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!


2 responses to this post.

  1. […] 24, 2008 Last time we derived the general solution of a linear, homogeneous PDE with constant coefficients: . […]


  2. […] found the general solutions for two types of linear, homogeneous, first order PDEs: those with constant coefficients, and those with varying coefficients. It turns out that you can also extend the method of […]


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: