Method of Characteristics: Relationship of constants

Question

Let us look at this simple $(1+1)$-dimensional transport equation:

$$\partial_t u+a\partial_x u = 0.$$

When solving this by the method of characteristics we obtain the following equation:

$$\frac{dt}{1}=\frac{dx}{a}=\frac{du}{0}.$$

From the last part, we find $u = c_1$ and from $dt = dx/a$ we obtain $x-at=c_2$.

To this point, I can explain everything that is happening, but now it is said that $c_1$ is a function of $c_2$, hence $u=c_1=F(x-at)$. I can apply this method, but I am not able to explain that why one constant should be related to the other constant. Is this just using the fact that we can express a constant as a function of another constant? This seems to be too much hand waving to me.

Answer 1

Think this way: Imagine you want to find one function $u=u(x,t)$ that it is constant through time along a path given by the curve $x=x(t)$, $i.e.$: $$\frac{du(x(t),t)}{dt}=\frac{\partial u}{\partial t}(x(t),t)+\frac{\partial u}{\partial x}(x(t),t)\frac{dx(t)}{dt}=0\tag{*}$$ This equation $(*)$ holds if, your PDE does: $$\frac{dx(t)}{dt}= a\tag{**}$$

Therefore the characteristic lines in which the solution of your PDE $u(x,t)$ is constant are $x-x_0=at$, $i.e$: $$u(x_0+at,t) = constant$$ This constant may be fixed without any loss of generality to $u(x_0,0)$. This fact shows that the function $u(x,t)$ can be expressed as a function of a unique parameter: $x_0$, and therefore, substituting the value of $x_0$ from the characteristic curve one has: $$u(x_0+at,t)=u(x_0)\rightarrow u(x,t)=u(x-at)$$

Answer 2

$$\partial_t u+a\partial_x u = 0.$$ The equations of the characteristic curves are derived from the system of ODEs : $$\frac{dt}{1}=\frac{dx}{a}=\frac{du}{0}.$$ A first family of characteristic curves comes from $\quad du=0 \quad\to\quad u=c_1$

A second family of characteristic curves comes from $\quad \frac{dt}{1}=\frac{dx}{a} \quad\to\quad x+a\,t=c_2$

Outside the characteristic curves, that is not indeed the constants which are related, but the functions. A general manner to express the relationship is on the form of implicit equation : $$\Phi(u,x+at)=0$$ with arbitrary differentiable function $\Phi$ of two variables.

Sometimes, but not always, the above implicit equation is solvable for one variable. This is the case here since $u$ doesn't appears in the second variable. So, in this favorable case, an equivalent manner to express the relationship is : $$u=F(x+at)$$ or equivalently $$x+at=G(u)$$ where $F$ is an arbitrary differentiable function and $G$ is it's inverse.