AMATH 731: Applied Functional Analysis Fall Sobolev spaces, weak solutions, Part II

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1 AMATH 731: Applied Functionl Anlysis Fll 217 Sobolev spces, wek solutions, Prt II (To ccompny Section 4.6 of the AMATH 731 Course Notes) In the previous hndout, we considered the following problem u (x)+g(x)u(x) = f(x), u() = u(b) =, (1) s motivtion for the study of wek solutions of DEs. Both sides of this eqution were multiplied by test function φ Cc (,b), with compct support in (,b): b b b u (x)φ(x) dx+ g(x)u(x)φ(x) dx = f(x)φ(x) dx. (2) Integrtion by prts then yielded the following eqution b b b u (x)φ (x) dx+ g(x)u(x)φ(x) dx = f(x)φ(x) dx, (3) where the boundry terms dispper due to the fct tht φ() = φ(b) =. Eq. (3) hs the form B(u,φ) = f,φ, (4) where, denotes the usul inner product on L 2 [,b] nd B(, ) denotes biliner form, i.e., (bounded) functionl tht is liner in ech of its two rguments. The gol is to determine wek solutions of Eq. (1) in terms of this eqution involving functionls. As will be seen below, we shll write tht, under pproprite conditions on B there exists unique function u tht stisfies the eqution B(u,v) = f,v (5) for ll v H, where H is n pproprite Hilbert spce. This function u will be clled the wek solution of Eq. (1). The so-clled Lx-Milgrm Theorem for bounded biliner functionls, to be discussed below, will gurntee the existence of such unique solution u. Riesz Representtion Theorem It is first instructive to recll this fundmentl result for bounded liner functionls (cf. Course Notes, p. 65). Theorem 1 Let F : H R be bounded liner functionl on Hilbert spce H. Then there exists unique z H so tht F(x) = x,z for ll x H. Moreover, F = z. Biliner forms nd the Lx-Milgrm Theorem Definition 1 A biliner form or functionl B on Hilbert spce H is mpping B : H H R such tht (x,y) is liner in ech of x,y H, i.e., for ll u 1,u 2,w H, nd c 1,c 2 R, B(c 1 u 1 +c 2 u 2,w) = c 1 B(u 1,w)+c 2 B(u 2,w), B(w,c 1 u 1 +c 2 u 2 ) = c 1 B(w,u 1 )+c 2 B(w,u 2 ). (6) 1

2 Theorem 2 (Lx-Milgrm) Let B : H H R be biliner functionl such tht the following conditions re stisfied: There exist constnts,b > such tht for ll u,v H, Finlly, let F : H R be bounded liner functionl on H. Then there exists unique element u H such tht B(u,v) b u v, (7) B(u,u) u 2. (8) B(u,v) = F(v), for ll v H. (9) Proof: 1. For ech fixed element u H, the mpping v B(u,v) is bounded liner functionl on H the boundedness follows from (7). From the Riesz Representtion Theorem, there exists unique element w H stisfying B(u,v) = w,v, v H. (1) This defines mpping A : u w so tht we shll write the bove s B(u,v) = Au,v, u,v H. (11) (Snek Preview/Spoiler Alert: Now go bck to Eq. (9) nd use the Riesz Representtion once gin to estblish tht for unique p H. From (11) nd (12) we hve tht Idelly, our solution is then given by But does this solution exist, i.e., does A 1 exist?) B(u,v) = F(v) = p,v (12) Au,v = p,v = Au = p. (13) u = A 1 p. (14) 2. We clim tht A : H H is bounded liner opertor: For c 1,c 2 R nd u 1,u 2 H, In ddition, implying tht Therefore, A is bounded. Also note tht A(c 1 u 1 +c 2 u 2 ),v = B(c 1 u 1 +c 2 u 2,v) (from 11) = c 1 B(u 1,v)+c 2 B(u 2,v) = c 1 Au 1,v +c 2 Au 2,v (from 11) = c 1 Au 1 +c 2 Au 2,v. (15) Au 2 = Au,Au = B(u,Au) b u Au, (16) Au b u, for ll u H. (17) u 2 B(u,u) = Au,u Au u (18) implying tht u Au, for ll u H. (19) 2

3 Therefore A is bounded from below. 3. We now show tht the rnge of A, R(A) is closed subspce of H. Let {v n } R(A) be Cuchy sequence with limit v H. Since ech v n R(A), there exists u n H such tht Au n = v n. From (19), u n u m Au n Au m = v n v m, (2) implying tht the sequence {u n } is lso Cuchy. Let u H denote the limit of this sequence. Since A is bounded, hence continuous, it follows tht Therefore v R(A), implying tht R(A) is closed. v = lim n v n = lim n Au n = A( lim n u n) = Au. (21) 4. From (19), A : H R(A) is bounded from below, implying tht n inverse opertor A 1 exists, i.e., A is one-to-one. 5. We now show tht R(A) = H. Assume tht R(A) H. Then, since R(A) is closed, there exists nonzero element w H such tht w R(A). But Aw R(A). This implies tht w 2 B(w,w) = Aw,w =, contrdicting (8). Therefore R(A) = H. 6. Now consider the given functionl F(v). From the Riesz Representtion Theorem, there exists p H such tht F(v) = p,v. From 4. nd 5., there exists u H such tht Au = p, i.e., u = A 1 p, so tht, from (11), B(u,v) = p,v = F(v). (22) 7. Finlly, we show tht there is t most one element u H stisfying (9). Assume the contrry, i.e., tht B(u,v) = F(v) nd B(ū,v) = F(v) for ll v H. Then, by linerity, B(u ū,v) = for v H. Set v = u ū to give, from (8), u ū 2 B(u ū,u ū) =. Therefore u = ū, nd the theorem is proved. Appliction to one-dimensionl boundry-vlue problem We first consider the following boundry vlue problem: u (x) = f(x), x 1, u() = u(1) =. (23) A physicl interprettion of this problem is tht u(x) is the trnsverse deflection (in the y-direction) of homogeneous string t point x under the influence of force f(x) (ctully force density more on this lter) cting in the y-direction. (We lso ssume the use of pproprite units tht simplify the form of the problem.) Note tht in this problem, the string is clmped t both ends, i.e., x = nd x = 1. Brief note on the origin of the bove eqution: Recll the 1D PDE for homogeneous vibrting string with n externl force f(x), 2 u t 2 = u c2 2 +f(x), x 1, (24) x2 where c 2 = T/ρ. (T is the tension in the string nd ρ the linel mss density (mss/unit length), both ssumed to be constnt.) For simplicity, we hve set c = 1. Eq. (23) corresponds to the stedystte, or time-independent, solution u(x,t) = u(x), i.e., u t =. 3

4 Using rguments similr to those in the previous hndout, the totl strin or elstic energy of the string is nd the totl work of the externl force is U = 1 2 W = The ssocited energy functionl J(u) for this problem is given by J(u) = 1 2 [u (x)] 2 dx, (25) f(x)u(x) dx. (26) [u (x)] 2 dx f(x)u(x) dx, (27) These three equtions hve the sme form s for the one-dimensionl rod problem exmined in the previous hndout. Following the sme type of vritionl method s before, one cn show tht the minimizer u of J(u) corresponds to the solution of (23): J(u +ǫv) = J(u )+ǫdj(u )v ǫ2 [v (x)] 2 dx. (28) But DJ(u ) = since u is solution to Eq. (23), implying tht J(u +ǫv) > J(u ). (29) Clssicl tretment nd its limittions In the clssicl picture indeed, the one employed erlier in this course one ssumes tht f C[,1] so tht u C 2 [,1]. The solution u of (23) my be expressed in terms of f using the Green s function ssocited with this boundry-vlue problem: where g(x,y) = u(x) = g(x, y)f(y) dy, (3) { y(1 x), y x 1, x(1 y), x y 1, Since f is ssumed to be C[,1], everything is nice here, nd u is twice differentible. But wht if f is not so nice? Well, if f were n L 1 function, the integrl would still mke sense nd u(x) would be defined. But wht bout u (x)? For exmple, wht bout the cse where f(x) is force concentrted t point? Such forces re often modelled with Dirc delt function, which corresponds to force of finite strength, sy A, (with A > corresponding to n upwrd-pointing force, A < to downwrd-pointing force) concentrted t point < < 1, implying tht the force density is infinite. Bypssing rigour for the moment, such force would be written s (31) f(x) = Aδ(x ). (32) Even in clssicl tretments, this expression for f(x) is usully inserted into into (3), keeping in mind the property tht, for ny v C[,1], The result is v(x)δ(x ) dx = v(). (33) u(x) = A g(x,y)δ(y ) dy = Ag(x,), (34) 4

5 implying tht u(x) = { Ax(1 ), x 1, A(1 x), x 1, Thus the grph of u(x) hs tringulr shpe, with vertex t x =, for which u() = A(1 ): F = Aδ(x )j (35) A(1 ) 1 The result, s is well known in undergrdute courses, is tht u(x) is not differentible t x = ; it is, however, piecewise differentible. In other words, we hve to move wy from the very nice spce u C 2 [,1]. From the discussion of the previous hndout, we see tht this solution for u(x) is well ccomodted in the energy spce, E R, or, equivlently, the Sobolev spce W 1,2, since [u (x)] 2 dx <. (36) Finlly, we mke comment regrding the reltionship of the Dirc delt function δ(x) nd the Green s function g(x, y) for this problem reltionship tht pplies to other similr boundry-vlue problems. For simplicity, let the mplitude of the point force be A = 1. Then, for point force f(x) locted t (,b), i.e., f(x) = δ(x ), the function u(x) = g(x,) the Green s function itself is seen to be the solution to the eqution u = δ(x ). (37) This result lso pplies to the problem of point electric chrges nd ssocited potentils. It is often stted in textbooks nd cn be proved rigorously using generlized functions nd ssocited wek derivtives see E. Zeidler, Applied Functionl Anlysis, Applictions to Mthemticl Physics, Springer-Verlg (1997), p Wek derivtive/sobolev spce tretment We now relx the restriction tht f C[,1] in (23) to f L 2 [,1] nd pply the method of wek derivtives in Sobolev spces to thisproblem. We ll work inthe spcewc 1,2 (, 1), the spce of functions with compct support on (,1) nd norm ( 1/2 u 1,2 = ([u(x)] 2 +[u (x)] 2 ) dx) (38) The most importnt spect of this spce is tht the (wek) derivtive u (x) is L 2 -integrble, cf. Eq. (36). Note: In the literture, the following nottion, W k,p c (D) is often denoted s W k,p (D). In ddition, the p = 2 Sobolev spces re often denoted s H k (D) = W k,2 (D), (39) 5

6 cknowledging tht these spces re Hilbert spces. Once gin, the superscript, i.e., H k (D), will be used to denote the subspce of functions with compct support on (,1). This is lso the nottion used in the AMATH 731 Course Notes. We dopt this nottion below. We now return to Eq. (23), u (x) = f(x), x 1, u() = u(1) =, (4) multiply both sides of it with n rbitrry element v H 1 (D) nd integrte by prts to give u (x)v (x) dx = f(x)v(x) dx, v H 1 (D), (41) where it is understood tht u (x) represents the generlized or wek derivtive of u(x). Eq. (41) hs the form B(u,v) = F(v), (42) where the biliner functionl B(u, v) nd the liner functionl F(v) re defined, respectively, s B(u,v) = u,v = F(v) = f,v = u (x)v (x) dx, f(x)v(x) dx. (43) We now seek to pply the Lx-Milgrm theorem to estblish the existence of unique solution u H 1 (D) to (41), hence to (23). First of ll, the liner functionl F(v) is bounded: ( ) 1/2 ( 1/2 F(v) = f(x)v(x) dx f(x) 2 dx v(x) dx) 2 <. (44) As for the biliner functionl B(u,v), it is bounded from bove: B(u,v) ( u (x) v (x) dx ) 1/2 ( ) 1/2 u (x) 2 dx v (x) 2 dx u H 1 v H 1, (45) thereby estblishing tht the condition (7) for the Lx-Milgrm Theorem is stisfied. Estblishing the second condition (8), i.e., bounding B(u,u) from below, is little trickier. We hve to resort to Poincré s Inequlity, cf. Theorem 4.9, AMATH 731 Course Notes, p. 68. Actully, the onedimensionl version of Corollry 4.2, on p. 69, is sufficient for this problem: There exists constnt c 1 > such tht u 2 H 1 (c 1 +1) u (x) 2 dx, for ll u H(,1). 1 (46) (In the 1D cse, the proof is quite simple, see below.) Since it follows tht B(u,u) = [u (x)] 2 dx, (47) 1 c 1 +1 u 2 H 1 B(u,u). (48) 6

7 Therefore, B stisfies both conditions of the Lx-Milgrm Theorem. We cn conclude tht there exists unique element u H 1 (,1) tht stisfies (42), therefore (41), nd therefore the boundry-vlue problem (23) in the generlized sense. Proof of Eq. (46): We first consider u C 1 (,1). From the Fundmentl Theorem of Clculus, for x (,1): x u(t)u (t) dt = 1 2 u(x)2 1 2 u()2 = 1 2 u(x)2. (49) But from the Cuchy-Schwrz inequlity, [ u(x)u (x) dx Combining these two results, we hve tht ] 1/2 [ 1/2 u(x) 2 dx u (x) dx] 2. (5) [ ] 1/2 [ 1/2 u(x) 2 2 u(x) 2 dx u (x) dx] 2. (51) This implies tht [ ] 1/2 [ 1/2 u(x) 2 dx 2 u(x) 2 dx u (x) dx] 2. (52) Squring both sides nd rerrnging yields Adding u(x) 2 dx 4 [u (x)] 2 dx. (53) u (x) 2 dx to both sides shows tht Eq. (46) is stisfied for ny u C 1 (,1), with c 1 = 4. For the cse u H 1(,1), there is sequence {u n} C 1(,1) such tht u n u in H 1-norm. Since Eq. (46) is stisfied by ll u n C 1 (,1), it will be stisfied for u. Finite elements nd the Ritz method We now outline procedure, bsed on the so-clled method of finite moments, nd the Ritz method, to provide pproximte solutions to the boundry-vlue problem (23). In wht follows, we outline the ppliction of the method to boundry-vlue problem over the generl intervl [, b], i.e., u() = u(b) =. Clerly, for the bove problem, =, b = 1. First, divide the intervl [,b] into n+1 equl subintervls using the prtition = < 1 < 2 < < n < n+1 = b, (54) where By definition, finite element j = j (b ). (55) n+1 is piecewise (tringulr-shped) liner function with e in : [,b] R, i = 1,2,,n, (56) e in ( i ) = 1, nd e in ( j ) =, for ll j i. (57) We define X n = spn{e 1n,e 2n,,e nn }. (58) 7

8 1 e 1n e 2n e 3n e nn n b Then u n X n if u n (x) = c in e in. (59) Note: Ech function e in stisfies the boundry condition e in () = e in (b) =, which implies tht u n () = u n (b) =, for ll u n X n. (6) The function u n X n is piecewise liner nd u n ( i ) = c in for ll i = 1,2,,n. Therefore the spce X n consists of ll piecewise liner functions with respect to the points, 1, 2,, n,b which stisfy the boundry-vlue condition u() = u(b) =. We now return to the energy functionl J(u) ssocited with this BVP, cf. Eq. (27), but now in the Sobolev spce H 1 (,b): J(u) = 1 2 b [u (x)] 2 dx f(x)u(x) dx, u H 1 (,b). (61) The minimiztion of this functionl with respect to functions u n X n is Ritz problem: min F(u n ). (62) u n X n This represents minimiztion problem with respect to the rel vribles c 1n,c 2n,,c nn in Eq. (59). If u n is solution to (62), then This produces the so-clled Ritz equtions: b u ne jn dx = c jn F(u n ) =, j = 1,2,,n. (63) Explicitly, we hve liner system of equtions in the unknowns c in : b For ech n, this liner system hs the form where e jn f dx, u n X n, j = 1,2,,n. (64) b b c in e ine jn dx = e jn f dx, j = 1,2,,n. (65) Ac = f, (66) ij = e i,e j, c i = c in, f i = e jn,f, 1 i,j n. (67) The mtrix A is quite concentrted ner the digonl, given tht the finite element e jn overlps only with itself nd its two immedite neighbours e j±1,n. 8

9 Proposition 1 (The Ritz method vi finite elements) Let f C[,b]. Then the bove Ritz method converges to the unique solution u of the boundry-vlue problem (23) in the sense of the Sobolev spce H 1 (,b) = W1,2 (, b), i.e. u u n 1,2 s n. (68) For proof of this proposition, s well s rigorous estimte of the error, see E. Zeidler, Applied Functionl Anlysis, Applictions to Mthemticl Physics, Springer-Verlg (1997). Wek solutions of PDEs Second-order elliptic PDEs Here we consider briefly boundry-vlue problems of the form Lu = f in D, u = on D, (69) where D is n open, bounded set of R n. Here, f : D R nd g : D R is given. L denotes second-order prtil differentil opertor, expressed in so-clled divergence form, Lu = xj ( ij (x) xi u)+ b i (x) xi u+c(x)u, (7) i,j=1 which is menble for tretments involving integrtion by prts, e.g., energy methods, wek solutions. The requirement tht u = on the boundry D is known s the Dirichlet boundry condition. Another clss of problems is s follows, Lu = in D, u = g on D. (71) This generlized Dirichlet problem in R 2, for which f =, is the subject of Question No. 6 in Problem Set 5 of the AMATH 751 Course Notes. Such boundry-vlue problem would rise when trying to find the electrosttic potentil u(x) inside chrge-free region D, produced by given chrge density g on the boundry D. With n eye to pplictions, we shll ssume tht ij = ji, 1 i,j n. (72) As well, we consider only elliptic prtil differentil opertors L, i.e., those for which the following condition holds: There exists constnt C > such tht ij (x)ξ i ξ j C ξ 2, (73) for (lmost) ll x D nd ll ξ R n. For L to be elliptic mens tht the symmetric n n mtrix A(x) is positive definite, with smllest eigenvlue λ C. Specil cse: ii = 1, ij = for i j, b i =, c = in (7), in which cse L = 2 =, the negtive Lplcin opertor. Physicl interprettion: Second-order elliptic PDEs re generliztions of Lplce s nd Poisson s equtions. Let us first review briefly some pplictions tht give rise to Lplce s eqution. Typiclly, 9

10 function u will denote the mount or density of some quntity (e.g., temperture, electrosttic potentil, chemicl concentrtion) in equilibrium. Then if V is n rbitrry subregion within D, with smooth boundry V, then the net flux of u through V is zero, i.e. F n ds =, (74) V where F denotes the flux density (discussed below) nd n the unit outer norml vector field to S. From the Divergence Theorem, F n ds = div F dx =, (75) implying tht V V div F = F = in D, (76) since V ws rbitrry. In mny pplictions, it is resonble ssumption tht the flux F is proportionl to the grdient u, but pointing in the opposite direction, since the flow will be from regions of higher concentrtion to those of lower concentrtion. In other words, Substitution into (76) yields Lplce s eqution F = K u, K >. (77) ( u) = 2 u =. (78) Some exmples: u flux lw in Eq. (77) chemicl concentrtion Fick s lw of diffusion temperture Fourier s lw of het conduction electrosttic potentil Ohm s lw of electricl conduction A clssicl exmple in electrosttics comes from the fundmentl eqution, div E(x) = ρ(x) ǫ, (79) where E(x) denotes the electric field t x R n due to chrge density ρ(x). (Here, ǫ is the permittivity of the vcuum.) Since E = V, (8) where V denotes the ssocited electrosttic potentil function, we hve 2 V = ρ ǫ, (81) or Poisson s eqution. Of course, in the bsence of chrge, this eqution reduces to Lplce s eqution. In the more generl cse, i.e., the opertor in Eq. (7), the second-order terms involving the ij represents diffusion within region D the coefficients ij describe the nisotropic, heterogeneous nture of the medium. The first-order terms involving the b i represent trnsport within D. The zeroth-order term cu describes locl cretion or depletion (for exmple, in chemicl pplictions, due to rections tht either produce or consume the chemicl). 1

11 In wht follows, it will be ssumed tht ij (x), b i (x), c(x) C(D). (82) This ssumption could be relxed even further to L (D). Furthermore, we ssume tht f L 2 (D). (83) We work in the Hilbert spce of functions H 1 (D) = W1,2 (D). We now multiply both sides of the eqution Lu = f by function v H 1 (D) nd integrte the first term by prts to obtin ij u xi v xj + b i u xi v +cuv dx = fv dx, (84) D i,j=1 D where Eq. (84) now hs the form u xi = xi u, etc.. (85) B(u,v) = F(v), for u,v H 1 (D), (86) where F(v) = f,v = fv dx (87) D is the liner functionl nd B(u,v) = ij u xi v xj + b i u xi v +cuv dx (88) D i,j=1 is the biliner form ssocited with the divergence form elliptic opertor L defined in (7). We now sy tht u H 1 (D) is wek solution of the boundry-vlue problem (69) if for ll v H 1 (D). B(u,v) = F(v) (89) Theorem 3 (Energy estimtes) There exist constnts α,β > nd γ such tht B(u,v) α u H 1 (D) v H 1 (D) (9) β u H 1 (D) B(u,u)+γ u L 2 (D). (91) Proof: See Prtil Differentil Equtions, by L.C. Evns, AMS (1998), pp Notethtifγ > intheboveestimtes, thenb(,)doesnotpreciselystisfythesecondhypothesis of the Lx-Milgrm theorem. A slight tinkering must be performed see the book by Adms, pp In the cse of the Lplcin opertor, the bove theorem (Energy estimtes) holds true for γ =. 11

12 Second-order prbolic PDEs We simply mention briefly tht the ide of wek solutions my be pplied PDEs tht involve time, often referred to s PDE evolution equtions. Second-order prbolic PDEs re generliztions of the het eqution, involving first-order time derivtive. (This is in contrst to hyperbolic equtions tht involve second-order time derivtives, e.g., the wve eqution.) In fct, the wek solution pproch provides the bsis of the so-clled Glerkin s method of computing pproximtions to these equtions. In wht follows, we ssume D to once gin be n open, bounded subset of R n nd define D T = D (,T] for some fixed time T >. We now consider the initil/boundry-vlue problem u t +Lu = f in D T, u = on D [,T], u = g on D {t = }. (92) L denotes for ech time t second-order prtil differentil opertor in divergence form, Lu = xj ( ij (x,t) xi u)+ b i (x,t) xi u+c(x,t)u, (93) i,j=1 We lso ssume tht the opertor L is uniformely elliptic for ech time t [,T], i.e., there exists constnt C > such tht (cf. Eq. (73), ij (x,t)ξ i ξ j C ξ 2. (94) In this cse, one sys tht the opertor +L is (uniformly) prbolic. t Specil cse: Once gin, ii = 1, ij = for i j, b i =, c = in (7), in which cse L = 2 =, the negtive Lplcin opertor, so tht the PDE u t 2 = f (95) becomes the het eqution, with source term f. In physicl pplictions, generl second-order prbolic equtions describe the time-evolution of quntity, e.g., chemicl concentrtion, within region D. Proceeding in mnner quite similr to tht for elliptic equtions, we ssume tht ij (x), b i (x), c(x) C(D), f L 2 (D T ), g L 2 (D). (96) We lso ssume tht ij = ji. As for the elliptic cse, we work in the Hilbert spce of functions H 1 (D) = W1,2 (D). Multiplying the opertor in (93) with v H 1 (D) nd integrting the first term by prts yields time-dependent biliner form B(u,v;t) = ij (,t)u xi v xj + b i (,t)u xi v +c(,t)uv dx, u,v H 1 (D). (97) D i,j=1 12

13 It is then tempting to write down n eqution involving time-vrying biliner nd liner functionls, B nd F, respectively, involving the function u = u(x,t). The stndrd procedure, however, is to ssocite with u mpping u : [,T] H 1 (D) (98) defined by [u(t)](x) := u(x,t) (x D, t T). (99) In other words, u will not be considered s function of x nd t but rther s mpping u of t into the spce H 1 (D) of functions of x. Similrily, one defines f : [,T] L 2 (D) (1) by [f(t)](x) := f(x,t) (x D, t T). (11) We now multiply the PDE u t + Lu = f by fixed function v H 1 (D) nd integrte by prts to produce u,v +B(u,v) = f,v, (12) for ech t T, where the prime represents differentition with respect to time. Definition 2 We sy tht function u L 2 (,T;H 1 (D) (13) is wek solution of the prbolic initil/boundry-vlue problem (92) provided tht for ech v H 1 (D) nd.e. time t T, nd u,v +B(u,v) = f,v, (14) u() = g. (15) (There is nother technicl point regrding the domin of definition of u but we omit it here see Adms, p. 352.) 13