Deriving general solution of a PDE $ \frac{ \partial u }{\partial x } + \frac{ \partial u }{\partial y } = \sqrt{ u } $.

Question

We have the PDE

$$ \frac{ \partial u }{\partial x } + \frac{ \partial u }{\partial y } = \sqrt{ u } $$

The characteristic curves are

$$ dx = dy = \frac{ d z }{\sqrt{z} } $$

Solving this system, we obtain that

$$ x = 2 \sqrt{z} + A $$

$$ y = 2 \sqrt{z} + B $$

Thus, we have the

$$ g(x,y,u) = x - 2 \sqrt{u} $$

$$ h(x,y,u) = y - 2\sqrt{u} $$

Thus, solution $u = u(x,y)$ satisfies

$$ F( x - 2\sqrt{u}, y - 2 \sqrt{u}) $$

where $F(g,h) $ is a smooth function satisfying $F_g^2 + F_h^2 \neq 0$.

Is this correct? How can I write a formula for $u $ explicitly? thanks

Answer 1

You treat your equation as a quasilinear, and so your solution is a way too complicated. Better, think of the equation as a semilinear one. It is known that if $(a,b) \ne (0,0),$ then the change of variables $$ \begin{cases} s = ax+by,\\ t= bx-ay \end{cases} $$ transforms the linear PDE $$ a u_x + bu_y=0 $$ to $$ (a^2+b^2) w_s=0. $$ In effect, the change of variables $$ \begin{cases} s = x+y,\\ t= x-y \end{cases} $$ transforms your PDE to \begin{equation*} \tag{$*$} 2 w_s =\sqrt{w} \end{equation*} which is reducible to the separable ODE $$ 2z'(s)=\sqrt{z(s)} \iff 2\sqrt{z(s)}-\frac s2 = C $$ where $C$ is an arbitrary constant. It follows that any solution of $(*)$ satisfies $$ 2\sqrt{w(s,t)}-\frac s2=f(t) $$ (why?) and hence any solution of the original PDE satisfes $$ 2\sqrt{u(x,y)}-\frac {x+y}2=f(x-y), $$ where $f$ is a continuously differentiable function (you can consult my book on PDEs to see what are the standard methods of solution of first-order semilinear and quasilinear PDEs.)