# Uniqueness: linear first order pde with constant coefficients

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 $$u(x,y)=f(bx-ay)$$?

• The second function need not be the same as the first, $u(x,y)=g(bx-ay)$. It only shares the characteristic lines. – Chrystomath Jan 24 at 10:06

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 $$f$$.