Equation 3.3.11, page 100 is the Fourier sin series of \(x\)\begin{equation} x=\sum _{n=1}^{\infty }B_{n}\sin \left ( n\frac{\pi }{L}x\right ) \qquad -L<x<L\tag{3.3.11} \end{equation} Where\begin{equation} B_{n}=\frac{2L}{n\pi }\left ( -1\right ) ^{n+1}\tag{3.3.12} \end{equation} The goal is to find the Fourier \(\cos \) series of \(x^{2}\). Since \(\int _{0}^{x}tdt=\frac{x^{2}}{2}\), then \(x^{2}=2\int _{0}^{x}tdt\). Hence from 3.3.11\[ x^{2}=2\int _{0}^{x}\left [ \sum _{n=1}^{\infty }B_{n}\sin \left ( n\frac{\pi }{L}t\right ) \right ] dt \] Interchanging the order of summation and integration the above becomes\begin{align} x^{2} & =2\sum _{n=1}^{\infty }\left ( B_{n}\int _{0}^{x}\sin \left ( n\frac{\pi }{L}t\right ) dt\right ) \nonumber \\ & =2\sum _{n=1}^{\infty }B_{n}\left ( \frac{-\cos \left ( n\frac{\pi }{L}t\right ) }{n\frac{\pi }{L}}\right ) _{0}^{x}\nonumber \\ & =\sum _{n=1}^{\infty }\frac{-2L}{n\pi }B_{n}\left [ \cos \left ( n\frac{\pi }{L}t\right ) \right ] _{0}^{x}\nonumber \\ & =\sum _{n=1}^{\infty }\frac{-2L}{n\pi }B_{n}\left [ \cos \left ( n\frac{\pi }{L}x\right ) -1\right ] \nonumber \\ & =\sum _{n=1}^{\infty }\left ( \frac{-2L}{n\pi }B_{n}\cos \left ( n\frac{\pi }{L}x\right ) +\frac{2L}{n\pi }B_{n}\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\frac{-2L}{n\pi }B_{n}\cos \left ( n\frac{\pi }{L}x\right ) +\sum _{n=1}^{\infty }B_{n}\frac{2L}{n\pi }\tag{1} \end{align}
But a Fourier \(\cos \) series has the form \begin{equation} x^{2}=A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\frac{\pi }{L}x\right ) \tag{2} \end{equation} Comparing (1) and (2) gives\[ A_{n}=\frac{-2L}{n\pi }B_{n}\] Using 3.3.12 for \(B_{n}\) the above becomes\begin{align*} A_{n} & =\frac{-2L}{n\pi }\frac{2L}{n\pi }\left ( -1\right ) ^{n+1}\\ & =\left ( -1\right ) ^{n}\left ( \frac{2L}{n\pi }\right ) ^{2} \end{align*}
And\begin{align*} A_{0} & =\sum _{n=1}^{\infty }B_{n}\frac{2L}{n\pi }\\ & =\sum _{n=1}^{\infty }\left ( \frac{2L}{n\pi }\left ( -1\right ) ^{n+1}\right ) \frac{2L}{n\pi }\\ & =\frac{4L^{2}}{\pi ^{2}}\sum _{n=1}^{\infty }\left ( -1\right ) ^{n+1}\frac{1}{n^{2}} \end{align*}
But \(\sum _{n=1}^{\infty }\left ( -1\right ) ^{n+1}\frac{1}{n^{2}}=\frac{\pi ^{2}}{12}\), hence the above becomes\begin{align*} A_{0} & =\frac{4L^{2}}{\pi ^{2}}\frac{\pi ^{2}}{12}\\ & =\frac{L^{2}}{3} \end{align*}
Summary The Fourier \(\cos \) series of \(x^{2}\) is\begin{align*} x^{2} & =A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\frac{\pi }{L}x\right ) \\ & =\frac{L^{2}}{3}+\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{2L}{n\pi }\right ) ^{2}\cos \left ( n\frac{\pi }{L}x\right ) \end{align*}
Since \[ x^{3}=3\int _{0}^{x}t^{2}dt \] Then, using result from part (a) for Fourier \(\cos \) series of \(t^{2}\) results in\begin{align*} x^{3} & =3\int _{0}^{x}\left [ A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\frac{\pi }{L}t\right ) \right ] dt\\ & =3\int _{0}^{x}\frac{L^{2}}{3}dt+3\int _{0}^{x}\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{2L}{n\pi }\right ) ^{2}\cos \left ( n\frac{\pi }{L}t\right ) dt\\ & =L^{2}\left ( t\right ) _{0}^{x}+3\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{2L}{n\pi }\right ) ^{2}\int _{0}^{x}\cos \left ( n\frac{\pi }{L}t\right ) dt\\ & =L^{2}x+3\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{2L}{n\pi }\right ) ^{2}\left [ \frac{\sin \left ( n\frac{\pi }{L}t\right ) }{n\frac{\pi }{L}}\right ] _{0}^{x}\\ & =L^{2}x+3\sum _{n=1}^{\infty }\frac{L}{n\pi }\left ( -1\right ) ^{n}\left ( \frac{2L}{n\pi }\right ) ^{2}\left [ \sin \left ( n\frac{\pi }{L}t\right ) \right ] _{0}^{x}\\ & =L^{2}x+\left ( 3\cdot 4\right ) \sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{L}{n\pi }\right ) ^{3}\sin \left ( n\frac{\pi }{L}x\right ) \end{align*}
Using 3.3.11 which is \(x=\sum _{n=1}^{\infty }B_{n}\sin \left ( n\frac{\pi }{L}x\right ) \), with \(B_{n}=\frac{2L}{n\pi }\left ( -1\right ) ^{n+1}\ \)the above becomes\[ x^{3}=L^{2}\sum _{n=1}^{\infty }\frac{2L}{n\pi }\left ( -1\right ) ^{n+1}\sin \left ( n\frac{\pi }{L}x\right ) +\left ( 3\cdot 4\right ) \sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{L}{n\pi }\right ) ^{3}\sin \left ( n\frac{\pi }{L}x\right ) \] Combining all above terms\[ x^{3}=\sum _{n=1}^{\infty }\left [ L^{2}\frac{2L}{n\pi }\left ( -1\right ) ^{n+1}+\left ( 3\cdot 4\right ) \left ( -1\right ) ^{n}\left ( \frac{L}{n\pi }\right ) ^{3}\right ] \sin \left ( n\frac{\pi }{L}x\right ) \] Will try to simplify more to obtain \(B_{n}\)\begin{align*} x^{3} & =\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{L^{3}}{n\pi }\left [ -2+\left ( 3\cdot 4\right ) \left ( \frac{1}{n\pi }\right ) ^{2}\right ] \sin \left ( n\frac{\pi }{L}x\right ) \\ & =\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}\frac{2L^{3}}{n\pi }\left [ -1+\left ( 3\times 2\right ) \left ( \frac{1}{n\pi }\right ) ^{2}\right ] \sin \left ( n\frac{\pi }{L}x\right ) \end{align*}
Comparing the above to the standard Fourier \(\sin \) series \(x^{3}=\sum _{n=1}^{\infty }B_{n}\sin \left ( n\frac{\pi }{L}x\right ) \) then the above is the required \(\sin \) series for \(x^{3}\) with\[ B_{n}=\left ( -1\right ) ^{n}\frac{2L^{3}}{n\pi }\left [ -1+\left ( 3\times 2\right ) \left ( \frac{1}{n\pi }\right ) ^{2}\right ] \sin \left ( n\frac{\pi }{L}x\right ) \] Expressing the above using \(B_{n}\) from \(x^{1}\) to help find recursive relation for next problem.
Will now use the notation \(^{i}B_{n}\) to mean the \(B_{n}\) for \(x^{i}\). Then since \(^{1}B_{n}=\frac{2L}{n\pi }\left ( -1\right ) ^{n+1}=\left ( -1\right ) ^{n}\left ( -\frac{2L}{n\pi }\right ) \) for \(x\), then, using \(^{3}B_{n}\) as the \(B_{n}\) for \(x^{3}\), the series for \(x^{3}\) can be written\begin{align*} x^{3} & =\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}L^{2}\left [ -\frac{2L}{n\pi }+6\left ( 2\frac{L}{n^{2}\pi ^{2}}\right ) \right ] \sin \left ( n\frac{\pi }{L}x\right ) \\ & =\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}L^{2}\left [ ^{1}B_{n}+6\left ( 2\frac{L}{n^{2}\pi ^{2}}\right ) \right ] \sin \left ( n\frac{\pi }{L}x\right ) \end{align*}
Where now \[ ^{3}B_{n}=\left ( -1\right ) ^{n}L^{2}\left [ B_{n}^{1}+6\left ( 2\frac{L}{n^{2}\pi ^{2}}\right ) \right ] \] The above will help in the next problem in order to find recursive relation.
Result from Last problem showed that\begin{align*} x & =\sum _{n=1}^{\infty }B_{n}^{1}\sin \left ( n\frac{\pi }{L}x\right ) \\ ^{1}B_{n} & =\left ( -1\right ) ^{n}\left ( -\frac{2L}{n\pi }\right ) \end{align*}
And\[ x^{3}=\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}L^{2}\left [ ^{1}B_{n}+\left ( 3\times 2\right ) \left ( 2\frac{L}{n^{2}\pi ^{2}}\right ) \right ] \sin \left ( n\frac{\pi }{L}x\right ) \] This suggests that\begin{align*} x^{5} & =\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}L^{2}\left [ ^{3}B_{n}+\left ( 5\times 4\times 3\times 2\right ) \left ( 2\frac{L}{n^{2}\pi ^{2}}\right ) \right ] \sin \left ( n\frac{\pi }{L}x\right ) \\ ^{3}B_{n} & =\left ( -1\right ) ^{n}L^{2}\left [ ^{1}B_{n}+6\left ( 2\frac{L}{n^{2}\pi ^{2}}\right ) \right ] \end{align*}
And in general\[ x^{m}=\sum _{n=1}^{\infty }\left ( -1\right ) ^{n}L^{2}\left [ ^{m-2}B_{n}+m!\left ( 2\frac{L}{n^{2}\pi ^{2}}\right ) \right ] \sin \left ( n\frac{\pi }{L}x\right ) \] Where\[ ^{m-2}B_{n}=\left ( -1\right ) ^{n}L^{2}\left [ ^{m-4}B_{n}+\left ( m-2\right ) !\left ( 2\frac{L}{n^{2}\pi ^{2}}\right ) \right ] \] The above is a recursive definition to find \(x^{m}\) Fourier series for \(m\) odd.
Equation 3.5.6 is\begin{equation} \frac{x^{2}}{2}=\frac{L}{2}x-\frac{4L^{2}}{\pi ^{3}}\left ( \sin \frac{\pi x}{L}+\frac{\sin \frac{3\pi x}{L}}{3^{3}}+\frac{\sin \frac{5\pi x}{L}}{5^{3}}+\frac{\sin \frac{7\pi x}{L}}{7^{3}}+\cdots \right ) \tag{3.5.6} \end{equation} Letting \(x=\frac{L}{2}\) in (3.5.6) gives\begin{align*} \frac{L^{2}}{8} & =\frac{L^{2}}{4}-\frac{4L^{2}}{\pi ^{3}}\left ( \sin \frac{\pi \frac{L}{2}}{L}+\frac{\sin \frac{3\pi \frac{L}{2}}{L}}{3^{3}}+\frac{\sin \frac{5\pi \frac{L}{2}}{L}}{5^{3}}+\frac{\sin \frac{7\pi \frac{L}{2}}{L}}{7^{3}}+\cdots \right ) \\ & =\frac{L^{2}}{4}-\frac{4L^{2}}{\pi ^{3}}\left ( \sin \frac{\pi }{2}+\frac{\sin 3\frac{\pi }{2}}{3^{3}}+\frac{\sin 5\frac{\pi }{2}}{5^{3}}+\frac{\sin 7\frac{\pi }{2}}{7^{3}}+\cdots \right ) \\ & =\frac{L^{2}}{4}-\frac{4L^{2}}{\pi ^{3}}\left ( 1-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}\cdots \right ) \end{align*}
Hence\begin{align*} \frac{L^{2}}{8}-\frac{L^{2}}{4} & =-\frac{4L^{2}}{\pi ^{3}}\left ( 1-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}\cdots \right ) \\ -\frac{L^{2}}{8} & =-\frac{4L^{2}}{\pi ^{3}}\left ( 1-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}\cdots \right ) \\ \frac{\pi ^{3}}{4\times 8} & =\left ( 1-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}\cdots \right ) \end{align*}
Or\[ \frac{\pi ^{3}}{32}=1-\frac{1}{3^{3}}+\frac{1}{5^{3}}-\frac{1}{7^{3}}\cdots \]
The function defined above is the Dirac delta function. (in the limit, as \(\Delta \rightarrow 0\)). Now\begin{align*} c_{n} & =\frac{1}{2L}\int _{-L}^{L}f\left ( x\right ) e^{in\frac{\pi }{L}x}dx\\ & =\frac{1}{2L}\int _{x_{0}}^{x_{0}+\Delta }\frac{1}{\Delta }e^{in\frac{\pi }{L}x}dx\\ & =\frac{1}{2L}\frac{1}{\Delta }\left [ \frac{e^{in\frac{\pi }{L}x}}{in\frac{\pi }{L}}\right ] _{x_{0}}^{x_{0}+\Delta }\\ & =\frac{1}{2L}\frac{L}{\Delta in\pi }\left [ e^{in\frac{\pi }{L}x}\right ] _{x_{0}}^{x_{0}+\Delta }\\ & =\frac{1}{i2n\Delta \pi }\left ( e^{in\frac{\pi }{L}\left ( x_{0}+\Delta \right ) }-e^{in\frac{\pi }{L}x_{0}}\right ) \end{align*}
Since \(\frac{e^{iz}-e^{-iz}}{2i}=\sin z\). The denominator above has \(2i\) in it. Factoring out \(e^{in\frac{\pi }{L}\left ( x_{0}+\frac{\Delta }{2}\right ) }\) from the above gives\begin{align*} c_{n} & =\frac{1}{i2n\Delta \pi }e^{in\frac{\pi }{L}\left ( x_{0}+\frac{\Delta }{2}\right ) }\left ( e^{in\frac{\pi }{L}\frac{\Delta }{2}}-e^{-in\frac{\pi }{L}\frac{\Delta }{2}}\right ) \\ & =\frac{1}{n\Delta \pi }e^{in\frac{\pi }{L}\left ( x_{0}+\frac{\Delta }{2}\right ) }\frac{\left ( e^{in\frac{\pi }{L}\frac{\Delta }{2}}-e^{-in\frac{\pi }{L}\frac{\Delta }{2}}\right ) }{i2} \end{align*}
Now the form is \(\sin \left ( z\right ) \) is obtained, hence it can be written as\[ c_{n}=\frac{e^{in\frac{\pi }{L}\left ( x_{0}+\frac{\Delta }{2}\right ) }}{n\Delta \pi }\sin \left ( n\frac{\pi }{L}\frac{\Delta }{2}\right ) \] Or\[ c_{n}=\frac{\cos \left ( n\frac{\pi }{L}\left ( x_{0}+\frac{\Delta }{2}\right ) \right ) +i\sin \left ( n\frac{\pi }{L}\left ( x_{0}+\frac{\Delta }{2}\right ) \right ) }{\Delta n\pi }\sin \left ( n\frac{\pi }{L}\frac{\Delta }{2}\right ) \]
Equation 4.2.7 is\begin{equation} \rho \left ( x\right ) \frac{\partial ^{2}u}{\partial t^{2}}=T_{0}\frac{\partial ^{2}u}{\partial x^{2}}+Q\left ( x,t\right ) \rho \left ( x\right ) \tag{4.2.7} \end{equation} Replacing \(Q\left ( x,t\right ) \) by \(-g\)\[ \rho \left ( x\right ) \frac{\partial ^{2}u}{\partial t^{2}}=T_{0}\frac{\partial ^{2}u}{\partial x^{2}}-g\rho \left ( x\right ) \] At equilibrium, the string is sagged but is not moving.
Therefore \(\frac{\partial ^{2}u_{E}}{\partial t^{2}}=0\). The above becomes\[ 0=T_{0}\frac{\partial ^{2}u_{E}}{\partial x^{2}}-g\rho \left ( x\right ) \] This is now partial differential equation in only \(x\). It becomes an ODE\[ \frac{d^{2}u_{E}}{dx^{2}}=\frac{g\rho \left ( x\right ) }{T_{0}}\] With boundary conditions \(u_{E}\left ( 0\right ) =0,u_{E}\left ( L\right ) =0\). By double integration the solution is found. Integrating once gives\[ \frac{du_{E}}{dx}=\int _{0}^{x}\frac{g\rho \left ( s\right ) }{T_{0}}ds+c_{1}\] Integrating again\begin{align} u_{E} & =\int _{0}^{x}\left ( \int _{0}^{s}\frac{g\rho \left ( z\right ) }{T_{0}}dz+c_{1}\right ) ds+c_{2}\nonumber \\ & =\int _{0}^{x}\left ( \int _{0}^{s}\frac{g\rho \left ( z\right ) }{T_{0}}dz\right ) ds+\int _{0}^{x}c_{1}ds+c_{2}\nonumber \\ & =\frac{g}{T_{0}}\int _{0}^{x}\int _{0}^{s}\rho \left ( z\right ) dzds+c_{1}x+c_{2} \tag{1} \end{align}
Equation (1) is the solution. Applying B.C. to find \(c_{1},c_{2}\). At \(x=0\) the above gives\[ 0=c_{2}\] The solution (1) becomes \begin{equation} u_{E}=\frac{g}{T_{0}}\int _{0}^{x}\int _{0}^{s}\rho \left ( z\right ) dzds+c_{1}x \tag{2} \end{equation} And at \(x=L\) the above becomes\begin{align*} 0 & =\frac{g}{T_{0}}\int _{0}^{L}\int _{0}^{s}\rho \left ( z\right ) dzds+c_{1}L\\ c_{1} & =\frac{-g}{LT_{0}}\int _{0}^{L}\int _{0}^{s}\rho \left ( z\right ) dzds \end{align*}
Substituting this into (2) gives the final solution\begin{equation} u_{E}=\frac{g}{T_{0}}\int _{0}^{x}\left ( \int _{0}^{s}\rho \left ( z\right ) dz\right ) ds+\left ( \frac{-g}{LT_{0}}\int _{0}^{L}\left ( \int _{0}^{s}\rho \left ( z\right ) dz\right ) ds\right ) x \tag{3} \end{equation} If the density was constant, (3) reduces to\begin{align*} u_{E} & =\frac{g\rho }{T_{0}}\int _{0}^{x}sds+\left ( \frac{-g\rho }{LT_{0}}\int _{0}^{L}sds\right ) x\\ & =\frac{g\rho }{T_{0}}\frac{x^{2}}{2}-\frac{g\rho }{LT_{0}}\frac{L^{2}}{2}x\\ & =\frac{g\rho }{T_{0}}\left ( \frac{x^{2}}{2}-\frac{L}{2}x\right ) \end{align*}
Here is a plot of the above function for \(g=9.8,L=1,T_{0}=1,\rho =0.1\) for verification.
Equation 4.2.9 is\begin{equation} \frac{\partial ^{2}u}{\partial t^{2}}=\frac{T_{0}}{\rho \left ( x\right ) }\frac{\partial ^{2}u}{\partial x^{2}} \tag{4.2.9} \end{equation} Since\begin{equation} \rho \left ( x\right ) \frac{\partial ^{2}u}{\partial t^{2}}=T_{0}\frac{\partial ^{2}u}{\partial x^{2}}+Q\left ( x,t\right ) \rho \left ( x\right ) \tag{1} \end{equation} And\begin{equation} \rho \left ( x\right ) \frac{\partial ^{2}u_{E}}{\partial t^{2}}=T_{0}\frac{\partial ^{2}u_{E}}{\partial x^{2}}+Q\left ( x,t\right ) \rho \left ( x\right ) \tag{2} \end{equation} Then by subtracting (2) from (1)\begin{align*} \rho \left ( x\right ) \frac{\partial ^{2}u}{\partial t^{2}}-\rho \left ( x\right ) \frac{\partial ^{2}u_{E}}{\partial t^{2}} & =T_{0}\frac{\partial ^{2}u}{\partial x^{2}}+Q\left ( x,t\right ) \rho \left ( x\right ) -T_{0}\frac{\partial ^{2}u_{E}}{\partial x^{2}}-Q\left ( x,t\right ) \rho \left ( x\right ) \\ \rho \left ( x\right ) \left ( \frac{\partial ^{2}u}{\partial t^{2}}-\frac{\partial ^{2}u_{E}}{\partial t^{2}}\right ) & =T_{0}\left ( \frac{\partial ^{2}u}{\partial x^{2}}-\frac{\partial ^{2}u_{E}}{\partial x^{2}}\right ) \end{align*}
Since \(v\left ( x,t\right ) =u\left ( x,t\right ) -u_{E}\left ( x,t\right ) \) then \(\frac{\partial ^{2}v}{\partial t^{2}}=\frac{\partial ^{2}u}{\partial t^{2}}-\frac{\partial ^{2}u_{E}}{\partial t^{2}}\) and \(\frac{\partial ^{2}v}{\partial x^{2}}=\frac{\partial ^{2}u}{\partial x^{2}}-\frac{\partial ^{2}u_{E}}{\partial x^{2}}\), therefore the above equation becomes\begin{align*} \rho \left ( x\right ) \frac{\partial ^{2}v}{\partial t^{2}} & =T_{0}\frac{\partial ^{2}v}{\partial x^{2}}\\ \frac{\partial ^{2}v}{\partial t^{2}} & =\frac{T_{0}}{\rho \left ( x\right ) }\frac{\partial ^{2}v}{\partial x^{2}}\\ & =c^{2}\frac{\partial ^{2}v}{\partial x^{2}} \end{align*}
Which is 4.2.9. QED.
Let us consider a small segment of the string of length \(\Delta x\) from \(x\) to \(x+\Delta x\). The mass of this segment is \(\rho \Delta x\), where \(\rho \) is density of the string per unit length, assumed here to be constant. Let the angle that the string makes with the horizontal at \(x\) and at \(x+\Delta x\) be \(\theta \left ( x,t\right ) \) and \(\theta \left ( x+\Delta x,t\right ) \) respectively. Since we are only interested in the vertical displacement \(u\left ( x,t\right ) \) of the string, the vertical force on this segment consists of two parts: Its weight (acting downwards) and the net tension resolved in the vertical direction. Let the total vertical force be \(F_{y}\). Therefore\[ F_{y}=\overset{\text{weight}}{\overbrace{-\rho \Delta xg}}+\overset{\text{net tension on segment in vertical direction}}{\overbrace{\left ( T\left ( x+\Delta x,t\right ) \sin \theta \left ( x+\Delta x,t\right ) -T\left ( x,t\right ) \sin \theta \left ( x,t\right ) \right ) }}\] Applying Newton’s second law in the vertical direction \(F_{y}=ma_{y}\) where \(a_{y}=\frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}\) and \(m=\rho \Delta x\), gives the equation of motion of the string segment in the vertical direction\[ \rho \Delta x\frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=-\rho \Delta xg+\left ( T\left ( x+\Delta x,t\right ) \sin \theta \left ( x+\Delta x,t\right ) -T\left ( x,t\right ) \sin \theta \left ( x,t\right ) \right ) \] Dividing both sides by \(\Delta x\)\[ \rho \frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=-\rho g+\frac{\left ( T\left ( x+\Delta x\right ) \sin \theta \left ( x+\Delta x,t\right ) -T\left ( x\right ) \sin \theta \left ( x,t\right ) \right ) }{\Delta x}\] Taking the limit \(\Delta x\rightarrow 0\)\[ \rho \frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=-\rho g+\frac{\partial }{\partial x}\left ( T\left ( x,t\right ) \sin \theta \left ( x,t\right ) \right ) \] Assuming small angles then \(\frac{\partial u}{\partial x}=\tan \theta =\frac{\sin \theta }{\cos \theta }\approx \sin \theta \), then we can replace \(\sin \theta \) in the above with \(\frac{\partial u}{\partial x}\) giving\[ \rho \frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=-\rho g+\frac{\partial }{\partial x}\left ( T\left ( x,t\right ) \frac{\partial u\left ( x,t\right ) }{\partial x}\right ) \] Assuming tension \(T\left ( x,t\right ) \) is constant, say \(T_{0}\) then the above becomes\begin{align*} \rho \frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}} & =-\rho g+T_{0}\frac{\partial }{\partial x}\left ( \frac{\partial u\left ( x,t\right ) }{\partial x}\right ) \\ \frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}} & =\frac{T_{0}}{\rho }\frac{\partial ^{2}u\left ( x,t\right ) }{\partial x^{2}}-\rho g \end{align*}
Setting \(\frac{T_{0}}{\rho }=c^{2}\) then the above becomes\[ \frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=c^{2}\frac{\partial ^{2}u\left ( x,t\right ) }{\partial x^{2}}-\rho g \] Note: In the above \(g\) (gravity acceleration) was used instead of \(Q\left ( x,t\right ) \) as in the book to represent the body forces. In other words, the above can also be written as\[ \frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=c^{2}\frac{\partial ^{2}u\left ( x,t\right ) }{\partial x^{2}}+\rho Q\left ( x,t\right ) \] This is the required PDE, assuming constant density, constant tension, small angles and small vertical displacement.
The natural frequencies of vibrating string of length \(L\) with fixed ends, is given by equation 4.4.11 in the book, which is the solution to the string wave equation\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }\sin \left ( n\frac{\pi }{L}x\right ) \left ( A_{n}\cos \left ( n\frac{\pi c}{L}t\right ) +B_{n}\sin \left ( n\frac{\pi c}{L}t\right ) \right ) \] The frequency of the time solution part of the PDE is given by the arguments of eigenfucntions \(A_{n}\cos \left ( n\frac{\pi c}{L}t\right ) +B_{n}\sin \left ( n\frac{\pi c}{L}t\right ) \). Therefore \(n\frac{\pi c}{L}\) represents the circular frequency \(\omega _{n}\). Comparing general form of \(\cos \omega t\) with \(\cos \left ( n\frac{\pi c}{L}t\right ) \) we see that each mode \(n\) has circular frequency given by \[ \omega _{n}\equiv n\frac{\pi c}{L}\] For \(n=1,2,3,\cdots \). In cycles per seconds (Hertz), and since \(\omega =2\pi f\), then \(2\pi f=n\frac{\pi c}{L}.\) Solving for \(f\) gives\begin{align*} f_{n} & =n\frac{\pi c}{2\pi L}\\ & =n\frac{c}{2L} \end{align*}
Where \(c=\sqrt{\frac{T_{0}}{\rho _{0}}}\) in all of the above.
Equation 4.4.11 above was for a string with fixed ends. Now the B.C. are different, so we need to solve the spatial equation again to find the new eigenvalues. Starting with \(u=X\left ( x\right ) T\left ( t\right ) \) and substituting this in the PDE \(\frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=c^{2}\frac{\partial ^{2}u\left ( x,t\right ) }{\partial x^{2}}\) with \(0<x<H\) gives\begin{align*} T^{\prime \prime }X & =c^{2}TX^{\prime \prime }\\ \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}
Where both sides are set equal to some constant \(-\lambda \). We now obtain the two ODE’s to solve. The spatial ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( H\right ) & =0 \end{align*}
And the time ODE is\[ T^{\prime \prime }+\lambda c^{2}T=0 \] The eigenvalues will always be positive for the wave equation. Taking \(\lambda >0\) the solution to the space ODE is\[ X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right ) \] Applying first B.C. gives\[ 0=A \] Hence \(X\left ( x\right ) =B\sin \left ( \sqrt{\lambda }x\right ) \) and \(X^{\prime }\left ( x\right ) =-B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) \). Applying second B.C. gives\[ 0=-B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }H\right ) \] Therefore for non-trivial solution, we want \(\sqrt{\lambda }H=\frac{n}{2}\pi \) for \(n=1,3,5,\cdots \) or written another way\[ \sqrt{\lambda }H=\left ( n-\frac{1}{2}\right ) \pi \qquad n=1,2,3,\cdots \] Therefore\[ \lambda _{n}=\left ( \left ( n-\frac{1}{2}\right ) \frac{\pi }{H}\right ) ^{2}\qquad n=1,2,3,\cdots \] These are the eigenvalues. Now that we know what \(\lambda _{n}\) is, we go back to the solution found before, which is\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }\sin \left ( \sqrt{\lambda _{n}}x\right ) \left ( A_{n}\cos \left ( \sqrt{\lambda _{n}}ct\right ) +B_{n}\sin \left ( \sqrt{\lambda _{n}}ct\right ) \right ) \] And see now that the circular frequency \(\omega _{n}\) is given by\begin{align*} \omega _{n} & =\sqrt{\lambda _{n}}c\\ & =\frac{\left ( n-\frac{1}{2}\right ) \pi }{H}c\qquad n=1,2,3,\cdots \end{align*}
In cycles per second, since \(\omega =2\pi f\) then\begin{align*} 2\pi f_{n} & =\frac{\left ( n-\frac{1}{2}\right ) \pi }{H}c\\ f_{n} & =\frac{\left ( n-\frac{1}{2}\right ) }{2H}c\qquad n=1,2,3,\cdots \end{align*}
The following are plots for \(n=1,2,3,4,5\) for \(t=0\cdots 3\) seconds by small time increments.
For part (a), the harmonics had circular frequency \(\omega _{n}=\frac{n\pi }{L}c\). Hence for odd \(n\), these will generate\begin{equation} \frac{\pi }{L}c,3\frac{\pi }{L}c,5\frac{\pi }{L}c,7\frac{\pi }{L}c,\cdots \tag{1} \end{equation} For part (b), \(\omega _{n}=\frac{\left ( n-\frac{1}{2}\right ) \pi }{H}c\). When \(H=\frac{L}{2}\), this becomes \(\omega _{n}=\frac{2\left ( n-\frac{1}{2}\right ) \pi }{L}c\). Looking at the first few modes gives\begin{align} & \frac{2\left ( 1-\frac{1}{2}\right ) \pi }{L}c,\frac{2\left ( 2-\frac{1}{2}\right ) \pi }{L}c,\frac{2\left ( 3-\frac{1}{2}\right ) \pi }{L}c,\frac{2\left ( 4-\frac{1}{2}\right ) \pi }{L}c,\cdots \nonumber \\ & \frac{\pi }{L}c,\frac{3\pi }{L}c,\frac{5\pi }{L}c,\frac{7\pi }{L}c,\cdots \tag{2} \end{align}
Comparing (1) and (2) we see they are the same. Which is what we asked to show.
\[ \rho _{0}\frac{\partial ^{2}u}{\partial t^{2}}=T_{0}\frac{\partial ^{2}u}{\partial x^{2}}-\beta \frac{\partial u}{\partial t}\] The term \(-\beta \frac{\partial u}{\partial t}\) is the force that acts on the spring segment due to damping. This is the Viscous damping force which is proportional to speed, where \(\beta \) represents viscous damping coefficient. This damping force always opposes the direction of the motion. Hence if \(\frac{\partial u}{\partial t}>0\) then \(-\beta \frac{\partial u}{\partial t}\) should come out to be negative. This occurs if \(\beta >0\). On the other hand, if \(\frac{\partial u}{\partial t}<0\) then \(-\beta \frac{\partial u}{\partial t}\) should now be positive. Which means again that \(\beta \) must be positive quantity. Hence only case were the damping force always opposes the motion of the string is when \(\beta >0\).
Starting with \(u=X\left ( x\right ) T\left ( t\right ) \) and substituting this in the above PDE with \(0<x<L\) gives\begin{align*} \rho _{0}T^{\prime \prime }X & =T_{0}TX^{\prime \prime }-\beta T^{\prime }X\\ \frac{\rho _{0}}{T_{0}}\frac{T^{\prime \prime }}{T}+\frac{\beta }{T_{0}}\frac{T^{\prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}
Hence we obtain two ODE’s. The space ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}
And the time ODE is\begin{align*} T^{\prime \prime }+c^{2}\beta T^{\prime }+c^{2}\lambda T & =0\\ T\left ( 0\right ) & =f\left ( x\right ) \\ T^{\prime }\left ( 0\right ) & =g\left ( x\right ) \end{align*}
The eigenvalues will always be positive for the wave equation. Hence taking \(\lambda >0\) the solution to the space ODE is\[ X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right ) \] Applying first B.C. gives\[ 0=A \] Hence \(X=B\sin \left ( \sqrt{\lambda }x\right ) \). Applying the second B.C. gives\[ 0=B\sin \left ( \sqrt{\lambda }L\right ) \] Therefore \begin{align*} \sqrt{\lambda }L & =n\pi \qquad n=1,2,3,\cdots \\ \lambda & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}
Hence the space solution is\begin{equation} X=\sum _{n=1}^{\infty }b_{n}\sin \left ( \frac{n\pi }{L}x\right ) \tag{1} \end{equation} Now we solve the time ODE. This is second order ODE, linear, with constant coefficients.\begin{align*} \frac{\rho _{0}}{T_{0}}\frac{T^{\prime \prime }}{T}+\frac{\beta }{T_{0}}\frac{T^{\prime }}{T} & =-\lambda \\ \frac{\rho _{0}}{T_{0}}T^{\prime \prime }+\frac{\beta }{T_{0}}T^{\prime }+\lambda T & =0\\ T^{\prime \prime }+\frac{\beta }{\rho _{0}}T^{\prime }+\frac{T_{0}}{\rho _{0}}\lambda T & =0 \end{align*}
Where in the above \(\lambda \equiv \lambda _{n}\) for \(n=1,2,3,\cdots \). The characteristic equation is \(r^{2}+c^{2}\beta r+c^{2}\lambda =0\). The roots are found from the quadratic formula\begin{align*} r_{1,2} & =\frac{-B\pm \sqrt{B^{2}-4AC}}{2A}\\ & =\frac{-\frac{\beta }{\rho _{0}}\pm \sqrt{\left ( \frac{\beta }{\rho _{0}}\right ) ^{2}-4\frac{T_{0}}{\rho _{0}}\lambda }}{2}\\ & =-\frac{\beta }{2\rho _{0}}\pm \frac{1}{2}\sqrt{\left ( \frac{\beta }{\rho _{0}}\right ) ^{2}-4\frac{T_{0}}{\rho _{0}}\lambda } \end{align*}
Replacing \(\lambda =\left ( \frac{n\pi }{L}\right ) ^{2}\), gives\begin{align*} r_{1,2} & =-\frac{\beta }{2\rho _{0}}\pm \frac{1}{2}\sqrt{\left ( \frac{\beta }{\rho _{0}}\right ) ^{2}-4\frac{T_{0}}{\rho _{0}}\left ( \frac{n\pi }{L}\right ) ^{2}}\\ & =-\frac{\beta }{2\rho _{0}}\pm \frac{1}{2}\sqrt{\frac{\beta ^{2}}{\rho _{0}^{2}}-4\frac{T_{0}}{\rho _{0}}\frac{n^{2}\pi ^{2}}{L^{2}}}\\ & =-\frac{\beta }{2\rho _{0}}\pm \frac{1}{2\rho _{0}}\sqrt{\beta ^{2}-n^{2}\left ( 4\rho _{0}T_{0}\frac{\pi ^{2}}{L^{2}}\right ) } \end{align*}
We are told that \(\beta ^{2}<4\rho _{0}T_{0}\frac{\pi ^{2}}{L^{2}}\), what this means is that \(\beta ^{2}-n^{2}\left ( 4\rho _{0}T_{0}\frac{\pi ^{2}}{L^{2}}\right ) <0\), since \(n^{2}>0\). This means we will get complex roots. Let \[ \Delta =n^{2}\left ( 4\rho _{0}T_{0}\frac{\pi ^{2}}{L^{2}}\right ) -\beta ^{2}\] Hence the roots can now be written as \[ r_{1,2}=-\frac{\beta }{2\rho _{0}}\pm \frac{i\sqrt{\Delta }}{2\rho _{0}}\] Therefore the time solution is\[ T_{n}\left ( t\right ) =e^{-\frac{\beta }{2\rho _{0}}t}\left ( A_{n}\cos \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) +B_{n}\sin \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) \right ) \] This is sinusoidal damped oscillation. Therefore\begin{equation} T\left ( t\right ) =\sum _{n=1}^{\infty }e^{-\frac{\beta }{2\rho _{0}}t}\left ( A_{n}\cos \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) +B_{n}\sin \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) \right ) \tag{2} \end{equation} Combining (1) and (2), gives the total solution\begin{equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{\beta }{2\rho _{0}}t}\left ( A_{n}\cos \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) +B_{n}\sin \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) \right ) \tag{3} \end{equation} Where \(b_{n}\) constants for space ODE merged with the constants \(A_{n},B_{n}\) for the time solution. Now we are ready to find \(A_{n},B_{n}\) from initial conditions. At \(t=0\)\[ f\left ( x\right ) =\sum _{n=1}^{\infty }\sin \left ( \frac{n\pi }{L}x\right ) A_{n}\] Multiplying both sides by \(\sin \left ( \frac{m\pi }{L}x\right ) \) and integrating gives\[ \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx=\int _{0}^{L}\sum _{n=1}^{\infty }\sin \left ( \frac{m\pi }{L}x\right ) \sin \left ( \frac{n\pi }{L}x\right ) A_{n}dx \] Changing the order of integration and summation\begin{align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx & =\sum _{n=1}^{\infty }A_{n}\int _{0}^{L}\sin \left ( \frac{m\pi }{L}x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =A_{m}\frac{L}{2} \end{align*}
Hence\[ A_{n}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx \] To find \(B_{n}\), we first take time derivative of the solution above in (3) which gives\begin{align*} \frac{\partial }{\partial t}u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{\beta }{2\rho _{0}}t}\left ( -\frac{\sqrt{\Delta }}{2\rho _{0}}A_{n}\sin \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) +B_{n}\frac{\sqrt{\Delta }}{2\rho _{0}}\cos \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) \right ) \\ & -\frac{\beta }{2\rho _{0}}\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{\beta }{2\rho _{0}}t}\left ( A_{n}\cos \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) +B_{n}\sin \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) \right ) \end{align*}
At \(t=0\), using the second initial condition gives\[ g\left ( x\right ) =\sum _{n=1}^{\infty }\sin \left ( \frac{n\pi }{L}x\right ) B_{n}\frac{\sqrt{\Delta }}{2\rho _{0}}-\frac{\beta }{2\rho _{0}}A_{n}\sin \left ( \frac{n\pi }{L}x\right ) \] Multiplying both sides by \(\sin \left ( \frac{m\pi }{L}x\right ) \) and integrating gives\[ \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx=\int _{0}^{L}\sum _{n=1}^{\infty }\sin \left ( \frac{m\pi }{L}x\right ) \sin \left ( \frac{n\pi }{L}x\right ) B_{n}\frac{\sqrt{\Delta }}{2\rho _{0}}dx-\sum _{n=1}^{\infty }\frac{\beta }{2\rho _{0}}A_{n}\sin \left ( \frac{m\pi }{L}x\right ) \sin \left ( \frac{n\pi }{L}x\right ) \] Changing the order of integration and summation\begin{align*} \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx & =\sum _{n=1}^{\infty }B_{n}\frac{\sqrt{\Delta }}{2\rho _{0}}\int _{0}^{L}\sin \left ( \frac{m\pi }{L}x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx-\sum _{n=1}^{\infty }\frac{\beta }{2\rho _{0}}A_{n}\int _{0}^{L}\sin \left ( \frac{m\pi }{L}x\right ) \sin \left ( \frac{n\pi }{L}x\right ) \\ & =B_{m}\frac{\sqrt{\Delta }}{2\rho _{0}}\frac{L}{2}-\frac{\beta }{2\rho _{0}}A_{n}\frac{L}{2}\\ & =\frac{L}{2}\left ( B_{m}\frac{\sqrt{\Delta }}{2\rho _{0}}-\frac{\beta }{2\rho _{0}}A_{n}\right ) \end{align*}
Hence\begin{align*} B_{m}\frac{\sqrt{\Delta }}{2\rho _{0}}-\frac{\beta }{2\rho _{0}}A_{n} & =\frac{2}{L}\int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx\\ B_{m} & =\left ( \frac{2}{L}\int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx+\frac{\beta }{2\rho _{0}}A_{n}\right ) \frac{2\rho _{0}}{\sqrt{\Delta }} \end{align*}
This completes the solution. Summary of solution\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\sin \left ( \frac{n\pi }{L}x\right ) e^{-\frac{\beta }{2\rho _{0}}t}\left ( A_{n}\cos \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) +B_{n}\sin \left ( \frac{\sqrt{\Delta }}{2\rho _{0}}t\right ) \right ) \\ A_{n} & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ B_{n} & =\left ( \frac{2}{L}\int _{0}^{L}g\left ( x\right ) \sin \left ( \frac{m\pi }{L}x\right ) dx+\frac{\beta }{2\rho _{0}}A_{n}\right ) \frac{2\rho _{0}}{\sqrt{\Delta }}\\ \Delta & =n^{2}\left ( 4\rho _{0}T_{0}\frac{\pi ^{2}}{L^{2}}\right ) -\beta ^{2} \end{align*}
\[ E=\frac{1}{2}\int _{0}^{L}\left ( \frac{\partial u}{\partial t}\right ) ^{2}dx+\frac{c^{2}}{2}\int _{0}^{L}\left ( \frac{\partial u}{\partial x}\right ) ^{2}dx \] Hence\[ \frac{dE}{dt}=\frac{1}{2}\frac{d}{dt}\int _{0}^{L}\left ( \frac{\partial u}{\partial t}\right ) ^{2}dx+\frac{c^{2}}{2}\frac{d}{dt}\int _{0}^{L}\left ( \frac{\partial u}{\partial x}\right ) ^{2}dx \] Moving \(\frac{d}{dt}\) inside the integral, it becomes partial derivative\begin{equation} \frac{dE}{dt}=\frac{1}{2}\int _{0}^{L}\frac{\partial }{\partial t}\left ( \frac{\partial u}{\partial t}\right ) ^{2}dx+\frac{c^{2}}{2}\int _{0}^{L}\frac{\partial }{\partial t}\left ( \frac{\partial u}{\partial x}\right ) ^{2}dx \tag{1} \end{equation} But \begin{equation} \frac{\partial }{\partial t}\left ( \frac{\partial u}{\partial t}\right ) ^{2}=\frac{\partial }{\partial t}\left ( \frac{\partial u}{\partial t}\frac{\partial u}{\partial t}\right ) =\frac{\partial ^{2}u}{\partial t^{2}}\frac{\partial u}{\partial t}+\frac{\partial u}{\partial t}\frac{\partial ^{2}u}{\partial t^{2}}=2\left ( \frac{\partial u}{\partial t}\frac{\partial ^{2}u}{\partial t^{2}}\right ) \tag{2} \end{equation} And\begin{equation} \frac{\partial }{\partial t}\left ( \frac{\partial u}{\partial x}\right ) ^{2}=\frac{\partial }{\partial t}\left ( \frac{\partial u}{\partial x}\frac{\partial u}{\partial x}\right ) =\frac{\partial ^{2}u}{\partial x\partial t}\frac{\partial u}{\partial x}+\frac{\partial u}{\partial x}\frac{\partial ^{2}u}{\partial x\partial t}=2\frac{\partial u}{\partial x}\frac{\partial ^{2}u}{\partial x\partial t} \tag{3} \end{equation} Substituting (2,3) into (1) gives\begin{align*} \frac{dE}{dt} & =\frac{1}{2}\int _{0}^{L}2\left ( \frac{\partial u}{\partial t}\frac{\partial ^{2}u}{\partial t^{2}}\right ) dx+\frac{c^{2}}{2}\int _{0}^{L}2\frac{\partial u}{\partial x}\frac{\partial ^{2}u}{\partial x\partial t}dx\\ & =\int _{0}^{L}\left ( \frac{\partial u}{\partial t}\frac{\partial ^{2}u}{\partial t^{2}}\right ) dx+c^{2}\int _{0}^{L}\frac{\partial u}{\partial x}\frac{\partial ^{2}u}{\partial x\partial t}dx \end{align*}
But \(\frac{\partial ^{2}u}{\partial t^{2}}=c^{2}\frac{\partial ^{2}u}{\partial x^{2}}\) then the above becomes\begin{align} \frac{dE}{dt} & =\int _{0}^{L}\left ( \frac{\partial u}{\partial t}\left [ c^{2}\frac{\partial ^{2}u}{\partial x^{2}}\right ] \right ) dx+c^{2}\int _{0}^{L}\frac{\partial u}{\partial x}\frac{\partial ^{2}u}{\partial x\partial t}dx\nonumber \\ & =c^{2}\int _{0}^{L}\left ( \frac{\partial u}{\partial t}\frac{\partial ^{2}u}{\partial x^{2}}\right ) dx+c^{2}\int _{0}^{L}\frac{\partial u}{\partial x}\frac{\partial ^{2}u}{\partial x\partial t}dx\nonumber \\ & =c^{2}\int _{0}^{L}\left ( \frac{\partial u}{\partial t}\frac{\partial ^{2}u}{\partial x^{2}}\right ) +\left ( \frac{\partial u}{\partial x}\frac{\partial ^{2}u}{\partial x\partial t}\right ) dx \tag{4} \end{align}
But since the integrand in (4) can also be written as\[ \frac{\partial }{\partial x}\left ( \frac{\partial u}{\partial t}\frac{\partial u}{\partial x}\right ) =\frac{\partial ^{2}u}{\partial x\partial t}\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial ^{2}u}{\partial x^{2}}\] Then (4) becomes\begin{align*} \frac{dE}{dt} & =c^{2}\int _{0}^{L}\frac{\partial }{\partial x}\left ( \frac{\partial u}{\partial t}\frac{\partial u}{\partial x}\right ) dx\\ & =c^{2}\left ( \frac{\partial u}{\partial t}\frac{\partial u}{\partial x}\right ) _{0}^{L} \end{align*}
Which is what we are asked to show. QED.