PDEs: Extending the Method of Characteristics

Last time we derived the general solution of a linear, homogeneous PDE with constant coefficients: c_2 u_x + c_1 u_t = 0. Physically, we can interpret such an equation as describing the flow of some chemical through water moving at a constant speed \frac{c_2}{c_1}. Suppose, however, that the water does not move at a constant speed. Suppose that c_1 and c_2 are themselves functions. In other words, suppose our PDE looks like this:

c_2 (x,t) u_x + c_1 (x,t) u_t = 0

It turns out that it’s not very difficult to extend the method of characteristics for this situation. Let’s graph it, though, to get a better idea of what we’re doing:

char2.jpg

As you can see, instead of characteristic lines we now have characteristic curves. Instead of the vector \textbf{s}=(c_1,c_2), we now have the vector-valued function \textbf{s}(x,t)=(c_1(x,t),c_2(x,t)). So pretty much everything is the same as before. Our solution u is still a function of the characteristics, and our job, as before, is to find the equation for these characteristics. The only difference is that we cannot eyeball the form of the characteristic equation as we did in the previous case. We have to actually calculate it.

What information do we have? Well, we know \textbf{s}(x,t), which describes each of the curves. From that, we can easily get that x' = \frac{c_2 (x,t)}{c_1 (x,t)}. Here we’re just extending the concept of the slope to account for a changing curve. So now we have an ODE which, hopefully, we’ll be able to solve. Suppose that we can solve it, and let \Lambda (x,t) be the solution (this is an implicit equation describing the characteristic curves). From our previous discussion, then, u(x,t) = f(C) = f(\Lambda (x,t) ).

Let’s do an example. Suppose we have the PDE x u_x + \frac{5}{t} u_t = 0. Let’s normalize it to get \frac{xt}{5}u_x + u_t = 0. From this, we get the ODE \frac{dx}{dt} = \frac{xt}{5}. You can solve this ODE using separation of variables. Rearrange the equation to get:

\int {\frac{dx}{x} = \frac{1}{5}\int {t\cdot dt}}

Solving, we get that \ln{x} = \frac{t^2}{10} + C, or C = \ln{x} - \frac{t^2}{10}. Thus u(x,t) = f(C) = f(\ln{x} - \frac{t^2}{10}).

As before, to solve for a specific solution, you need an initial condition u(x,0) = g(x). Suppose that u(x,0) = x^2. Plugging t=0 into the solution, we get u(x,0) = f(\ln{x} ) = x^2. Make a substitution \alpha = \ln{x} and solve for x. We get x = e^{\alpha}. This transforms our equation into the form f(\alpha ) = e^{2\alpha}.

Well here, \alpha is simply a dummy variable. We already know that u(x,t) = f(C), for some C. So just let C=\alpha. Then u(x,t)=f(C)=e^{2C}= \exp\left(2\ln{x} - \frac{t^2}{5}\right)= x^2 \exp\left(-\frac{t^2}{5}\right). Plug this into the PDE to see that it in indeed a solution.

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One response to this post.

  1. […] 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 characteristics for linear, inhomogeneous, […]

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