6.2: New Variables

In order to understand the solution in all mathematical details we make a change of variables

$$w = x + ct,\quad z = x - ct . \nonumber$$

We write $u(x,t)=\bar u(w,z)$. We find

$$\begin{align} \dfrac{\partial}{\partial x} u & = \dfrac{\partial}{\partial w}{\bar u}\dfrac{\partial}{\partial x} w + \dfrac{\partial}{\partial z} {\bar u}\dfrac{\partial}{\partial x} z = \dfrac{\partial}{\partial w}{\bar u} + \dfrac{\partial}{\partial z} {\bar u}, \nonumber\\[4pt] \dfrac{\partial^2}{\partial x^2} u & = \dfrac{\partial^2}{\partial w^2}{\bar u}+2 \dfrac{\partial^2}{\partial w \partial z} {\bar u}+ \dfrac{\partial}{\partial z} {\bar u}, \nonumber\\[4pt] \dfrac{\partial}{\partial t} u & = \dfrac{\partial}{\partial w}{\bar u}\dfrac{\partial}{\partial t} w + \dfrac{\partial}{\partial z} {\bar u}\dfrac{\partial}{\partial t} z = c\left(\dfrac{\partial}{\partial w}{\bar u} - \dfrac{\partial}{\partial z} {\bar u}\right), \nonumber\\[4pt] \dfrac{\partial^2}{\partial t^2} u & = c^2\left(\dfrac{\partial^2}{\partial w^2}{\bar u}-2 \dfrac{\partial^2}{\partial w \partial z} {\bar u}+ \dfrac{\partial}{\partial z} {\bar u} \right)\end{align} \nonumber$$

We thus conclude that

$$\dfrac{\partial^2}{\partial x^2} u(x,t) - \frac{1}{c^2} \dfrac{\partial^2}{\partial t^2} u(x,t) = 4\dfrac{\partial^2}{\partial w \partial z} {\bar u} = 0 \nonumber$$

An equation of the type $\dfrac{\partial^2}{\partial w \partial z} {\bar u} = 0$ can easily be solved by subsequent integration with respect to $z$ and $w$. First solve for the $z$ dependence,

$$\dfrac{\partial}{\partial w}{\bar u} = \Phi(w), \label{eq5}$$

where $\Phi$ is any function of $w$ only. Now solve this equation for the $w$ dependence,

$$\bar u(w,z) = \int \Phi(w) dw = F(w) + G(z) \nonumber$$ In other words, with $F$ and $G$ arbitrary functions.

6.2.1: Infinite String

Equation $\ref{eq5}$ is quite useful in practical applications. Let us first look at how to use this when we have an infinite system (no limits on $x$). Assume that we are treating a problem with initial conditions

$$u(x,0) = f(x),\;\;\dfrac{\partial}{\partial t} u(x,0) = g(x). \nonumber$$

Let me assume $f(\pm\infty)=0$. I shall assume this also holds for $F$ and $G$ (we don’t have to, but this removes some arbitrary constants that don’t play a rôle in $u$). We find

$$\begin{align} F(x)+G(x) &= f(x),\nonumber\\[4pt] c(F'(x)-G'(x)) & = g(x).\end{align} \nonumber$$

The last equation can be massaged a bit to give

$$F(x)-G(x) = \underbrace{\frac{1}{c}\int_0^x g(y) dy}_{=\Gamma(x)} + C \nonumber$$

Note that $\Gamma$ is the integral over $g$. So $\Gamma$ will always be a continuous function, even if $g$ is not!

And in the end we have

$$\begin{align} F(x) &= \dfrac{1}{2} \left[f(x) +\Gamma(x)+C\right]\nonumber\\[4pt] G(x) &= \dfrac{1}{2} \left[f(x) -\Gamma(x)-C\right] \end{align} \nonumber$$

Suppose we choose (for simplicity we take $c=1 \text{m/s}$)

$$f(x) = \begin{cases} x+1 & \text{if $-1<x<0$} \\ 1-x & \text{if $0<x<1$} \\ 0 & \text{elsewhere} \end{cases} . \nonumber$$

and $g(x)=0$. The solution is then simply given by

$$u(x,t) = \dfrac{1}{2} \left[f(x+t)+f(x-t)\right]. \label{eq.11}$$ This can easily be solved graphically, as shown in Figure $\PageIndex{1}$.

Figure $\PageIndex{1}$: The graphical form of ($\ref{eq.11}$), for (from left to right) $t=0s$,$t=0.5s$ and $t=1s$. The dashed lines are $\dfrac{1}{2} f(x+t)$ (leftward moving wave) and $\dfrac{1}{2} f(x-t)$ (rightward moving wave). The solid line is the sum of these two, and thus the solution $u$.

6.2.2: Finite String

The case of a finite string is more complex. There we encounter the problem that even though $f$ and $g$ are only known for $0<x<a$, $x\pm ct$ can take any value from $-\infty$ to $\infty$. So we have to figure out a way to continue the function beyond the length of the string. The way to do that depends on the kind of boundary conditions: Here we shall only consider a string fixed at its ends.

$$\begin{align} u(0,t) &=u(a,t)=0,\nonumber\\[4pt] u(x,0)&=f(x)\\[4pt] \dfrac{\partial}{\partial t} u (x,0) &= g(x).\end{align} \nonumber$$

Initially we can follow the approach for the infinite string as sketched above, and we find that

$$\begin{align} F(x) &= \dfrac{1}{2} \left[f(x) +\Gamma(x)+C\right], \nonumber\\[4pt] G(x) &= \dfrac{1}{2} \left[f(x) -\Gamma(x)-C\right]. \end{align} \nonumber$$

Look at the boundary condition $u(0,t)=0$. It shows that

$$\dfrac{1}{2} [f(ct)+f(-ct)]+\dfrac{1}{2} [\Gamma(ct)-\Gamma(-ct)]=0. \nonumber$$

Now we understand that $f$ and $\Gamma$ are completely arbitrary functions – we can pick any form for the initial conditions we want. Thus the relation found above can only hold when both terms are zero

$$\begin{align} f(x)&=-f(-x),\nonumber\\[4pt] \Gamma(x)&=\Gamma(x).\label{eq:refl1}\end{align}$$

Now apply the other boundary condition, and find

$$\begin{align} f(a+x)&=-f(a-x),\nonumber\\[4pt] \Gamma(a+x)&=\Gamma(a-x).\label{eq:refl2}\end{align}$$

Figure $\PageIndex{2}$: A schematic representation of the reflection conditions in Equations $\ref{eq:refl1}$ and $\ref{eq:refl2}$. The dashed line represents $f$ and the dotted line $\Gamma$.

The reflection conditions for $f$ and $\Gamma$ are similar to those for sines and cosines, and as we can see from Figure $\PageIndex{2}$ both $f$ and $\Gamma$ have period $2a$.