using change of coordinates to solve a PDE

Question

Let's say I have a simple PDE like $u_x + u_y = 2$ that I want to solve. I know that, among other methods, I could use a change of coordinates approach by setting $r = x$ and $t = x - y$, but I'm not sure exactly where to go from here.

In my textbook there is the following discussion:

But I do not understand exactly how $u_x$ ends up becoming $au_{x'} + bu_{y'}$ and how $u_y$ ends up becoming $bu_{x'} - au_{y'}$. I don't even see where the sum comes from. I know this is probably a really dumb question, but to me it seems like the math used here is different than what I've been taught all my life.

Answer 1

It is the application of chain rule in multivariate calculus. You can find proofs in many calculus textbooks, but the intuition behind it is, well-behaved bivariate functions looks like a flat plane when we zoom in very closely around a point. Instead of tangent lines, we have tangent planes. We want the derivative to reflect the infinitesimal changes in both axes, so we have two terms, with each term looking like the univariate "high school calculus" chain rule, only with partial differentials.

Applying the chain rule for $u_x$ and $u_y$; we should get two equations with two terms each.

Note that from (3), $\partial x'/\partial x=a$, and $\partial u/\partial x' = u_{x'}$ by change of notation. Do the same thing for all other terms.

Answer 2

Change of variables $\quad\begin{cases}r=x \\ t=x-y \end{cases} \quad$ in the PDE :$\quad \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=2$

It is easy to obtain the relationship between $\frac{\partial u}{\partial x}$ , $\frac{\partial u}{\partial y}$ and $\frac{\partial u}{\partial r}$ , $\frac{\partial u}{\partial t}$ from the fundamental identity :

$$\text{Total differential : } \quad du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy = \frac{\partial u}{\partial r}dr+\frac{\partial u}{\partial t}dt $$

$\quad\begin{cases}r=x \quad\to\quad dr=dx \\ t=x-y \quad\to\quad dt=dx-dy \end{cases} \quad $ that we put into the above equation.

$$\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy = \frac{\partial u}{\partial r}dx+\frac{\partial u}{\partial t}(dx-dy) $$

$$\left(\frac{\partial u}{\partial x}-\frac{\partial u}{\partial r}-\frac{\partial u}{\partial t}\right)dx+\left(\frac{\partial u}{\partial y}+\frac{\partial u}{\partial t} \right)dy =0 $$ This implies : $$\begin{cases} \frac{\partial u}{\partial x}-\frac{\partial u}{\partial r}-\frac{\partial u}{\partial t}=0 \\ \frac{\partial u}{\partial y}+\frac{\partial u}{\partial t}=0 \end{cases} \quad\to\quad \begin{cases} \frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}+\frac{\partial u}{\partial t} \\ \frac{\partial u}{\partial y}=-\frac{\partial u}{\partial t} \end{cases}$$

Then, putting them into the PDE :

$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=2=\left(\frac{\partial u}{\partial r}+\frac{\partial u}{\partial t}\right)+\left(-\frac{\partial u}{\partial t}\right)\quad$ and after simplification : $$\frac{\partial u}{\partial r}=2$$ Integrating with respect to $r$ : $$u=2r+f(t)$$ where $f$ is any differentiable function.

Finally, comming back to the initial variables : $$u=2x+f(x-y)$$