I have the following multivariable function:

$u = x + y + F(xy)$

where F is an arbitrary differentiable function. I would like to construct a first order PDE with u(x,y) as the following solution. Initially I thought of taking $u_x$ and $u_y$ which is easy for the $x+y$ term of $u$. However, what can I do about the $F(xy)$ term? I know that it initially turns out as a solution to characteristic equations, but how do I reverse engineer a first order PDE that has $F(xy)$ as a solution? Would it require me to arbitrarily define a function of $F(xy)$ such as $F(xy) = xy$ or $F(xy) = sin(xy)$, etc. then to take arbitrary derivatives of $u_x$ and $u_y$?

  • $\begingroup$ Do you know anything about $\dfrac{\partial F}{\partial x}$ or $\dfrac{\partial F}{\partial y}$? If not, we can do a little with the chain rule, but not much. $\endgroup$ – Eric Towers Apr 19 '16 at 3:18

If you know the "method of characteristics", take it backwards :

We know the explicit solution : $\quad u-x-y=F(xy)\quad $ any differenciable function $F$ of one variable.

The implicit solution is : $\quad \Phi(xy\:,\:u-x-y)=0\quad $ any differentiable function $\Phi$ of two variables.

So, the characteristic equations are : $\begin{cases} xy=c_1 \\ u-x-y=c_2 \end{cases}\quad\to\quad \begin{cases} x\:dy+y\:dx=0 \\ du=dx+dy \end{cases}$

From $x\:dy+y\:dx=0$ : $$\frac{dx}{x}=\frac{dy}{-y}=\frac{dx+dy}{x-y}$$ And with $du=dx+dy$ : $$\frac{dx}{x}=\frac{dy}{-y}=\frac{u}{x-y}$$ The corresponding PDE is : $$x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}=x-y$$

  • $\begingroup$ Ah, going in reverse makes a lot of sense. Thank you very much! $\endgroup$ – Alvin Nunez Apr 21 '16 at 0:10

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.