Solve PDE (using change of variable)

Question

I am wondering why we can substitute $u_{x}(ax+by,bx-ay)$ and $u_{y}(ax+by,bx-ay)$ to the original PDE equation $au_x(x,y)+bu_y(x,y)=0$.

Why is the case that $au_{x}(ax+by,bx-ay)+bu_{y}(ax+by,bx-ay)=0$?.

I thought after change of variable, we have different coordinate system.

I find another way to solve this problem is to define another fucntion $v$ such that $u(x,y)=v(ax+by,bx-ay)=v(x',y')$. By defining in such way, I find it more intuitive. Anyone could explain the equivalence between these two methods to solve the problem?

or we just abuse the notation $u$ in the above?

Answer 1

Your original PDE is \begin{equation} au_{x} + bu_{y} = 0 \end{equation}

where both a,b $\not =$ 0

Now you consider the transformations:

\begin{equation} x'= ax +by \\ y' = bx - ay \end{equation}

You impose: \begin{equation} u(x,y) = u(x',y') \end{equation}

Now consider: \begin{equation} \frac{\partial }{\partial x}(u(x,y)) = \frac{\partial }{\partial x}(u(x',y')) \\ \\ u_{x} = \frac{\partial u}{\partial x'}\frac{\partial x'}{\partial x} + \frac{\partial u}{\partial y'}\frac{\partial y'}{\partial x} \end{equation}

Similarly you get: \begin{equation} \frac{\partial }{\partial y}(u(x,y)) = \frac{\partial }{\partial y}(u(x',y')) \\ \\ u_{y} = \frac{\partial u}{\partial y'}\frac{\partial y'}{\partial y} + \frac{\partial u}{\partial x'}\frac{\partial x'}{\partial y} \end{equation}

Now plugging these into the PDE: \begin{equation} au_{x} + bu_{y} = 0 \\ a(\frac{\partial u}{\partial x'}\frac{\partial x'}{\partial x} + \frac{\partial u}{\partial y'}\frac{\partial y'}{\partial x} ) + b( \frac{\partial u}{\partial y'}\frac{\partial y'}{\partial y} + \frac{\partial u}{\partial x'}\frac{\partial x'}{\partial y}) = 0 \end{equation}

Now from the relations: \begin{equation} x'= ax +by \\ y' = bx - ay \end{equation}

You can calculate: \begin{equation} \frac{\partial x'}{\partial x} = a \\ \frac{\partial y'}{\partial x} = b \\ \frac{\partial x'}{\partial y} = b \\ \frac{\partial y'}{\partial y} = -a \end{equation}

Plugging in these relations you get: \begin{equation} a(au_{x'} + bu_{y'}) + b(bu_{x'}-au_{y'}) = 0 \\ ({a}^2+{b}^2)u_{x'} = 0 \end{equation}

Now we assumed: Both a,b $\not =$ 0. Hence, \begin{equation} {a}^2+{b}^2 \not = 0 \end{equation}

Therefore: \begin{equation} u_{x'} = 0 \end{equation}

Now integrating this w.r.t x' \begin{equation} u(x',y') = f(y') \\ \end{equation}

But, \begin{equation} y' = bx - ay \\ \end{equation}

Therefore:

\begin{equation} u(x,y) = f(bx - ay) \\ \end{equation}

We could've also just said \begin{equation} u(x,y) = v(x',y') \\ \end{equation}

It's the same thing.

Answer 2

The change of variable is well defined if the map given by $(x,y)=(au+bv,bu-av)$ is bijective. That is the determinant is not null, i.e., $$a^2+b^2\neq 0.$$