Consider the simple wave function

$$\frac{\partial^2 u}{\partial t^2}-C^2\frac{\partial^2 u}{\partial x^2}=0$$

Using the characteristic coordinates $\xi = x + Ct$ and $\eta = x - Ct$ we transform this equation into

$$\frac{\partial^2u}{\partial\xi\partial\eta} = 0$$

Integrating with respect to $\eta$ we have

$$\frac{\partial u}{\partial\xi} = f_1(\xi)$$

and integrating with respect to $\xi$ we have

$$u(\xi, \eta) = \xi f_1(\xi) + f_2(\eta)$$

Hence the solution in terms of $x$ and $t$ is

$$u(x,t) = (x+Ct)f_1(x+Ct) + f_2(x-Ct)$$

Where $f_1, f_2$ are arbitrary. Now the solution given is

$$u(x,t) = f_1(x+Ct) + f_2(x-Ct)$$

Is this simply down to re-defining the arbitrary function $f_1(\xi) = \xi f_1(\xi)$?

  • 1
    $\begingroup$ D'Alembert. $\endgroup$ – Did Mar 31 '16 at 15:10

In general $\int f_1(\xi) d \xi$ is not $\xi f_1(\xi)$, it is some antiderivative of $f_1$, say $F_1(\xi)$, plus a function which doesn't depend on $\xi$.

Incidentally, I think it is a bit clearer what is going on if instead of actually changing coordinates, you just do operator factorization like:

$$\left (\frac{\partial}{\partial t}+C\frac{\partial}{\partial x} \right ) \left (\frac{\partial}{\partial t}-C\frac{\partial}{\partial x} \right )u=0$$

Then this says that the quantity $\frac{\partial u}{\partial t}-C\frac{\partial u}{\partial x}$ is being transported to the right at velocity $C$. (If $C<0$ then it's really to the left; you get the idea.) You may solve this homogeneous transport equation by characteristics, obtaining some solution $v(x,t)$. Then to find $u$ you solve the inhomogeneous transport equation $\frac{\partial u}{\partial t}-C\frac{\partial u}{\partial x}=v(x,t)$.


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