Prove that the problem has a unique solution.

Question

Consider the Cauchy Problem $a\dfrac{\partial u}{\partial x}+b\dfrac{\partial u}{\partial y}=1;u(x,y)=x$ on $ax+by=1;a^2+b^2\neq 0$

Then prove that the problem has a unique solution.

By Lagrange Equations:

$\dfrac{{\operatorname {dx}}}{a}=\dfrac{{\operatorname {dy}}}{b}=\dfrac{{\operatorname {du}}}{1}\implies x=au+c_1;y=bu+c_2$

So we get $y-bu=f(x-au)$. I cant proceed further,neither I can use the boundary conditions.Please help

Answer 1

You can written as $bu - y = f(bx - ay)$ and the boundary condition is given that $u(x,y) = x $ on $u(x,y)= x$ on $ax + by =1$

Therefore $bu(x , \frac{1-ax}{b}) = bx = \frac{1-ax}{b} + f(bx - a \frac{1-ax}{b})$ , we get

$f(\frac{b^2x + a^2x - a}{b}) = b^2x + ax -1$

put $z = \frac{b^2x +a^2x -a}{b}$ and got the function $f$