My Professor's notes had this problem:
Solve the PDE $u_x + u_y = 0$ in the domain $y > \phi(x)$, $x \in \mathbb{R}$ given that $u = g(x)$ on the curve $y = \phi(x)$, where $\phi(x) = \frac{x}{1 + |x|}$.
Her notes say that the general solution is $u(x, y) = A(x – y)$, where $A()$ is an arbitrary function. However, if I try to solve this myself, then I am unable to get this.
My class only started learning this, so please forgive me for any errors.
So based on what we have learned I think that the first two characteristic equations are $\dfrac{dx}{dt} = 1$ and $\dfrac{dy}{dt} = 1$. If we use separation of variables we get $x(t) = t + C_1$ and $y(t) = t + C_2$. One key point we were told here is that $C_1$ and $C_2$ are constants that are the same for any single characteristic curve but change between different characteristic curves.
I then used change of variables to get $x(0) = s = C_1$ and $y(0) = \dfrac{s}{1 + |s|} = C_2$ which must mean that the change of variables for the map $(s, t) \to (x, y)$ is $x(s, t) = t + s$ and $y(s, t) = t + \dfrac{s}{1 + |s|}$.
Ok, I think I have done everything up to now correctly.
I now solve the last characteristic equation $\dfrac{du}{dt} = 0$. If we again use separation of variables then we get $u(x,y) = C_3$.
But now I wonder how she got $u(x, y) = A(x – y)$ as the general solution? And where did the function $A$ come from?
Best Answer
Let $p:=x-y$ and $q:=x+y$. Then, $x=\dfrac{1}{2}q+\dfrac{1}{2}p$ and $y=\dfrac{1}{2}q-\dfrac{1}{2}p$. The two new variables $p$ and $q$ are independent: $$\frac{\partial p}{\partial q}=\frac{\partial{x}}{\partial{q}}\,\left(\frac{\partial p}{\partial x}\right)+\frac{\partial{y}}{\partial{q}}\,\left(\frac{\partial p}{\partial y}\right)=\left(\frac12\right)\cdot 1+\left(\frac12\right)\cdot(-1)=0$$ and $$\frac{\partial q}{\partial p}=\frac{\partial{x}}{\partial{p}}\,\left(\frac{\partial q}{\partial x}\right)+\frac{\partial{y}}{\partial{p}}\,\left(\frac{\partial p}{\partial y}\right)=\left(\frac12\right)\cdot 1+\left(-\frac12\right)\cdot1=0\,.$$ That is, if a function $U(p,q)$ satisfies $\dfrac{\partial U}{\partial q}=0$, then $U(p,q)=A(p)$ for some function $A$.
Now, let $U(p,q)$ denote $u\left(\frac{1}{2}q+\frac{1}{2}p,\frac{1}{2}q-\frac{1}{2}p\right)=u(x,y)$ in your question. Since $$\frac{\partial U}{\partial q}=\frac{\partial x}{\partial q}\,\left(\frac{\partial U}{\partial x}\right)+\frac{\partial y}{\partial q}\,\left(\frac{\partial U}{\partial y}\right)=\frac{1}{2}\,\left(\frac{\partial u}{\partial x}\right)+\frac{1}{2}\,\left(\frac{\partial u}{\partial y}\right)=\frac12\,\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\right)=0\,,$$ we have that $U(p,q)=A(p)$ for some function $A$. Thus, $$u(x,y)=A(x-y)\,.$$