Solve PDE (using change of variable) 
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?
 A: 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.
A: 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., $$a^2+b^2\neq 0.$$
