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:
- blew it along the way
- I'm missing some clever trick that should make solving a lot easier
- Am not applying the method of characteristics well
Please share your insights with me.
Many thanks!
Best Answer
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.