The operator \(Lu=-u''\) on \(L^2(-\infty ,\infty )\) has a Green’s function defined by \(-G''-\lambda G=\delta (x-\xi )\) given by \[ G(x,\xi ;\lambda )=-\frac {e^{i\sqrt {\lambda }|x-\xi |}}{2i\sqrt {\lambda }}. \] Assume the representation \[ -\frac {1}{2\pi i}\int _{C_\infty }G(x,\xi ;\lambda )\,d\lambda =\delta (x-\xi ) \] and show how it leads to the Fourier Transform Theorem that \[ f(x)=\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{-i\mu x}F(\mu )\,d\mu , \qquad F(\mu )=\int _{-\infty }^{\infty }e^{i\mu x}f(x)\,dx. \]
Question
Math 6420 • PDE
Fall 2010 • Question 9