I'm having trouble with solving the quasilinear PDE

$$\begin{cases} u_x+u_y&=2\sqrt{u}, \\ u(x,x)&=g(x) \end{cases}$$

via method of characteristics as in this paper.

My attempt:

First I started by formulating the ODEs

$$\begin{cases} \dot{x}&=1 \\ \dot{y}&=1 \\ \dot{z}&=2\sqrt{z}. \end{cases}$$

Solving those and applying the initial conditions gives me

$$\begin{cases} x&=t+x_0 \\ y&=t+x_0 \\ z&=\frac{1}{4}(2t+z_0)^2=\frac{1}{4}(2t+2\sqrt{g(x_0)})^2. \end{cases}$$

But how do I eliminate $t$ and $x_0$ if $x=y$?


1 Answer 1


The system $$\left\lbrace \begin{aligned} t+x_0 = x\\ t+x_0=y \end{aligned}\right. $$ has no solution $(t, x_0)$ if $x≠y$, and it has infinitely many solutions if $x=y$. Therefore, we can't express the solution uniquely in terms of $x$, $y$. Situation is somewhat similar to this post and related ones.

The Lagrange-Charpit equations $$ \frac{dx}{1} = \frac{dy}{1} = \frac{du}{2\sqrt u} $$ provides the characteristic families $x-y = c_1$ and $x - \sqrt{u} = c_2$. The second characteristic family might be rewritten $u = (x - c_2)^2$). General solutions read $$ u = \big(x - f(x-y)\big)^2 $$ where $c_2 = f(c_1)$ involves an arbitrary function $f$. If $x=y$, we find $u = (x - f(0))^2$ which needs to equal $g(x)$ according to the boundary condition. Two cases arise:

  • if $g(x) = (x - c_3)^2$, then we obtain an infinity of solutions under the constraint $f(0) = c_3$;
  • else, no function $f$ matches the boundary condition: there is no solution.

Again, we can't express the solution uniquely in terms of $x$, $y$.

  • 1
    $\begingroup$ Thanks! I got to that but computed $dz/dx=2\sqrt{z}$, which led me to the solution $u(x,y)=\frac{1}{4}(2x+f(y-x))^2$, with the condition $f(0)=2(\sqrt{g(x)}-x)$. $\endgroup$ Jun 2, 2020 at 5:55
  • 1
    $\begingroup$ @HannahBloom it's pretty much the same as in my answer! $\endgroup$
    – EditPiAf
    Jun 3, 2020 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.