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.


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.

  • $\begingroup$ 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

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