Ngô Quốc Anh

March 26, 2009

Solving initial value problem for heat equation via Fourier transform

Filed under: Các Bài Tập Nhỏ, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 21:05

Followed by Solving initial value problem for wave equation via Fourier transform, in this topic I will show you how to solve initial value problem for heat equation via Fourier transform.

Following is the problem

\displaystyle\left\{\begin{gathered} {u_t} = {c^2}{u_{xx}}, \quad x \in \mathbb{R}, \hfill \\ u(x,0) = \varphi (x). \hfill \\ \end{gathered}\right.

From the equation by taking Fourier transform to the both sides, we obtain

\displaystyle\widehat{{u_t}\left( {\eta ,t} \right)} - c^2 {\left( {i\eta } \right)^2}\widehat{u\left( {\eta ,t} \right)} = 0.

This is an ODE, the solution is given by

\displaystyle\widehat{u\left( {\eta ,t} \right)} = C(\eta)e^{-c^2 \eta^2 t}.

From the initial date we get

\displaystyle\widehat{u\left( {\eta ,0} \right)} = \widehat{\varphi \left( \eta \right)}

which implies that

\displaystyle\widehat{\varphi \left( \eta \right)} = C(\eta).

Thus, we obtain

\displaystyle\widehat{u\left( {\eta ,t} \right)} = \widehat{\varphi \left( \eta \right)}e^{-c^2 \eta^2 t}.

Finally, we obtain by taking the inverse Fourier transform

\displaystyle u\left( {x,t} \right) = \frac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^\infty {{e^{i\eta x}}\widehat u\left( {\eta ,t} \right)d\eta } = \frac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^\infty {{e^{i\eta x - {c^2}{\eta ^2}t}}\hat \varphi \left( \eta \right)d\eta }.

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