Outline. Math Partial Differential Equations. Fourier Transforms for PDEs. Joseph M. Mahaffy,

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1 Outline Math 53 - Partial Differential Equations s for PDEs Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu Department of Mathematics and Statistics Dynamical Systems Group Computational Sciences Research Center San Diego State University San Diego, CA Inverse Spring 9 Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (/) Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (/) Heat Equation: Consider the PDE on an infinite domain: with IC Additionally, the IC satisfies: The BCs are u t k u, t >, < <, u(, ) f(). lim. ± lim u(, t). ± As before, we use separation of variables: From the PDE, we obtain φ()g (t) kφ ()g(t), u(, t) φ()g(t). The eigenvalue problem becomes: or d φ + λφ, φ(± ) <. d g kg φ φ λ. This has bounded solutions for λ. In particular, if λ ω, then φ() c cos(ω) + c sin(ω). Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (3/) Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (4/)

2 From before, any λ solves the eigenvalue problem, so we obtain a continuous spectrum for λ. The solution to the t-dependent equation is: g(t) e kωt. Superposition principle: Since the eigenvalues form a continuous spectrum, the superposition principle requires integration over the continuous spectrum, rather than an infinite sum. The solution becomes: u(, t) [A(ω) cos(ω) + B(ω) sin(ω) e kωt dω. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (5/) It remains to show this satisfies the IC, so u(, ) [A(ω) cos(ω) + B(ω) sin(ω) dω. It remains to show there eist A(ω) and B(ω), which are valid for most functions, f(). Comple eponentials: Recall that Euler s formula gives: cos(ω) eiω + e iω and sin(ω) eiω e iω, i so comple solutions are linear combinations of comple eponentials. An alternate way to write the solution is: u(, t) c(ω)e iω e kωt dω. Convention uses e iω with ω being the wave number. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (6/) : Consider the periodic function:, < <, f (), < <,, < <, f ( + n) f () for all integers n, creating a -periodic function. : The limiting case is given by: {, < <,, otherwise. The functions above are even, so the Fourier series contains only cosine terms, so f () a + n a n cos ( ) The Fourier coefficients are: f () f 4 () f 8 () a d and a n cos ( ) d nπ sin ( ) nπ Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (7/) Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (8/)

3 Fourier coefficients: The sequence of Fourier coefficients is called the amplitude spectrum of f () because a n is the maimum amplitude of a n cos ( ).5 n n 3 nπ n 5 n n n 6 nπ n n n 4 n nπ n n The amplitude spectrum becomes denser as increases. Thus, the discrete system approaches the continuous system. : The Fourier series for f () is f () a + n With the Fourier coefficient formulas, f () (a n cos(ω n ) + b n sin(ω n )), ω n nπ f (v) dv + + sin(ω n ) Define ω ω n+ ω n (n+)π n f (v) sin(ω n v) dv [ cos(ω n ) f (v) cos(ω n v) dv. nπ π, so ω π. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (9/) Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (/) to With the information above and for all finite, we have f () f (v) dv + [ cos(ω n ) ω π n + sin(ω n ) ω f (v) sin(ω n v) dv. et, then lim f (). It is plausible (assuming f() absolutely integrable) that [ cos(ω) f(v) cos(ωv) dv + sin(ω) π Definition (Absolutely Integrable) A function f() is absolutely integrable if the limits eists for lim a a f() d + lim b b f() d. f (v) cos(ω n v) dv f(v) sin(ωv) dv dω. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (/) In the limiting case, the Fourier series naturally transformed to the Fourier Integral: π A(ω) π [ cos(ω) f(v) cos(ωv) dv + sin(ω) f(v) sin(ωv) dv dω, [A(ω) cos(ω) + B(ω) sin(ω) dω, () Theorem () f(v) cos(ωv) dv and B(ω) π f(v) sin(ωv) dv. If f() is piecewise smooth in every finite interval and f() is absolutely integrable, then f() can be represented by a Fourier integral (3). At a point of discontinuity the value of the Fourier integral equals the midpoint of the left and right hand limits of f() at that point. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (/)

4 - Pair Inverse : Consider the eample: {, <,, >. The Fourier integral representation is: and A(ω) π B(ω) π π [A(ω) cos(ω) + B(ω) sin(ω) dω, f(v) cos(ωv) dv π f(v) sin(ωv) dv π cos(ω) sin(ω) ω cos(ωv) dv sin(ω) πω sin(ωv) dv., <, dω,,, >. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (3/) We omitted the comple Fourier series earlier, but it satifies: f( + ) + f( ) The function f() is -periodic with n c n e i/ c n f()e i/ d. For periodic functions, < <, the allowable wave numbers ω are ω nπ π n, the wave lengths are, which are integral partitions of the region length n. The distance between successive values of the wave number is: ω ω n+ ω n (n + )π nπ π Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (4/) Inverse Inverse It follows that the comple Fourier series can be written: ( f( + ) + f( ) ) ω f(s)e iωs ds e iω. π n is the limiting form as. The values ω are the square root of the eigenvalues, and as, the eigenvalues get closer together, approaching a continuum. Definition ( Identity) f( + ) + f( ) [ f(s)e iωs ds e iω dω. π Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (5/) Definition () It follows that F (ω) π f( + ) + f( ) f(s)e iωs ds. F (ω)e iω dω. Note: Different authors do different things with the watch the definitions carefully. π factor, so If f() is continuous, then the Fourier integral representation of f() is F (ω)e iω dω. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (6/)

5 Inverse Inverse Inverse Definition ( Pair) If f() is continuous, then the Fourier integral representation of f() is F (ω)e iω dω, F (ω) f(s)e iωs ds π The two equations above are called the pair. This relationship shows that f() is composed of waves e iω for all wave numbers and wave lengths. The pair is valid provided f() is absolutely integrable: f() d <. Generally, we also want f() to also be piecewise smooth, but this condition can be relaed. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (7/) The Gaussian function often arises from the diffusion operator: G(ω) e αω. The function whose Fourier transform is G(ω): or g() G(ω)e iω dω g() α α.5.5 ω.5.5 π α e /4α. e αω e iω dω Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (8/) Inverse It follows that the inverse of a Gaussian is itself a Gaussian: Table F (ω)e iω dω F (ω) π f()e iω d e β F (ω) 4πβ e ω /4β π α e /4α Derivation: Consider: The derivative is: g() e αω e iω dω. dg d iωe αω e iω dω. F (ω) e αω Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (9/) Inverse Integration by parts with vanishing at the endpoints: dg d α The solution of this ODE is e αω e iω dω α g(). g() g()e /4α, g() et z αω (or dz αdω), so g() α π e z dz α. e αω dω. Note: There are two ways to solve this (one involves comple variables): ( e dz) z e ( +y ) d dy π re r dr dθ π. Joseph M. Mahaffy, jmahaffy@mail.sdsu.edu PDEs - s (/)