# 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?

• I think this is related to the fact that the solutions of transport equation are constant along characteristics.. Jul 28, 2020 at 14:02
• yes, because u is constant along the characteristics line, we define our new coordinate system in a such way that $x'=ax+by$ is parallel to the characteristic line. However, I am wondering what is the justification for the substitution. Jul 28, 2020 at 14:09
• As users are submitting close votes, you should explain where you found this problem and why it is important to you. If you don't give more information then your question might be closed. Jul 29, 2020 at 3:14

Your original PDE is $$$$au_{x} + bu_{y} = 0$$$$

where both a,b $$\not =$$ 0

Now you consider the transformations:

$$$$x'= ax +by \\ y' = bx - ay$$$$

You impose: $$$$u(x,y) = u(x',y')$$$$

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

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

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

Now from the relations: $$$$x'= ax +by \\ y' = bx - ay$$$$

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

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

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

Therefore: $$$$u_{x'} = 0$$$$

Now integrating this w.r.t x' $$$$u(x',y') = f(y') \\$$$$

But, $$$$y' = bx - ay \\$$$$

Therefore:

$$$$u(x,y) = f(bx - ay) \\$$$$

We could've also just said $$$$u(x,y) = v(x',y') \\$$$$

It's the same thing.

• What if we have a bounded domain $U$, $u(x,y)=u(x’,y’)$ maybe not be well defined. Since $x’=ax+by$ Also, u is a unique solution for the PDE problem. If we define like $u(x,y)=u(x’,y’)$. Do we abuse the notation u here? Jul 30, 2020 at 2:35
• I think it is a bit of notation abuse since my PDE professor always used u(x,y) = v(x',y'). As far as domain issues go, I think this shouldn't be a problem. Our solution is always a set of curves in the 2d plane dependent upon the initial condition we give to this problem. I see no reason why we couldn't do a domain extension if needed and then restrict back, uniqueness will still be maintained.
– Dk65
Jul 30, 2020 at 2:47
• Thank you! It clears my confusion. Jul 30, 2020 at 2:50

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., $$a^2+b^2\neq 0.$$

• I have edited the problem. I hope my quetion will make more sense now. Jul 29, 2020 at 7:42