Problem: consider the following PDE: $$-u_t=\mbox{sign}(u) u_x+ \frac{1}{2}u_{xx},$$ with some boundary condition $u(T,x)=\delta_a(x)-\delta_{-a}(x)$, $a>0$ fixed, being $\mbox{sign}(u)\in \{-1,1\}$ the sign of $u$ and $\delta$ the Dirac-function. The final condition could also be taken general $u(T,x)=\Phi(x)$ with $\Phi$ regular if it helps.

Here, $u$ is a function $u:[0,T]\times \mathbb{R}\rightarrow \mathbb{R}$.

Observe that if we change $\mbox{sign}(u)$ by simply a function $f(x)$ of $x$, this would make things easier or even if $f$ was constant, a semi-explicit solution using series could be found. However, $\mbox{sign}(u)$ is in some sense "simple" since it just changes sign from -1 to 1 and so on.

Question: Could there be a chance to find a (semi-explicit) solution or at least, properties of the sign of $u$? i.e. the regions where $u$ is positive, negative, etc. Is there any trick one could use for this specific kind of PDE?

Thanks for any tips or ideas!

  • $\begingroup$ For $u(T,x)=\delta_y(x)$ the solution $u$ will be the fundamental solution of the corresponding parabolic equation $u_t=u_x+ \frac{1}{2}u_{xx}$ with unversed time since it is positive for $t>0$. $\endgroup$ – Andrew May 13 '16 at 20:05
  • $\begingroup$ Yes you are right, I thought about that as well, if you start positive then you remain positive and then the equation can be reduced to the heat equation by tranforming the function. Nevertheless, I have a more complex starting condition, namely $u(T,x)=\delta_{a}-\delta_{-a} $, $a>0$ fixed. Just imagine we approach this initial condition by a function $\Phi$ which is positive on the right, negative on the left and $\Phi(0)=0$. Would a similar argument work then as well? $\endgroup$ – Martingalo May 13 '16 at 20:58
  • $\begingroup$ Derivatives $u_{xx}$ and $u_t$ cannot be both continuous at points where $u=0$ and $u_x\ne0$. How the equation is understood then? $\endgroup$ – Andrew May 13 '16 at 21:43
  • $\begingroup$ Hopefully $u=0$ on a null measure set and the equation might be understood in a weak sense indeed, due to the discontinuities popping up from $\mbox{sign}(u)$. $\endgroup$ – Martingalo May 13 '16 at 21:45
  • $\begingroup$ Let $G(x,y,t)$ be the Green's function of the first BVP $u_t=u_x+ \frac{1}{2}u_{xx}$, $x>0$, $u|_{x=0}=0$. Put $u(x,t)=G(x,a,t)$ for $x\ge0$ and Put $u(x,t)=-G(-x,a,t)$ for $x<0$. Then $u$ satisfy the initial condition. Also $u$ is positive for $x>0$, $t>0$; $u_{xx}$ and $u_t$ change sign after the mapping $x\to-x$, but $u_x(-x,t)=u(x,t)$. So $u$ satisfy the equation when $x\ne0$. But $u_{xx}$ is not continuous on the line $x=0$. $\endgroup$ – Andrew May 13 '16 at 22:09

Green's function generally is defined in the corresponding domain only. So what I meant can be written as $u(t,x)=G^+(x,a,t)1_{\{x>0\}}-G^-(-x,a,t)1_{\{x<0\}}$, where $G^+(x,a,t)$ is the solution of $−u_t=u_x+1/2u_{xx}$, $x>0$ and $u(T,x)=δ_a$, $u|_{x=0}=0$; $G^−(x,a,t)$ is the solution of $−u_t=-u_x+1/2u_{xx}$ and $u(T,x)=δ_a$, $u|_{x=0}=0$.

The solution $u$ and its derivatives $u_x$ and $u_t$ are continuous for $t<T$ but, as I've mentioned in the comments, $u_{xx}$ is not continuous on the line $x=0$.


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.