I want to solve the following PDE:

$$\begin{align} u_{tt}&=c^2u_{xx}-\gamma u_x, \quad 0<x<1, \quad t>0,\\ \\ u(0,t)&=u(1,t)=0, \\ u(x,t=0)&=x(1-x),\\ u_t(x,t=0)&=0. \end{align}$$

Though I see the literature has examples with the damped wave equation, I don't have an explanation for the $-\gamma u_x$ term. How would I solve it?

  • $\begingroup$ The "simplest" way would be to do it on the Fourier side. Taking the Fourier series expansion in the spatial interval $(0,1)$ you rewrite the equation as $$\hat{u}_{tt} = (-c^2 \xi^2 - i\gamma \xi) \hat{u}$$ which will allow you to write the solution in Fourier representation. $\endgroup$ – Willie Wong Feb 14 '14 at 10:06
  • $\begingroup$ @ Willie Wong : You are right, I deleted my first answer because there was a big mistake in it. $\endgroup$ – JJacquelin Feb 14 '14 at 10:30

One can, as I mentioned in the comment above, do the problem by Fourier series.

One can however to a funny change of variables to convert the equation to one that is perhaps better studied.

We observe the following fact:

$$ (c \partial_x - \frac{\gamma}{2c})(c\partial_x - \frac{\gamma}{2c}) = c^2 \partial^2_{xx} - \gamma \partial_x + \frac{\gamma}{4c^2} $$

which means that your equation can be re-written as

$$ u_{tt} = (c \partial_x - \frac{\gamma}{2c})(c\partial_x - \frac{\gamma}{2c})u - \frac{\gamma}{4c^2}u $$

Now, we have that

$$ c \partial_x u - \frac{\gamma}{2c} u = c e^{\gamma x / (2c^2)} \partial_x \left( e^{- \gamma x/(2c^2)} u\right) $$

which rewrites the equation as

$$u_{tt} = e^{\gamma x / (2c^2)} c^2 \partial^2_{xx} \left(e^{-\gamma x /(2c^2)}u\right) - \frac{\gamma}{4c^2}u $$

So we make the substitution $v(x,t) = e^{-\gamma x / (2c^2)} u(x,t)$. The equation for $v$ is simply the following Klein-Gordon equation

$$ v_{tt} = c^2 v_{xx} - \frac{\gamma}{4c^2} v $$

with the boundary conditions

$$ v(0,t) = v(1,t) = 0 $$

and the initial values

$$ \begin{align} v_t(x,0) &= 0 \\ v(x,0) &= e^{-\gamma x / (2c^2)} x(1-x) \end{align}$$


If $\gamma \neq \gamma(u)$, or even simpler, $\gamma$ is a constant, then your problem can be solved using separation of varaiables since the PDE is linear. Furthermore, the boundary conditions are homogenous, as well as the PDE.

Set $u(t,x) = P(x)Q(t)$, with $P\neq0, Q\neq 0$. Plug it into the equation to obtain:

$$PQ'' = c^2 P''Q - \gamma P' Q,$$ which can be rewritten as:

$$\frac{\frac{c^2}{\gamma} P'' - P'}{P} = \frac{Q'}{\gamma Q} = \lambda, $$ where $\lambda$ is some real constant, which in general can be zero, positive or negative. The equation written above leads you to two problems, one for $P(x)$ and another for $Q(t)$. Let's solve the problem for $P$ (you may solve the problem for $Q$ as well, but it's not relevant).

The boundary conditions for $P$ are given by the original boundary conditions for $u$, indeed: $P(0) = P(1) = 0$; so the problem for $P$ becomes:

$$P''- \beta P'- \beta \lambda P = 0, \quad 0 < x < 1, \quad P(0) = P(1) = 0,$$ where $\beta = \frac{\gamma}{c^2}$. The solution of this equation depends on the value of $\lambda$ (in general it'll be: $P(x) = Ae^{s_1x} + Be^{s_2 x}$, where $s_i$ are the two -different- roots of $s^2 - \beta s - \beta \lambda$.), whether it's positive, zero or negative.

Solve for the so called eigenfunctions, $P_n(x)$, and expand the solution in terms of them as follows:

$$u(t,x) = \sum^\infty_n P_n(x) Q_n(t),$$

plug this into the original PDE and solve for $Q_n(t)$ by applying the prescribed initial conditions.

Can you take it from here?


  • 1
    $\begingroup$ This is the case when the involved differential operator $L=c^2\partial^2_x−\gamma\partial_x$ cannot be selfadjoint with boundary conditions of any type. Of course, separation of variables can be employed, but it is not recommended here because it requires far more advanced knowledge in the spectral theory of differential operators. First, one must check the operator's being normal, which is the case. Second, find the operator's biorthogonal basis, and know what to do with such basis. The best way here is to avoid it using a routine substitution employed by Willie Wong (see below). $\endgroup$ – mkl314 Feb 15 '14 at 0:07

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