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

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