5
$\begingroup$

We have the following PDE: \begin{equation} \frac{\partial p(x,t)}{\partial t}= - a\frac{\partial p(x,t)}{\partial x} + \frac{D}{2} \frac{ \partial^2 p(x,t) }{\partial x^2}, \quad0<x<L, \quad t>0 \end{equation} where $a$ and $D$ are constants. We also have the following initial and boundary conditions: \begin{equation} p(x,0)=0, \hspace{5mm} a p(x,t)- \frac{D}{2} \frac{\partial p(x,t)}{\partial x} = f(t) \quad at \quad x=0, \hspace{5mm} \frac{\partial^2 p(x,t)}{\partial x^2}=0 \quad at \quad x=L \end{equation}

The expression, $\frac{\partial^2 p(x,t)}{\partial x^2}=0$, is not a standard well-known boundary condition and makes the PDE difficult to solve. I would be grateful if anyone have any idea or comment for analytically solving this PDE.

$\endgroup$
3
$\begingroup$

You can reduce this problem to a more familiar one with the following trick. Consider the problem $$ q_t(x,t) = -au_x(x,t) + \frac{D}{2}q_{xx}(x,t) \quad (0 < t, 0 < x < 2L)\\ q(x,0) = 0, \quad aq(0,t) - \frac{D}{2}q_x(0,t) = f(t) = -aq(2L,t) - \frac{D}{2}q_x(2L,t) $$ and solve it with a standard method. The solution will have a specific symmetry about the line $\{x = L\}$ by uniqueness, namely it will satisfy $q(x,t) = -q(2L-x,t)$. This implies that $q_{xx}(L,t) = 0$ for all $t$.

Then set $p(x,t) = q(x,t)$ for $t >0, 0 < x < L$ and you are done.

$\endgroup$
  • $\begingroup$ Thanks for your answer. What if I just put $\frac{\partial^2 p(x,t)}{\partial x^2}=0$ into the PDE to get a first order boundary condition? And which method do you recommend for the reduced problem of yours? Duhamel's theorem? $\endgroup$ – Hossein Jul 30 '15 at 8:07
  • $\begingroup$ @HosseinHosseini - you could try to use separation of variables. $\endgroup$ – Hans Engler Jul 31 '15 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.