u t = u p (t)q(x) = p(t) q(x) p (t) p(t) for some λ. = λ = q(x) q(x)

Transcription

1 . Separation of variables / Fourier series The following notes were authored initially by Jason Murphy, and subsequently edited and expanded by Mihaela Ifrim, Daniel Tataru, and myself. We turn to the study of initial value problems with boundary conditions: (t, x) (0, ) Ω. The problems have three components: (i) the equation (PDE) (ii) the initial conditions (IC) (t = 0) (iii) the boundary conditions (BC) (x Ω) We focus on the following boundary conditions: Dirichlet: u(t, x) is specified for x Ω Neumann: u(t, x) n(x) is specified for x Ω We focus on homogeneous BCs (that is, u = 0 or u n = 0). The idea of separation of variables is to look for solutions of the form u(t, x) = p(t)q(x). Example. Consider the heat equation (with some BCs): Thus we look for non-trivial p, q so that u t = u p (t)q(x) = p(t) q(x) p (t) p(t) p (t) p(t) In particular, we look for λ and q such that = λ = q(x) q(x) q = λq. This is the eigenfunction/eigenvalue problem for. for some λ. = q(x). q(x) Linear algebra review : Let X be a vector space and A : X X a linear transformation. A nonzero vector v X is an eigenvector of A if there exists λ C such that Av = λv. We call λ the eigenvalue associated to v. However there is an additional twist: the eigenfunction must satisfy the BCs for u to be a solution. Depending on the BCs you may find different eigenvalues/eigenfunctions. Now suppose q λ is an eigenfunction of with eigenvalue λ (satisfying the BCs). Then we get a solution u λ (t, x) = p λ (t)q λ (x) where p λ (t) = e λt. Note u λ satisfies the PDE and BCs. What about the IC, say u(0) = f? In general we won t have f = q λ for any λ. Key observation: λ c λu λ still solves the PDE and BCs. Thus IF we could write then we could solve the PDE/BC/IC by writing f(x) = λ c λ q λ (x), u(t, x) = λ c λ p λ (t)q λ (x). We are led to the following problems:. Solve the eigenvalue problem for with prescribed BCs.

2 2 2. Write functions satisfying BCs as a linear combination of eigenfunctions. If we can solve and 2, then we can use separation of variables to solve the PDE/BC/IC. We restrict attention to the case Ω = [0, L]. A direct computation for Problem (homework ) yields: (i) (Dirichlet) eigenvalues λ n = ( nπ L )2, eigenfunctions q n (x) = sin( nπ L x) (n > 0) (ii) (Neumann) eigenvalues are λ n = ( nπ L )2, eigenfunctions q n (x) = cos( nπ L x) (n 0) We discuss Problem 2. In fact, we will show: Theorem. If F : [, L] C is periodic (that is, F () = F (L)) and smooth, then we can write F (x) in a Fourier series: F (x) = c n e inπ L x. In fact, this will imply (see homework): (i) if f : (0, L) satisfies homogeneous Dirichlet BCs, then f(x) = n= c n sin( nπ L x) (ii) if f : (0, L) satisfies homogeneous Neumann BCs, then f(x) = n=0 c n cos( nπ L x). For (i)/(ii), extend f to (, L) by odd/even reflection and show that the Fourier expansion reduces to a sine/cosine series. We turn to the Theorem. Linear algebra review 2: Let X be a vector space over C. An inner product on X is a function b : X X C such that b(v, w) = b(w, v) b(αv + βw, z) = αb(v, z) + βb(w, z) b(v, v) 0 b(v, v) = 0 = v = 0 Note that the first two properties imply that b(v, αw + βz) = αb(v, w) + βb(v, z). Thus b is linear with respect to the first variable and antilinear with respect to the second. We use the notation v, w = b(v, w). We call v, w orthogonal if v, w = 0. We write v w. Given an inner product,, the induced norm of a vector v X is v := v, v. The norm can be interpreted as a notion of length of a vector. Lemma.. The norm satisfies the following properties: Homogeneity: cv = c v for all c C. Cauchy-Schwartz inequality: v, w v w. Triangle inequality: v + w v + w. Pythagorean theorem: If v and w are orthogonal, then v + w 2 = x 2 + y 2. A set {v n } n Z X is called orthonormal if v n, v m = { n = m 0 n m. A set {v n } X is called an orthonormal basis if it is orthonormal and forms a Schauder basis, i.e. for each f X there exist unique constants c n C such that lim f N N c n v n = 0.

3 3 If this limit holds, we write in fact orthonormality would then imply To see this, fix m, and for each ε > 0 write f = f = c n v n ; c n = f, v n. c n v n + r N, r N < ε, N > m. Taking the inner product of both sides with v m and applying Cauchy-Schwarz, we conclude that f, v m c m = r N, v m r N v m < ε for any ε > 0. Thus If {v n } is an orthonormal basis then for any f X we have f = n f, v n v n. Theorem.2 (Bessel s inequality/parseval s identity). If {v n } n Z is any orthonormal set (not necessarily a basis), then for any f X one has f, v n 2 f 2. (Bessel s inequality) Moreover, equality holds iff {v n } is also a basis (Parseval s identity). Proof. For each N > 0, we observe that 0 f f, v n v n 2 = f 2 f, v n 2, which yields the inequality. Further {v n } is a basis, then upon taking N we obtain 0 lim which is Parseval s identity. N f f, v n v n 2 = f 2 f, v n 2 = 0, Example.. Finite dimensional example. Let X = C n, and let z, w := n j= z iw i. Then, is an inner product, with n z 2 = z, z = z j 2. An orthonormal basis is given by vectors Example.2. The main example. Let For f, g X, define The corresponding norm is j= e = (, 0, 0,... ), e 2 = (0,, 0,... ),..., e n = (..., 0, ). X = {F : [, L] C : F is periodic and continuous}. f, g := L f(x)g(x) dx. ( f := L f(x) 2 dx) /2.

4 4 We reformulate the theorem in terms of vector spaces and inner products. Let e n (x) := e inπx L. A simple computation shows that the set {e n } n Z is orthonormal: e n, e m = L e i(n m)πx L dx = For a function f : [, L] C and an integer n Z, write ˆf(n) := f, e n = L { 0, n m,, n = m. f(x)e inπx L The sequence { ˆf(n)} n Z are the (complex) Fourier coefficients of f. Define the partial Fourier sums S N f by dx. S N f(x) = ˆf(n)e n (x) = f, e n e n (x). Theorem reformulated. The set {e n } n Z, e n (x) = e iπnx L, is an orthonormal basis for X. Thus for every continuous function f : [, L] C with f(l) = f(), the Fourier partial sums S N f = N ˆf(n)e n converge to f in the mean squared sense: ( lim f S Nf = lim N N L f S N f 2 dx) /2 = 0. Further, if f is differentiable at a point x, then S N f(x) converges to f(x): lim f(x) S Nf(x) = 0 for each x (, L). N Finally, if f is smooth on [, L] and its derivatives f (j) are also periodic, then S N f converges to f uniformly on [, L]: lim N max f(x) S N f(x) = 0. x [,L] As an immediate corollary of the first part and Parseval s identity, f 2 = L f(x) 2 dx = In particular, we derive the iemann-lebesgue lemma: lim ˆf(n) = 0 if f X. n ˆf(n) 2 for all f X. emarks. The first part of the theorem and the corollary hold in much greater generality: if f : [, L] C is merely square integrable in the sense that f 2 is integrable on [, L] (i.e. L f 2 dx is well-defined and finite), then f S N f 0 as N. If f X is merely continuous, f S N f 0 does not imply that f(x) S N f(x) 0 for each x; in general, convergence of a sequence of functions in the mean squared sense does not imply converge pointwise. Uniform convergence is the strongest of the three modes of convergence and implies both pointwise convergence and mean squared convergence. Indeed, if f(x) S N f(x) < ε for all x in [, L], then ( L ) ( f(x) S N f(x) 2 dx Proof. We already showed that {e n } n Z are orthonormal. L ) ε 2 dx ε.

5 5 Fix f X. We can write where Then S N f(x) = = D N (x) := Define the Cesáro means where F N (x) = N N L f, e n e n (x) = ( f(y) = f D N (x), ) e n (x y) dy ( L ) f(y)e n ( y) dy ( ) (N+ e n (x) = 2 sin )πx L sin ( ) πx = Dirichlet kernel. σ N f(x) = f N n=0 σ N f(x) := N ( N D n (x) = N N n=0 N n=0 S N f(x). D N ) (x) = f F N, ( ( sin Nπx sin ( π Observe that the Fejér kernel is nonnegative and also that L F N (x) dx = L F n (x) dx = N In fact, F N behaves essentially as an approximate identity: x) )) 2 = Fejér kernel. N n=0 L D n (x) dx =. Theorem.3 (Fejér s theorem). If f X is continuous and -periodic, σ N f converges to f uniformly in [, L]. Proof of Fejér. For fixed f X, choose M > 0 such that f(x) M for all M. We start by proving pointwise convergence, i.e. that σ N f(x) f(x) for each x. Fix x. The continuity of f at x means that for each ε > 0, there exists δ > 0 such that f(x) f(y) < ε whenever x y < δ. Write σ N f(x) f(x) = L [f(x y) f(x)]f N (y) dy. We break the integral into the regions y < δ and δ y L. The integral for small y will be estimated using the continuity of f: f(x y) f(x) F N (y) dy ε F N (y) dy ε. y <δ y <δ For the second integral exploits the fact that most of the mass of F N is concentrated in a O(N ) neighborhood of y = 0. Observe that sin ( π y) c y for some c > 0 whenever y L (so that the argument of sin lies in the interval [ π/2, π/2]), so F N (y) C N y for some constant C, and 2 f(x y) f(x) F N (y) dy M C 2CM dy L N y 2 LδN ε δ y L whenever N 2CM Lεδ. Overall, we conclude that for each ε > 0, σ N f(x) f(x) 2ε for all N sufficiently large. δ y L

6 6 To prove uniform convergence, we just need the property, proved in Math 04, that functions in X are in fact uniformly continuous. This means that for each ε > 0 there exists δ > 0 such that f(x) f(y) for all x, y such that x y < δ. The same argument as above yields, for each ε > 0, a lower bound for N which is independent of x, so that σ N f(x) f(x) 2ε for all x. Now we are ready to show that S N f converges to f in the mean squared sense. For each ε > 0, by Fejér s theorem there exists M such that σ M f(x) f(x) < ε for all x. The computation above shows that σ M f f < ε as well. Then for any N > M, f S N f = (f σ M f) S N (f σ M f) + σ M f S N σ M f. Since σ M f = M m= M a me m (x) for some constants a m C, and a m = σ M f, e m, we see the last two terms cancel. On the other hand, by Bessel s inequality Altogether we see that S N (f σ M f) f σ M f < ε. f S N f < 2ε for all N > M. Using the iemann Lebesgue lemma and a more careful analysis of the Dirichlet kernel, one can show that if f is differentiable at x 0 then S N f(x 0 ) f(x 0 ) pointwise. 2. Fourier transform The theory of Fourier series says that for f : [, L] C periodic and smooth we can write f as a linear combination of waves of frequencies n, f(x) = where the Fourier coefficients are given by f, e n = f, e n e inπ L x, ( ) L f(x)e inπ L x dx. The Fourier transform extends these ideas to the case L : for f : C we define f : C (formally) by f(ξ) = f(x)e ixξ dx. Thus f(ξ) is the Fourier coefficient at a frequency ξ. Question. Can we recover f from f? Do we have an analogue of ( )? Suppose f : C has compact support, say f(x) = 0 for x > M. Choose L > M. Then L f, e n = f(x)e inπ L x dx = π L 2π f(x)e inπ L x dx = π nπ L f( L ). Thus for fixed x, we have f(x) = π L inπ nπ f( L )e L x. Let us write ε = π L and G(y) = f(y)e iyx and send L (that is, ε 0). Then f(x) = εg(εn) G(ξ) dξ = f(ξ)e ixξ dy.

7 7 We arrive (formally) at the Fourier inversion formula f(x) = f(ξ)e ixξ dξ, f(ξ) (2π) = /2 We turn now to the details. Definition. (Schwartz space) f(x)e ixξ dx. S() = {f C () : x k f (l) (x) is bounded for all k, l 0.} If f S then f is absolutely integrable, and so f is a bounded function f : C. By definition, a function f : C is absolutely integrable if f dx := lim f dx <. M [ M,M] In fact we will show f S. Lemma. Let f S(). if g(x) = f (x) then ĝ(ξ) = iξ f(ξ) if g(x) = ixf(x) then ĝ(ξ) = d dξ f(ξ) ix0ξ if g(x) = f(x x 0 ) then ĝ(ξ) = e f(ξ) if g(x) = e ixξ0 f(x) then ĝ(ξ) = f(ξ ξ 0 ) if g(x) = f(λx) for λ > 0, then ĝ(ξ) = f( ξ λ λ ). Proof. If g(x) = f (x), then If g(x) = ixf(x), then ĝ(ξ) = ĝ(ξ) = = [ = iξ e ixξ f (x) dx ] d dx (e ixξ )f(x) dx + [f(x)e ixξ ] x= e ixξ f(x) dx = iξ f(ξ). ixf(x)e ixξ dx = f(x) d dξ (e ixξ ) dx = d dξ f(ξ). emark: We can now see the connection of the F to PDE: it interchanges derivatives and multiplication by x. The last property is one manifestation of the uncertainty principle: the narrower f is (taking λ large), the wider its Fourier transform becomes, and vice versa. Heuristically, this happens because sharply peaked functions contain lots of high frequency/short wavelength components, so its frequency profile (given by its Fourier transform) will have long tails. On the other hand, if f is smooth and spread out in space, it has mostly long wavelengths so its Fourier transform will concentrate at low frequencies. Proposition. If f S then f S. Proof. Using the lemma, for any f S and any k, l 0 we have ξ k( )l d dξ f(ξ) = ĝ(ξ), where g = ( d i dx )k ( ix) l f S. As g S, we have ĝ is bounded. We conclude f S. Thus if we write Ff = f, we have F : S S. In fact: Theorem. F : S S is a bijection. More precisely, if we define F : S S by F f(ξ) = ˇf(ξ) = e ixξ f(x) dx,

8 8 then FF = F F = I. Thus, the Fourier inversion formula holds: f S f(x) = FF f(x) = e ixξ ˆf(ξ) dξ. Lemma. Let f(x) = e x2 /2. Then f S and f = f. Proof. (You check that f S.) We have Thus so Now we note f (x) = xf(x) = i ( ixf(x)) iξ f(ξ) = i d dξ f(ξ). = f(ξ) = e ξ2 /2 f(0). f(0) = d dξ f = ξ f, e x2 /2 dx =. We define K(x) = (2π) /2 e x2 /2. For ε > 0 we define K ε (x) = ε K( x ε ). Then K ε form approximate identities as ε 0, Using (iii): if G ε (x) = K(εx), then Ĝε = K ε. Proof of Theorem. We first show the inversion formula. Let f S, x, and ε > 0. Let g(y) = f(x y). f K ε (x) = f(x y)k ε (y) dy = g(y)ĝε(y) dy = ĝ(y)g ε (y) dy = e ixy f( y)k(εy) dy = f(y)e ixy K( εy) dy. Now send ε 0. Then f K ε (x) f(x), while K( εy) K(0) = (2π) /2. Thus f(x) = f(y)e ixy dy. We can now see that F is a bijection: we define F : S S by F g(x) = g(ξ)e ixξ dξ. The Fourier inversion formula says that F F = Id on S. Combining this with the fact that F f(y) = Ff( y), we can see that F F = Id. We conclude that F = F and F is a bijection on S. We next prove the Plancherel theorem for the Fourier transform. As before, we introduce the L 2 inner product and norm on S() by defining f, g = f(x)g(x) dx for f, g S, f = ( /2 f, f = f(x) dx) 2. Theorem. (Plancherel) For f, g S we have f, g = f, ĝ. In particular f = f. emark. This says that F is a unitary transformation. Proof of Theorem. Essentially the same computation as in the proof the multiplication formula shows that Ff, g = f, F g for all f, g S.

9 9 Hence by the inversion formula we have f, g = f, F Fg = Ff, Fg. /2 Lemma. F(f g)(ξ) = (2π) f(ξ)ĝ(ξ). Proof. F(f g)(ξ) = (2π) /2 (f g)(x)e ixξ dx ( ) = (2π) /2 g(y)e iyξ f(x y)e i(x y)ξ dx dy = f(ξ)ĝ(ξ). We next discuss the Fourier transform in higher dimensions. Let d. A multi-index α is an element of N d, where N = {0,, 2,... }. For α = (α,... α d ), we define α = d i= α i, x α = x α xα d d, α f = Schwartz space: S( d ) = {f C ( d ) : x α β f is bounded for all α, β} For f S we define f(ξ) = (2π) d/2 d f(x)e ix ξ dx. α f x α xα d d Then F is a bijection on S (in fact unitary) and the Fourier inversion formula holds: f(x) = (2π) d/2 d f(ξ)e ix ξ dξ. We have the Fourier transform F : S( d ) S( d ): Ff(ξ) = f(ξ) = (2π) d/2 d f(x)e ix ξ dx. This is a bijection, and F : S( d ) S( d ) is given by In particular we have the Fourier inversion formula: F f(x) = ˇf(x) = (2π) d/2 d f(x)e ix ξ dx. f(x) = (2π) d/2 d f(ξ)e ix ξ dξ. We have also the Plancherel theorem f, g = f, ĝ, f, g = f(x)g(x) dx d and the convolution identities F(f g)(ξ) = (2π) d/2 f(ξ)ĝ(ξ), F ( f ĝ)(x) = (2π) d/2 (f g)(x). Let us finally discuss the application of the Fourier transform to PDEs. Similar to the d case we have the following: if g(x) = α f(x) then ĝ(ξ) = (iξ) α f(ξ) if g(x) = ( ix) α f(x) then ĝ(ξ) = α f(ξ) Here α = (α,..., α d ) N d, with α = d i= α i, x α = x α xα d d, α f = Example. Consider α = (, 0,..., 0). We have F( f x )(ξ) = iξ f(ξ). α f x α. xα d d.

10 0 Similarly with α = (2, 0,..., 0) we get Thus we find F( 2 f x 2 )(ξ) = ξ 2 f(ξ). F( f)(ξ) = ξ 2 f(ξ). Example. (Laplace/Poisson equation) Consider the equation u = f on d. Taking the Fourier transform, we find Thus the solution is given by F( u)(ξ) = f(ξ) ξ 2 û(ξ) = f(ξ) û(ξ) = f(ξ) ξ 2. Evidently the fundamental solution is given by u(x) = F [ f(ξ) ξ 2 ](x) = (2π) d/2 [f F ( ξ 2 )](x). Φ(x) = (2π) d/2 F ( ξ 2 )(x). However, notice that ξ is not a Schwartz function. Thus to make sense of this we need to extend our 2 theory of the Fourier transform. Example 2. (Heat equation) { ut u = 0 (t, x) (0, ) d u(0, x) = f(x) x d. We apply the Fourier transform in the x variables only. We find For each ξ, this is an ODE in t that we can solve: Thus û t (t, ξ) = F( u)(t, ξ) û t (t, ξ) = ξ 2 û(t, ξ). û(t, ξ) = û(0, ξ)e t ξ 2 = e t ξ 2 f(ξ). u(t, x) = F [ f e t ξ 2 ](x) = (2π) d/2 [f F (e t ξ 2 )](x). Evidently (2π) d/2 F (e t ξ 2 )(x) is the fundamental solution, and hence we must have (2π) d/2 F (e t ξ 2 )(x) = (4πt) d/2 e x 2 /2t. In fact, recall that we showed F(e x2 /2 )(ξ) = e ξ2 /2 in d =. In the same way we have F(e x 2 /2 )(ξ) = e ξ 2 /2 in dimension d. Thus F(e x 2 /2t )(ξ) = t d/2 e t ξ 2 /2 We conclude F (e t ξ 2 /2 )(x) = t d/2 e x 2 /2t, which gives ( ). Example 3. (Wave equation) { utt u = 0 (t, x) (0, ) d Taking the Fourier transform in the x variables: For each ξ this is an ODE that we can solve: Imposing the initial conditions: while u(0, x) = f(x), u t (0, x) = g(x) x d. û tt (t, ξ) = ξ 2 û(t, ξ). û(t, ξ) = A(ξ) cos(t ξ ) + B(ξ) sin(t ξ ). f(ξ) = û(0, ξ) = A(ξ), û t (t, ξ) = ξ A(ξ) sin(t ξ ) + ξ B(ξ) cos(t ξ ), ( )

11 so Thus Defining we get ĝ(ξ) = û t (0, ξ) = ξ B(ξ). û(t, ξ) = f(ξ) cos(t ξ ) + ĝ(ξ) sin(t ξ ) ξ [ ] = f(ξ) sin(t ξ ) t ξ W (t, x) = (2π) d/2 F [ sin(t ξ ) ] ξ (x), u(t, x) = [f t W (t, )](x) + [f W (t, )](x). We call W the fundamental solution of the wave equation. + ĝ(ξ) sin(t ξ ). ξ Once again, however, sin(t ξ ) ξ is not a Schwartz function, and hence to make sense of W we need to extend our theory of the Fourier transform. 3. Tempered distributions and the Fourier Transform ecall that a distribution on d is a linear functional F : D( d ) C on the space of test functions which is continuous in the following sense: if ϕ n, ϕ D( d ) are test functions such that ϕ n, ϕ are all supported in some compact set K d, and sup x α (ϕ n ϕ)(x) 0 for each multiindex α, then F (ϕ n ) F (ϕ). The space of distributions on d is denoted by D ( d ). In the language of linear algebra, D is the dual space of D. Intuitively, we can think of a distribution as an object that does not necessarily have a well-defined value at each point x d, but does admit local averages ; for each test function ϕ D we get a measurement F (ϕ) which can be thought of as integrating F against ϕ. If ϕ dx =, then we can regard F (ϕ) as a d weighted average. ecall that when F = l f is the distribution associated to a continuous function f, then F (ϕ) = f(x)ϕ(x) dx. d A tempered distribution on d is a linear functional F : S( d ) C on the space of Schwartz functions, which is continuous in the following sense: if ϕ n, ϕ S( d ) are Schwartz functions such that sup x α β (ϕ n ϕ)(x) 0 for all multindices α, β 0. x then F (ϕ n ) F (ϕ). The set of tempered distributions is a vector space, denoted by S ( d ), which can be thought of as the dual space of S( d ). Just as each test function u D defines an associated distribution l u D, we can embed S into S : for u S we can define l u S by l u (f) = u(x)f(x) dx. d The map u l u is injective, since given l u we can recover the value of u at any point x by testing l u against an approximate identity centered at x (see HW2). Thus, a Schwartz function u and its associated distribution l u are really just different ways of thinking about the same objects. Convention: if u S, we identify u with the distribution l u (ϕ) = u(x)ϕ(x) dx, and use the notation u in both contexts. If necessary, we shall indicate whether we are thinking of u as a function (in terms of its pointwise values) or as a distribution. In fact, we can make this identification u l u S for any absolutely integrable function u, which includes many more functions than just Schwartz functions. Since test functions (smooth with compact support) are also Schwartz, we see that tempered distributions are also distributions: D S S D. However, not every distribution is tempered.

12 2 Example 3.. For any N > 0, u(x) = x N is a tempered distribution on. Indeed, if ϕ S(), then u(ϕ) = x N ϕ(x) dx is well-defined; since for each M > 0 we have a bound ϕ(x) C M ( + x ) M which guarantees the convergence of the integral (for instance by choosing M = N + 2). Example 3.2. u(x) = e x2 is a distribution on, since u(ϕ) = e x2 ϕ(x) dx converges for any ϕ D(). However, it is not a tempered distribution since the integral may not converge if ϕ is merely Schwartz and not compactly supported. For example, ϕ(x) := e x2 /2 is Schwartz but u(x)ϕ(x) dx = ex2 /2 dx does not converge. These two examples illustrate what it means for a distribution to be tempered : since tempered distributions are tested against which merely decay faster than any power of x, they are not allowed to grow arbitrarily fast as x. As with ordinary distributions, we use integration by parts to motivate the definition of derivatives of tempered distributions: α u(x)f(x) dx = ( ) α d u(x) α f(x) dx d for u, f S. So we define: α u(f) := ( ) α u( α f). We can define we can define convolution f u S for f S and u S in two essentially equivalent ways. First, if f, u, g S then (f u)(x) g(x) dx = u(y)f(x y)g(x) dy dx d d d ( ) = u(y) f(y x)g(x) dx dy ( f(x) := f( x)) d d = u(y)( f g)(y) dy. d Thus we define f u as the tempered distribution (f u)(g) = u( f g). Alternatively, we can define f u as a function in a way that parallels the usual convolution of two functions. If f, u S, then f u(x) = u f(x) = u(y)f(x y) dy = u(y)τ x f(y) dy = u(τx f), where τ x f(y) = f(y x) = f(x y). Now if f S and u S, we define f u = u f as the function u f(x) = f u(x) := u(τ x f). One can show that the function f u from the second definition agrees with the tempered distribution f u from the first definition in the sense that they yield the same output when tested against Schwartz functions. Whereas we could multiply general distributions by arbitrary smooth functions, we only multiply tempered distributions by smooth functions that don t grow too quickly.

13 3 A smooth function f is tempered if for each multiindex α 0, there exists N > 0 such that α f(x) C N ( + x ) N. For u S and a tempered smooth function f, we define fu S by [fu](ϕ) = u(fϕ). Just as we can translate functions, if x 0 d, we define the translation of f S by τ x0 f(ϕ) := f(τ x0 ϕ). We now turn to the theory of the Fourier transform on tempered distributions. ecall that for u, f S we have the multiplication identity: û(x)f(x) dx = u(x) f(x) dx. d d Moreover for f S we have f S. This motivates the following: Definition. For u S we define Fu = û : S( d ) C by We also write û = Fu. û(f) = u( f). As F : S S is linear and continuous, we find that û S. Thus F : S S. Moreover, F is continuous on S in the following sense: if u n S is a sequence of tempered distributions which converges to some u S, then û n (ϕ) := u n ( ϕ) u( ϕ) =: û(ϕ) for any ϕ S. This continuity property allows us to evaluate some Fourier transforms of some u S by approximation arguments (see HW0). Similarly we can define F : S S by F u(f) = u(f f). Then F û(f) = û(f f) = u(ff f) = u(f) = F û = u. Similarly FF (u) = u. Thus F = F and F is a bijection on S. Moreover, all of the nice algebraic properties of F continue to hold on S. A. α u = (iξ) α û B. ( ix)α u = α û C. F(f u) = (2π) d/2 fû Moral. You can safely do formal computations with the Fourier transform even if you are not working with Schwartz functions. Proof: for (A) we have: α u(f) = α u( f) = ( ) α u( α f) For (B) we have: = ( ) α u( ( ix) α f) = û((ix) α f) = (ix) α û(f) ( ix) α u(f) = ( ix) α u( f) = ( ) α u((ix) α f) = ( ) α u( α f) = ( ) α û( α f) = α û(f).

14 4 For (C) we first note that for f, g S: F ( f ĝ) = (2π) d/2 fg }{{} as before We also note that for f S: Thus F f = f = = f ĝ = (2π) d/2 F(fg). f = f. F(f u)(g) = (f u)(ĝ) = u( f ĝ) = u( f ĝ) = (2π) d/2 u(f( fg)) = (2π) d/2 û( fg) = (2π) d/2 [ fû](g). Example. Consider δ 0 S defined by δ 0 (f) = f(0). Then δ 0 = (2π) d/2 as a distribution: δ 0 (f) = δ 0 ( f) = f(0) = (2π) d/2 d f(x) dx. So the Fourier transform of δ 0 is a constant. More generally, using (A) we see that the polynomials are given by the Fourier transforms of δ 0 and its derivatives. 4. Fourier Multipliers Definition 4.. Let K S ( d ) be a tempered distribution. The linear map T K S S defined by T K f = f K is called a convolution operator. The distribution K is the convolution kernel of T K. emark. By the discussion following the definition of convolution, T K f can be regarded as either a function or as a tempered distribution, and those interpretations are equivalent. Example 4.. If K S is a function, then T K f is given by the integral T K f(x) = K(x y)f(y) dy. d Definition 4.2. Let m be a tempered distribution on d. The map m(d) : S S defined by i.e. m(d)f = F (m ˆf), is a Fourier multiplier. m(d)f = m ˆf, emark 4.3. Fourier multipliers work by reshaping the frequency profile of the input function. In the language of signal processing, a multiplier m(d) such that m is supported for small or large ξ would be called a low-pass filter or high-pass filter, respectively. Since m(d)f = F (m f) = (2π) d/2 f ˇm, f K = (2π) d/2 K f, we see that convolution operators are also Fourier multipliers and vice versa. Example 4.2. For t > 0, define the map e t f by the rule ê t f(ξ) = e t ξ 2 ˆf(ξ). Thus e t = m(d) is a Fourier multiplier which multiplies the Fourier transform of the input by m(ξ) = e t ξ 2. Since ˇm(x) = (2t) d/2 e x 2 /4t, we also have the convolution representation e t f = (2π) d/2 f ˇm = (2π) d/2 f ˇm(x) = f K t, where K t (x) = (4πt) d/2 e x 2 /4t is the fundamental solution of the heat equation.

15 5 The frequency-space representation of e t makes clear why the solutions to the heat equation are smoother than the initial data. If f is rough then its Fourier transform f(ξ) (which measures its frequency profile ) will decay very slowly as ξ, since it takes lots of high-frequency waves to build sharp spikes in f. For any t > 0, the multiplier e t ξ 2 is rapidly decreasing to 0 as ξ, so e t suppresses high frequency components of its input, thereby smoothing it out. For example, if f = δ 0 then δ 0 = (2π) d/2 (that is, waves at all frequencies contribute equally to build up an infinitely tall and narrow spike at x = 0). For any t > 0, the Fourier transform êt δ 0 = (2π) d/2 e t ξ 2 has Gaussian decay, and indeed (e t δ 0 )(x) = (4πt) d/2 e x 2 /4t is smooth. Example 4.3. Earlier we derived the Fourier transform of the solution to the homogeneous wave equation { ( 2 t )u = 0, (t, x) (0, ) d, finding that u(0, x) = g(x), t u(0, x) = h(x), û(t, ξ) = cos(t ξ )ĝ( ξ ) + sin(t ξ ) ĥ(ξ). ξ Thus u(t, x) can be written in terms of the Fourier multipliers u(t, x) = [cos(t D )g](x) + [ sin(t D ) h](x). D In Problem 4 of HW0, you derive the convolution kernels of the multipliers cos(t D ) and sin(t D ) D case d =. in the