1
$\begingroup$

I have a system of PDEs

$\frac{\partial f}{\partial x}=1-f$

$\frac{\partial f}{\partial y}=-f$

$f(x,0)=1-e^{-x}$

$f(0,y)=0$

$x>0, y>0$

I've tried to integrate it manually, but the solution I've received is wrong $(1-e^{-x})\cdot e^{-y}$

How to solve this may be with some computer algebra system?

$\endgroup$
3
$\begingroup$

By a bootstrapping argument, you get that, if a solution $f$ exists, then $f \in \mathcal{C}^\infty(\mathbb{R}^2,\mathbb{R})$. Therefore, Schwarz's theorem holds, and you get $\frac {\partial^2 f} {\partial x \partial y} = \frac {\partial^2 f} {\partial y \partial x}$.

In your example, you get $$ \frac {\partial} {\partial y} \left( \frac {\partial f} {\partial x} \right) = \frac {\partial} {\partial y} (1-f) = -\frac {\partial f} {\partial y} = f $$ and $$ \frac {\partial} {\partial x} \left( \frac {\partial f} {\partial y} \right) = \frac {\partial} {\partial x} (-f) = -\frac {\partial f} {\partial x} = f-1. $$ Therefore, you get $f(x,y) = f(x,y) - 1$ for all $(x,y)$, which means that your system unfortunately has no solution.


This reasoning is supported by trying to solve the system.

The equation $\frac {\partial f} {\partial x} + f = 1$ yields $f(x,y) = 1 + g(y) e^{-x}$. Evaluating at $x=0$ yields $g(y) = -1$. Therefore, $f(x,y) = 1 + e^{-x}$, which is a contradiction since $\frac {\partial f} {\partial y} + f$ should vanish.

$\endgroup$

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.