The Fourier Transform Method R. C. Trinity University Partial Differential Equations April 22, 2014
Recall The Fourier transform The Fourier transform of a piecewise smooth f L 1 (R) is ˆf(ω) = F(f)(ω) = 1 f(x)e iωx dx, and f can be recovered from ˆf via the inverse Fourier transform Remarks: f(x) = F 1 (ˆf)(x) = 1 ˆf(ω)e iωx dω. See Appendix B1 for a table of Fourier transform pairs. The Fourier transform can help solve boundary value problems with unbounded domains.
Fourier transforms of two-variable functions If u(x,t) is defined for < x <, we define its Fourier transform in x to be û(ω,t) = F(u(x,t))(ω) = 1 u(x,t)e iωx dx. Because the Fourier transform treats t as a constant, we have ( n ) u F x n = (iω) n F(u) = (iω) n û and ( n ) u F t n = 1 = n t n ( 1 n u t n(x,t)e iωx dx u(x,t)e iωx dx ) = n t nf(u) = n û t n.
Example Solve the 1-D heat equation on an infinite rod, u t = c 2 u xx, < x <, t > 0, u(x,0) = f(x). We take the Fourier transform (in x) on both sides to get û t = c 2 (iω) 2 û = c 2 ω 2 û û(ω,0) = ˆf(ω). Since there is only a t derivative, we solve as though ω were a constant: û(ω,t) = A(ω)e c2 ω 2 t ˆf(ω) = û(ω,0) = A(ω).
To solve for u, we invert the Fourier transform, obtaining u(x,t) = 1 = 1 û(ω,t)e iωx dω ˆf(ω)e c2 ω 2t e iωx dω. This expresses the solution in terms of the Fourier transform of the initial temperature distribution f(x). Remark: By writing ˆf as an integral, reversing the order of integration, and using the (inverse) Fourier transform of e c2 ω 2t, one can show that in fact u(x,t) = 1 2c πt f(s)e (x s)2 /4c 2t ds.
Example Solve the boundary value problem u t = tu xx, < x <, t > 0, u(x,0) = f(x), which models the temperature in an infinitely long rod with variable thermal diffusivity. Taking the Fourier transform (in x) on both sides yields û t = t(iω) 2 û = tω 2 û, û(ω,0) = ˆf(ω). The ODE in t is separable, with solution û(ω,t) = A(ω)e t2 ω 2 /2 ˆf(ω) = û(ω,0) = A(ω).
As before, Fourier inversion gives u(x,t) = 1 ˆf(ω)e t2 ω 2 /2 e iωx dω. In comparison with the preceding example, this decays more rapidly as t increases. This is is physically reasonable, since the thermal diffusivity is increasing with t. Remark: Notice that this is the solution of the previous example, with t 2 /2 replacing c 2 t. Using the previous remark, this means u(x,t) = 1 t f(s)e (x s)2 /2t 2 ds.
Example Solve the third order mixed derivative boundary value problem u tt = u xxt, < x <, t > 0, u(x,0) = f(x), u t (x,0) = g(x). Taking the Fourier transform (in x) on both sides yields û tt = (iω) 2 û t = ω 2 û t, û(ω,0) = ˆf(ω), û t (ω,0) = ĝ(ω) Solving the ODE it t for û t gives û t (ω,t) = A(ω)e ω2 t û(ω,t) = A(ω) ω 2 e ω2t +B(ω) = A(ω)e ω2t +B(ω).
Imposing the initial conditions we find that ˆf(ω) = û(ω,0) = A(ω)+B(ω) ĝ(ω) = û t (ω,0) = ω 2 A(ω) A(ω) = ĝ(ω) B(ω) = ω 2 ˆf(ω)+ ĝ(ω) ω 2. Plugging these into û and applying Fourier inversion yields u(x,t) = 1 ( ĝ(ω) ω 2 e ω2t + ˆf(ω)+ ĝ(ω) ) ω 2 e iωx dω = 1 ( ˆf(ω)+ ĝ(ω) ) ω 2 (1 e ω2t ) e iωx dω = f(x)+ 1 ĝ(ω) ω 2 (1 e ω2t )e iωx dω.
Example Solve the boundary value problem t 2 u x u t = 0, < x <, t > 0, u(x,0) = f(x), and express the solution explicitly in terms of f. Taking the Fourier transform (in x) on both sides yields t 2 (iω)û û t = 0, û(ω,0) = ˆf(ω). The ODE in t is separable, with solution û(ω,t) = A(ω)e it3 ω/3 ˆf(ω) = û(ω,0) = A(ω).
Using Fourier inversion leads to u(x,t) = 1 = 1 = f (x + t3 3 ˆf(ω)e it3 ω/3 e iωx dω ˆf(ω)e iω(x+t3 /3) dω ). Remark: This particular problem is amenable to the method of characteristics, although the Fourier transform method may seem somewhat more straightforward.
Example Solve the Dirichlet problem in the upper half-plane 2 u = u xx +u yy = 0, < x <, y > 0, u(x,0) = f(x), which models the steady state temperature in a semi-infinite plate. Taking the Fourier transform (in x) on both sides yields (iω) 2 û +û yy = û yy ω 2 û = 0, û(ω,0) = ˆf(ω). The ODE in y has characteristic equation r 2 ω 2 = 0 r = ±ω û(ω,y) = A(ω)e ωy +B(ω)e ωy.
We now require that û(ω,y) remain bounded as y. Consequently, } ω > 0 A(ω) = 0 û(ω,y) = C(ω)e y ω ω < 0 B(ω) = 0 Fourier inversion then gives the result u(x,y) = 1 = 1 ˆf(ω) = û(ω,0) = C(ω) ˆf(ω)e y ω e iωx dω f(s)e iω(s x) ds e y ω dω }{{} sub. z=s x = 1 f(x +z)e iωz dz e y ω dω.
Now reverse the order of integration to obtain u(x,y) = 1 = 1 f(x +z) e y ω e iωz dωdz f(x +z)f ω (e y ω )(z)dz, where we have written F ω to indicate the Fourier transform in ω. Recall that 2 F(e x 1 ) = π 1+ω 2 F(g(ax)) = 1 ( ω ) a ĝ a F(e a x ) = 2 π a a 2 +ω 2.
Setting a = y and ω = z, it finally follows that u(x,y) = y π f(x +z) y 2 +z 2 dz }{{} sub. s=x+z = y π which is known as the Poisson integral formula. f(s) y 2 +(x s) 2 ds, Remark: The simplification technique above can be generalized by introducing the convolution of functions g and h: (g h)(x) = 1 See Sections 7.2, 7.4 and 7.5. g(x t)h(t)dt.