g i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f

Transcription

1 1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where is some open bounded domin in R n. The condition on u on the boundry is clled irichlet boundry condition. We shll ssume tht f L 2 Omeg). This eqution cn be solved explicitely for limited number of situtions where the domin hs symmetry, like bll or hlf spce. In generl to infer the existence of solution is difficult problem. In order to hrness the power of functionl nlysis for proving the existence of solution one relxes the problem. Recll the definition of H 1 () form the exercises. These were ll functions f L 2 ( with the property tht there exist function gf i L2 () so tht for ll φ C c f φ x i dx = g i fφdx (). It ws shown in the exercises tht H 1 () endowed with the inner product n fgdx + gf i gi hdx (2) is Hilbert spce. We shll, henceforth, buse nottion nd write i=1 g i f(x) = f x i (x). Note tht since is bounded the constnt function is certinly in H 1 (). Wht bout functions tht vnish on the boundry. Note tht ny function ψ Cc () is in H 1 (). in fct, the eke derivtive equls the usul derivtive in this cse (why?). To cpture functions in H 1 () tht vnish on the boundry of we introduce new spce. efinition 1.1. The spce H() 1 is the closure of the set Cc () in the norm of H 1 (). It is Hilbert spce with the inner product (2). Almost by definition this spce is Hilbert spce. If f n is Cuchy sequence in H() 1 then it converges in H 1 () to some element f. Since ech f n H() 1 there exists φ n Cc () so tht Hence f n φ n H 1 () < 1 n. f φ n H 1 () f f n H 1 () + f n φ n H 1 () < f f n H 1 () + 1 n which tends to zero s n. Hence f H 1 (). efinition 1.2. u H() 1 is wek solution of (1) if for every v H() 1 we hve tht u vdx = fvdx (3) 1

2 2 Note tht we hve dropped the complex conjugtion since we shll be deling with rel vlued functions. Also note tht the derivtives re ll in the wek sense. Suppose tht u is twice differentible solution of (1). Then we cn integrte by prts nd obtin [ u f]vdx = from which we conclude tht u = f. The pth to this stge is, however, thorny nd we shll give some indiction how to procedd t the end An importnt inequlity. Let f be smooth function of the intervll [, ] nd ssume tht f() = f() =. Our gol is to get lower bound on the rtio f (x) 2 dx f(x) 2 dx. (4) We know from Fourier nlysis tht ny smooth function cn be expnded in Fourier series of the type πkx f(x) = c k 2 sin( ) The functions πkx 2 sin( ) form n orthonorml system nd hence we hve tht f(x) 2 dx = c k 2. The function f is given by f (x) = π nd once more using orthogonlity we find tht kc k 2 cos( πkx ) k 2 c k 2. Hence our rtio (4) cn be expressed s k2 c k 2 c k 2. Clerly by replcing k 2 by its smlles vlue 1 we find tht k2 c k 2 c k 2 with equlity if nd only if c k = for ll k > 1. Thus, we proved

3 Theorem 1.3 (Wirtinger s inequlity). For ny smooth function f on the intervl [, ] with f() = f() = we hve tht f (x) 2 dx f(x) 2 dx with equlity only if f is multiple of sinx). Here is nother mybe even simpler proof for this inequlity. Since f() = f() = we my write f(x) = h(x)g(x) where h(x) = sinx ) nd g(x) some function which hs compct support. Now f (x) = h (x)g(x) + h(x)g (x) so tht h (x g(x dx + h(x g (x dx + h(x)h (x)[g(x ] dx. Integrting the lst term by prts nd noting tht the boundry terms vnish we get Since we find = h (x g(x dx + h(x g (x dx h(x g (x dx h (x) = h(x) h(x)h (x)[g(x ]dx. [h(x)h (x)] [g(x ]dx. h(x g (x dx + h(x [g(x ]dx f(x dx. This theorem hs n interesting consequence. Let R n be n open set nd ssume tht fits into slb, i.e., between two prllel n 1 dimensionl plnes. enote by the infimum of the distnces of the pirs of plnes so tht fits between them. Theorem 1.4. Let u C c (). Then u(x) 2 dx u(x) 2 dx. Proof. let u Cc (). The support C, i.e., the closure of the set where the function does not vnish is compct set, by ssumption. Hence it hs dimeter d < (why?). Pick ny < δ < d There exist two plnes distnce d + δ pprt so tht C fits between these two plnes (why?). By rotting the function we cn ssume tht these two plnes re perpendiculr to the x-xis nd by trnslting the function w cn ssum tht one of the plnes psses through the origin nd the other through the point (d + δ,,..., ). We estimte d+δ [ d+δ u 2 dx = u 2 dx u ] 2 dx 1 dx 2 dx n x 1 R n 1 R n 1 Using Wirtinger s inequlity we find for every fixed x 2,..., x n d+δ u ( π d+δ 2 dx 1 u 2 dx 1 x 1 d + δ 3

4 4 which implies the result. As n immedite corollry we hve Theorem 1.5. Let u be ny function in H(). 1 Then u(x) 2 dx u(x) 2 dx. In prticulr u(x) 2 dx is norm which is equivlent with u H 1 () nd H() 1 with inner product (u, v) = u vdx is Hilbert spce. Proof. Let u H(). 1 Then there exists sequence of function φ n Cc () so tht u φ n H 1 () s n. Hence u φ 2 dx nd s n. Hence u 2 dx = lim φ n n 2 dx u φ n 2 dx Further u 2 dx u 2 H 1 () = u(x) 2 dx + lim φ n (x) 2 dx = u(x) 2 dx. n ( ( u 2 dx + 1) u 2 dx π nd hence the norms re equivlent. The finl sttement is immedite Existence of wek solution. Theorem 1.6. There exists unique function u H() 1 tht stisfies fvdx = u vdx for ll v H 1 (). Proof. This will fll out from the Riesz representtion theorem. Consider the liner functionl v fvdx nd note tht it is bounded on H() 1 since fvdx ( f 2 dx) 1/2 ( v 2 dx) 1/2

5 nd since v H 1 () we hve tht ( v 2 dx) 1/2 π ( v 2 dx) 1/2. By the Riesz representtion theorem there exists unique u H() 1 so tht fvdx = u vdx which proves the existence of the wek solution. Now the hrd work strts. It would be nice to be ble to integrte by prts nd to obtin [ u f]vdx = for ll v H 1 (). At lest we could conclude tht u = f pointwise lmost everywhere in Note, however, tht the solution u is in H 1 () only nd we hve no clue in wht sense one should interpret the second derivtive of u. It is lso interesting tht t this stge one hs only (3) to work with nd this leds to the theory of elliptic regulrity which ws developped round 195. First, in the sme wy we introduced H 1 (), one introduces the higher Sobolev spces H k (). Then one proves tht function tht is in high enough Soboleve spce H k () is in fct continuously differentible. In prticulr function tht is in H k () for ll k is C in the interior. One cn lso prove tht function tht is in H k () for ll k, is smooth on the boundry provided tht the boundry itself is smooth. This kind of nlysis hs nothing to do with the PE per se. In first step one proves tht the wek solution u is in H 2 (U) where is n open subset of whose closure is compct. If f H k () one cn ctully prove tht u H k+2 (U). This step is clled interior regulrity. The wy to do this is by crefully chosen test functions. The next problem is the boundry. One would like to conclude tht the function u vnishes on the boundry but this is tricky since ny function in Sobolev spce is defined only lmost everywhere nd since the boundry hs mesure zero one hs to crefully define wht one mens by boundry vlue. This kind of theorems re known s trce theorems. Now, if the boundry is smooth one proceeds to prove higher regulrity of u which, ssuming tht f nd re smooth, tht u C () nd the eqution (1) holds in the usul sense. 5