# Solving system of PDEs

I am stuck on the following problem: Solve for $f(x,y)$, where:

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

$\frac{\partial f}{\partial x} = \frac{1}{2}xy$

My original strategy was to integrate the first equation giving:

$\frac{y^2}{2} + g(x)$, where $g(x)$ is a constant function, and then differentiating this and setting it equal to the second equation giving:

$g'(x) = \frac{1}{2}xy$, which I integrate again to give:

$g(x) = \frac{1}{4}x^2y + h(y)$, giving:

$f(x,y) = \frac{y^2}{2} + \frac{1}{4}x^2y + h(y)$,

but, where do you go from here? Any suggestions would be appreciated!

Thanks!

• You didn't carry the negative sign when differentiating wrt x, also the last step should contain an arbitrary constant – Triatticus Jul 8 '16 at 3:31
• Triatticus - Hi. Thanks. I have updated it now. I don't know where one would go from here! – Thomas Moore Jul 8 '16 at 3:37
• Are you sure about the values of $f_y$ and $f_x$? Something seems to be inconsistent. – okrzysik Jul 8 '16 at 3:38
• You don't need a constant dependent on y you've already eliminated that choice from the start, after the last step the only constant necessary is independent of x and y so C. Your answer is correct if it just has a C at the end – Triatticus Jul 8 '16 at 3:39
• Since $\dfrac{\partial^2 f}{\partial x\partial y} \ne \dfrac{\partial^2 f}{\partial y\partial x}$, there is no (reasonable) solution. – Ted Shifrin Jul 8 '16 at 3:52

## 1 Answer

Not in any strong sense, suppose your function $f$ is twice times continuously differentiable, you can use Schwarz's theorem, i.e. \begin{equation} \frac{\partial^{2}f(x,y)}{\partial x\partial y}\equiv \frac{\partial^{2}f(x,y)}{\partial y\partial x} \end{equation} the partial derivatives of second order commute, and would end up with the condition \begin{equation}\frac{1}{2}x\equiv 0\end{equation} which is impossible. Notice, that your question is strongly related to the question if there exists a potential with gradient $(\tfrac{1}{2}xy,y)$ and then the condition I gave would be called the integrability condition...

• Hello. Thank you. That is much clearer now! – Thomas Moore Jul 8 '16 at 12:52