Consider the equation $u_x + u_y = \sqrt{u} $
Derive the general solution
Observe that the trivial solution $u(x,y) ≡ 0 $ is not covered by the
general solution.
This is my solution:
Characteristic equation;
${\frac {dx}{1}}$=${\frac {dy}{1}}$=${\frac {du}{\sqrt{u}}}$
solving the first equality
$x-y=c_1$
$dx={\frac {du}{\sqrt{u}}}$
$c_2=2\sqrt{u}-x$
$F(c_1,c_2)=0$ then $c_2=f(c_1)$
$F(x-y,2\sqrt{u}-x)=0$
$2\sqrt{u}-x=f(x-y)$
then
$u={\frac {(x+f(x-y))^2}{4}}$
is my solution true? i am not sure and i don't understand how to observe that trivial condition is not covered by the general solution. i would appreciate it if you help.
Best Answer
The derivation and the proposed solution look correct. Indeed, we have $$ u_x+u_y=\frac{x+f(x-y)}{2}=\sqrt{u} $$ for $x+f(x-y)\geq 0$. Note that $u\equiv 0$ would imply $x+f(x-y) = 0$ for all such $x$, $y$, which is impossible.