Consider the linear PDE

$y u_x + x u_y = 0$.

Applying the method of characteristics, we find that a general solution is given by

$u(x,y) = f(x^2 - y^2) $

for some function $f$. My question is: can we state that $all$ solutions are of this form?

Does the method of characteristic uniquely determine the general solution of a linear PDE such as the one above? If so,why?


There are more solutions, which a different formulation of the "method of characteristics" may have found. Any solution to this linear transport PDE has to be constant along characteristics, but this does not reduce to $u=f(x^2-y^2)$.

Let $u$ be any solution. The characteristic curves (which can be defined independently of $u$) solve $$ \partial_s \binom{X}{Y} = \binom{Y}{X},$$ And by chain rule $$ \partial_s \big( u(X,Y) \big) = u_x(X,Y)X_s + u_y(X,Y) Y_s = Y u_x(X,Y) + Xu_y(X,Y) = 0. $$

Having a look at a plot of the characteristics, which are (as you correctly said) the level sets of $x^2 - y^2$,

enter image description here

We see that the characteristic curves $x^2 - y^2 = c$ cover the whole of $\mathbb R^2$, so that every value of $u$ is determined as long as we give enough initial data. But this plot also shows us another thing: every level set except for $x^2 - y^2 = 0$ is made up of a pair of connected components, and when you prescribe $u(x,y) = f(x^2-y^2)$, you are forcing $u$ to be the same value on each such pair of these components. This is not necessary.

One way to obtain all possible solutions is to prescribe the values of $u$ on both the $x$ and $y$-axes, with a compatability condition for the characteristic point at $(x,y)=(0,0)$: $$ u(x,0) = f_1(x), \\ y(0,y) = f_2(y), \\ f_1(0) = f_2(0).$$


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