FOURIER METHODS AND DISTRIBUTIONS: SOLUTIONS

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1 Centre for Mathematical Sciences Mathematics, Faculty of Science FOURIER METHODS AND DISTRIBUTIONS: SOLUTIONS. We make the Ansatz u(x, y) = ϕ(x)ψ(y) and look for a solution which satisfies the boundary conditions u(, y) = u(, y) =. Substituting u into Laplace s equation gives and hence ϕ (x)ψ(y) + ϕ(x)ψ (y) = ϕ (x) ϕ(x) = ψ (y) ψ(y) = λ, for some constant λ R (assuming ϕ(x)ψ(y) ). Hence, ϕ = λϕ and ψ = λψ The boundary conditions on the lateral sides give ϕ() = ϕ() =, whence λ = (kπ) 2 for some k =,2,3,... and ϕ(x) = Asin(kπx) for some A R. This in turn gives ψ(y) = B cosh(kπy) +C sinh(kπy) for some constants B and C. Now we look for a solution satisfying the boundary conditions u y (x,) = and u(x,) = x 2 x by making the Ansatz u(x,y) = k= (B k cosh(kπy) +C k sinh(kπy))sin(kπx) (A can be swallowed into B and C). Evaluating at y = gives u y (x,) = k= so that C k = for each k. Evaluating at y = gives u(x,) = k= kπc k sin(kπx) =, B k cosh(kπ)sin(kπx) = x 2 x. In order to find B k, we expand the function x 2 x in a sine series. We have that where c k = 2 = 4 k 2 π 2 x 2 x = k= (x 2 x)sin(kπx)dx = 2 kπ sin(kπx) dx = c k sin(kπx), { 8 k 3 π 3, (2x )cos(kπx)dx k odd,, k even. Please, turn over!

2 It follows that u(x,y) = j= 8 sin((2 j + )πx)cosh((2 j + )πy) (2 j + ) 3 π 3. cosh((2 j + )π) The calculations above were formal, but using Weierstrass M-test, one sees that the series converges uniformly on the whole square and that one can differentiate it once termwise on the square [,] [,] and infinitely many times on the square [,] [,) (note that cosh((2 j + )πy)/ cosh((2 j + )π) exponentially fast for y [, )). Thus the series defines a function u C ([,] [,)) C ([,] [,]) which solves Laplace s equation and the boundary conditions. Note that we cannot hope to do much better since if u were C 2 on the whole closed square, we would get the contradiction u xx (,) = 2 = u yy (,) =, where the boundary conditions and Laplace s equation have been used. 2. We first write Laplace s equation in polar coordinates: 2 u r 2 + u r r + 2 u r 2 θ 2 =. Making the Ansatz u(r,θ) = ϕ(r)ψ(θ) leads to the equations and ψ (θ) = λψ(θ) r 2 ϕ + rϕ = λϕ. Assuming that u is -periodic in θ gives λ = k 2, k Z, and ψ k (θ) = e ikθ (this is a convenient way of making sure that each solution is only counted once). Inserting λ = k 2 in the equation for ψ gives r 2 ϕ + rϕ = k 2 ϕ. This is an Euler equation with linearly independent solutions r k, r k if k and, logr if k =. Hence, if we require u to be bounded at the origin, we obtain the solutions u k (r,θ) = r k e ikθ, k Z. We now try to find a solution of Laplace s equation satisying the boundary condition u(,θ) = g(θ) by making the Ansatz This gives so that Hence, Moreover, u(r,θ) = k= k= r k e ik(θ s) = u(r,θ) = k= g(θ) = u(,θ) = c k r k e ikθ. k= c k e ikθ, c k = g(θ)e ikθ dθ. π ( π ) g(s)e iks ds r k e ikθ = π π = k= (re i(θ s) ) k + k= (re i(θ s) ) k = r 2 re i(θ s) 2 = r2 e is re iθ 2. g(s) k= r k e ik(θ s) ds. re i(θ s) + rei(θ s) re i(θ s)

3 Hence, and u(r,θ) = π r 2 g(s) e is re iθ 2, r 2 e is re iθ 2 = K(reiθ,e is ), where K is the Poisson kernel for the unit disc (see Evans Section 2.2.4). The denominator can also be written ( 2r cos(θ s) + r 2 ). 3. Making the Ansatz u(x,t) = ϕ(x)ψ(t) leads to iϕ(x)ψ (t) + ϕ (x)ψ(t) = and hence ϕ (x) ϕ(x) = iψ (t) ψ(t) = λ. The periodic boundary conditions give ϕ(x) = Ae ikx and λ = k 2, k Z (where we ve allowed negative k but haven t written the solutions e ikx just like in the previous exercise in order to count each solution just once). We also obtain ψ(t) = Be ik2t. We now make the Ansatz u(x,t) = k= c k e ik2t e ikx in order to find a solution which also satisfies the initial condition. This gives g(x) = u(x,) = k= c k e ikx so that c k = g(x)e ikx dx π are the usual complex Fourier coefficients of g. Assuming that g is -periodic and smooth, one can integrate by parts arbitrarily many times in the formula for c k and obtain that c k C N ( + k ) N for any N, where C N is a constant depending on N. From this it is easily seen that u defined as above indeed is a solution of the initial/boundary-value problem for the Schrödinger equation. Note also that u(,t) 2 L 2 ( π,π) = k= c k e ik2t 2 = k= c k 2 = g 2 L 2 ( π,π) by Parseval s formula. Alternatively, one can deduce this using the energy method assuming only that u is a smooth solution of the initial/boundary-value problem. The advantage of the latter is that it also gives uniqueness (without assuming that u is given by the above Fourier series formula). 4. Alternative : We use an energy method. Note that d dt u 2 (x,t)dx = = = = u(x,t)u t (x,t)dx u(x,t)u xx (x,t)dx u 2 x(x,t)dx + [u(x,t)u x (x,t)] x=π x= u 2 x(x,t)dx Please, turn over!

4 since u x (,t) = u x (π,t) =. It follows that u 2 (x,t)dx u 2 (,t)dx = for all t [,T ]. Hence, u(x,t). Alternative 2: We use a maximum principle argument. Extend u to an even function on [ π,π] [,T ] by setting u(x,t) = u( x,t) for x [ π,] and then to a -periodic function on R [,T ]. The extended function (also denoted u) belongs to C 2 (R (,T ]) C(R [,T ]) and solves the heat equation with u(x,) =. Since u is bounded (so that u(x,t) Ae a x 2 for some A,a > ), one can in fact deduce directly from Theorem 7 in Evans Section 2.3 that u vanishes. However, one can also see this from the usual maximum principle by regarding the domain [ π, π] with periodic boundary conditions as a compact manifold without boundary (the unit circle). This means that the points ( π,t) and (π,t) for t (,T ) can be regard as interior points. Similarly, the points ( π,t ) and (π,t ) can be regarded as interior points relative to the set R {T }. The details are left to the reader. 5. Note that e x L (R) so that the Fourier transform is well-defined in the classical sense (as an integral). We have F (e x ) = e ixξ e x dx = = ( iξ + + iξ ) = 2 + ξ 2. Fourier s inversion formula tells us that ( ) e x = F 2 + ξ 2 e ( iξ )x dx + e (+iξ )x dx where the right hand side can be interpreted either in the distributional sense or using the L 2 - definition of the inverse Fourier transform. However, since 2 belongs to L (R), the +ξ 2 right hand side also makes sense as a classical Fourier integral. This means that we have the equality e x = 2 + ξ 2 eixξ dξ for each x. Evaluating at x = gives + x 2 dx = + ξ 2 dξ = π (this can of course also be deduced by using the primitive function arctan x). Using Plancherel s theorem, we get that e 2 x dx = 4 ( + ξ 2 ) 2 dξ and simplifying this gives ( + x 2 ) 2 dx = π e 2x dx = π Assuming that u is sufficiently regular, we get that û (ξ,y) = ξ 2 û(ξ,y),

5 where the primes denote derivation with respect to y and the hats Fourier transformation in x. From this we get that û(ξ,y) = a(ξ )e ξ y ξ y + b(ξ )e for some a(ξ ) and b(ξ ). In order to obtain a bounded solution for y > we set b(ξ ) =, in which case a(ξ ) = ĝ(ξ ). Thus, û(ξ,y) = ĝ(ξ )e ξ y, which in turn gives u(x,y) = F (ĝ(ξ )e ξ y ) = g F (e ξ y ). By the previous exercise, we have F (e ξ y )(x) = F (e ξ y )( x) = y F (e ξ ) and hence u(x,y) = π ( x ) = 2 y y( + x2 ) = 2y y x 2 + y 2 2 y (x s) 2 + y 2 g(s)ds. This coincides with Poisson s formula for the upper half plane in Evans. 7. The smoothness follows e.g. from the integral representation ix ξ t ξ 2 u(x,t) = () n/2 ĝ(ξ )dξ R n e Note that we can differentiate under the integral sign as many times as we like since g, and therefore also ĝ, belongs to S (R n ). The boundedness also follows from this formula, or from the solution formula involving the fundamental solution and g(x). The uniqueness follows from Evans, Section 2.3, since u is assumed to be bounded. It s clear that u(,) = g belongs to S (R n ). For t >, û(ξ,t) = e t ξ 2 ĝ(ξ ) is a product of two Schwartz functions and therefore also a Schwartz function. But then u(,t) = F (û(ξ,t)) S (R n ). Finally, u(,t) α,β = sup x R n x α D β x u(x,t) () n/2 F (x α D β x u(x,t)) L (R n ) where the Fourier transformation is with respect to x and can be estimated by a constant multiple of F (x α D β x u(x,t)) = D α ξ (ξ β û(ξ,t)) ( + ξ ) α + β D γ ĝ(ξ ) ξ γ α for t [,T ] (this is not optimal). For each γ, we can estimate the L -norm by ( ) ( + ξ α + β ) D γ ĝ(ξ ) dξ ( + ξ ) n dξ ( + ξ ) α + β +n+ D γ R n ξ R n ξ ĝ(ξ ) L (R n ) C ( + ξ ) α + β +n+ D γ ξ ĝ(ξ ) L (R n ) with C = R n(+ ξ ) n dξ. Finally, we can in a similar way estimate (+ ξ ) α + β +n+ D γ ξ ĝ(ξ ) L (R n ) by a finite number of semi-norms of g. Please, turn over!

6 4 8. Denote the terms in the second series in (.9) by u j (x,t). Since u j L (R 2 ), the π(2 j+) 2 series converges uniformly on R 2 to a continuous function by Weierstrass M-test. Moreover, each function u j (x,t) is a classical solution of the wave equation. It follows by linearity that each partial sum n s n (x,t) = u j (x,t) j= solves the wave equation and that s n (x,t) converges uniformly on R 2 (and hence also in D (R 2 )) to u(x,t). But then for each ϕ D(R 2 ) we have that ( 2 t c 2 2 x )u,ϕ = u,( 2 t c 2 2 x )ϕ = lim s n,( 2 n t c 2 x 2 )ϕ = lim s n,( 2 n t c 2 x 2 )ϕ = lim ( 2 n t c 2 x 2 )s n,ϕ =. In other words, u solves the wave equation 2 t u = c 2 2 x u in the sense of distributions. 9. Suppose that n and C exist. If ϕ m ϕ in S (R d ), then ϕ m ϕ α,β as m for each α and β in N d. Hence u,ϕ m ϕ C ϕ m ϕ n. It follows that u is continuous. Suppose that there are no such numbers. This implies that for each n there is a ϕ n S (R d ) such that u,ϕ n > n ϕ n n. In particular, ϕ n (otherwise both sides would be zero). Now let ψ n = ϕ n /(n ϕ n n ). Then for fixed α and β in N d we have that ψ n α,β ψ n n = n for n α + β. Hence, ψ n in S (R d ). On the other hand u,ψ n > for all n, so u,ψ n. This means that u is not continuous.. a) Taking the Fourier transform of both sides (assuming that u S (R)), we get that (ξ 2 + )û =. Thus and is a particular solution. û = + ξ 2 ( ) u p (x) = F + ξ 2 = 2 e x

7 b) Since Ae x + Be x is a classical solution of the homogeneous equation u + u =, it is clear that u = u p + Ae x + Be x solves the inhomogeneous equation u + u = δ. This u belongs to S (R) if and only if A = B = since e x is exponentially growing as x and e x is exponentially growing as x. Indeed, we can test u against a sequence of nonnegative test functions in D(R) which converges to a test function ϕ in S (R) which vanishes for negative x and equals e x/2 as x + in order to deduce that A =. A similar argument shows that B =. Note that this exercise shows that we will in general get different answers if we consider fundamental solutions in S and D.. We have that ( t 2 x 2 )Φ,ϕ = Φ,( t 2 x 2 )ϕ = 2 = 2 x (ϕ tt (x,t) ϕ xx (x,t))dt dx (ϕ x (s, s) ϕ t (s, s))ds 2 (ϕ x (s,s) + ϕ t (s,s))ds (integrate the first term with respect to t and the second with respect to x and rearrange the terms). Note that ϕ has compact support, so that we don t have to worry about the convergence of these integrals. The last expression can be rewritten as 2 which concludes the proof. 2. It suffices to consider d ds ϕ(s, s)ds 2 d ds ϕ(s,s) = ϕ(,) = δ (,),ϕ, u, u 2,ψ(x,y) since the other version can be treated using symmetry. First note that ψ(x, ) D(R n 2) for every fixed x R n so that the function x u 2,ψ(x,y) is well-defined. Also notice that ψ(x, ) if x is sufficiently large. Hence u 2,ψ(x,y) = for such x, meaning that x u 2,ψ(x,y) has compact support. To show that the function x u 2,ψ(x,y) is continuous, we simply note that ψ(x,y) ψ(x,y) in D(R n 2) as x x (check the details!). Similarly, (ψ(x + he j,y) ψ(x,y))/h x j ψ(x,y) in D(R n 2) as h by the mean value theorem. Hence x j u 2,ψ(x,y) = u 2, x j ψ(x,y) and by the previous argument, the right-hand side is continuous. It follows that u 2,ψ(x,y) C (R n ). Iterating this, one finds that u 2,ψ(x,y) C (R n ) and hence also in D(R n ). The definition u u 2,ψ = u, u 2,ψ(x,y) therefore makes sense. If ψ = ϕ ϕ 2 one obtains u u 2,ϕ ϕ 2 = u, u 2,ϕ (x)ϕ 2 (y) = u,ϕ (x) u 2,ϕ 2 (y) = u,ϕ u 2,ϕ We assume throughout that n 2 (otherwise the definitions lack meaning). After a change of variables, we can assume that the principal symbol has the form p 2 (ξ ) = n σ j ξ j 2 j= Please, turn over!

8 with σ j {,,}. If the operator is hyperbolic according to Definition 4., it follows by Example 4. that it is also strictly hyperbolic in some direction in the sense of Definition 4.9. Let n [ξ,η] = σ j ξ j η j j= so that [ξ,ξ ] = p 2 (ξ ). If P(D) is strictly hyperbolic in the direction ν in the sense of Definition 4.9, then p 2 (ν) = [ν,ν] and in particular all σ j can t be zero. Moreover, ν j for some j such that σ j. Expanding p 2 (ξ + τν) = [ξ,ξ ] + 2τ[ξ,ν] + τ 2 [ν,ν] = we see that no σ j can be zero. Indeed, say for simplicity that σ and ν, while σ 2 =. Then we could take ξ 2 and ξ j = for all other j, giving p 2 (ξ + τν) = τ 2 [ν,ν] with the double zero τ =. It s also clear that not all σ j can have the same sign, since then the equation p 2 (ξ +τν) = would have no real zeros for ξ not parallel to ν. If n = 2 or n = 3 it now follows that P(D) is hyperbolic in the sense of Definition 4.. Finally, assume that n 4 and that at least two σ j are positive and at least two negative. Without loss of generality we can assume that [ν,ν] >, ν and σ = σ 2 =. Otherwise, we change P(D) to P(D) and relabel the coordinates (note that this would not work for n 3). Choosing ξ = (ξ,ξ 2,,...,), with ν ξ + ν 2 ξ 2 =, we then obtain p 2 (ξ + τν) = [ξ,ξ ] + τ 2 [ν,ν] > for all real τ so that P(D) is not hyperbolic according to Definition 4.9. In order to find the characteristic and strictly hyperbolic directions, we assume first that the equation has been transformed into the above form with σ = = σ n = and σ n =, so that the corresponding principal symbol is given by n p 2 (ξ ) = j= ξ 2 j ξ 2 n. Note that this is a quadratic form with n positive squares and one negative square. Sylvester s law of inertia implies that any subspace on which p is negative definite must have dimension one. The same is true for subspaces on which p is negative semidefinite (in fact, it must be negative definite on such a subspace). Write ξ = (ξ,ξ n ) and define [ξ,η] as above, [ξ,η] = ξ η ξ n η n. The condition that ξ is characteristic can be written [ξ,ξ ] =, that is ξ n = ± ξ. This defines a cone in R n. In order for ν to be strictly hyperbolic, we first of all need it to be noncharacteristic, that is, [ν,ν]. This means that ν lies in the set {ξ R n : ξ n ± ξ }. In the case n = 2, this consists of four connected components. In the case n 3 there are only three connected components (draw a picture!). Due to this, the characterization of strict hyperbolicity depends on whether n = 2 or 3.

9 For strict hyperbolicity we also need the equation [ξ + τν,ξ + τν] = to only have simple real roots for every ξ not parallel to ν. Expanding this, we obtain the quadratic equation [ξ,ξ ] + 2τ[ν,ξ ] + τ 2 [ν,ν] =, where [ν,ν]. The discriminant of this second degree equation is given (up to a constant factor) by [ν,ξ ] 2 [ν,ν][ξ,ξ ]. In order for there to be only simple real roots (that is two different real roots), we need the inequality [ν,ν][ξ,ξ ] < [ν,ξ ] 2 for all ξ R n not parallel to ν. This can be seen as a reverse Cauchy-Schwarz inequality. Case. Assume first that n 3. If [ν,ν] > we can choose ξ with [ξ,ξ ] > and [ν,ξ ] = (take ξ n = and ν ξ = ). Hence, ν is not strictly hyperbolic (this is the same argument used above when showing that Definition 4.9 implies Definition 4.). Assume next that [ν,ν] <. If [ξ,ξ ] > we are done, so we can assume that [ξ,ξ ]. We begin by considering the case [ξ,ξ ] <. We claim that the reverse Cauchy-Schwarz inequality holds in this case. The basic idea is to use the projection onto ξ. Let e = [ξ,ξ ] ξ, so that [e,e] =, and set Then ν = [ν,e]e, ν = ν ν. [ν,e] = [ν,e], [ν,e] = [ν,e] [ν,e] =. It follows that [ν,ν ] =. We also have that [ν,ν ] >. Otherwise, we would have a subspace of dimension two (spanned by e and ν ) on which [, ] is negative semidefinite, contradicting Sylvester s law of inertia. We can therefore write ν = λξ +ν where [ξ,ν ] = and [ν,ν ]. Then so since [ξ,ξ ] <, while and hence [ν,ν] = λ 2 [ξ,ξ ] + [ν,ν ] > λ 2 [ξ,ξ ], [ν,ν][ξ,ξ ] < λ 2 [ξ,ξ ] 2, [ν,ξ ] 2 = λ 2 [ξ,ξ ] 2, [ν,ν][ξ,ξ ] < [ξ,ξ ] 2. If on the other hand [ξ,ξ ] =, then we cannot have [ν,ξ ] = since then the quadratic form would be negative semidefinite on the subspace spanned by ξ and ν. But this also implies that [ν,ν][ξ,ξ ] < [ξ,ξ ] 2. Please, turn over!

10 This concludes the proof for n 3. Geometrically, the condition is that ν lies in the forward cone {ξ R n : ξ n > ξ } or the backward cone {ξ R n : ξ n < ξ }. Case 2. Assume that n = 2. In this case, all non-characteristic directions are hyperbolic. To see this, we repeat the last steps of the above proof in the case that [ν,ν] <. Since the condition of hyperbolicity is invariant under the transformation ν = ν ν 2, it also holds if [ν,ν] > (what goes wrong in the counterproof is that it s no longer possible to take ξ 2 = and ν ξ = since we only have two dimensions to play with).