I need help this: Solve the following initial value problem
$u_x+u_y=u^2,~~ u(x,-x)=x$
Here what I did: The characteristic equations: $~dx/dt=1,~~ dy/dt=1~$ and $~du/dt=u^2~.$
- $du/u^2= dx$ Integrate: $-1/u+ C_1=x$ ; $C_1 =1/u+ x$; $ u=1/(C_1-x)$
- $dx = dy$ Integrate: $y + C_2=x$ ; $C_2 = x - y$
Therefore, the complete integral of the equation:
$F ( 1/u + x,x - y) = 0$ or in another form:
$u =1/(C_1-x)=1/(f(C_2)-x)= ( f (x- y)- x )^{-1}$ where $f$ - some differentiable function.
The initial condition $u(x,-x) = x$;
$( f (x+x)- x )^{-1}=x$;
$1=x(f(2x)-x)\implies f(2x)=(1+x^2)/x$
But I don't know what do now...
Thanks
So you get, $$-1/u+ C_1=x\implies C_1=x+1/u\\~~~~~~~~~~~~~~~~~~~~~~\text{and}~~~~~ C_2 = x - y$$ Hence the solution of the given PDE is of the form $~x+1/u=f(x - y)~,$ where $~f~$ is arbitrary function.
Now initial condition gives, $~u(x,-x)=x~,$ so the above solution becomes $$x+\dfrac 1x=f(2x)~~~~~\text{i.e.,}~~~ f(y)=\dfrac y2+\dfrac 2y$$ $($Putting $y=2x)$
So $$x+\dfrac 1u=f(x - y)\implies x+\dfrac 1u=\dfrac {x-y}2+\dfrac 2{x-y}\implies \dfrac 1u=\dfrac {x-y}2+\dfrac 2{x-y}-x$$ $$\implies \dfrac 1u=\dfrac{y^2-x^2+4}{2(x-y)}\implies (y^2-x^2+4)u=2(x-y)$$ This is the required solution of the given initial valued PDE.