Non uniqueness of a solution of a first order PDE: Exercise 9.23 (b) John Lee "Introduction to Smooth Manifolds"

Question

This question is part of exercise 9.23 of "Introduction to smooth manifold" of John Lee second edition. Prove the following equation has a global solution which is not unique $$ \frac{\partial}{\partial x}u + y \, \frac{\partial}{\partial y}u=0 \\ u(x,1)=e^{-x} $$ I have used the method of characteristics as specified by the hint, but I did not find any reason of why this Cauchy problem could have a non unique solution. Here's what I have done $$ \begin{cases} x'(t,s)=1 \\ y'(t,s)=y(t,s) \\ x(t,s)=s \\ y(t,s)=1 \end{cases}, $$ then I imposed the boundary condition and this brought me to $u(x,y)=y\cdot e^{-x}.$ But then how could I proceed to build another solution? Any suggestions? Could it be a misprint? In another post the author has pointed out a mistake in one of the questions of problem 9.23, could this be another error, or am I missing something? I have already checked the correction list of the book and there was no error mentioned about this question.

Answer 1

The characteristic passing through the point $(x_0,y_0)$ is $y=y_0 \, e^{x-x_0}$, so the characteristics passing through the points $(x_0,1)$ (where $u$ is given) fill out the upper half plane $y > 0$. Thus, they don't tell you anything about what $u$ is supposed to be in the lower half plane $y \le 0$, except that for a global solution you probably want $u$ to be nice (say class $C^1$) everywhere, and in particular across the line $y=0$. The general solution is $u(x,y)=f(y \, e^{-x})$ where $f(t)$ is some $C^1$-function, and to satisfy the given condition you have to take $f(t)=t$ for $t>0$, but you can continue this in any $C^1$ way that you like to $t \le 0$. For example, you may take $f(t)=t$ in that region too, but also for example $f(t)=t+t^2$.