Soving PDE $u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0$

Question

I have some problems with solving PDEs.

1. \begin{cases} \ u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0 \\[2ex] u(x,x)=1,\quad u(1,y)=y \end{cases}

What I've done: $$u(x,y)=\frac{1}{x^2}v(x,y)$$ $$u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0 \Rightarrow v_{xx}-v_{yy}=0$$ $$v(x,y)=C_1(x+y)+C_2(x-y)$$ $$u=\frac{1}{x^2}(C_1(x+y)+C_2(x-y))$$ From $u(x,x)=1, u(1,y)=y$ we get $$\begin{cases} \ C_1(2x)+C_2(0)=1 \\[2ex] C_1(1+y)+C_2(1-y)=y \end{cases}$$

And this is where I stopped. No idea how to solve such a system. Any ideas?

2. \begin{cases} \ u_{xy}-\frac{1}{x-y}(u_x-u_y)=1 \\[2ex] u(x,x)=0,\quad u(2,y)=2+2y+\frac{1}{2}y^2 \end{cases}

$$u(x,y)=\frac{1}{x-y}v(x,y)$$ $$u_{xy}-\frac{1}{x-y}(u_x-u_y)=1 \Rightarrow v_{xy}=x-y$$ $$v=\frac{1}{2}x^2y-\frac{1}{2}y^2x+C_1(x)+C_2(y)$$ $$u=\frac{1}{x-y} \left(\frac{1}{2}x^2y-\frac{1}{2}y^2x+C_1(x)+C_2(y)\right)$$ $$\begin{cases} \ C_1(x)+C_2(-x)=x^3 \\[2ex] C_1(2)+C_2(y)=4-\frac{y^3}{4} \end{cases}$$

And again the same problem. What's wrong? Thank you.

Answer 1

The equation $C_1(2x)+C_2(0)=1$ shows that $C_1(2x)$ doesn't vary with $x$, i.e. $C_1$ is a constant function, say $C_1(x) = c.$

Inserting this in $C_1(1+y)+C_2(1-y)=y$ we get $C_2(1-y) = y-c$ i.e. $C_2(y) = (1-y)-c = (1-c)-y.$

Thus, $$ \begin{cases} C_1(x) = c \\ C_2(x) = (1-c)-y \end{cases} $$

Answer 2

In fact your equations are $$\begin{cases} \ C_1(2x)+C_2(0)=x^2 \\[2ex] C_1(1+y)+C_2(1-y)=y \end{cases}$$which after solving lead to$$\begin{cases} C_1(x)={x^2\over 4}+a \\ C_2(x)=-{x^2\over 4}-a \end{cases}$$and the final answer becomes $$u(x,y)={y\over x}$$