Let's say we have the heat equation $\partial_t u(x,t) - \alpha \Delta_x u(x,t)=0$. I wanted just to look at it in one dimension so the equation becomes $\partial_t u(x,t) - \alpha \partial_{xx} u(x,t)=0$. I was wondering how I can solve the equation if $u(x+2\pi,t)=u(x,t)$ with knowing that $u(x,t)=(4\pi\alpha t)^{-\frac{1}{2}}\int_{-\infty}^{\infty}\exp{(-\frac{\lvert x-y\rvert^2}{4\alpha t})}\phi(y)dy$, where $u(x,0)=\phi(x)$. I dont know how to use the periodic property for solving the integral, because as much as I know I can then just trim it into infnitely many integrals which range from $0$ to $2\pi$ for example. Can someone maybe give me a hint for this one?
Periodic heat equation
0
$\begingroup$
integration
periodic-functions
heat-equation
1 Answers
0
The eigenvalues of the second derivative with periodic b.c. on $[0,2\pi]$ are known. Pick an orthonormal basis of eigenvectors for these eigenvalues and then you can express the solution as a Fourier series.