24 d’Alembert’s Solution to wave PDE

(added December 13, 2018)

The PDE is

\begin{equation} \frac {\partial ^{2}\psi }{\partial t^{2}}=c^{2}\frac {\partial ^{2}\psi }{\partial x^{2}} \tag {1}\end{equation}

Let

\begin{align*} u & =x-ct\\ v & =x+ct \end{align*}

Then

\begin{align} \frac {\partial \psi }{\partial t} & =\frac {\partial \psi }{\partial u}\frac {\partial u}{\partial t}+\frac {\partial \psi }{\partial v}\frac {\partial v}{\partial t}\nonumber \\ & =-c\frac {\partial \psi }{\partial u}+c\frac {\partial \psi }{\partial v} \tag {2}\end{align}

And

\begin{align} \frac {\partial \psi }{\partial x} & =\frac {\partial \psi }{\partial u}\frac {\partial u}{\partial x}+\frac {\partial \psi }{\partial v}\frac {\partial v}{\partial x}\nonumber \\ & =\frac {\partial \psi }{\partial u}+\frac {\partial \psi }{\partial v} \tag {3}\end{align}

Then, from (2)

\begin{align} \frac {\partial ^{2}\psi }{\partial t^{2}} & =-c\left ( \frac {\partial ^{2}\psi }{\partial u^{2}}\frac {\partial u}{\partial t}+\frac {\partial ^{2}\psi }{\partial u\partial v}\frac {\partial v}{\partial t}\right ) +c\left ( \frac {\partial ^{2}\psi }{\partial v^{2}}\frac {\partial v}{\partial t}+\frac {\partial ^{2}\psi }{\partial v\partial u}\frac {\partial u}{\partial t}\right ) \nonumber \\ & =-c\left ( -c\frac {\partial ^{2}\psi }{\partial u^{2}}+c\frac {\partial ^{2}\psi }{\partial u\partial v}\right ) +c\left ( c\frac {\partial ^{2}\psi }{\partial v^{2}}-c\frac {\partial ^{2}\psi }{\partial v\partial u}\right ) \nonumber \\ & =c^{2}\frac {\partial ^{2}\psi }{\partial u^{2}}-c^{2}\frac {\partial ^{2}\psi }{\partial u\partial v}+c^{2}\frac {\partial ^{2}\psi }{\partial v^{2}}-c^{2}\frac {\partial ^{2}\psi }{\partial v\partial u}\nonumber \\ & =c^{2}\frac {\partial ^{2}\psi }{\partial u^{2}}+c^{2}\frac {\partial ^{2}\psi }{\partial v^{2}}-2c^{2}\frac {\partial ^{2}\psi }{\partial v\partial u} \tag {4}\end{align}

And from (3)

\begin{align} \frac {\partial ^{2}\psi }{\partial x^{2}} & =\left ( \frac {\partial ^{2}\psi }{\partial u^{2}}\frac {\partial u}{\partial x}+\frac {\partial ^{2}\psi }{\partial u\partial v}\frac {\partial v}{\partial x}\right ) +\left ( \frac {\partial ^{2}\psi }{\partial v^{2}}\frac {\partial v}{\partial x}+\frac {\partial ^{2}\psi }{\partial v\partial u}\frac {\partial u}{\partial x}\right ) \nonumber \\ & =\left ( \frac {\partial ^{2}\psi }{\partial u^{2}}+\frac {\partial ^{2}\psi }{\partial u\partial v}\right ) +\left ( \frac {\partial ^{2}\psi }{\partial v^{2}}+\frac {\partial ^{2}\psi }{\partial v\partial u}\right ) \nonumber \\ & =\frac {\partial ^{2}\psi }{\partial u^{2}}+\frac {\partial ^{2}\psi }{\partial v^{2}}+2\frac {\partial ^{2}\psi }{\partial v\partial u} \tag {5}\end{align}

Substituting (4,5) into (1) gives

\begin{align*} -2c^{2}\frac {\partial ^{2}\psi }{\partial v\partial u} & =2c^{2}\frac {\partial ^{2}\psi }{\partial v\partial u}\\ -4c^{2}\frac {\partial ^{2}\psi }{\partial v\partial u} & =0 \end{align*}

Since \(c\neq 0\) then

\[ \frac {\partial ^{2}\psi }{\partial v\partial u}=0 \]

Integrating w.r.t \(v\) gives

\[ \frac {\partial \psi }{\partial u}=f\left ( u\right ) \]

Integrating w.r.t \(u\)

\[ \psi \left ( x,t\right ) =F\left ( u\right ) +G\left ( v\right ) \]

Therefore

\begin{equation} \psi \left ( x,t\right ) =F\left ( x-ct\right ) +G\left ( x+ct\right ) \tag {6}\end{equation}

The functions \(F\left ( x,t\right ) ,G\left ( x,t\right ) \) are arbitrary functions found from initial and boundary conditions if given. Let initial conditions be

\begin{align*} \psi \left ( x,0\right ) & =f_{0}\left ( x\right ) \\ \frac {\partial }{\partial t}\psi \left ( x,0\right ) & =g_{0}\left ( x\right ) \end{align*}

Where the first condition above is the shape of the string at time \(t=0\) and the second condition is the initial velocity.

Applying first condition to (6) gives

\begin{equation} f_{0}\left ( x\right ) =F\left ( x\right ) +G\left ( x\right ) \tag {7}\end{equation}

Applying the second condition gives

\begin{align} g_{0}\left ( x\right ) & =\left [ \frac {\partial }{\partial t}F\left ( x-ct\right ) \right ] _{t=0}+\left [ \frac {\partial }{\partial t}G\left ( x+ct\right ) \right ] _{t=0}\nonumber \\ & =\left [ \frac {dF\left ( x-ct\right ) }{d\left ( x-ct\right ) }\frac {\partial \left ( x-ct\right ) }{\partial t}\right ] _{t=0}+\left [ \frac {dG\left ( x+ct\right ) }{d\left ( x+ct\right ) }\frac {\partial \left ( x+ct\right ) }{\partial t}\right ] _{t=0}\nonumber \\ & =\left [ -c\frac {dF\left ( x-ct\right ) }{d\left ( x-ct\right ) }\right ] _{t=0}+\left [ c\frac {dG\left ( x+ct\right ) }{d\left ( x+ct\right ) }\right ] _{t=0}\nonumber \\ & =-c\frac {dF\left ( x\right ) }{dx}+c\frac {dG\left ( x\right ) }{dx} \tag {8}\end{align}

Now we have two equations (7,8) and two unknowns \(F,G\) to solve for. But the (8) has derivatives of \(F,G\,\). So to make it easier to solve, we integrate (8) w.r.t. to obtain

\begin{equation} \int ^{x}g_{0}\left ( s\right ) ds=-cF\left ( x\right ) +cG\left ( x\right ) \tag {9}\end{equation}

So we will use (9) instead of (8) with (7) to solve for \(F,G\). From (7)

\begin{equation} F\left ( x\right ) =f_{0}\left ( x\right ) -G\left ( x\right ) \tag {10}\end{equation}

Substituting (10) in (9) gives

\begin{align} \int ^{x}g_{0}\left ( s\right ) ds & =-c\left ( f_{0}\left ( x\right ) -G\left ( x\right ) \right ) +cG\left ( x\right ) \nonumber \\ & =-cf_{0}\left ( x\right ) +2cG\left ( x\right ) \nonumber \\ G\left ( x\right ) & =\frac {\left ( \int ^{x}g_{0}\left ( s\right ) ds\right ) +cf_{0}\left ( x\right ) }{2c}\nonumber \\ & =\frac {1}{2c}\left ( \int ^{x}g_{0}\left ( s\right ) ds+cf_{0}\left ( x\right ) \right ) \tag {11}\end{align}

Using the above back in (10) gives \(F\left ( x\right ) \) as

\begin{equation} F\left ( x\right ) =f_{0}\left ( x\right ) -\frac {1}{2c}\left ( \int ^{x}g_{0}\left ( s\right ) ds+cf_{0}\left ( x\right ) \right ) \tag {12}\end{equation}

Using (11,12) in (6) gives the final solution

\begin{align*} \psi \left ( x,t\right ) & =F\left ( x-ct\right ) +G\left ( x+ct\right ) \\ & =f_{0}\left ( x-ct\right ) -\frac {1}{2c}\left ( \int ^{x-ct}g_{0}\left ( s\right ) ds+cf_{0}\left ( x-ct\right ) \right ) +\frac {1}{2c}\left ( \int ^{x}g_{0}\left ( s\right ) ds+cf_{0}\left ( x\right ) \right ) \\ & =f_{0}\left ( x-ct\right ) -\frac {1}{2c}\int ^{x-ct}g_{0}\left ( s\right ) ds-\frac {1}{2}f_{0}\left ( x-ct\right ) +\frac {1}{2c}\int ^{x+ct}g_{0}\left ( s\right ) ds+\frac {1}{2}f_{0}\left ( x+ct\right ) \\ & =\frac {1}{2}\left ( f_{0}\left ( x-ct\right ) +f_{0}\left ( x-ct\right ) \right ) +\frac {1}{2c}\int _{x-ct}^{x+ct}g_{0}\left ( s\right ) ds \end{align*}

The above is the final solution. So if we are given initial position and initial velocity of the string as function of \(x\), we can find exact solution to the wave PDE.