Complete the plucked string problem to get equation 4.0
Solution
Here we start with the solution given in 4.8
\[ y_{0}={\displaystyle \sum \limits _{n=1}^{\infty }} b_{n}\sin \left (\frac {n\pi x}{L}\right ) =f\relax (x) \quad (1) \]
Where \(f\relax (x) \) represents the initial position (shape) of the string.
Now need to find \(b_{n}\)
First need to define \(f\relax (x) \), from diagram we see that from \(x=0\) to \(x=L/2\) the slope is \(\frac {h}{L/2}=\frac {2h}{L}\) hence from equation of line we get \(\ y=\frac {2h}{L}x\)
From \(x=L/2\) to \(x=L\), slope is \(-\frac {2h}{L}\), so \(y=h-\frac {2h}{L}\left ( x-\frac {L}{2}\right ) =h-\frac {2h}{L}x+h=2h-\frac {2h}{L}x=\ 2\left ( h-\frac {hx}{L}\right )\)
so we have \[ f\relax (x) =\left \{ \ \begin {array} [c]{lll}\frac {2h}{L}x & & 0\leq x\leq \frac {L}{2}\\ 2\left (h-\frac {hx}{L}\right ) & & \frac {L}{2}<x\leq L \end {array} \right . \]
so from \(\relax (1) \) we get, after applying inner product w.r.t. \(\sin \left (\frac {n\pi x}{L}\right ) \)
\begin {align*} b_{n}{\displaystyle \int \limits _{0}^{L}} \sin ^{2}\left (\frac {n\pi x}{L}\right ) & ={\displaystyle \int \limits _{0}^{L}} f\relax (x) \sin \left (\frac {n\pi x}{L}\right ) \ dx\\ b_{n}\ \frac {L}{2} & ={\displaystyle \int \limits _{0}^{\frac {L}{2}}} f\relax (x) \sin \left (\frac {n\pi x}{L}\right ) \ dx+{\displaystyle \int \limits _{\frac {L}{2}}^{L}} f\relax (x) \sin \left (\frac {n\pi x}{L}\right ) \ dx\\ b_{n}\ \frac {L}{2} & ={\displaystyle \int \limits _{0}^{\frac {L}{2}}} \frac {2h}{L}x\ \sin \left (\frac {n\pi x}{L}\right ) \ dx+{\displaystyle \int \limits _{\frac {L}{2}}^{L}} 2\left (h-\frac {hx}{L}\right ) \sin \left (\frac {n\pi x}{L}\right ) \ dx\\ b_{n}\ \frac {L}{2} & =\frac {2h}{L}{\displaystyle \int \limits _{0}^{\frac {L}{2}}} \ x\ \sin \left (\frac {n\pi x}{L}\right ) \ dx+{\displaystyle \int \limits _{\frac {L}{2}}^{L}} 2h\ \sin \left (\frac {n\pi x}{L}\right ) \ dx-{\displaystyle \int \limits _{\frac {L}{2}}^{L}} \frac {2hx}{L}\sin \left (\frac {n\pi x}{L}\right ) \ dx\\ b_{n}\ \frac {L}{2} & =\frac {2h}{L}{\displaystyle \int \limits _{0}^{\frac {L}{2}}} \ x\ \sin \left (\frac {n\pi x}{L}\right ) \ dx+2h{\displaystyle \int \limits _{\frac {L}{2}}^{L}} \sin \left (\frac {n\pi x}{L}\right ) \ dx-\frac {2h}{L}{\displaystyle \int \limits _{\frac {L}{2}}^{L}} x\sin \left (\frac {n\pi x}{L}\right ) \ dx\\ b_{n}\ \frac {L}{2} & =\frac {16\ h\ L\cos \left (\frac {n\pi }{4}\right ) \sin \left (\frac {n\pi }{4}\right ) ^{3}}{n^{2}\pi ^{2}}\\ b_{n} & =\frac {32\ h\ \cos \left (\frac {n\pi }{4}\right ) \sin \left ( \frac {n\pi }{4}\right ) ^{3}}{n^{2}\pi ^{2}} \end {align*}
so
\[ b_{n}\ \ =\frac {32\ h\ L\cos \left (\frac {n\pi }{4}\right ) \sin \left ( \frac {n\pi }{4}\right ) ^{3}}{n^{2}\pi ^{2}}\]
Looking at few values of n to see the pattern
\begin {align*} b_{n} & =\frac {32\ h\ \cos \left (\frac {\pi }{4}\right ) \sin \left (\frac {\pi }{4}\right ) ^{3}}{1^{2}\pi ^{2}},\frac {32\ h\ \cos \left (\frac {2\pi }{4}\right ) \sin \left (\frac {2\pi }{4}\right ) ^{3}}{2^{2}\pi ^{2}},\frac {32\ h\ \cos \left (\frac {3\pi }{4}\right ) \sin \left (\frac {3\pi }{4}\right ) ^{3}}{3^{2}\pi ^{2}},...\\ & =\ \frac {8h}{\pi ^{2}},0,-\frac {8h}{9\ \pi ^{2}},0,\frac {8h}{25\ \pi ^{2}},\dots \\ & =\frac {8h}{\pi ^{2}}\left (1,0,-\frac {1}{9},0,\frac {1}{25},\dots \right ) \end {align*}
Notice that we have terms for only odd n.
Now, substituting the above in the general solution given in equation 4.7 in book, which is
\[ y={\displaystyle \sum \limits _{n=1}^{\infty }} b_{n}\sin \left (\frac {n\pi x}{L}\right ) \cos \left (\frac {n\pi vt}{L}\right ) \]
Gives
\begin {align*} y & =\ \frac {8h}{\pi ^{2}}\left (\sin \left (\frac {\pi x}{L}\right ) \cos \left (\frac {\pi vt}{L}\right ) +0+-\frac {1}{9}\sin \left (\frac {3\pi x}{L}\right ) \cos \left (\frac {3\pi vt}{L}\right ) +0+\frac {1}{25}\sin \left ( \frac {5\pi x}{L}\right ) \cos \left (\frac {5\pi vt}{L}\right ) +...\right ) \\ y & =\frac {8h}{\pi ^{2}}\left (\sin \left (\frac {\pi x}{L}\right ) \cos \left (\frac {\pi vt}{L}\right ) -\frac {1}{9}\sin \left (\frac {3\pi x}{L}\right ) \cos \left (\frac {3\pi vt}{L}\right ) +\frac {1}{25}\sin \left ( \frac {5\pi x}{L}\right ) \cos \left (\frac {5\pi vt}{L}\right ) +...\right ) \end {align*}
The above is the result we are asked to show.
A string of length L has zero initial velocity and a displacement \(y_{0}\left ( x\right ) \) as shown. Find the displacement as a function of x and t.
Solution
The PDE that governs this problem is the wave equation \(\nabla ^{2}y=\frac {1}{v^{2}}\frac {\partial ^{2}y}{\partial t^{2}}\)
The candidate solutions are
\[ y=\left \{ \begin {array}[c]{l} \sin (kx)\,\sin (\omega t) \\ \sin (kx)\,\cos (\omega t) \\ \cos (kx)\,\sin (\omega t) \\ \cos (kx)\,\cos (\omega t) \end {array} \right . \]
where \(\omega =kv\) and \(k=\frac {2\pi }{\lambda }\) where \(\lambda \) is the wave length
Now we discard solutions that contains \(\cos kx\) since the string is fixed at \(x=0\).
So we are left with
\[ y=\left \{ \begin {array} [c]{l}\sin (kx)\,\sin (\omega t) \\ \sin (kx)\,\cos (\omega t) \end {array} \right . \]
Now, \(y=0\) at \(x=L\) then from \(\sin kx=0\) or \(\sin kL=0\ \)we need \(k=\frac {n\pi }{L}\)
Hence solutions become
\[ y=\left \{ \begin {array} [c]{l}\sin (\frac {n\pi }{L}x)\, \sin (\frac {n\pi }{L}vt) \\ \sin (\frac {n\pi }{L}x)\, \cos (\frac {n\pi }{L}vt) \end {array} \right . \]
Applying initial conditions, which says that at time \(t=0\), velocity is zero.
Hence from above, after taking \(\frac {\partial y}{\partial t}\), we get
\[ \frac {\partial y}{\partial t}=\left \{ \begin {array}[c]{l}\frac {n\pi v}{L}\ \sin (\frac {n\pi }{L}x)\ \cos (\frac {n\pi v}{L}t) \\ -\frac {n\pi v}{L}\ \sin (\frac {n\pi }{L}x)\ \sin (\frac {n\pi v}{L}t) \end {array} \right . \]
For the above to be zero at \(t=0\) then we discard first solution above with \(\cos t\) in it. Hence final general solution is now
\[ y=\left \{ {}\right . \begin {array} [c]{l}\sin (\frac {n\pi }{L}x)\ \cos (\frac {n\pi }{L}vt) \end {array} \]
A general solution is a linear combination of the above solutions, hence
\[ y={\displaystyle \sum \limits _{n=1}^{\infty }} \ b_{n}\ \sin \left (\frac {n\pi }{L}x\right )\ \cos \left (\frac {n\pi }{L}vt\right ) \qquad (1) \]
To find \(b_{n}\), we apply the second initial condition, which is \(y=y_{0}=f\relax (x) \)
(Notice that we use two initial conditions, i.e. at time t=0 we are looking at speed and position, this is because we started with a PDE with \(\frac {\partial ^{2}y}{\partial t^{2}}\) in it, which is a second order in t.)
At t=0, (1) becomes
\[ y={\displaystyle \sum \limits _{n=1}^{\infty }} \ b_{n} \sin \left (\frac {n\pi }{L}x\right ) = f(x) \qquad (2) \]
To find \(f\relax (x)\) from diagram, we see that for \(0\leq x\leq \frac {L}{4}\), \(y=x\frac {h}{L/4}=\frac {4h}{L}x\)
For \(\frac {L}{4}<x\leq \frac {L}{2}\), \(y=-\left (x-\frac {L}{4}\right ) \frac {h}{L/4}+h=-\left (x-\frac {L}{4}\right ) \frac {4h}{L}+h=-x\frac {4h}{L}+\frac {L}{4}\frac {4h}{L}+h=-x\frac {4h}{L}+2h\)
For \(\frac {L}{2}<x\leq L\), \(y=0\)
Hence
\[ y=\left \{ \begin {array} [c]{lll}\frac {4h}{L}x & & 0\leq x\leq \frac {L}{4}\\ 2h-x\frac {4h}{L} & & \frac {L}{4}<x\leq \frac {L}{2}\\ 0 & & \frac {L}{2}<x\leq L \end {array} \right . \]
Do the inner product on both sides of equation (2) w.r.t. \(\sin \frac {n\pi }{L}x\)
\begin {align*} \ \ b_{n}\int _{0}^{L}\sin ^{2}\frac {n\pi }{L}x\ \ dx & =\int _{0}^{L}f\left ( x\right ) \ \sin \frac {n\pi }{L}x\ \ dx\ \ \\ b_{n}\frac {L}{2} & =\ \int _{0}^{\frac {L}{4}}f\relax (x) \ \sin \frac {n\pi }{L}x\ dx\ +\ \int _{\frac {L}{4}}^{\frac {L}{2}}f\relax (x) \ \sin \frac {n\pi }{L}x\ dx+\int _{\frac {L}{2}}^{L}f\relax (x) \ \sin \frac {n\pi }{L}x\ dx\\ & =\int _{0}^{\frac {L}{4}}\frac {4h}{L}x\ \sin \frac {n\pi }{L}x\ dx+\int _{\frac {L}{4}}^{\frac {L}{2}}\left (2h-x\frac {4h}{L}\right ) \ \sin \frac {n\pi }{L}x\ dx+\int _{\frac {L}{2}}^{L}0\ \sin \frac {n\pi }{L}x\ dx\\ & =\int _{0}^{\frac {L}{4}}\frac {4h}{L}x\ \sin \frac {n\pi }{L}x\ dx+\int _{\frac {L}{4}}^{\frac {L}{2}}2h\sin \left (\frac {n\pi }{L}x\right ) -x\frac {4h}{L}\sin \left (\frac {n\pi }{L}x\right ) \ \ dx\\ & =\int _{0}^{\frac {L}{4}}\frac {4h}{L}x\ \sin \frac {n\pi }{L}x\ dx+\int _{\frac {L}{4}}^{\frac {L}{2}}2h\sin \left (\frac {n\pi }{L}x\right ) dx-\int _{\frac {L}{4}}^{\frac {L}{2}}x\frac {4h}{L}\sin \left (\frac {n\pi }{L}x\right ) \ \ dx\\ & =\frac {4h}{L}\int _{0}^{\frac {L}{4}}x\ \sin \frac {n\pi }{L}x\ dx+2h\int _{\frac {L}{4}}^{\frac {L}{2}}\sin \left (\frac {n\pi }{L}x\right ) dx-\frac {4h}{L}\int _{\frac {L}{4}}^{\frac {L}{2}}x\sin \left (\frac {n\pi }{L}x\right ) \ \ dx\\ b_{n} & =\ \frac {8h}{n^{2}\pi ^{2}}\left (2\sin \left (\frac {n\pi }{4}\right ) -\sin \frac {n\pi }{2}\right ) \end {align*}
Looking at few values of \(b_{n}\)
\begin {align*} b_{n} & =\frac {8h}{1^{2}\pi ^{2}}\left (2\sin \left (\frac {\pi }{4}\right ) -\sin \frac {\pi }{2}\right ) ,\frac {8h}{2^{2}\pi ^{2}}\left (2\sin \left ( \frac {2\pi }{4}\right ) -\sin \frac {2\pi }{2}\right ) ,\frac {8h}{3^{2}\pi ^{2}}\left (2\sin \left (\frac {3\pi }{4}\right ) -\sin \frac {3\pi }{2}\right ) ,\dots \\ & =\frac {8h}{\pi ^{2}}\left [ \left (2\sin \left (\frac {\pi }{4}\right ) -\sin \frac {\pi }{2}\right ) ,\frac {1}{2^{2}}\left (2\sin \frac {2\pi }{4}-\sin \frac {2\pi }{2}\right ) ,\frac {1}{3^{2}}\left (2\sin \frac {3\pi }{4}-\sin \frac {3\pi }{2}\right ) ,\dots \right ] \\ & =\frac {8h}{\pi ^{2}}\left [ \frac {1}{n^{2}}\left \{ \left (2\sin \left ( \frac {\pi }{4}\right ) -\sin \frac {\pi }{2}\right ) ,\left (2\sin \frac {2\pi }{4}-\sin \frac {2\pi }{2}\right ) ,\left (2\sin \frac {3\pi }{4}-\sin \frac {3\pi }{2}\right ) ,\dots \right \} \right ] \\ & =\frac {8h}{\pi ^{2}} \left [ \frac {1}{n^{2}}\left (2\sin \left (\frac {n\pi }{4}\right ) -\sin \frac {n\pi }{2}\right ) \right ] \end {align*}
Hence from equation (1) above, we get
\begin {align*} y & ={\displaystyle \sum \limits _{n=1}^{\infty }} \ b_{n}\sin \frac {n\pi }{L}x\cos \frac {n\pi }{L}vt\quad \\ & ={\displaystyle \sum \limits _{n=1}^{\infty }} \ \frac {8h}{\pi ^{2}}\left [ \frac {1}{n^{2}}\left (2\sin \left (\frac {n\pi }{4}\right ) -\sin \frac {n\pi }{2}\right ) \right ] \ \sin \frac {n\pi }{L}x\cos \frac {n\pi }{L}vt \quad \\ & =\frac {8h}{\pi ^{2}}{\displaystyle \sum \limits _{n=1}^{\infty }} B_{n}\ \sin \frac {n\pi }{L}x\cos \frac {n\pi }{L}vt\ \end {align*}
Where \[ B_{n}=\frac {1}{n^{2}}\left (2\ \sin \left (\frac {n\pi }{4}\right ) -\sin \frac {n\pi }{2}\right ) \]
The above is the result required to show.
A string of length L is initially stretched straight, its ends are fixed for all time t. At time t=0 its points are given the velocity \(V\relax (x) =\left (\frac {\partial y}{\partial t}\right ) _{t=0}\ \)as shown in diagram below. Determine the shape of the string at time t.
Solution
The PDE that governs this problem is the wave equation \(\nabla ^{2}y=\frac {1}{v^{2}}\frac {\partial ^{2}y}{\partial t^{2}}\)
The candidate solutions are
\[ y=\left \{ \begin {array}[c]{l} \sin (kx)\,\sin (\omega t) \\ \sin (kx)\,\cos (\omega t) \\ \cos (kx)\,\sin (\omega t) \\ \cos (kx)\,\cos (\omega t) \end {array} \right . \]
Where \(\omega =kv\) and \(k=\frac {2\pi }{\lambda }\) where \(\lambda \) is the wave length
Now we discard solutions that contains \(\cos kx\) since the string is fixed at \(x=0\).
So we are left with
\[ y=\left \{ \begin {array} [c]{l}\sin (kx)\,\sin (\omega t) \\ \sin (kx)\,\cos (\omega t) \end {array} \right . \]
Now, \(y=0\) at \(x=L\) then from \(\sin kx=0\) or \(\sin kL=0\ \)we need \(k=\frac {n\pi }{L}\)
Hence solutions become
\[ y=\left \{ \begin {array} [c]{lll}\sin \frac {n\pi }{L}x\sin \frac {n\pi }{L}vt & & \\ \sin \frac {n\pi }{L}x\cos \frac {n\pi }{L}vt & & \end {array} \right . \]
Applying initial conditions, which says that at time \(t=0\), velocity is given by \(V\relax (x) \)
Hence from above, after taking \(\frac {\partial y}{\partial t}\), we get
\[ \frac {\partial y}{\partial t}=\left \{ \begin {array} [c]{lll}\frac {n\pi v}{L}\ \sin \frac {n\pi }{L}x\cos \frac {n\pi v}{L}t & & \\ -\frac {n\pi v}{L}\sin \frac {n\pi }{L}x\sin \frac {n\pi v}{L}t & & \end {array} \right . \]
For the above we discard velocity solution above with \(\sin t\) in it since that will give zero velocity at time t=0, which is not the case here. Hence we discard y solution with \(\cos t\) in it, then the final general solution for y is now
\[ y=\begin {array} [c]{lll}\sin \frac {n\pi }{L}x\sin \frac {n\pi }{L}vt & & \end {array} \]
A general solution is a linear combination of the above solutions, hence
\[ y={\displaystyle \sum \limits _{n=1}^{\infty }} \ b_{n}\ \sin \frac {n\pi x}{L}\ \sin \frac {n\pi vt}{L} \qquad (1) \]
To find \(b_{n}\), we apply the velocity initial condition. Hence differentiate equation (1) and set t=0, we have
\[ \frac {\partial y}{\partial t}={\displaystyle \sum \limits _{n=1}^{\infty }} \ b_{n}\ \frac {n\pi v}{L}\sin \frac {n\pi x}{L}\cos \frac {n\pi vt}{L} \]
Setting t=0
\[ \frac {\partial y}{\partial t}={\displaystyle \sum \limits _{n=1}^{\infty }} \ b_{n}\ \frac {n\pi v}{L}\sin \frac {n\pi x}{L}=V_{t=0}\qquad (2) \]
Now to find \(V_{t=0}\). From diagram, we see that for \(0\leq x\leq \frac {L}{2}-w\), \(V_{t=0}=0\)
For \(\frac {L}{2}-w<x\leq \frac {L}{2}+w\),\(V_{t=0}=h\)
For \(\frac {L}{2}+w<x\leq L\), \(V_{t=0}=0\)
Hence
\[ V_{t=0}=\left \{ \begin {array} [c]{lll}0 & & 0\leq x\leq \frac {L}{2}-w\\ h & & \frac {L}{2}-w<x\leq \frac {L}{2}+w\\ 0 & & \frac {L}{2}+w<x\leq L \end {array} \right . \]
Do the inner product on both sides of equation (2) w.r.t. \(\sin \frac {n\pi }{L}x\)
\begin {align*} \ \ b_{n}\frac {n\pi v}{L}\int _{0}^{L}\sin ^{2}\frac {n\pi }{L}x\ \ dx & =\int _{0}^{L}V\relax (x) \ \sin \frac {n\pi x}{L}\ \ dx\ \ \\ b_{n}\frac {n\pi v}{2} & =\ \int _{0}^{\frac {L}{2}-w}0\ \sin \frac {n\pi }{L}x\ dx\ +\ \int _{\frac {L}{2}-w}^{\frac {L}{2}+w}h\ \sin \frac {n\pi x}{L}\ dx+\int _{\frac {L}{2}+w}^{L}0\ \sin \frac {n\pi }{L}x\ dx\\ b_{n}\frac {n\pi v}{2} & =\ \int _{\frac {L}{2}-w}^{\frac {L}{2}+w}h\ \sin \frac {n\pi x}{L}\ dx\\ b_{n}\frac {n\pi v}{2} & =\ -h\ \frac {L}{n\pi }\left [ \cos \frac {n\pi x}{L}\right ] _{\frac {L}{2}-w}^{\frac {L}{2}+w}\\ b_{n}\frac {n\pi v}{2} & =\ -h\ \frac {L}{n\pi }\left [ \cos \frac {n\pi \left ( \frac {L}{2}+w\right ) }{L}-\cos \frac {n\pi \left (\frac {L}{2}-w\right ) }{L}\right ] \\ b_{n}\frac {n\pi v}{2} & =\ -h\ \frac {L}{n\pi }\left [ \cos \left (\frac {n\pi }{2}+\frac {n\pi w}{L}\right ) -\cos \left (\frac {n\pi }{2}-\frac {n\pi w}{L}\right ) \right ] \\ b_{n} & =\ -\frac {2hL}{n^{2}\pi ^{2}v}\left [ \cos \left (\frac {n\pi }{2}+\frac {n\pi w}{L}\right ) -\cos \left (\frac {n\pi }{2}-\frac {n\pi w}{L}\right ) \right ] \end {align*}
But \(\cos \left (a+b\right ) =\cos \relax (a) \cos \relax (b) -\sin \relax (a) \sin \relax (b) \)
and \(\cos \left (a-b\right ) =\cos \relax (a) \cos \relax (b) +\sin \relax (a) \sin \relax (b) \)
Let \(a=\frac {n\pi }{2},b=\frac {n\pi w}{L}\)
Hence \(b_{n}\) becomes
\begin {align*} b_{n} & =-\frac {2hL}{n^{2}\pi ^{2}v}\left [ \cos \left (a+b\right ) -\cos \left (a-b\right ) \right ] \\ & =-\frac {2hL}{n^{2}\pi ^{2}v}\left [ \cos \relax (a) \cos \left ( b\right ) -\sin \relax (a) \sin \relax (b) -\left \{ \cos \left ( a\right ) \cos \relax (b) +\sin \relax (a) \sin \relax (b) \right \} \right ] \\ & =-\frac {2hL}{n^{2}\pi ^{2}v}\left [ \cos \relax (a) \cos \left ( b\right ) -\sin \relax (a) \sin \relax (b) -\cos \relax (a) \cos \relax (b) -\sin \relax (a) \sin \relax (b) \right ] \\ & =-\frac {2hL}{n^{2}\pi ^{2}v}\left [ -\sin \relax (a) \sin \left ( b\right ) \ -\sin \relax (a) \sin \relax (b) \right ] \\ & =\frac {4hL}{n^{2}\pi ^{2}v}\sin \relax (a) \sin \relax (b) \ \\ & =\frac {4hL}{n^{2}\pi ^{2}v}\sin \left (\frac {n\pi }{2}\right ) \sin \left ( \frac {n\pi w}{L}\right ) \ \end {align*}
For even \(n\), the term \(\sin \left (\frac {n\pi }{2}\right ) \) is zero. For \(n\) odd \(\sin \left (\frac {n\pi }{2}\right ) =1\) when \(n=1,5,9,\dots \) and \(\sin \left (\frac {n\pi }{2}\right ) =-1\) when \(n=3,7,11,\dots \)Hence
\[ b_{n}=A\relax (n) \frac {4hL}{n^{2}\pi ^{2}v}\sin \left (\frac {n\pi w}{L}\right )\qquad n=1,3,5,7,\dots \]
And \(A\relax (n) \) is a function which returns \(1\) when \(n=1,5,9,..\) and returns \(-1\) when \(n=3,7,11,\dots \)
Hence now we have \(b_{n}\) we can substitute in (1)
\begin {align*} y & ={\displaystyle \sum \limits _{n=1}^{\infty }} \ b_{n}\ \sin \frac {n\pi x}{L}\ \sin \frac {n\pi vt}{L}\ \ \ \ \ \\ y & ={\displaystyle \sum \limits _{n\ odd}^{\infty }} \ A\relax (n) \frac {4hL}{n^{2}\pi ^{2}v}\ \sin \left (\frac {n\pi w}{L}\right ) \ \left [ \ \sin \frac {n\pi x}{L}\ \sin \frac {n\pi vt}{L}\right ] \ \ \ \ \ \ \ \\ y & =\ \ \frac {4hL}{\pi ^{2}v}{\displaystyle \sum \limits _{n\ odd}^{\infty }} A\relax (n) \ \frac {1}{n^{2}}\ \sin \left (\frac {n\pi w}{L}\right ) \ \left [ \ \sin \frac {n\pi x}{L}\ \sin \frac {n\pi vt}{L}\right ] \end {align*}
Which is the general solution. Looking at few expanded terms in the series we get
\(y=\frac {4hL}{\pi ^{2}v}\ \left \{ \sin \left (\frac {\pi w}{L}\right ) \ \sin \frac {\pi x}{L}\ \sin \frac {\pi vt}{L}-\frac {1}{9}\ \sin \left ( \frac {3\pi w}{L}\right ) \ \ \sin \frac {3\pi x}{L}\ \sin \frac {3\pi vt}{L}+\ \frac {1}{25}\ \sin \left (\frac {5\pi w}{L}\right ) \ \sin \frac {5\pi x}{L}\ \sin \frac {n\pi vt}{L}\right \} \)
Which is the result required.
Compute numerically the coefficients \(\ c_{m}=\frac {200}{k_{m}J_{1}\left ( k_{m}\right ) }\) for the first 3 terms of the series \(u={\displaystyle \sum \limits _{m=1}^{\infty }} c_{m}J_{0}\left (k_{m}r\right ) e^{-k_{m}z}\) for the steady state temp. in a solid semi-infinite cylinder when \(u=0\) at \(r=1\) and \(u=100\) at \(z=0.\) find \(u\) at \(r=1/2,z=1\)
Solution
Here, we are looking at the solution for temp. inside a semi-infinite cylinder. This solution is for the case of a uniform temp. distribution on the boundary \(z=0\) is given by \(u\) equation shown above. Note that in the expression \(c_{m}=\frac {200}{k_{m}J_{1}\left (k_{m}\right ) }\), the \(k_{m}\) are the zeros of \(J_{0}\) not \(J_{1}.\)
Need to find \(c_{1},c_{2},c_{3}\) where \(c_{1}=\frac {200}{k_{1}J_{1}\left (k_{1}\right ) }\)
To find \(k_{1}\) and \(J_{1}\left (k_{1}\right ) \) I used mathematica.
I plotted \(J_{0}\relax (x) \) to see where the zeros are located first
So I see there is a zero near 2,5, and 9. I use mathematica to find these:
Now I need to find \(J_{1}\left (k_{m}\right ) \). This is the result for 3 terms:
Hence, now the \(c_{m}\) terms can be found:
\begin {align*} c_{1} & =\frac {200}{k_{1}J_{1}\left (k_{1}\right ) }=\frac {200}{\left ( 2.404\right ) \left (0.519\right ) }=\allowbreak 160.\,\allowbreak 30\\ c_{2} & =\frac {200}{k_{2}J_{1}\left (k_{2}\right ) }=\frac {200}{\left ( 5.52\right ) \left (-0.34\right ) }=-106.\,\allowbreak 56\\ c_{3} & =\frac {200}{k_{3}J_{1}\left (k_{3}\right ) }=\frac {200}{\left ( 8.65\right ) \left (0.27\right ) }=\allowbreak 85.\,\allowbreak 635 \end {align*}
Evaluating \(u={\displaystyle \sum \limits _{m=1}^{\infty }} c_{m}J_{0}\left (k_{m}r\right ) e^{-k_{m}z}\) for the first 3 terms when \(r=1/2,z=1\)
\begin {align*} u & =c_{1}J_{0}\left (k_{1}r\right ) e^{-k_{1}z}+c_{2}J_{0} \left (k_{2}r\right ) e^{-k_{2}z}+c_{3}J_{0}\left (k_{3}r\right ) e^{-k_{3}z}\\ & =c_{1}J_{0}\left (k_{1}\frac {1}{2}\right ) e^{-k_{1}}+c_{2}J_{0}\left ( k_{2}\frac {1}{2}\right ) e^{-k_{2}}+c_{3}J_{0}\left (k_{3}\frac {1}{2}\right )e^{-k_{3}}\\ & =\left (160.30\right ) J_{0}\left (2.404\frac {1}{2}\right ) e^{-2.404}-\left (106.56\right ) J_{0}\left (5.52\frac {1}{2}\right ) e^{-5.52}+\left (85.635\right ) J_{0}\left (8.65\frac {1}{2}\right ) e^{-8.65}\\ & =\left (160.30\right ) J_{0}\left (1.202\right ) e^{-2.404}-\left (106.56\right ) J_{0}\left (2.76\right ) e^{-5.52}+\left (85.635\right ) J_{0}\left (4.325\right ) e^{-8.65} \end {align*}
Mathematica was used to evaluate \(J_{0}\) values above.
Hence
\begin {align*} u & =\left (160.30\right ) \left (0.67\right ) e^{-2.404}-\left (106.56\right ) \left (-0.168\right ) e^{^{-5.52}}+\left (85.635\right ) \left (-0.356\right ) e^{-8.65}\\ u & =9.704\,3+7.171\,3\times 10^{-2}-5.3389\times 10^{-3}\\ u & =9.7707\ \text {degrees} \end {align*}
Find the solution for the steady state temp. distribution in a solid semi-infinite cylinder if the boundary temp. are \(u=0\) at \(r=1\) and \(u=y=r\sin \theta \) at \(z=0\).
Solution
The candidate solutions are given by the solution to the Laplace equation in cylindrical coordinates which are
\[ u=\left \{ \begin {array} [c]{lll}J_{n}\left (k\ r\right ) \sin \left (n\theta \right ) e^{-k\ z} & & (1)\\ & & \\ J_{n}\left (k\ r\right ) \cos \left (n\theta \right ) e^{-k\ z} & & (2) \end {array} \right . \]
Where \(k\) is a zero of \(J_{n}\) (This is because we have used the B.C. of \(u=0\) at \(r=1\) to determine that the \(k^{\prime }s\) have to be the zeros of \(J_{n}\)) when deriving the above solutions. See book page 560.
From boundary conditions we want \(u=r\sin \theta \) when \(z=0\), hence we need to keep the solution (1) above, with \(n=1\). Hence a solution is \[ u=J_{1}\left (k\ r\right ) \sin \left (\theta \right ) e^{-k\ z}\qquad (3) \]
A general solution is a linear series combinations (eigenfunctions) of (3), each eigenfunction for each of the zeros of \(J_{1}\). Call these zeros \(k_{m}\)
\[ u=\sum _{m=1}^{\infty }c_{m}\ J_{1}\left (k_{m}\ r\right ) \sin \left ( \theta \right ) e^{-k_{m}\ z}\qquad (4) \]
We now apply B.C. at \(z=0\) to find \(c_{m}\). From (4) when \(z=0\) \[ r\sin \theta =\sum _{m=1}^{\infty }c_{m}\ J_{1}\left (k_{m}\ r\right ) \sin \left (\theta \right ) \qquad (5) \]
We use (5) to find \(c_{m}\) and then substitute into (4) to obtain the final solution.
To find \(c_{m}\) from (5), take the inner product of each side with respect to \(rJ_{1}\left (k_{u}\ r\right ) \) from \(r=0\) to \(r=1\)
\begin {align*} \int _{0}^{1}r\sin \theta \left [ rJ_{1}\left (k_{u}\ r\right ) \right ] \ dr & =\sum _{m=1}^{\infty }c_{m}\ \left (\int _{0}^{1}J_{1}\left (k_{m}\ r\right ) \sin \left (\theta \right ) \left [ rJ_{1}\left (k_{u}\ r\right ) \right ] \ dr\right ) \\ \sin \theta \int _{0}^{1}r^{2}J_{1}\left (k_{u}\ r\right ) \ dr & =\sum _{m=1}^{\infty }c_{m}\ \sin \left (\theta \right ) \left (\int _{0}^{1}J_{1}\left (k_{m}\ r\right ) \left [ rJ_{1}\left (k_{u}\ r\right ) \right ] \ dr\right ) \end {align*}
Dividing each side by \(\sin \theta \)
\[ \int _{0}^{1}r^{2}J_{1}\left (k_{u}\ r\right ) \ dr=\sum _{m=1}^{\infty }c_{m}\ \left (\int _{0}^{1}J_{1}\left (k_{m}\ r\right ) \left [ rJ_{1}\left ( k_{u}\ r\right ) \right ] \ dr\right ) \]
From orthogonality of Bessel function, we know that
\[ \int _{0}^{1}J_{p}\left (k_{m}\ r\right ) r J_{p}\left (k_{u}\ r\right ) dr=0 \]
If \(m\neq u\). Hence in above equation all terms on the right drop except for one when \(u=m\). We get
\[ \int _{0}^{1}r^{2}J_{1}\left (k_{m}\ r\right ) \ dr=\ c_{m}\ \int _{0}^{1}r\ \ J_{1}\left (k_{m}\ r\right ) J_{1}\left (k_{m}\ r\right ) dr \]
Or
\[ c_{m}=\frac {\int _{0}^{1}r^{2}J_{1}\left (k_{m}\ r\right ) \ dr}{\int _{0}^{1}r\ \ J_{1}\left (k_{m}\ r\right ) \ \ J_{1}\left (k_{m}\ r\right ) \ dr}\qquad (6) \]
The integral in the denominator above is found from equation 19.10 in text on page 523 which gives \[ \int _{0}^{1}r\ \ J_{1}\left (k_{m}\ r\right ) \ \ J_{1}\left (k_{m}\ r\right ) \ dr=\frac {1}{2}\left [ J_{2}\left (k_{m}\right ) \right ] ^{2}\qquad (7) \]
Now, we need to find the integral of the numerator in equation (6).
Using equation 15.1 in text, page 514, which says
\[ \frac {d}{dx}\left [ x^{p}J_{p}\relax (x) \right ] =x^{p}J_{p-1}\left ( x\right ) \]
Putting \(p=2\) above, and letting \(x=k_{m}r\) gives
\begin {align*} \frac {1}{k_{m}}\frac {d}{dr}\left [ \left (k_{m}r\right ) ^{2}J_{2}\left ( k_{m}r\right ) \right ] & =\left (k_{m}r\right ) ^{2}J_{1}\left ( k_{m}r\right ) \\ \frac {1}{k_{m}}\frac {d}{dr}\left [ k_{m}^{2}r^{2}J_{2}\left (k_{m}r\right ) \right ] & =k_{m}^{2}r^{2}J_{1}\left (k_{m}r\right ) \\ \frac {1}{k_{m}}\frac {d}{dr}\left [ r^{2}J_{2}\left (k_{m}r\right ) \right ] & =r^{2}J_{1}\left (k_{m}r\right ) \end {align*}
Integrating each side w.r.t \(r\) from \(0\dots 1\)
\begin {align*} \frac {1}{k_{m}}\int _{0}^{1}\frac {d}{dr}\left [ r^{2}J_{2}\left ( k_{m}r\right ) \right ] \ dr & =\int _{0}^{1}r^{2}J_{1}\left (k_{m}r\right ) \ dr\\ \ \frac {1}{k_{m}}\ \left [ r^{2}J_{2}\left (k_{m}r\right ) \right ] _{0}^{1} & =\int _{0}^{1}r^{2}J_{1}\left (k_{m}r\right )\, dr\\ \frac {1}{k_{m}}\ \left [ J_{2}\left (k_{m}\right ) -0\right ] & =\int _{0}^{1}r^{2}J_{1}\left (k_{m}r\right ) \ dr\\ \frac {1}{k_{m}}\ J_{2}\left (k_{m}\right ) & =\int _{0}^{1}r^{2}J_{1}\left ( k_{m}r\right )\,dr\qquad (8) \end {align*}
Substituting (7) and (8) into (6)
\begin {align*} c_{m} & =\frac {\int _{0}^{1}r^{2}J_{1}\left (k_{m}\ r\right ) \ dr}{\int _{0}^{1}r\ \ J_{1}\left (k_{m}\ r\right ) \ \ J_{1}\left (k_{m}\ r\right ) \ dr}\ \ \ \ \\ \ & =\frac {\frac {1}{k_{m}}\ J_{2}\left (k_{m}\right ) }{\frac {1}{2}\left [ J_{2}\left (k_{m}\right ) \right ] ^{2}\ }\\ & =\frac {2\ }{k_{m}\ J_{2}\left (k_{m}\right )} \end {align*}
Substituting this into (4) above, we get
\begin {align*} u & =\sum _{m=1}^{\infty }c_{m}\ J_{1}\left (k_{m}\ r\right ) \sin \left ( \theta \right ) e^{-k_{m}\ z}\\ u & =\sum _{m=1}^{\infty }\frac {2\ }{k_{m}\ J_{2}\left (k_{m}\right ) \ }\ J_{1}\left (k_{m}\ r\right ) \ \sin \left (\theta \right ) \ e^{-k_{m}\ z} \end {align*}
where \(k_{m}\) are zeros of \(J_{1}\)
The above is the result we are asked to show.
A flat circular plate of radius 1 is initially at temp. 100\(^{0}\). From \(t=0\) on, the circumference of the plate is held at 0\(^{0}.\) Find the time-dependent temp distribution \(u\left (r,\theta ,t\right ) \)
Solution
First convert heat equation from Cartesian coordinates to polar.
heat equation in 2D Cartesian is \[ \nabla ^{2}u=\frac {\partial ^{2}}{\partial x^{2}}u+\frac {\partial ^{2}}{\partial x^{2}}u=\frac {1}{\alpha ^{2}}\frac {\partial }{\partial t}u \]
First need to express Laplacian operator \(\nabla ^{2}\) in polar coordinates: \begin {align*} x &= r\cos \theta \\ y &=r\sin \theta \\ \end {align*}
Hence
\begin {align} \frac {\partial x}{\partial r} & =\cos \theta \tag {A}\\ \frac {\partial y}{\partial r} & =\sin \theta \nonumber \end {align}
And
\begin {align} \frac {\partial x}{\partial \theta } & =-r\ \sin \theta \tag {B}\\ \frac {\partial y}{\partial \theta } & =r\ \cos \theta \nonumber \end {align}
From geometry, we also know that \begin {align*} r &=\sqrt {x^{2}+y^{2}}\\ \theta &=\arctan \frac {y}{x}\\ \end {align*}
The above 2 relations imply
\(\frac {\partial }{\partial r}=\frac {\partial x}{\partial r}\frac {\partial }{\partial x}+\frac {\partial y}{\partial r}\frac {\partial }{\partial y}\ \) and \(\frac {\partial }{\partial \theta }=\frac {\partial x}{\partial \theta }\frac {\partial }{\partial x}+\frac {\partial y}{\partial \theta }\frac {\partial }{\partial y}\)
Hence we can express the above, using equations (A) and (B) as follows:
\begin {align*} \frac {\partial }{\partial r} & =\frac {\partial x}{\partial r}\frac {\partial }{\partial x}+\frac {\partial y}{\partial r}\frac {\partial }{\partial y}\\ & =\cos \theta \frac {\partial }{\partial x}+\sin \theta \frac {\partial }{\partial y}\ \ \ \ \ \ \ \end {align*}
Multiply each side by \(r\)
\begin {align} r\frac {\partial }{\partial r} & =r\cos \theta \frac {\partial }{\partial x}+r\sin \theta \frac {\partial }{\partial y}\nonumber \\ & =x\ \frac {\partial }{\partial x}+y\frac {\partial }{\partial y}\qquad \tag {1} \end {align}
Squaring each sides of (1) gives
\begin {align} r\frac {\partial }{\partial r}\left (\ r\frac {\partial }{\partial r}\right ) & =\left (x\ \frac {\partial }{\partial x}+y\frac {\partial }{\partial y}\right ) ^{2}\nonumber \\ r\left (r\frac {\partial ^{2}}{\partial r^{2}}+\frac {\partial }{\partial r}\right ) & =\ \ x\frac {\partial }{\partial x}\ x\frac {\partial }{\partial x}+y\frac {\partial }{\partial y}\ y\frac {\partial }{\partial y}+2x\frac {\partial }{\partial x}\ y\frac {\partial }{\partial y}\ \nonumber \\ r^{2}\frac {\partial ^{2}}{\partial r^{2}}+r\frac {\partial }{\partial r} & =x\frac {\partial }{\partial x}\left (\ x\frac {\partial }{\partial x}\right ) +y\frac {\partial }{\partial y}\left (\ y\frac {\partial }{\partial y}\right ) +2x\frac {\partial }{\partial x}\ \left (y\frac {\partial }{\partial y}\right ) \nonumber \\ & =\ x\left (x\frac {\partial ^{2}}{\partial x^{2}}+\overbrace {\frac {\partial }{\partial x}x}^{=1}\frac {\partial }{\partial x}\right ) +y\left ( y\frac {\partial ^{2}}{\partial y^{2}}+\overbrace {\frac {\partial }{\partial y}y}^{=1}\frac {\partial }{\partial y}\right ) +2x\left (y\frac {\partial ^{2}}{\partial x\partial y}+\overbrace {\frac {\partial }{\partial x}y}^{=0}\frac {\partial }{\partial y}\right ) \nonumber \\ & =\ x^{2}\frac {\partial ^{2}}{\partial x^{2}}+x\frac {\partial }{\partial x}+y^{2}\frac {\partial ^{2}}{\partial y^{2}}+y\frac {\partial }{\partial y}+2xy\frac {\partial ^{2}}{\partial x\partial y}\tag {2} \end {align}
Notice that when manipulating of differential operators, \(x\frac {\partial }{\partial x}\neq \frac {\partial }{\partial x}x\). Similarly
\begin {align} \frac {\partial }{\partial \theta } & =\frac {\partial x}{\partial \theta }\frac {\partial }{\partial x}+\frac {\partial y}{\partial \theta }\frac {\partial }{\partial y}\nonumber \\ & =-r\ \sin \theta \frac {\partial }{\partial x}+r\ \cos \theta \frac {\partial }{\partial y} \nonumber \\ & =-y\frac {\partial }{\partial x}+x\frac {\partial }{\partial y}\tag {3} \end {align}
Squaring each side of (3) gives
\begin {align} \left (\frac {\partial }{\partial \theta }\right ) ^{2} & =\left ( -y\frac {\partial }{\partial x}+x\frac {\partial }{\partial y}\right ) ^{2}\nonumber \\ \frac {\partial }{\partial \theta }\frac {\partial }{\partial \theta } & =-y\frac {\partial }{\partial x}\left (-y\frac {\partial }{\partial x}\right ) +x\frac {\partial }{\partial y}\left (x\frac {\partial }{\partial y}\right ) -y\frac {\partial }{\partial x}\left (x\frac {\partial }{\partial y}\right ) +x\frac {\partial }{\partial y}\left (-y\frac {\partial }{\partial x}\right ) \nonumber \\ \frac {\partial ^{2}}{\partial \theta ^{2}} & =-y\ \left (-y\frac {\partial ^{2}}{\partial x^{2}}+\overbrace {\frac {\partial }{\partial x}\left (-y\right ) }^{=0}\frac {\partial }{\partial x}\right ) +x\left (x\frac {\partial ^{2}}{\partial y^{2}}+\overbrace {\frac {\partial }{\partial y}\relax (x) }^{=0}\frac {\partial }{\partial y}\right ) \nonumber \\ & -y\left (x\frac {\partial ^{2}}{\partial y\partial x}+\overbrace {\frac {\partial }{\partial x}x}^{=1}\frac {\partial }{\partial y}\right ) +x\left ( -y\frac {\partial ^{2}}{\partial x\partial y}+\overbrace {\frac {\partial }{\partial y}y}^{=1}\frac {\partial }{\partial x}\right ) \nonumber \\ & =y^{2}\frac {\partial ^{2}}{\partial x^{2}}+x^{2}\frac {\partial ^{2}}{\partial y^{2}}-yx\frac {\partial ^{2}}{\partial y\partial x}-y\ \frac {\partial }{\partial y}-xy\frac {\partial ^{2}}{\partial x\partial y}-x\frac {\partial }{\partial x}\nonumber \\ & =y^{2}\frac {\partial ^{2}}{\partial x^{2}}+x^{2}\frac {\partial ^{2}}{\partial y^{2}}-2yx\frac {\partial ^{2}}{\partial y\partial x}-y\ \frac {\partial }{\partial y}-x\frac {\partial }{\partial x}\tag {4} \end {align}
Adding equation (2) and (4) and carry cancellations
\(r^{2}\frac {\partial ^{2}}{\partial r^{2}}+r\frac {\partial }{\partial r}+\frac {\partial ^{2}}{\partial \theta ^{2}}=\left (x^{2}\frac {\partial ^{2}}{\partial x^{2}}+x\frac {\partial }{\partial x}+y^{2}\frac {\partial ^{2}}{\partial y^{2}}+y\frac {\partial }{\partial y}+2xy\frac {\partial ^{2}}{\partial x\partial y}\right ) +\left (y^{2}\frac {\partial ^{2}}{\partial x^{2}}+x^{2}\frac {\partial ^{2}}{\partial y^{2}}-2yx\frac {\partial ^{2}}{\partial y\partial x}-y\ \frac {\partial }{\partial y}-x\frac {\partial }{\partial x}\right ) \)
\(r^{2}\frac {\partial ^{2}}{\partial r^{2}}+r\frac {\partial }{\partial r}+\frac {\partial ^{2}}{\partial \theta ^{2}}=\left (x^{2}\frac {\partial ^{2}}{\partial x^{2}}+y^{2}\frac {\partial ^{2}}{\partial y^{2}}\right ) +\left ( y^{2}\frac {\partial ^{2}}{\partial x^{2}}+x^{2}\frac {\partial ^{2}}{\partial y^{2}}\ \right ) \)
Hence we get
\begin {align*} r^{2}\frac {\partial ^{2}}{\partial r^{2}}+r\frac {\partial }{\partial r}+\frac {\partial ^{2}}{\partial \theta ^{2}} & =x^{2}\frac {\partial ^{2}}{\partial x^{2}}+y^{2}\frac {\partial ^{2}}{\partial y^{2}}+y^{2}\frac {\partial ^{2}}{\partial x^{2}}+x^{2}\frac {\partial ^{2}}{\partial y^{2}}\ \\ & =\left (x^{2}+y^{2}\right ) \left (\frac {\partial ^{2}}{\partial x^{2}}+\frac {\partial ^{2}}{\partial y^{2}}\right ) \end {align*}
Dividing by \(r^{2}\)
\[ \frac {\partial ^{2}}{\partial r^{2}}+\frac {1}{r}\frac {\partial }{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}}{\partial \theta ^{2}}=\frac {\left ( x^{2}+y^{2}\right ) }{r^{2}}\left (\frac {\partial ^{2}}{\partial x^{2}}+\frac {\partial ^{2}}{\partial y^{2}}\right ) \ \]
But \(r^{2}=x^{2}+y^{2}\) hence
\[ \frac {\partial ^{2}}{\partial r^{2}}+\frac {1}{r}\frac {\partial }{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}}{\partial \theta ^{2}}=\ \left ( \frac {\partial ^{2}}{\partial x^{2}}+\frac {\partial ^{2}}{\partial y^{2}}\right ) \ =\nabla ^{2}\]
Now that we have the Laplacian in polar coordinates, we can solve the problem by applying separation of variables on the heat PDE expressed in polar coordinates.
\begin {equation} \frac {\partial ^{2}}{\partial r^{2}}u+\frac {1}{r}\frac {\partial }{\partial r}u+\frac {1}{r^{2}}\frac {\partial ^{2}}{\partial \theta ^{2}}u=\frac {1}{\alpha ^{2}}\frac {\partial }{\partial t}u\tag {5} \end {equation}
Let solution \(u\left (r,\theta ,t\right ) \) be a linear combination of functions each depends on only \(r,\theta ,\) or \(t\)
\begin {equation} u\left (r,\theta ,t\right ) =R\relax (r) \Theta \left (\theta \right ) T\relax (t) \tag {6} \end {equation}
Substitute (6) in (5). First evaluate the various derivatives:
\[ \frac {\partial }{\partial r}u=\Theta \left (\theta \right ) T\relax (t) \frac {\partial }{\partial r}R\relax (r) \]
\[ \frac {\partial ^{2}}{\partial r^{2}}u=\Theta \left (\theta \right ) T\left ( t\right ) \frac {\partial ^{2}}{\partial r^{2}}R\relax (r) \]
\[ \frac {\partial }{\partial \theta }u=R\relax (r) T\relax (t) \frac {\partial }{\partial \theta }\Theta \left (\theta \right ) \]
\[ \frac {\partial ^{2}}{\partial \theta ^{2}}u=R\relax (r) T\relax (t) \frac {\partial ^{2}}{\partial \theta ^{2}}\Theta \left (\theta \right ) \]
\[ \frac {\partial }{\partial t}u=R\relax (r) \Theta \left (\theta \right ) \frac {\partial }{\partial t}T\relax (t) \]
Hence equation (5) becomes
\begin {align*} \frac {\partial ^{2}}{\partial r^{2}}u+\frac {1}{r}\frac {\partial }{\partial r}u+\frac {1}{r^{2}}\frac {\partial ^{2}}{\partial \theta ^{2}}u & =\frac {1}{\alpha ^{2}}\frac {\partial }{\partial t}u\\ \Theta \left (\theta \right ) T\relax (t) \frac {d^{2}}{dr^{2}}R\left ( r\right ) +\frac {1}{r}\Theta \left (\theta \right ) T\relax (t) \frac {d}{dr}R\relax (r) +\frac {1}{r^{2}}R\relax (r) T\left ( t\right ) \frac {d^{2}}{d\theta ^{2}}\Theta \left (\theta \right ) & =\frac {1}{\alpha ^{2}}R\relax (r) \Theta \left (\theta \right ) \frac {d}{dt}T\relax (t) \end {align*}
Divide by \(R\relax (r) \Theta \left (\theta \right ) T\relax (t) \)
\begin {align*} \ \frac {1}{R\relax (r) }\frac {d^{2}}{dr^{2}}R\relax (r) +\frac {1}{r}\frac {1}{R\relax (r) }\frac {d}{dr}R\relax (r) +\frac {1}{r^{2}}\frac {1}{\Theta \left (\theta \right ) }\frac {d^{2}}{d\theta ^{2}}\Theta \left (\theta \right ) & =\frac {1}{\alpha ^{2}}\ \frac {1}{T\relax (t) }\frac {d}{dt}T\relax (t) \\ \frac {1}{R\relax (r) }\left [ \frac {d^{2}}{dr^{2}}R\relax (r) +\frac {1}{r}\frac {d}{dr}R\relax (r) \right ] +\frac {1}{r^{2}}\frac {1}{\Theta \left (\theta \right ) }\frac {d^{2}}{d\theta ^{2}}\Theta \left ( \theta \right ) & =\frac {1}{\alpha ^{2}}\ \frac {1}{T\relax (t) }\frac {d}{dt}T\relax (t) \end {align*}
We notice that the RHS depends only on \(t\) and the LHS depends only on \(r,\theta \) and they equal to each others, hence they both must be constant. Let this constant be \(-k^{2}\)
Hence
\begin {align} \frac {1}{\alpha ^{2}}\ \frac {1}{T\relax (t) }\frac {d}{dt}T\left ( t\right ) & =-k^{2}\tag {7}\\ \frac {1}{R\relax (r) }\left [ \frac {d^{2}}{dr^{2}}R\relax (r) +\frac {1}{r}\frac {d}{dr}R\relax (r) \right ] +\frac {1}{r^{2}}\frac {1}{\Theta \left (\theta \right ) }\frac {d^{2}}{d\theta ^{2}}\Theta \left ( \theta \right ) & =-k^{2}\tag {8} \end {align}
equation (7) is a linear first order ODE with constant coeff. \(\frac {d}{dt}T\relax (t) =-\alpha ^{2}T\relax (t) k^{2}\) or \(\frac {dT\relax (t) }{T\relax (t) }=-\alpha ^{2}k^{2}dt\)
Integrating to solve gives \begin {align*} \int \frac {dT\relax (t) }{T\relax (t) } &=\int -\alpha ^{2}k^{2}dt\\ \ln T\relax (t) &=-\alpha ^{2}k^{2}t\\ \end {align*}
or\begin {equation} T\relax (t) =e^{-\alpha ^{2}k^{2}t}\tag {9} \end {equation} Looking at equation (8). Multiply each sides by \(r^{2}\) we get\begin {align} \ \frac {r^{2}}{R\relax (r) }\left [ \frac {d^{2}}{dr^{2}}R\left ( r\right ) +\frac {1}{r}\frac {d}{dr}R\relax (r) \right ] +\frac {1}{\Theta \left (\theta \right ) }\frac {d^{2}}{d\theta ^{2}}\Theta \left ( \theta \right ) & =-r^{2}k^{2}\nonumber \\ \frac {r^{2}}{R\relax (r) }\left [ \frac {d^{2}}{dr^{2}}R\left ( r\right ) +\frac {1}{r}\frac {d}{dr}R\relax (r) \right ] +r^{2}k+\frac {1}{\Theta \left (\theta \right ) }\frac {d^{2}}{d\theta ^{2}}\Theta \left ( \theta \right ) & =0\nonumber \\ r^{2}\left (\frac {\ 1}{R\relax (r) }\left [ \frac {d^{2}}{dr^{2}}R\relax (r) +\frac {1}{r}\frac {d}{dr}R\relax (r) \right ] +k^{2}\right ) +\frac {1}{\Theta \left (\theta \right ) }\frac {d^{2}}{d\theta ^{2}}\Theta \left (\theta \right ) & =0\tag {10} \end {align}
The second term depends only on \(\theta \) and the first term depends only on \(r\) and they are equal, hence they must be both constant. Let this constant be \(-n^{2}\) hence
\begin {align*} \frac {1}{\Theta \left (\theta \right ) }\frac {d^{2}}{d\theta ^{2}}\Theta \left ( \theta \right ) & =-n^{2}\\ \ \frac {d^{2}}{d\theta ^{2}}\Theta \left (\theta \right ) & =-n^{2}\Theta \left (\theta \right ) \end {align*}
This is a second order linear ODE with constant coeff. Solution is
\begin {equation} \Theta \left (\theta \right ) =\left \{ \begin {array} [c]{lll}\sin n\theta & & \\ & & \\ \cos n\theta & & \end {array} \right . \tag {11} \end {equation}
From (10) we now have
\begin {align} r^{2}\left (\frac {\ 1}{R\relax (r) }\left [ \frac {d^{2}}{dr^{2}}R\relax (r) +\frac {1}{r}\frac {d}{dr}R\relax (r) \right ] +k^{2}\right ) -n^{2} & =0\nonumber \\ \frac {\ r^{2}}{R\relax (r) }\left [ \frac {d^{2}}{dr^{2}}R\left ( r\right ) +\frac {1}{r}\frac {d}{dr}R\relax (r) \right ] +r^{2}k^{2}-n^{2} & =0\nonumber \\ \ r^{2}\left [ \frac {d^{2}}{dr^{2}}R\relax (r) +\frac {1}{r}\frac {d}{dr}R\relax (r) \right ] +\left (r^{2}k^{2}-n^{2}\right ) R\left ( r\right ) & =0\nonumber \\ r^{2}\frac {d^{2}}{dr^{2}}R\relax (r) +r\frac {d}{dr}R\relax (r) +\left (r^{2}k^{2}-n^{2}\right ) R\relax (r) & =0\tag {12} \end {align}
Equation (12) is the Bessel D.E., its solutions are \(J_{n}\left (kr\right ) \) and \(N_{n}\left (kr\right ) \) . As described on book on page 560, we can not use the \(N_{n}\left (kr\right ) \) solution since plate contains the origin and \(N_{n}\relax (0) \) is not defined. So we use solution \(R\left ( r\right ) =J_{n}\left (kr\right ) .\) From boundary conditions, we want solution to be zero at \(r=1\), hence we want \(J_{n}\relax (k) =0\), hence the k’s are the zeros of \(J_{n}\)
Putting these solutions together, we get from (6)
\begin {align*} u\left (r,\theta ,t\right ) & =R\relax (r) \Theta \left ( \theta \right ) T\relax (t) \\ & =\left \{ \begin {array} [c]{lll}J_{n}\left (kr\right ) \sin n\theta e^{-\alpha ^{2}k^{2}t} & & \\ & & \\ J_{n}\left (kr\right ) \cos n\theta e^{-\alpha ^{2}k^{2}t} & & \end {array} \right . \end {align*}
From symmetry of plate, the solution can not depend on the angle \(\theta \), hence let \(n=0\) and so as not to get \(u=0\), we must pick the solution with \(\cos n\theta \) term. Hence our solution now is
\[ u\left (r,t\right ) =J_{0}\left (kr\right ) \ e^{-\alpha ^{2}k^{2}t}\]
Where \(k\) is a zero of \(J_{0}\)
The general solution is a linear combination of this eigenfunction for all zeros of \(J_{0}\), hence
\begin {equation} u\left (r,t\right ) =\sum _{m=1}^{\infty }c_{m}\ J_{0}\left (k_{m}r\right ) \ e^{-\alpha ^{2}k_{m}^{2}\ t}\tag {13} \end {equation}
We find \(c_{m}\) by using initial condition. When \(t=0\) \(,\) temp. was 100\(^{0} \) hence
\[ 100=\sum _{m=1}^{\infty }c_{m}\ J_{0}\left (k_{m}r\right ) \ \ \]
Applying inner product w.r.t. \(rJ_{0}\left (k_{u}r\right ) \) from \(0\dots 1\)
\begin {align*} \int _{0}^{1}100\ rJ_{0}\left (k_{u}r\right ) \ dr & =\int _{0}^{1}\left ( \sum _{m=1}^{\infty }c_{m}\ J_{0}\left (k_{m}r\right ) \right ) \ rJ_{0}\left ( k_{u}r\right ) \ dr\\ 100\int _{0}^{1}\ rJ_{0}\left (k_{u}r\right ) \ dr & =\sum _{m=1}^{\infty }c_{m}\int _{0}^{1}J_{0}\left (k_{m}r\right ) \ rJ_{0}\left (k_{u}r\right ) \ dr \end {align*}
From orthogonality of \(J_{0}\left (k_{m}r\right ) \ \)and \(J_{0}\left ( k_{u}r\right ) \), all terms drop expect when \(m=u\)
\[ 100\int _{0}^{1}\ rJ_{0}\left (k_{u}r\right ) \ dr=\ c_{u}\int _{0}^{1}\ \ r \left [ J_{0}\left (k_{u}r\right ) \right ] ^{2}\ dr \]
From here we can follow the book on page 561 to get \[ c_{m}=\frac {200}{k_{m}J_{1}\left (k_{m}\right ) }\]
Substitute this in equation 13
\begin {align*} u\left (r,t\right ) & =\sum _{m=1}^{\infty }c_{m}\ J_{0}\left (k_{m}r\right ) \ e^{-\alpha ^{2}k_{m}^{2}\ t}\\ & =\sum _{m=1}^{\infty }\frac {200}{k_{m}J_{1}\left (k_{m}\right ) }\ J_{0}\left (k_{m}r\right ) \ e^{-\alpha ^{2}k_{m}^{2}\ t}\\ & =200\sum _{m=1}^{\infty }\frac {1}{k_{m}J_{1}\left (k_{m}\right ) }\ J_{0}\left (k_{m}r\right ) \ e^{-\alpha ^{2}k_{m}^{2}\ t} \end {align*}
Where \(k_{m}\) are zeros of J\(_{0}\)
Notice that final solution does not depend on \(\theta \)
Solve \begin {align*} r\frac {d}{dr}\left (r\frac {dR}{dr}\right ) &=n^{2}R\\ \frac {d}{dr}\left (r^{2}\frac {dR}{dr}\right ) &=l\left (l+1\right ) R\\ \end {align*}
Solution
First equation, use power series method.
\begin {align*} r\frac {d}{dr}\left (r\frac {dR}{dr}\right ) & =n^{2}R\\ r\ \left (r\frac {d^{2}R}{dr^{2}}+\frac {dR}{dr}\right ) -n^{2}R & =0\\ r^{2}\frac {d^{2}R}{dr^{2}}+r\frac {dR}{dr}-n^{2}R & =0 \end {align*}
Let \(R=a_{0}r^{s}+a_{1}r^{s+1}+a_{2}r^{s+2}+a_{3}r^{s+3}+a_{4}r^{s+4}+\cdots \) then
\begin {align*} R & =a_{0}r^{s}+a_{1}r^{s+1}+a_{2}r^{s+2}+a_{3}r^{s+3}+a_{4}r^{s+4}+\cdots \\ -n^{2}R & =-n^{2}a_{0}r^{s}-n^{2}a_{1}r^{s+1}-n^{2}a_{2}r^{s+2}-n^{2}a_{3}r^{s+3}-n^{2}a_{4}r^{s+4}-\cdots \\ \frac {dR}{dr} & =s\ a_{0}r^{s-1}+\left (s+1\right ) \ a_{1}r^{s}+\left ( s+2\right ) \ a_{2}r^{s+1}+\left (s+3\right ) \ a_{3}r^{s+2}+\ \cdots \\ r\frac {dR}{dr} & =s\ a_{0}r^{s}+\left (s+1\right ) \ a_{1}r^{s+1}+\left ( s+2\right ) \ a_{2}r^{s+2}+\left (s+3\right ) \ a_{3}r^{s+3}+\ \cdots \\ \frac {d^{2}R}{dr^{2}} & =\ \left (s-1\right ) s\ a_{0}r^{s-2}+s\left ( s+1\right ) \ a_{1}r^{s-1}+\left (s+1\right ) \left (s+2\right ) \ a_{2}r^{s}+\left (s+2\right ) \left (s+3\right ) \ a_{3}r^{s+1}+\ \cdots \\ r^{2}\frac {d^{2}R}{dr^{2}} & =\ \left (s-1\right ) s\ a_{0}r^{s}+s\left ( s+1\right ) \ a_{1}r^{s+1}+\left (s+1\right ) \left (s+2\right ) \ a_{2}r^{s+2}+\left (s+2\right ) \left (s+3\right ) \ a_{3}r^{s+3}+\ \cdots \end {align*}
Table is
| \(r^{s}\) | \(r^{s+1}\) | \(r^{s+2}\) | \(r^{s+m}\) | |
| \(-n^{2}R\) | \(-n^{2}a_{0}\) | \(-n^{2}a_{1}\) | \(-n^{2}a_{2}\) | \(-n^{2}\ a_{m}\) |
| \(r\frac {dR}{dr}\) | \(s\ a_{0}\) | \(\left (s+1\right ) \ a_{1}\) | \(\left ( s+2\right ) \ a_{2}\) | \(\ \left (s+m\right ) a_{m}\) |
| \(r^{2}\frac {d^{2}R}{dr^{2}}\) | \(\ \left (s-1\right ) s\ a_{0}\) | \(s\left ( s+1\right ) \ a_{1}\) | \(\left (s+1\right ) \left (s+2\right ) \ a_{2}\) | \(\left (s+m-1\right ) \left (s+m\right ) \ a_{m}\) |
Hence, from first column we see , and since \(a_{0}\) \(\neq 0\) we solve for \(s\)
\begin {align*} -n^{2}a_{0}+s\ a_{0}+\left (s-1\right ) s\ a_{0} & =0\\ a_{0}\left (-n^{2}+s+\left (s-1\right ) s\right ) & =0\\ -n^{2}+s+\left (s-1\right ) s & =0\\ -n^{2}+\ s^{2} & =0\\ s & =\pm n \end {align*}
We see from second column, \(a_{1}\left (-n^{2}+\left (s+1\right ) +s^{2}+s\right ) =0\) or \(a_{1}\left (-s^{2}+2s+1+s^{2}\right ) =0\), hence \(a_{1}\left (2s+1\right ) =0\)
For \(a_{1}\neq 0\) then \(s=-\frac {1}{2}\), this means \(n\) is not an integer since \(s=\pm n\). hence \(a_{1}\) must be zero.
The same applies to all \(a_{m}\) , \(m>0\) Hence solution contains only \(a_{0}\)
\begin {align*} R & =a_{0}r^{\pm n}\\ R & =\left \{ \begin {array} [c]{lll}a_{0}r^{-n} & & \\ & & \\ a_{0}r^{+n} & & \end {array} \right . \end {align*}
For some constant \(a_{0}\). This solution is when \(n\neq 0\)
If \(n=0\), table is
| \(r^{s}\) | \(r^{s+1}\) | \(r^{s+2}\) | \(r^{s+m}\) | |
| \(-n^{2}R\) | \(0\) | \(0\) | \(0\) | \(0\) |
| \(r\frac {dR}{dr}\) | \(s\ a_{0}\) | \(\left (s+1\right ) \ a_{1}\) | \(\left ( s+2\right ) \ a_{2}\) | \(\ \left (s+m\right ) a_{m}\) |
| \(r^{2}\frac {d^{2}R}{dr^{2}}\) | \(\ \left (s-1\right ) s\ a_{0}\) | \(s\left ( s+1\right ) \ a_{1}\) | \(\left (s+1\right ) \left (s+2\right ) \ a_{2}\) | \(\left (s+m-1\right ) \left (s+m\right ) \ a_{m}\) |
From first column:
\begin {align*} s a_{0}+s^{2}a_{0}-sa_{0}&=0\\ a_{0}\left (s+s^{2}-s\right ) &=0\\ s^{2}&=0\\ s&=0\\ \end {align*}
And all other \(a^{\prime }s\) are zero. Hence \(R=a_{0}\) or \(R\) is constant.
Now for the second ODE
\begin {align*} \frac {d}{dr}\left (r^{2}\frac {dR}{dr}\right ) & =l\left (l+1\right ) R\\ r^{2}\frac {d^{2}R}{dr^{2}}+2r\frac {dR}{dr}-l\left (l+1\right ) R & =0 \end {align*}
Let \(R=a_{0}r^{s}+a_{1}r^{s+1}+a_{2}r^{s+2}+a_{3}r^{s+3}+a_{4}r^{s+4}+\cdots \) then
\begin {align*} R & =a_{0}r^{s}+a_{1}r^{s+1}+a_{2}r^{s+2}+a_{3}r^{s+3}+a_{4}r^{s+4}+\cdots \\ -l\left (l+1\right ) R & =-l\left (l+1\right ) a_{0}r^{s}-l\left ( l+1\right ) a_{1}r^{s+1}-l\left (l+1\right ) a_{2}r^{s+2}-l\left ( l+1\right ) a_{3}r^{s+3}-l\left (l+1\right ) a_{4}r^{s+4}-\cdots \\ \frac {dR}{dr} & =s\ a_{0}r^{s-1}+\left (s+1\right ) \ a_{1}r^{s}+\left ( s+2\right ) \ a_{2}r^{s+1}+\left (s+3\right ) \ a_{3}r^{s+2}+\ \cdots \\ 2r\frac {dR}{dr} & =2s\ a_{0}r^{s}+2\left (s+1\right ) \ a_{1}r^{s+1}+2\left (s+2\right ) \ a_{2}r^{s+2}+2\left (s+3\right ) \ a_{3}r^{s+3}+\ \cdots \\ \frac {d^{2}R}{dr^{2}} & =\ \left (s-1\right ) s\ a_{0}r^{s-2}+s\left ( s+1\right ) \ a_{1}r^{s-1}+\left (s+1\right ) \left (s+2\right ) \ a_{2}r^{s}+\left (s+2\right ) \left (s+3\right ) \ a_{3}r^{s+1}+\ \cdots \\ r^{2}\frac {d^{2}R}{dr^{2}} & =\ \left (s-1\right ) s\ a_{0}r^{s}+s\left ( s+1\right ) \ a_{1}r^{s+1}+\left (s+1\right ) \left (s+2\right ) \ a_{2}r^{s+2}+\left (s+2\right ) \left (s+3\right ) \ a_{3}r^{s+3}+\ \cdots \end {align*}
Table is
| \(r^{s}\) | \(r^{s+1}\) | \(r^{s+2}\) | \(r^{s+m}\) | |
| \(-n^{2}R\) | \(-l\left (l+1\right ) a_{0}\) | \(-l\left (l+1\right ) a_{1}\) | \(-l\left (l+1\right ) a_{2}\) | \(-l\left (l+1\right ) \ a_{m}\) |
| \(2r\frac {dR}{dr}\) | \(2s\ a_{0}\) | \(2\left (s+1\right ) \ a_{1}\) | \(2\left ( s+2\right ) \ a_{2}\) | \(\ 2\left (s+m\right ) a_{m}\) |
| \(r^{2}\frac {d^{2}R}{dr^{2}}\) | \(\ \left (s-1\right ) s\ a_{0}\) | \(s\left ( s+1\right ) \ a_{1}\) | \(\left (s+1\right ) \left (s+2\right ) \ a_{2}\) | \(\left (s+m-1\right ) \left (s+m\right ) \ a_{m}\) |
From first column: \begin {align*} -l\left (l+1\right ) a_{0}+2s\ a_{0}+\left (s-1\right ) s\ a_{0}&=0\\ a_{0}\left (-l\left (l+1\right ) +2s\ +\left (s-1\right ) s\ \right ) &=0\\ -l\left (l+1\right ) +2s\ +\left (s-1\right ) s&=0\\ -l\left (l+1\right ) +s\ +s^{2}&=0\\ \left (s-l\right ) \left (s-\left (-l-1\right ) \right ) &=0\\ \end {align*}
Hence \(s=l\) or \(s=-l-1\).
We also see that all other \(a^{\prime }s\) will be zero, since recursive formula has only \(a_{m}\) in it and no other \(a\). Hence
\begin {align*} R & =a_{0}r^{s}\\ R & =\left \{ \begin {array} [c]{lll}a_{0}r^{l} & & \\ & & \\ a_{0}r^{-l-1} & & \end {array} \right . \end {align*}
For some constant \(a_{0}\)