quasi linear partial differential equation problem

Question

I need to solve to find infinity number of solutions for the PDE: $$\begin{cases}\frac{u_x}{y}-u_y=u\quad y>0\\ u\left(x,e^{-x}\right)=0\end{cases}$$ what I have done so far is: $$\begin{cases}x'(t)=\frac{1}{y}\\y'(t)=-1\\u'(t)=u\end{cases}\rightarrow x(t,s)=\begin{cases}x(t,s)=C_1(s)-ln(C_2(s)-t)\\ y(t,s)=C_2(s)-t\\ y(t,s)=C_3(s)e^t\end{cases}$$ How do I proceed from this point?

Answer 1

$$\frac{u_x}{y}-u_y=u$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{1/y}=\frac{dy}{-1}=\frac{du}{u}$$ A first characteristic equation comes from solving $\frac{dx}{1/y}=\frac{dy}{-1}\quad\implies\quad dx+\frac{dy}{y}=0$ : $$e^x y=c_1$$ Note : you correctly wrote $x(t,s)=C_1(s)-\ln(C_2(s)-t)$ and $y(t,s)=C_2(s)-t$ . Eliminating $t$ leads to $x=C_1-\ln(y)$ which is equivalent to the above equation $e^x y=c_1$ with $c_1=e^{C_1}$ .

A second characteristic equation comes from solving $\frac{dy}{-1}=\frac{du}{u}$ : $$e^y u=c_2$$

Note : You corectly wrote $y(t,s)=C_2(s)-t$ . But there is a typo in your writting $y(t,s)=C_3(s)e^t$ which should be $u(t,s)=C_3(s)e^t$ . Eliminating $t$ leads to $u=C_3 e^{C_2-y}\quad\implies\quad e^y u=C_3 e^{C_2}$ . This is equivalent to the above equation $e^y u=c_2$ with $c_2=C_3 e^{C_2}$ .

The general solution of The PDE is : $$e^y u=F(e^x y)$$ Where $F$ is an arbitrary function. $$\boxed{u(x,y)=e^{-y} F(e^x y)}$$

Case of condition : $u(x,e^{-x})=0$ $$0=u(x,e^{-x})=e^{-(e^{-x})} F(e^x e^{-x})=e^{-(e^{-x})}F(1)$$ This implies $$F(1)=0$$ They are infinity many functions which are equal to $0$ at argument $=1$ . Thus the answer to the question is : They are infinity many solutions to the PDE in case of the particular specified condition. $$\boxed{u(x,y)=e^{-y} F(e^x y)\quad\text{with arbitrary function } F \text{ such as } F(1)=0.}$$

Note : This conclusion was expected because the condition $u(x,e^{-x})=0$ is definded on the curve $y=e^{-x}$ that is $e^xy=1$ which is a characteristic curve (case $c_1=1$ ). It is known that the solution is not unique when the condition is defined on a characteristic curve.

Answer 2

Recall the idea behind what you've done: we want to find a family of curves $\{C_s\}_s$ such that

1. every point in $\mathbb{R}^2$ lies on exactly one $C_s$,
2. as you travel along $C_s$, $u$ changes in a predictable and simple-to-formulate manner (that is, $\frac{d}{dt}(u(C_s(t)))=(\text{nice formula})$), and
3. for each $s$, your initial condition $u(x,e^{-x})=0$ specifies the value of $u(C_s(t))$ for exactly one value of $t$ (call it $t_s$).

This reduces an a priori 2D PDE into a 1D ODE, because we can compute $u(a,b)$ by

• first figuring out $s$ such that $(a,b)$ lies on $C_s$ and then
• integrating along $C_s$ from $C_s(t_s)$ to $(a,b)$.

You've shown (2) (it turns out $\frac{d}{dt}(u(C_s(t)))=A_se^t$), and (1) turns out to be true. But (3) is violated really badly.

Seeing whether (1) and (3) hold is tricky, because your formulas for $C_s$ are super ugly. As it turns out, there's an easier way to describe the characteristic curves.

Notice that $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-1}{\frac{1}{y}}=-y$$ so that each characteristic curve is $$y(x)=Ae^{-x}$$ In particular, $x\mapsto(x,e^{-x})$ is one such curve.

I leave it to you to now concoct a different solution for every $C^1$ function $s\mapsto u(s)$ that vanishes when $s=0$.