijmj-(r).

(5.2.4)

Proof: Since w* (r) is a monotonically decreasing function (as we have done several times before, we write, by abuse of notation, w*(x) = w*(\x\)), its derivative exists almost everywhere. If this derivative vanishes for some r, then (5.2.4) is trivially true. If the derivative is non-zero, then, clearly, Br = {w* > c] for some c > 0. Thus, it is enough to prove (5.2.4) when Br is a level set of w*. It follows immediately from the definition that w satisfies the system of equations: Wi = 0

on dU

i 1 < i < m.

(5.2.5)

Let c > 0 be less than the maximum value of w^ and set S \ c = {wi > c}. Let rc be the radius of H* c . Then,

-NujNr^wf(rc)

=

- j ^

°£d*

- /n«,c Aw*dx

= J ^ \Vw*\da

= HT=i /oi. c

a

ijf3(wj)dx

<J2T=iavIniJj(wj)dxThe inequality in the second line above comes from the classical isoperimetric inequality, as in the proof of the Polya - Szego theorem (cf. Step 2 of the proof of Theorem 2.3.1). The last inequality above uses the fact that, when fj is monotonically increasing, then, (fj(u)j))* = fj{w*) and the property (1.3.2). The result now follows for any Br which is of the form fi*c and the proof is complete. • Theorem 5.2.1 [Kesavan and Pacella (1999)] Let Q, C RN be a bounded domain. Let fi : ]R —> R, 1 < i < m be a family of continuous, positive and monotonically increasing functions. Let Fi be the primitive of fi such that

134

Symmetrization

and

Applications

Fi{0) = 0 for 1 < i < m. Let A = (aij) be a symmetric matrix of order m with non-negative constant coefficients. Let u e (HQ(Q) nC(fi)) m be a solution of (5.2.1) and let w be given by (5.2.2). Then m

m

Y^aijmiirij

„

\x\N-2Fi(w*(x))dx

> 4N(N-l)uN^2

(5.2.6)

where, for 1 < i < m, fhi =

I fi(u)i)dx. (5.2.7) Jn Further, equality is achieved in this inequality if, and only if, Q is a ball and all the Wi are radially symmetric and radially decreasing. Proof: Let Mi and rrii be given by (5.2.3) for 1 < i < m. Then, Ml(r)

= NtJNr^FiiwKr))

and m{(r) =

^ ^ " V t K W ) .

Differentiating M[(r) once again with respect to r and using (5.2.4), we get m

M/'(r) > N(N -l)uNrN-2Ft(wUr))

-

ft(w;(r))J2alJmJ(r). j=i

Multiplying throughout by Nuj^r1^-1 first derivatives of Mi and m^, we get

and using the expressions for the 771

NuN^^M'^r)

N 2

-Y,aijmj(r)m,i(r).

> N(N -l)uNr - Ml{r) 3=1

We sum the above inequalities over all 1 < i < m and integrate over the interval [0, R] where R is the radius of fi*. On one hand, since M[(R) = NtJNRN~lFi(0) = 0, we get, on integrating the left-hand side by parts, pR

\ Jo

t-R

NuNrN-xM'l{r)dr

= -N(N-l)uN

Jo

rN~2Ml(r)dr

which, except for a change of sign, is exactly the first integral on the righthand side. On the other hand, since a^ = a^ we also have J2 oymj(r)mJ(r) -

-—

J^

a^mii^m^r)

Nonlinear Problems

135

Since t h e rrii are all monotonic increasing, we thus deduce t h a t rJV-2F<«(r))NwjvrJV-1dr. i,j = l

i=l ^°

Now m^tf)

=

/

fi{w*)dx

— I fi(wi)dx

= m*.

T h e integral on t h e right-hand side of t h e preceding inequality is precisely the expression, on passing to polar coordinates, for t h e integral on t h e right-hand side of (5.2.6) and so t h e proof t h a t inequality is complete. If equality is achieved in (5.2.6), then, retracing t h e argument above, we see t h a t for almost every 0 < r < R, we have equality in (5.2.4). Hence, for almost every value c > 0, /

\Vw*\da

=

[

\Vwi\da.

Integrating this over t h e range of values c of Wi, we get, using t h e co-area formula, /

Jn*

\Vw*\2dx

=

\Vwi\2dx

/

Jo.

for each 1 < i < m. Since t h e right-hand side of (5.2.4) is positive, it follows t h a t , in the case of equality in (5.2.6), t h a t w* (r) < 0 for almost every r. Thus by t h e theorem of Brothers and Ziemer (cf. Theorem 2.3.3), we deduce t h a t (up t o a translation of t h e origin) Q, — Q* and that, for all 1 < % < m , W{ — w*. This completes t h e proof. • Corollary 5.2.1 then

With the hypotheses m

and notations m

Y j aijfhiThj

> 871"^

as above, if N — 2,

.

/

Fi(vji)dx.

P r o o f : W h e n TV = 2, t h e right-hand side of (5.2.6) reduces to m

,,

8TTV /

i=i Jn*

m

Fi{w*)dx

-

„

8TTV / i=1

Jn

F^w^dx

(5.2.8)

136

Symmetrization

and

Applications

using the properties of the Schwarz symmetrization and so we get (5.2.8).

R e m a r k 5.2.1 Let B ~ (bij(x)) be a matrix of order N such that, for all £ G R ^ and for almost all x G H, we have K|2 < B(x)U

< m2-

(5.2.9)

Let u = (ui) G (HQ($l)nC(£l))m be a positive solution vector to the system: mil

-div(BVm) = fi(ZT-i «««*) Ui = 0

\,l
on dQ, J '

Let Wi be defined by (5.2.2). If c > 0 is a value of Wi, then, -BVwtM

= -Vwt.B^

= -^(BI/.I/) > - ^ of ov where v is the unit outward normal on dH^c, thanks to the condition (5.2.9). Now the proof of Lemma 5.2.1 will go through mutatis mutandis and so Theorem 5.2.1 is still valid. Observe, further, since the isoperimetric inequality (5.2.6) only involves the Wi, it is still true if w = (wi) satisfies -div(B,Vt«() = £ ™ , o y / i t o ) i n n \ i < j < Wi = 0 on dCl J ' — —

m

where all the Z?j, 1 < i < m, satisfy (5.2.9).

•

We now establish a Pohozaev type identity for solutions of the system (5.2.1). T h e o r e m 5.2.2 Let u be a solution vector of the system (5.2.1). Let w be given by (5.2.2). Then

Y ] (2-N) i^i L

f Vui.Vwidx Jn

+ 2N [ Fi(wi)dx Vn

Proof: We start with the identity div((z.Vuj)Vui) = Vui.(l + (x.V))Vuj + (a;.Vuj)A^

(5.2.10)

Nonlinear

137

Problems

Integrating this by parts over Q, we get dUi V7 dui •da X.VUj x.VUj-^-

f j

'an =

/ Vui.Vujdx+ Jn

/ Y^ -rr^Xki:—^—dx + i dx ^ifjiidxi k9xi Jn

(x.VuAAuidx.

Integrating the middle term on the right-hand side twice by parts and collecting the terms, we get du' / (X.VUJ)-^ J an

du' + (x.Vui)—j- -

= (2-7V) / Vui.Vujdx+

(x.u^Vui.Vuj) da

j [(x.Vuj)&Ui +

(x.Vui)&Uj]dx.

As already remarked earlier, since the Ui and u>i vanish on <9f2, we have Vui = jfif-v and Vwi = ^ u on d$l. In view of this, the left-hand side of the above relation can be simplified as /

Jdn

(x.i/)———-da.

du du

Multiplying the resulting identity by a%j and summing over i and j , we get, using the symmetry of A, „

m

dui

duj

/ fa-") V] ai du du f

N

f

(2-TV) / Vui.Vwidx in

ft

+ 2 \ Y^ ,/n~

Xk^Auidx dxk

The second term on the right-hand side can be computed, by integration by parts, as follows: m

r

N

p. ^fc^—lAuidx -

i=1-•«*=!

oxk

m

~ N

n

- 2 V / 1=1 Jil fc=l n» N m

y^xk—-f(u>i)dx

e y e r dxp.k

xk {Fi{wi))dx

W^ ^

u m

„

Fi(wi)dx.

Symmetrization

138

and

Applications

There are no boundary integrals since Wi vanishes on dQ, and Ft(0) = 0. The identity (5.2.10) now follows immediately. • We now consider the case when Q is a ball in the plane M2. Theorem 5.2.3 Let Q — BR C R 2 ; the ball with centre at the origin and of radius R. Let fi : IR —*• IR be positive and monotonically increasing continuous functions for 1 < i < m. Let A be a symmetric matrix of order m with non-negative coefficients. Assume, further, that A is positive definite. Let u — (ui) G (HQ(£1) D C(Q))m be a positive solution vector to (5.2.1). Then m is radially symmetric and radially decreasing for each 1 < i < m. Proof: Let w be given by (5.2.2). Observe that v — x/R, where u is the unit outward normal on the boundary. Thus x.v — R for x e dQ. Then, since TV = 2, the Pohozaev identity (5.2.10) reads as

R

= 4 E / ^\{wi)dx.

E ^ ^

(5.2.11)

Let us set 1

f

dui

Since — Aui — fi(wi), it follows, from Green's formula, that ffii =

l
-2-KRJJLU

(5.2.12;

Since A is positive definite, we have

P

= iJ /

m

Q

ai

o

m

2^ i7hJ~fr7

da

~

27rR

1^ ^jtoto-

Thus, in view of (5.2.12), we have f -ft /

v~^ dui du _ > Oii— T^dcr > —- >

_ _ aumim-i

Nonlinear Problems

139

Thus, in view of (5.2.11), we get \_] aijihifrij

< 871-yj /

Fi(u)i)dx.

But this is exactly the inequality (5.2.8) but in the opposite direction. Thus it follows that we have equality in (5.2.8) and so, by Theorem 5.3.1, all the Wi and hence, all the Ui, are radial and radially decreasing. • R e m a r k 5.2.2 The fact that the differential operator was linear and the structure of the nonlinearities helped us to define the auxiliary functions Wi from the solutions Ui and derive the Payne - Rayner type inequality. If we tried to imitate this procedure for the p-Laplacian, we would encounter a term of the form m

Y2 aijmj

(r)mi(r)

i,j=X

which cannot be explicitly integrated in closed form. The positive deflniteness was needed to obtain a suitable lower bound for the boundary integral in Pohozaev's identity, in order to deduce the symmetry result. • A similar problem was studied by [Chanillo and Kiessling (1995)]. In their case fi(t) = exp(t) for all 1 < i < m. They not only assume that the matrix is symmetric and with non-negative coefficients, but also that the row sums are equal to unity. We do not need this hypothesis here. On the other hand, since their problem is posed on all of M2, they are able to do away with the boundary term in Pohozaev's identity and so they do not need the additional hypothesis of positive definiteness of the matrix A. 5.3

Comments

In the preceding sections, we proved Payne - Rayner type isoperimetric inequalities for positive solutions of certain nonlinear problems. Then, using a Pohozaev type identity, we deduced the radial symmetry of positive solutions in a ball. As already mentioned, the idea of such a proof goes back to [Lions (1981b)] who considered the solution of a semilinear equation involving the Laplacian in a ball in the plane. In that paper, it is implicitly assumed that equality in the isoperimetric inequality occurs only in the spherically symmetric situation. This was rigorously proved in the general

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case of the p-Laplacian by [Kesavan and Pacella (1994)], who also found the correct generalization of Lions' result to all dimensions. Of course, as already observed, the symmetry result here depends heavily on the fact that the nonlinearity is positive. Again, for the p-Laplacian, this is quite reasonable. However, for the Laplacian, any C1 function / will do. The study of symmetry properties of solutions based on symmetry properties of the domain first started with the work of [Gidas, Ni and Nirenberg (1979)] (though the ideas were already present in earlier works of Alexandrov and of Serrin). It was further refined and improved by [Berestycki and Nirenberg (1991)]. These depend heavily on the availability of maximum principles and the method is generally known as the method of moving planes. Strong maximum principles are generally not available for degenerate operators like the p-Laplacian. While symmetry results using variants of the moving plane method do exist (cf. (Grossi, Kesavan, Pacella and Ramaswamy (1998)], [Damascelli, Pacella and Ramaswamy (1999)] and [Damascelli and Pacella (2000)], for instance), mention must be made of the work of [Brock (2000)], who uses symmetrization techniques to prove such results. He has developed a procedure known as continuous Steiner symmetrization and has applied it to obtain symmetry results for solutions of the p-Laplacian. There are several other applications of symmetrization to nonlinear problems. Of course, the study of free boundary value problems, like the obstacle problem (cf. Section 3.4) can already be thought of as falling into this category. We also refer the reader to the works of [Bandle and Sperb (1983)], [Mossino (1979)] and [Mossino and Temam (1981)] for isoperimetric inequalities for free boundary value problems occuring in models describing the physics of plasmas. See also [Bandle (1976b)] for an isoperimetric inequality for solutions of a nonlinear eigenvalue problem. These are just a few instances of applications of symmetrization and the literature is vast, rich and varied. Needless to say that there is a wealth of literature on isoperimetric inequalities for linear and nonlinear problems using techniques other than symmetrization or techniques depending on other types of symmetrization. References to such works could be found in those cited in this text.

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Index

Bessel functions, 86, 90, 103 modified, 106 zeroes of, 86 Bramble - Payne inequality, 120 Brunn - Minkowski inequality, 24 buckling load, 116

Faber-Krahn inequality, 85 Fleming - Rischell theorem, 28 fundamental tone, 106 gap inequality, 101 Hardy - Littlewood inequality, 11, 14, 16 Hausdorff dimension, 22 Hausdorff measures, 21

Cabre's lemma, 25 Chiti's theorem, 97 co-area formula, 28, 31 co-area regularity, 40 conjecture Polya - Szego, 116 PPW, 99 Rayleigh (membranes), 85 Rayleigh (plates), 106 Saint Venant, 73

isoperimetric inequality, 20, 23 isospectral domains, 88 level set, 1 lower contact set, 24 Minkowski content, 21 moving planes, method of, 24, 140

de Giorgi perimeter, 22 Dido's problem, 19 Dirichlet principle, 71 distribution function, 1

Polya - Szego theorem, 35 Payne's inequality, 78, 118 Payne-Rayner inequality, 125 Payne-Sperb-Stakgold inequality, 126 Pohozaev's identity, 130, 136

electrostatic capacity, 70 electrostatic potential, 70 equality case Faber - Krahn inequality, 87 Payne - Rayner inequality, 128 Talenti's theorem, 53 isoperimetric inequality, 26 Polya - Szego inequality, 41 equimeasurable, 4

Rayleigh quotient, 84, 90, 107, 117 rearrangement spherically symmetric decreasing, 13, 17 increasing, 16 unidimensional 147

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decreasing, 2 increasing, 16 Riesz' inequality, 18, 39 Schwarz symmetrization, 13 Sobolev constant, 43, 44 Sobolev's inequality, 43 Szego-Weinberger inequality, 91 Talenti's theorem, 48 torsional rigidity, 73, 76 vanishing at infinity, 17 warping function, 73, 74

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