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Let us say I find the characteristic lines of some easy PDE $a U_x + b U_y = 0$ to be $bx-ay=c$, where $b, a, c$ are constants. Now, we say the solution must be constant along those lines, so it HAS to be a function of $(bx-ay)$. We can write $U(x,y)=f(bx-ay)$ then.

Here's my question. If I have another function $u(x,y)$ that is a solution and is constant along those same lines, but does not have the same fixed value along each line as $f(bx-ay)$, what says that it HAS to be able to be written in the form $u(x,y)=f(bx-ay)$?

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    $\begingroup$ The second function need not be the same as the first, $u(x,y)=g(bx-ay)$. It only shares the characteristic lines. $\endgroup$ – Chrystomath Jan 24 at 10:06
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The fact that $U$ must be constant along the lines $bx-ay = c$ is only a consequence of the method of characteristics. To determine the values $U = f(c)$ along those lines in a unique way, we need to specify some boundary conditions (or boundary data).

For example, let us assume that $U$ is known along the line $Ax+By=C$. For any point $(x,y)$, we construct the intersection $(x_0,y_0)$ of the line $Ax+By=C$ with the characteristic curve: \begin{aligned} bx_0-ay_0 &= c = bx-ay\\ Ax_0+By_0 &= C \end{aligned} The determinant of this system is $\Delta = aA + bB$, and $\Delta\neq 0$ leads to $$ x_0 = \frac{aC + Bc}{aA + bB} ,\quad y_0 = \frac{bC - Ac}{aA + bB} $$ Existence and uniqueness is only guaranteed if $\Delta \neq 0$, i.e. if the characteristic lines are not parallel to the line $Ax+By=C$ where $U$ is known. The function $f$ defining the solution as $U(x,y) = f(bx-ay)$ is related to the boundary data. Another set of boundary data would provide another such function $f$.

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