2

I am struggling to understand conceptually why we are allowed to fix a variable to solve PDE's and still get out a general solution, take the following example for instance.

Say we have a partial differential equation of the form

$$ U_{x} + 2U_{y} + (2x-y)U = 2x^2 + 3xy - 2y^2$$ we can use the following change of coordinates $$\widetilde{x}=x+2y \\ \widetilde{y}=2x-y$$

and we are left with

$$5U_\widetilde{x} + \widetilde{y}U=\widetilde{x}\widetilde{y}$$

$\textbf{We can now solve this by fixing $\widetilde{y}$ to get}$ $$U(\widetilde{x},\widetilde{y})= (\widetilde{x} - \frac{5}{\widetilde{y}})+\exp(\frac{-\widetilde{x} \widetilde{y}}{5})f(\widetilde{y})$$

Whice when reverted back to original coordinates gives

$$U(x,y) = (x+2y - \frac{5}{2x-y})+e^{-x-2y}g(2x-y)$$

Where I struggle conceptually is trying to understand why we are allowed to fix $\widetilde{y}$? Further why is the solution then applicable to all $\widetilde{y}$ how comes the solution isn't limited to only fixed $\widetilde{y}$ values,since that's what we used to get the solution?

seraphimk
  • 573

1 Answers1

3

Strictly speaking you're not fixing $y$, but you're using an integrating factor to integrate a partial derivative with respect to $x$.

If you multiply both sides of the equation by $\frac{e^{\tilde{x} \tilde{y}}}{5}$, the PDE becomes: \begin{equation} \dfrac{\partial }{\partial \tilde x} \left(U e^{\frac{\tilde x \tilde y}{5}} \right) = \frac{1}{5} \tilde x \tilde y e^{\frac{\tilde x \tilde y}{5}} = \dfrac{\partial }{\partial \tilde x} \left( \left( \tilde x -\frac{5}{\tilde y} \right)e^{\frac{\tilde x \tilde y}{5}} \right), \end{equation} that is \begin{equation} \dfrac{\partial }{\partial \tilde x} \left(\left(U-\tilde x + \frac{5}{\tilde y} \right) e^{\frac{\tilde x \tilde y}{5}} \right) = 0. \end{equation} So you have a function whose partial derivative WRT $\tilde x$ is zero, so no matter how much you change $\tilde x$, the function will remain the same if $\tilde y$ is unchanged. This means that the argument inside the differential operator can only be a function of $\tilde y$. A priori this is unknown and you may call it $f(\tilde y)$ - it is then determined by initial/boundary data.

spetrevski
  • 1,243
  • so the point of "fixing" a variable is to use the integrating factor to try and get the function in a form such that it is only a function of one variable? Also, technically should we not need two functions i.e. one where we have "fixed" y and one where we have "fixed" x – seraphimk May 13 '20 at 17:49
  • 1
    Yes, the point is to be able to write the expression as $\partial / \partial x (...) = 0$ or $\partial / \partial y (...)=0$ which forces the expression in the parentheses to be a function of the other variable only. You're not "fixing" anything - the expression I wrote holds for any $x$ and any $y$. It only helps you deduce something in its fullest generality. – spetrevski May 13 '20 at 18:21