## PDEs: Solving Simple PDEs Using Characteristics

Last time we derived a simple PDE describing the process of non-diffusive advection in a constantly flowing stream of water. That PDE looked like this: $v\cdot u_x + u_t = 0$, where $v$ was the constant speed of the water. The solution of this PDE is a function of the form $u(x,t)$.

This PDE follows a more general form that looks like this: $c_2\cdot u_x + c_1\cdot u_t = 0$, where $c_1$ and $c_2$ are constants. If you’ve studied multivariable calculus, you might recognize this kind of equation. It is, in fact, a directional derivative.

We can rewrite $c_2 u_x + c_1 u_t = \nabla u \cdot \textbf{s} = 0$, where $\textbf{s} = (c_1, c_2)$. We can read this equation as “the gradient (or derivative) of $u(x,t)$ in the direction of $\textbf{s}$ is zero”. Which means that in the direction $\textbf{s}$, the solution $u$ attains the same constant value. It may be easier to see by graphing it:

If you think of $u$ as being a surface graphed above the plane of the image above, then on each of the light grey lines, $u$ has the same “height” above the screen. These lines are called characteristics. There’s no reason to suppose, however, that $u$ doesn’t vary across the different characteristics. As long as we follow one particular characteristic, however, $u$ will remain the same.

The mathematical interpretation of this line of reasoning is that $u$ is a function of the characteristics, since each characteristic determines a specific value of the solution. Therefore if we let $C$ represent the value of the characteristic, then $u = f(C)$, for some function $f$. From the graph, we can tell that the characteristics are in fact lines. If we let $b$ be the value of the x-intercept, then the equation for the characteristic looks like: $x = \frac{c_2}{c_1} t + b$. Well, $C$ is a constant value, so we need to rearrange this expression to: $c_1 x - c_2 t = c_1 b$. Since $c_1 b$ is constant for each line, we can let that be our $C$.

Therefore we conclude that $u = f(C) = f(c_1 x - c_2 t)$, for all functions $f$. We can actually rewrite this as $u = f(x-\frac{c_2}{c_1} t)$ by just manipulating the equation at the end of the preceding paragraph. This tells us that the solution $u$ is moving in the positive $x$ direction, which matches well with our understanding of the advective equation. Note that to actually solve for specific solutions we need an initial condition of the form $u(x,0)=g(x)$. Otherwise there’s no way to tell what the function $f$ actually is!