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Next: Separation of Variables Up: Partial Differential Equations of Previous: Modelling: Derivation of the

D'Alembert's solution of the Wave Equation

A particularly neat solution to the wave equation, that is valid when the string is so long that it may be approximated by one of infinite length, was obtained by d'Alembert. The idea is to change coordinates from $x$ and $t$ to $\xi$ and $\eta$ in order to simplify the equation. Anticipating the final result, we choose the following linear transformation

$\displaystyle \xi = x-ct \qquad {\rm and} \qquad \eta =x +ct.$      

Thus, $u(x,t)=u(\xi, \eta) = u(\xi(x,t), \eta (x,t))$ and we must use the chain rule to express derivatives in terms of $x$ and $t$ as derivatives in terms of $\xi$ and $\eta$. Hence,
$\displaystyle {\partial u \over \partial x}$ $\textstyle =$ $\displaystyle {\partial \xi \over \partial x} {\partial u
\over \partial \xi} +...
...tial \eta} = {\partial u \over \partial \xi} + {\partial u \over \partial
\eta}$  
$\displaystyle {\partial u \over \partial t}$ $\textstyle =$ $\displaystyle {\partial \xi \over \partial t} {\partial u
\over \partial \xi} +...
...\eta} = - c{\partial u \over \partial \xi} + c{\partial u \over \partial
\eta}.$  

The second derivatives require a bit of care.

$\displaystyle {\partial^{2} u \over \partial x} = {\partial \over \partial x} (...
...tial \eta} ({\partial u \over \partial \xi} +
{\partial u \over \partial \eta})$      

Thus,
$\displaystyle {\partial^{2} u \over \partial x^{2}} = {\partial^{2} u \over \pa...
...2}u \over \partial \xi \partial \eta} + {\partial^{2}u \over
\partial \eta^{2}}$      

and in a similar manner
$\displaystyle {\partial^{2} u \over \partial t^{2}} = c^{2} {\partial^{2} u \ov...
... \partial \xi \partial \eta} + c^{2}
{\partial^{2} u \over \partial \eta^{2}} .$      

Thus, (2.6) becomes
  $\textstyle \,$ $\displaystyle {\partial^{2} u \over \partial \xi^{2}} - 2 {\partial^{2}u \over ...
...{2}u \over \partial \xi \partial \eta} +
{\partial^{2}u \over \partial \xi^{2}}$  
  $\textstyle \,$ $\displaystyle \Rightarrow {\partial^{2} u \over \partial \xi \partial \eta} = 0 .$  

This equation is much simpler and can be solved by direct integration. First of all integrate with respect to $\xi$ to give

$\displaystyle {\partial u \over \partial \eta} = g(\eta)$      

where $g(\eta)$ is an arbitrary function of $\eta$. Then integrate with respect to $\eta$ to obtain
$\displaystyle u (\xi, \eta) = F(\xi) + G(\eta)$      

where $F(\xi)$ is an arbitrary function of $\xi$ and $G(\eta) = \int g(\eta)d
\eta$.

Finally, we replace $\xi$ and $\eta$ by their expressions in terms of $x$ and $t$ to obtain

$\displaystyle u(x,t) = F(x-ct)+G(x+ct).$     (2.7)

Example 2. .14Verify that $\sin (x-ct) + \sin (x+ct)$ is a solution to the wave equation. Here $F(\xi) = \sin \xi$ and $G(\eta) = \sin \eta$.

Thus

\begin{eqnarray*}
{\partial u \over \partial x} &=& \cos (x-ct) + \cos (x+ct), \...
...l t^{2}} &=& - c^{2} \sin (x-ct) - c^{2} \sin (x+ct).
\nonumber
\end{eqnarray*}



Therefore,

\begin{displaymath}
{\partial^{2} u \over \partial x^{2}} = {1 \over c^{2}}
{\partial^{2} u \over \partial t^{2}}.
\end{displaymath}

Note that we may use the trigonometric identities for $\sin (A \pm B)$ to obtain

$\displaystyle \sin (x-ct) + \sin (x+ct) = 2 \sin x \cos ct$      

and we remark that this is the product of a function of $x$ with a function of $t$. This result will be used later.

D'Alembert's solution involves two arbitrary functions that are determined (normally) by two initial conditions. If the initial conditions are

$\displaystyle u(x,o)$ $\textstyle =$ $\displaystyle f(x)$  
$\displaystyle {\partial u \over \partial t} (x,o)$ $\textstyle =$ $\displaystyle h(x).$ (2.8)

Then, from (2.7), we have
\begin{displaymath}
F(x) +G(x) = f(x).
\end{displaymath} (2.9)

Note that $F$ is a function of $\xi =x-ct$ but $\xi = x$ at $t=0$. The tricky part is calculating ${\partial u \over \partial t}$
$\displaystyle {\partial u \over \partial t}$ $\textstyle =$ $\displaystyle {\partial \xi \over \partial t}{dF(\xi)
\over d\xi} + {\partial \eta \over \partial t}{dG \over d \eta}
(\eta)$  
$\displaystyle {\partial u \over \partial t}$ $\textstyle =$ $\displaystyle -c {dF(\xi) \over d \xi} + c {dG \over d \eta}
(\eta)$  

but at $t=0, \xi = x$ and $\eta = x$. Thus, the functional form of $F$ and $G$ are obtained by replacing $\xi$ by $x$ and $\eta = x$.

Example 2. .15If, for example, $ u(x,t) = \sin (\xi) = \sin (x-ct)$, then

$\displaystyle \begin{array}{ll}
u(x,t)=F(\xi)= \sin \xi & u(x,0)=F(x)= \sin x\\...
...& {\partial u
\over \partial t} (x,0) =-cF^{\prime} (x) = -c \cos x
\end{array}$      

The initial speed, that is $\partial u / \partial t$ evaluated at $t=0$, is then

$\displaystyle -cF^{\prime}(x)+cG^{\prime}(x)=h(x).$      

Integrating with respect to $x$, as both $F$ and $G$ are functions of $x$ when $t=0$, we get
\begin{displaymath}
-F(x) +G(x) = {1 \over c} H(x) = {1 \over c} \int h(x) dx
\end{displaymath} (2.10)

Subtracting (2.10) from (2.9) gives
$\displaystyle F(x)$ $\textstyle =$ $\displaystyle {1 \over 2} \{ f(x) - {1 \over c} H(x) \}$  
$\displaystyle F(\xi)$ $\textstyle =$ $\displaystyle {1 \over 2} \{ f(\xi) - {1 \over c} H {\xi} \} = {1 \over 2} \{ f
(x-ct)-{1 \over c} H(x-ct) \}$  

Adding (2.9) and (2.10) gives

$\displaystyle G(x)$ $\textstyle =$ $\displaystyle {1 \over 2} \{ f(x) + {1 \over c} H(x) \}$  
$\displaystyle \Rightarrow G(\eta)$ $\textstyle =$ $\displaystyle {1 \over 2} \{ f(\eta) + {1 \over c} H(\eta) \} = {1
\over 2} \{ f(x+ct) + {1 \over c} H(x+ct) \}$  

Hence, d'Alembert's solution that satisfies the initial conditions (2.8) is

$\displaystyle u(x,t)$ $\textstyle =$ $\displaystyle {1 \over 2} \left \{ f(x-ct)+f(x+ct) +{1 \over c} \left (H(x+ct) -
H(x-ct)\right ) \right \}$  
  $\textstyle =$ $\displaystyle {1 \over 2} \left \{ f(x-ct) + f(x+ct) + {1 \over c}
\int^{x+ct}_{x-ct} h(s) ds \right \}.$  



Example 2. .16$f(x-ct)$ can be thought of as a ``shape", defined by the function $f(\xi)$, moving to the right. We can see this if we consider the value $\xi = 0$. Say

$\displaystyle f(0) =1.$      


 Then $f=1$ at 		$x=0$ when 		$ct=0$ 

$x=1$ when $ct=1$
$x=2$ when $ct=2$

Thus, we are moving to the right to larger values of $x$. Similarily $G(x+ct)$ corresponds to a wave propagating to the left.

Example 2. .17 $F(x) = a \cos kx \quad G(x) = 0$ Thus, the solution to the wave equation is

$\displaystyle u= a \cos k(x-ct)$      

corresponding to a wave travelling to the right (increasing $x$).


Here 		 $a$ is the amplitude, 

$k$ is the wavenumber - $k= {2 \pi \over \lambda}$,
$\lambda$ is the wavelength - distance from one crest to the next,
$kc$ is the angular frequency $kc = 2\pi / \tau$,
$\tau$ is the period of the wave - time from one crest to the next to pass afixed point.

Figure 2.2:
\includegraphics [scale=0.7]{fig2.ps}

Example 2. .18 $F(x)= a \sin kx \qquad G(x) = a \sin kx$ This gives the solution

$\displaystyle u$ $\textstyle =$ $\displaystyle a \{ \sin k (x-ct)+ \sin k (x+ct) \},$  
  $\textstyle =$ $\displaystyle a \{ \sin kx \cos kct + \cos kx \sin kct + \sin kx \cos kct - \cos kx \sin
kct \},$  
  $\textstyle =$ $\displaystyle 2a \sin kx \cos kct.$  

The solution is a standing wave as shown in Figure 2.2. The maximum and minimum displacements are at $x=0, \pm {\pi \over k}, \ldots,
\pm {N \pi \over k}$ and the nodes (where $u=0$) are at $x= \pm {\pi \over 2k},
\ldots , \pm (N+ {\textstyle{1 \over 2}}) {\pi \over K}.$

Example 2. .19

Figure 2.3:
\includegraphics [scale=0.7]{fig3.ps}

Consider the case of the string initially at rest with the above initial displacement. Thus,
$\displaystyle u(x,0)=f(x) = \left\{ \begin{array}{cc}
0 & \vert x\vert \geq L, ...
... -L \leq x \leq 0, \\
\delta (1-x/L) & 0 \leq x \leq L,\\
\end{array} \right.$      

where $\delta$ is a constant. In Figure 2.3, $\delta = 2$.
$\displaystyle {\partial u \over \partial t} (x,0) = 0 .$      

From d'Alembert's solution we have
$\displaystyle F(x) + G(x)$ $\textstyle =$ $\displaystyle f(x)$  
$\displaystyle -c F^{\prime} (x) + cG^{\prime} (x)$ $\textstyle =$ $\displaystyle 0 \Rightarrow F(x) = G(x) + C,$  

where $C$ is a constant. Adding the two solutions we have
$\displaystyle F(x)$ $\textstyle =$ $\displaystyle {\textstyle{1 \over 2}} f(x) + {1 \over 2}C,$  
$\displaystyle G(x)$ $\textstyle =$ $\displaystyle {1\over 2} f(x) - {1\over 2} C$  

Hence, the solution is
$\displaystyle u(x,t) = {\textstyle{1 \over 2}} f(x-ct) + {\textstyle{1 \over 2}} f(x+ct),$      

and the arbitrary constant, $C$, does not appear in the solution. Thus, we may, in fact, take $C=0$ without any loss of generality. So we have 2 equal triangles of height $\delta /2$ going to the right and the left. This is shown in Figure 2.4.

Figure 2.4: The solution obtained in Example 19 is shown for $\delta = 2$ and $ c = 1$ at various times.
\includegraphics [scale=0.7]{fig4.ps}


next up previous
Next: Separation of Variables Up: Partial Differential Equations of Previous: Modelling: Derivation of the
Prof. Alan Hood
2000-10-30