AD-A092 684 WISCONSIN UNIV MADISON DEPT OF MATHEMATICS F/6 12/1ON NONTRIVIAL SOLUTIONS OF A SEMILINEAR WAVE EQUATION(U)1 979 P m RABINOWITZ N0OQIW 76-C 0300

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MICROCOPY RESOLUTION TEST CHARTNATIONAL BUREAU OF STANDARDS 1963 A

LEVEL 1---- On nontrivial solutions of a semilinear wave equation

Paul H./Rabinowitz

Mathematics Department / D T IUniversity of Wisconsin ECTE

Madison, Wisconsin 53706

E

>The question of the existence of nontrivial time periodic solutions of

autonomous or forced semilinear wave equations has been the object of consider-

able recent interestJ "Td ",d-h-u:n ,,- '-

(1.1) ft) - U , + f (x .u) = 0, 0 < x< I

(or its analogue where f also depends on t in a time periodic fashion)

together with boundary conditions in x and periodicity conditions in t In

particular the following result was proved in [11, Theorem 3.37 and Corollary

4.14]:

Theorem 1.2: Let f e C(0,A ] X R,R) and satisfy

(f 1 ) f (x, 0) = 0 and f (x, r) is strictly monotone Increasing in r

(f 2 ) f(x,r) = o(Irl) at r= 0

(f 3 ) There are constants r> 0 and g > 2 such that E" 79

This researeh was sponsored in part by the Office of Naval Research under

Contract o4 , N00014-76-C-0300V Reproduction in whole or in part Is

permitted for any purpose of the United States v)nmer,".. f r%,

.Jd 031J

0 < F(x.r) : rf(xr)

for Ir Z r and x E [0,11 where

rF(x,r)= J f(x,s) ds

0

Then for any T which is a rational multiple of I , equation (1.1) possesses

a nontrivial continuous weak solution u satisfying

(13) u(O t) = 0 U(1,ot)

(1.3)

u(x, t+T)= u(x,t)

k kFurthermore f E C implies u e C

As part of the proof of Theorem 1. 2, it was shown that the functional

(1.4) I(u)= fT [1(utu z0 0 x-Yxu]dd

defined on the class of functions satisfying (1.3) (and of which (1.1) is

formally the Euler equation) has a positive critical value. Therefore (f,)

and the form of I imply that ut t 0 for the corresponding critical point u.

Thus u is nonconstant and must depend explicitly on t . It was further

observed in [11] (Theorem 5.24 and Remark 5.25) that if g satisfies

(fl) - (f 3 ) the equation

(1.5) utt- Uxx- g(x,u)= 0, 0< x<

together with (1.3). also possesses a nontrivial weak solution. Indeed the

arguments of Theorem 1.2 go through with minor modifications to establish

this fact. However the functional one studies for this case is

3

(16 [ u (ux - t ) Glx, u)] dx dt0 0 xU

where G is the primitive of g Again the positivity of (u) for a critical

point u implies u is nonconstant but we can no longer conclude that u

depends explicitly on t . In fact it is known [13,14] that as a consequence

of (f.) - (f 3 ) , the ordinary differential equation boundary value problem

(1.7) d- d- -uA = 9

dx, g(x,u), u(O) = 0 = u(I)

has an unbounded sequence of solutions which can be characterized by the

number of zeros they possess in (0,I) .

Our goal in this paper is to show that if (f 3 ) is strengthened somewhat,

(1.5), (1.3) possesses infinitely many time dependent solutions. More

precisely we will prove: Acesslom

Theorem 1.8: Let g a C([O,l] X R R) and suppose g satisfiesTABInounce4

(f 1 )- (f.) and ustifi0a

13) There is a constant I > 0 such that

0 < gF(x r) S r f(x, r) Avilmd ..' Avail and/az'

Diet., epee al

forall rX0 I

Then for any Tit IQ thereisa k0 t IN such that for all k a k0

(1.5), (1.3) possesses a solution uk which is kT periodic in t andauk Mroe"- k 0 Moreover infinitely many of the functions uk are distinct.

. . . ...at. . .- . . . . . . . . . l l E [III . . T _ . . . . . . . . . . . . . . . . .

4

Remark 1.9: We have no estimate for the size of k 0 and do not know if the

result is false in general for k0 = I . Note also that since (1. 5) is an

autonomous equation with respect to t , whenever u(x, t) is a solution,

so is u(x, t+e) for any 0 e R . The above statement about the Uk'S being

distinct means in particular that they do not differ by merely a translation in time.

The proof of Theorem 1.8 draws on several results from [Il] and ideas

from [12] . For convenience we will take I = w and T Zw . Choosing

k e IN, we seek a solution uk of (1.5) which is Zk periodic in t

and k X 0. Making the change of time scale r = t/k , the period becomesat

Zw again and the problem to be solved is

U rTr - k2 (Uxx + g(x. U)) = 0 0 < x< W

(1.10) U(O,- r) = 0= U(A , r)

U(x, T+Zw) U(x, T)

with U(x, ) = u(x, t)

For the convenience of the reader and to set the stage for a key estimate,

the argument used in [11] to establish the existence of nontrivial solutions of

(1.10) will be sketched quickly. Solutions are obtained by an approximation

argument. To begin (1.10) is modified in two ways. The wave operator

- k 8 a possesses in infinite dimensional null space in the class of

functions satisfying the boundary and periodicity conditions of (1.10) and given

by

N= span (sinJx sin kJt, sin Jxcos kJt I J i IN)

The fact that N is infinite dimensional complicates the analysis of (1.10)

and to introduce some compactness to the problem in N, we perturb (1.10)

by adding a 1 term to the left hand side of the equation. Here p > 0

and V is the (L orthogonal) projection of U onto N . A second difficulty

in treating (1.10) arises due to the unrestricted rate of growth of g(x, r) as

Ir I -- . . We get around this by truncating g . More precisely g(x, r) is

replaced by gK(x, r) which coincides with g for Ir r K and grows

cubically at oo [11] . Thus (1.10) is replaced by the modified problem

U + VT- k (Uxx+gK(x'U))= 0, 0 < x<

(1.11) U(O, T) = 0 U(1, T)

I U(x, T+Z1r) U(x, T)

where gKsatisfies (f), (f 2 ), (f 3 ) with a new constant 1j = min (g,4) .

Letting GK denote the primitive of gK - in a formal fashion (1. 11)

can be interpreted as the Euler equation arising from the functional

(1.12) J(U;k,3,K)f= l-Ux- U - T V - G K(XU)IxdT0 0

Let

Em = span {sinjxsinnT, sinjxcos nT 1 0-s j, n s m)

The strategy pursued in [11] was to find a critical point Umk of 11 Em

let m -. w, and then let 1--0 to get a solution Uk of (1.10) with g

replaced by gK Then L° " bounds for Uk independent of K show if

we choose K(k) sufficiently large, gK(x, Uk) = g(x, Uk) so (1.10) obtains.

6

A separate comparison argument Is required to prove that Uk " 0

The first step in carrying out the details of the above argument involves

obtaining an upper bound Mk for Cmk- J(Umk; kf3,K) with Mk

independent of m, 13, and K • For the current problem which also depends

on k, it Is crucial to know the behavior of Mk as a function of k . Thus

we will take a closer look at cmk and use a variant of an argument of [12]

By Lemma 1.13 of [11] , cmk can be characterized in a minimax fashion.

We will not write down this characterization explicitly but will note a consequence

of it which in turn provides an upper bound for cmk . Set

0mk = span {sinjx sinn , sinJxcos nrI 0 <sj, n- sm and na- >J k }

and

Pk = ak sin x sin (k-1) r

where a- so I k =1 . Set = mk (D span k " Then by

Lemma 1.13 of [11]

(1.13) 0 < cmk s max J(u; k,PK)u E Tmk

Inequality (1.13) will lead to a suitable choice for Mk . Note that by (f 3 )

(or even (f 3 ) ), there are constants a,, a. >- 0 such that

(1.14) GK(xr) aaIIl - a2

for all r e R, x t [0,r] . Consequently J-w as u- in Tmk

(under I • IL2 ) so there is a point z a Zmk at which the maximum in (1.13)

is achieved. Writing

_ _ _ _ _

7

(1.1) z-- l l + 6k

where 0 6 1 mk' = 1, and Y + 62 = I and substituting (1.15)

into (1.13) gives

(1.16) k fGK(x, z) dxdt if f (k' zxzdxd

Lobnn 0014 Ln 11) hwta

(1.17) k2 (a, lz" -a 3 )S k 11z 112

LL

Applying the Holder inequality yields

(1.18) IIz B A

where A is a constant independent of m, k, , K . Hence by (1.13),

(1.18), and the form of J ,

(1.19) Cmk s Mk

for a constant M independent of m, k, p, K.

Letting m-- w and then 1-. 0, and formalizing what we have just

shown gives:

Lemma .ZO: Under the hypotheses of Theorem 1.8 (with I= w and T = Zw),

for all k u IN, there exists a solution Uk of (1.10) satisfying

tiI~

r

8

(1.21) ck J(Uk; k,0,K) - Mk

with M independent of k + K.8 uk

It remains to show that for all k sufficiently large. U - 0

and infinitely many of the functions Uk( x, t) = Uk(x, T) are distinct. If

Uk is independent of r for any subsequence of k's tending to oo

Uk = Uk(x) is a classical solution of (1.7). Thus by (1.21) with K= K(k)

suitably large,d~ k 2

1.22c = ° 2k 2 fo! [.-.-2 - G(x,Uk)] dx

By (1.7),

(1.23) f i- I dx= f U (x foUrxU dx

Combining (1.21) - (1.23) yields

(1.24) [ U -(xU) -G(x,Uk)] dx-0

as k- o along this subsequence. Moreover by (f3)

rt Io' T I"(1.25) [ [Ukg(x, Uk) - G(xUk)] dx> (-_=)Ukg(xUk)dx.

001 dU k

Thus Ukg(x Uk)-- 0 in LI . From (1.23) again we conclude that d- 0kg~x Uk)- 0dxin L which easily implies Uk- O in Lf • By (fZ), for any

e > 0, there is a 6 > 0 such that Ir s 6 implies Ig(x,r)I s er

Choosing F-<1 and k large enough so that IUkl s 6, (1.23) then

shows

1 9

dU dUk dUk 2(1.26) IU S 7II k < - ll

'd 2 ~~ kII 2 ~~X dx"2

a contradiction. Consequently Uk depends on r for all large k

To prove the second assertion of Theorem 1.8, suppose two functions

Uk(x, r), Uj(x, r) correspond to the same function of x, t modulo a

translation in time (keeping Remark 1.9 in mind). ThusUkk( /k) U(x,t) and U (X.) Uj(xj,-)= U(xt+e)U. x' T) = kx O Uxt

for some e E R or Uk(X, )= U(x,kr), U (x 5 )= U(xJT+8) • Since

U must be both 2w k and 2w j periodic in t, letting a denote the

greatest common divisor of j and k, we have j = aJ k = ak and

U has period 27a in t . Furthermore

(1.27) ckf f - kx k kG(x, Uk )] d d0 02w k fw 2 t-kf I (U- u, U G(x,U)jdxdt

0 02 Z-ra ow 1 2 2 k2k 2a -f [Y(u x -Ut) G (x, U)] dx dt a b

and similarly

(1.z8) cj Lba

Consequently if there were a sequence of solutions Uk of (1.10)k

corresponding to the same function U (up to a translation in t ), by

(1.27) - (1.28) we have

zq

and ck -0 o like ki along this sequence contrary to (1.19). Thus at most(12)k k

10

finitely many functions Uk(x0 r) correspond to the same solution Uk(X, t)

of (1.5), (1.3) and infinitely many of the functions uk must be time

dependent solutions of (1.5), (1.3). The proof of Theorem 1.8 is complete.

Remark 1.30. Both the existence assertions from [11] and the arguments given

above use hypothesis (f 2 ) which requires that g vanish more rapidly than

linearly at 0 However this condition can be weakened. The simplest such

generalization would be to replace g(x,r) by ar + g(x,r) with a a

constant and for this case we have:

Theorem 1.31: Let g satisfy (f 1 ), (f 2 ), (f 3 ) and let a > 0 . Then for

all T e IQ , there exists a k0 e IN such that for all k - k 0 , the problemI0utt- uxx- au- g(x,u)= 0 0 < x<

(1.32) u(O, t) = 0 = u(1, t)

u(x, t+kT)= u(x,t)

aukhas a continuous weak solution uk which is k T periodic in t and - k 0

Moreover infinitely many of these functions are distinct.

Proof: For convenience we again take I = r , T = 2i . It was shown in

[11] that Theorem 1.2 carries over to (1.32) for a > 0 . It is also easy to see

that the argument of Lemma 1.20 will give (I.Z1) for this setting. Likewise

(1.27) - (1.29) are unaffected by the a term. Thus we get Theorem 1.29

provided that we can show Uk(x, r) depends on r for all large k . If not,

the analogues of (1.22) - (1.23) here are

2

I dUk(1.33) ck -Z,, k' fo I:L - - "a Uk - U(x, Uk] dC k 0 2 d x 2 k) d

and2

(1.34) T l-kI dx f (aU? + UgX,U)) dx.(134 dxk k x )0 0

Thus (1.19), (1.33) - (1.34), and (f 3 ) show that Ukg(xUk)--0 in L

as k-oo as in (1.24) - (1.25). Since

(1.35) lig(xUk)ll 1 " max Ig(xr)I + IIUkg( , Uk)IIk IL0O x s , Irl -l L

and the right hand side of (1.35) is uniformly bounded in k it follows from

d2 U(1.7) that the functions _L are uniformly bounded in L . The boundary

dx2

conditions Uk(O) = 0 = Uk(7) imply that there is xk G (Or) such that

dUk

Xk) = Hence

dUk x dz Uk(t)

S=Xk

which implies that

d~k d Uk 1

(1.36) - Lk d dx L

dUkThus the functions Uk , -- are bounded in L and by (1.7) again, so

are k . It follows that a subsequence of Uk converges (in 2 •

to a solution U of (1.7) as k-o. But (fl) and 1 LUk(xoUk'I -0

as k--o imply U = O.

Next observe that (1. 7) can be written as

(1.37) Uk(x)= f H(x, y) (a Uk(Y) + G (y. Uk(y))) dy0

12

where H is the Green's function for under the boundary condition

dx2

U(0) = 0 = U(vr) • Dividing (1.37) by J1UkII I gives,

(1.38) U(1= k (U k(Y) G(x, Uk(y))

S = 11 1 0 1H(x , Y ) (a C I + .... dy .

By (f 2 ) , the arguments of the integral operator are uniformly bounded in C

1 2Hence since this operator is compact from C to C , by (f 2 ) again a

subsequence of Uk/ I1 Uk II cI converge to V satisfying II VII C1 = I and

(1.39) V(x) = af H(x,y) V(y) dy0

or equivalently

(1.40) -V" = aV 0 < x< w ; V(0)= 0 = V(r) •

If a is not an eigenvalue of - d under these boundary conditions

we have a contradiction and the proof is complete. Thus suppose a is an

eigenvalue. Consider the eigenvalue problems :

(1.41) -z" = Xaz, 0 < x < Tr ; z(0) 0= z(r)

(1.42) -y" (a +g(xk}y, 0 < x < -f ;y(0)= 0 = ylr)

where ip is C 1 on [0,r] . Let X (resp. ,j(q)) denote the jth

eigenvalue of (1.41) (resp. (1.42)), the eigenvalues being ordered according

to increasing magnitude. As is well known any eigenfunction corresponding

to Xm or m(p) belongs to

r '3

Sm= (q E C ([0,i], ) q(o)= 0= q(), 0q has exactly M-1

zeros in (0, r), and 0'/O at all zeros of 9' in [Ow]}

(Indeed the eigenvalues of (1.41) are Xm = m2 a - and corresponding eigen-

functions are multiples of sin mx) Since g(x,9)q 2-1 0 via (fl) , we

have X. Ij(9) for all j E IN and V E C, 0 viaa standard

comparison theorem [15, Chapter 6J . By (1.40), 1 is an eigenvalue of

(1.41), say I= Xm and V E Sm . Thus gm(p) I and since Sm is open

(in the CI topology) and U k/UkU I - V in C 1 along some subsequence,C

it follows that Uk/ 11Uk l C 1 and therefore Uk belongs to Sm for all large

k in this subsequence. Writing (1.7) as

(1.43) ~ U" g(x, Uk )(1.43 (a+ k Uk 0 < x< W U k(0)= 0= Uk

we see m(Uk) I By (fl) again, g(xU k )U kI > 0 except at the m+I

zeros of Uk . An examination of the proof of the Sturm Comparison Theorem

[16, p. 208-209] then shows Uk has a zero between each pair of successive

zeros of V. Consequently Uk E Sm+I , a contradiction. Thus Theorem 1.31

is established.

Remark 1.44: In [5], Brezis, Coron, and Nirenberg study (1.1), (1.3)

replacing (f 3 ) by

) -rf(r)- F(r)- z 3f(r)l -Y

and

(f 5 ) f(r)/r - as Irl -

14

(and with no analogue of (fZ)). If we use (f 4 ) - (f 5 ) with x dependent f

in place of (f 3 ), it is not difficult to see that the proof of [11] carries over for

this case as does Lemma 1.ZO and (.Z7) - (1.29). Thus we obtain a

variant of Theorem 1.8 for this case once it is established that Uk(x, )

depends on for large k To do this, we argue as in the proof of Theorem 1.8.

Assume (f 4 ) holds with y = 0 . Then by (l.Z5) and (f 4 ), Ig(x, Uk)I 1 - 0

as k- o . This in turn implies IlUkl LO -0 via (1.7) and (1.36).

Hence (1.26) again provides a contradiction.

It is also possible for us to drop (f.) and even the requirement that

f(x, 0) = 0 in (fh) but then a new existence mechanism is required and we shall

not carry out the details here.

15

References

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(3] Bahri, A. and H. Brezis, Periodic solutions of a nonlinear wave

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[4] Brezis, H. and J. M. Coron, Periodic solutions of nonlinear wave

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[51 Brezis, H., J. M. Coron, and L. Nirenberg, Free vibrations for a

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in honor of Erich H. Rothe, Academic Press, (1978), 161-177.

__________

16

[II] Rabinowitz, P. H., Free vibrations for a semilinear wave equation,

Comm. Pure Appl. Math. 31, (1978), 31-68.

[12] Rabinowitz, P. H., Subharmonic solutions of a forced wave equation,

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(13] Turner, R. E. L., Superlinear Sturm-Liouville problems, J. Diff. Eq.,

13, (1973), 157-171.

[14] Rabinowitz, P. H., Some aspects of nonlinear elgenvalue problems.

Rocky Mtn. Math. j., 3, (1973), 161-202.

[15] Courant, R. and D. Hilbert, Methods of Mathematical Physics,

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[16] Coddington, E. A. and N. Levinson, Theory of Ordinary Differential

Equations, McGraw-Hill (1955).

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