Solving a qusilinear PDE - am I missing a clever trick?

Question

I'm more or less familiar with the characteristic method for solving a quasilinear PDE. I've tried to apply it to the following problem, but I having difficulties: $$ u(x+u)u_x - y(y+u)u_y = 0 \\ u(1, t) = \sqrt{t} $$

When writing down the characteristic equations:

$$ x_s = u(x+u) \\ y_s = -y(y+u) \\ u_s = 0 $$ (t stands for initial curve parameterization, s stands for characteristic parameterization) I do manage to find that:

$$ u(s) = C_1 (t) $$ And substituting the initial condition provides:$u = \sqrt{t}$. I find that suspicious, since I think this implies that $u=\sqrt{y}$ which does not satisfy the equation. Since I had no better idea, I was trying to solve the other two equations (note that u is constant with respect to s): $$ ln(x+u) = us + C_2(t) \\ \frac{1}{u}(ln(y) - ln(y+u)) = -s + C_3(t) $$

Which I can not seem to infer u or t in terms of x, y . Anyway, I think I've either:

1. blew it along the way
2. I'm missing some clever trick that should make solving a lot easier
3. Am not applying the method of characteristics well

Please share your insights with me. Many thanks!

Answer 1

You almost had it!

You're right that $u=\sqrt{t}$, but what that means is that $u$ is constant along the characteristic curve corresponding to the parameter value $t$, so you're wrong in thinking that it implies that $u(x,y)=\sqrt{y}$.

Anyway, finishing the calculation that you started, we find that the characteristic curve going through the point $(x,y,u)=(1,t,\sqrt{t})$ when $s=0$ is $$ \begin{pmatrix} x \\ y \\ u \end{pmatrix} = \begin{pmatrix} (1+\sqrt{t}) \, e^{s \sqrt{t}} - \sqrt{t} \\[1ex] \dfrac{t}{(1+\sqrt{t}) \, e^{s \sqrt{t}} - \sqrt{t}} \\[1ex] \sqrt{t} \end{pmatrix} , $$ where somewhat miraculously the relation $y=t/x$ happens to hold. In other words, the characteristic curve corresponding to the parameter value $t>0$ is nothing but the curve $xy=t$, or more precisely it's the branch of that curve which lies in the first quadrant $x>0$, $y > 0$. And since $u$ is constant on each such curve, we find that $$ u(x,y) = \sqrt{\strut xy} ,\qquad x>0 ,\quad y>0 . $$ Remark 1: The given initial data only specifies $u=\sqrt{t}$ at the points $(x,y)=(1,t)$ where $t \ge 0$, because of the square root, and the characteristic for $t=0$ is simply the $x$-axis, where $u=\sqrt{xy}$ isn't differentiable, so we need to exclude $t=0$ and keep only $t>0$.

Remark 2: You can (and should) check directly that $u=\sqrt{xy}$ indeed satisfies the PDE and the initial condition.

Answer 2

$$u(x+u)u_x - y(y+u)u_y = 0$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{u(x+u)}=\frac{dy}{-y(y+u)}=\frac{du}{0}=ds$$ A first characteristic equation comes from $du=0$ : $$u=c_1$$ A second characteristic equation comes from solving $\frac{dx}{c_1(x+c_1)}=\frac{dy}{-y(y+c_1)}$

$\frac{dx}{dy}+\frac{c_1}{y(y+c_1)}x=-\frac{(c_1)^2}{y(y+c_1)}\quad$ is a first order linear ODE. Solving it for $x(y)$ leads to :

$x=\frac{(c_1)^2+c_2(c_1+y)}{y}$

$$\frac{xy-(c_1)^2}{c_1+y}=c_2$$

The general solution of the form of implicit equation $c_2=F(c_1)$ , with $c_1=u$ is : $$\boxed{\frac{xy-u^2}{u+y}=F(u)}$$ $F$ is an arbitrary function (to be determined according to the specified condition).

Condition : $u(1,t)=\sqrt{t}$

$\frac{t-(\sqrt{t})^2}{\sqrt{t}+t}=F(\sqrt{t})=0$

The function $F=0$ is known now. We put it into the above general solution. $$\frac{xy}{u+y}-\frac{u^2}{u+y}=0\quad\implies\quad u^2=xy$$ $$\boxed{u(x,y)=\sqrt{xy}}$$ The root $_:-\sqrt{xy}\:$ is excluded because the condition implies $u\geq 0$.