I need some help this exercise:

I have to solve diffusion equation $u_t = Du_{xx}$ with periodic boundary conditions

$u(t,-l) = u(t,l)$ and $u_x(t,-l) = u_x(t,l) $.

The initial value should be a function $f(x) = H \sin(2\pi\nu x)$, and $l=\pi$.

I have two questions:

  1. Is it possible to solve this problem?
  2. When,yes - How to solve it?

Thank you for your help in advance.

  • $\begingroup$ Something is wrong with your formulation. Periodic boundary conditions on $(-\pi,\pi)$ look like $$u(t,-\pi)=u(t,\pi),\quad u_x(t,-\pi)=u_x(t,\pi),$$ and generally not equal to zero! For the 2nd order differential operator, boundary conditions indicated in your post are generally known as Cauchy conditions, while imposing such conditions on the whole boundary (i.e. at $x=\pm\pi$) is known to make the problem overdetermined, i.e. ill-posed. $\endgroup$ – mkl314 Jun 12 '14 at 12:30
  • $\begingroup$ I see... thank you. I changed that. But what if I want the flux to be 0 on the boundary? $\endgroup$ – FloS Jun 12 '14 at 13:42
  • $\begingroup$ Zero flux on the boundary implies $u_x(t,-\pi)=u_x(t,\pi)=0$. The latter is enough for the problem to be well-posed without any other boundary conditions. $\endgroup$ – mkl314 Jun 12 '14 at 14:00
  • $\begingroup$ you mean neumann boundary conditions. when i take them i'm violating my boundary conditions with my initial condition because $f(x)=\sin(x) \rightarrow f'(x)=-\cos(x) \rightarrow f(0)=0$. Or is this wrong? $\endgroup$ – FloS Jun 12 '14 at 15:09
  • $\begingroup$ So, the enigmatic $\nu=\frac{1}{2\pi}$. Of course, initial data $f(x)=\sin{(x)}$ violates compatibility conditions $f'(-\pi)=f'(\pi)=0$. Solution with such initial data cannot be classical. More precisely, the property of a classical solution $u\in C\bigl([0,\infty)\times [-\pi,\pi]\bigr)$ will be lost, though it will stay in $C^{\infty}\bigl((0,\infty)\times (-\pi,\pi)\bigr)$ as a weak solution represented by an infinite not uniformly convergent Fourier series. $\endgroup$ – mkl314 Jun 12 '14 at 17:44

You could solve this problem using the method of separation of variables.

Then the solution has the form: $u(t,x)=X(x)T(t)$.

Replace this at the problem, and then you have to solve two problems, one of $X$ and one of $T$.

  • $\begingroup$ My actual problem is not to solve the equation by separation mehtod, my problem is to fit the initial condition without violating the boundary conditions. $\endgroup$ – FloS Jun 12 '14 at 10:29

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