Solution to first order partial Cauchy problem?

Question

Solve the Cauchy Problem :

$$(x+y)u_{x} + (x-y)u_y = 1$$ with $ \; \; u(1,y)=\frac{1}{\sqrt{2}}$

My attempt:

We have the following characteristic equations :

$$\frac{dx}{x+y} = \frac{dy}{x-y} = \frac{du}{1}$$

Solving

$$\frac{dx}{x+y} = \frac{dy}{x-y} \Rightarrow xdx - ydy = xdy + ydy \Rightarrow x^2 - y^2 -2xy = c_1 $$ where $c_1$ is a constant

I am unable to find another pair of characteristic equations to solve.

In order to reduce it to canonical form, if I use transformation $\phi(x,y)= x^2 - y^2 -2xy $ and $\gamma(x,y)= y$

I am getting $(x+y)u_{\gamma} = 1$ which is again of no help.

How do I proceed?

Answer 1

Your first characteristic equation is correct : $$x^2-y^2-2xy=c_1$$ From this : $\quad x-y=\pm\sqrt{c_1+2y^2}$

Second characteristic, from $\quad\frac{dy}{x-y}=\frac{du}{1}=\frac{dy}{\pm\sqrt{c_1+2y^2}}$

$u-\int\frac{dy}{\pm\sqrt{c_1+2y^2}}=c_2$

$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2(c_1+2y^2)}+2y\bigg|=c_2$

$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2(x^2-y^2-2xy+2y^2)}+2y\bigg|=c_2$

$$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2}(x-y)+2y\bigg|=c_2$$

Answer 2

From $$dx = (x+y)du\\dy = (x-y)du$$ using the eigenvalues $\pm \sqrt 2$ and eigenvectors of the symmetric matrix $\pmatrix{1&1\\1&-1}$ one can consider the combination $$ d((1\pm\sqrt2)x+y)=((1\pm\sqrt2)(x+y)+(x-y))du=\pm\sqrt2((1\pm\sqrt2)x+y)du $$ so that $$ (1\pm\sqrt2)x+y)=c_{2/3}e^{\pm\sqrt 2\,u}. $$