First order Cauchy problem

Question

I'm trying to solve the PDE \begin{align*} au_x + (bx+cu-1)u_y = d \\ u(x,0) = 0. \end{align*}

So far, I got the solution for the case $cd-ab \neq 0$ as \begin{align*} u(x,y) = \frac{d}{cd-ab}\left( 1-bx-\sqrt{(1-bx)^2+2(cd+ab)y} \right) \end{align*} Now, I'm stuck on the case $ab = cd$. Can you point me the way to go?

Answer 1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=a$ , letting $x(0)=0$ , we have $x=at$

$\dfrac{du}{dt}=d$ , letting $u(0)=u_0$ , we have $u=dt+u_0=\dfrac{dx}{a}+u_0$

$\dfrac{dy}{dt}=bx+cu-1=(ab+cd)t+cu_0-1$ , letting $y(0)=f(u_0)$ , we have $y=\dfrac{(ab+cd)t^2}{2}+(cu_0-1)t+f(u_0)=\dfrac{(ab-cd)x^2}{2a^2}+\dfrac{x(cu-1)}{a}+f\left(u-\dfrac{dx}{a}\right)$

$u(x,0)=0$ , i.e. $y(x,u=0)=0$ :

$0=\dfrac{(ab-cd)x^2}{2a^2}-\dfrac{x}{a}+f\left(-\dfrac{dx}{a}\right)$

$f\left(-\dfrac{dx}{a}\right)=-\dfrac{(ab-cd)x^2}{2a^2}+\dfrac{x}{a}$

$f(x)=-\dfrac{(ab-cd)x^2}{2d^2}-\dfrac{x}{d}$ when $d\neq0$

$\therefore y=\dfrac{(ab-cd)x^2}{2a^2}+\dfrac{x(cu-1)}{a}-\dfrac{ab-cd}{2d^2}\left(u-\dfrac{dx}{a}\right)^2-\dfrac{1}{d}\left(u-\dfrac{dx}{a}\right)$ when $d\neq0$

Answer 2

Solve the equation by the method of characteristics.

The characteristic ODE are as follows.

$\dfrac{d x}{ d t} =a$, $\dfrac{d u}{ d t} = d$, $\dfrac{d y}{ d t} =b x +c u -1$,

with initial conditions $x(0) = x_0$, $u(0) = 0 $, $y(0) = 0$ at $t=0$.

By solving the first two characteristic ODE, there are

$x = x_0 + a t $ and $u = d t$.

By substituting $x,u$ into the third characteristic ODE, there is

$\dfrac{d y}{ d t} = b x_0 -1 +(ab +cd) t $, with $y(0) = 0$.

The solution is $y = (b x_0 - 1) t + \dfrac{1}{2} (ab + cd) t^2$.

By substituting $x_0$ into $y$, there is

$y = (b(x - at) - 1) t + \dfrac{1}{2} (ab + cd) t^2 = (b x - 1) t + \dfrac{1}{2} (cd - ab) t^2 $, which is

$\dfrac{1}{2} (cd - ab) t^2 + (b x - 1) t - y =0$.

(1) If $cd \neq ab $, then $t = \dfrac{1-bx \pm \sqrt {(bx -1)^2 + 2 (cd - ab) y}}{cd - ab}$.

Notice that $y = 0$ at $t = 0$, $ t = \dfrac{1-bx - \sqrt {(1-bx)^2 + 2 (cd - ab) y}}{cd - ab}$.

Thus, the solution of the PDE is $u = \dfrac{d}{cd -ab} \Bigg(1-bx - \sqrt {(1-bx)^2 + 2 (cd - ab) y} \Bigg)$.

(2) If $cd = ab $, then $t = \dfrac{ y }{bx - 1}$.

Thus, the solution of the PDE is $u = \dfrac{ d y }{bx - 1}$.

Remark: Note that the solution is not defined on the whole plane and the regularity of the solution needs further discussion.