\[ \rho \frac{\partial ^{2}u}{\partial t^{2}}=T_{0}\frac{\partial ^{2}u}{\partial t^{2}}+\alpha u+\beta \frac{\partial u}{\partial t}\] The PDE equation represents the vertical displacement \(u\left ( x,t\right ) \) of the string as a function of time and horizontal position. This is 1D wave equation. The term \(\beta \frac{\partial u}{\partial t}\) represents the damping force (can be due to motion of the string in air or fluid). The damping coefficient \(\beta \) must be negative to make \(\beta \frac{\partial u}{\partial t}\) opposite to direction of motion. Damping force is proportional to velocity and acts opposite to direction of motion.
The term \(\alpha u\) represents the stiffness in the system. This is a restoring force, and acts also opposite to direction of motion and is proportional to current displacement from equilibrium position. Hence \(\alpha <0\) also.
Let \(u=X\left ( x\right ) T\left ( t\right ) \). Substituting this into the above PDE gives\[ \rho T^{\prime \prime }X=T_{0}X^{\prime \prime }T+\alpha XT+\beta T^{\prime }X \] Dividing by \(XT\neq 0\)\begin{align*} \rho \frac{T^{\prime \prime }}{T} & =T_{0}\frac{X^{\prime \prime }}{X}+\alpha +\beta \frac{T^{\prime }}{T}\\ \rho \frac{T^{\prime \prime }}{T}-\beta \frac{T^{\prime }}{T} & =T_{0}\frac{X^{\prime \prime }}{X}+\alpha \end{align*}
To make each side depends on one variable only, we move \(\rho \left ( x\right ) ,\beta \left ( x\right ) \) to the right side since these depends on \(x\). Then dividing by \(\rho \left ( x\right ) \) gives\[ \frac{T^{\prime \prime }}{T}-\frac{\beta }{\rho }\frac{T^{\prime }}{T}=T_{0}\frac{X^{\prime \prime }}{\rho X}+\frac{\alpha }{\rho }\] If \(\frac{\beta \left ( x\right ) }{\rho \left ( x\right ) }=c\) is constant, then we see the equations have now been separated, since \(\frac{\beta \left ( x\right ) }{\rho \left ( x\right ) }\) do not depend on \(x\) any more and the above becomes\[ \frac{T^{\prime \prime }}{T}-c\frac{T^{\prime }}{T}=T_{0}\frac{X^{\prime \prime }}{\rho X}+\frac{\alpha \left ( x\right ) }{\rho \left ( x\right ) }\] Now we can say that both side is equal to some constant \(-\lambda \) giving the two ODE’s\begin{align*} \frac{T^{\prime \prime }}{T}-c\frac{T^{\prime }}{T} & =-\lambda \\ T_{0}\frac{X^{\prime \prime }}{\rho X}+\frac{\alpha }{\rho } & =-\lambda \end{align*}
Or\begin{align*} T^{\prime \prime }-cT^{\prime }+\lambda T & =0\\ X^{\prime \prime }+X\left ( \frac{\alpha }{T_{0}}+\lambda \frac{\rho }{T_{0}}\right ) & =0 \end{align*}
From above, the spatial ODE is
\begin{equation} X^{\prime \prime }+X\left ( \frac{\alpha }{T_{0}}+\lambda \frac{\rho }{T_{0}}\right ) =0 \tag{1} \end{equation} Comparing to regular Sturm Liouville (RSL) form, which is\begin{align} \frac{d}{dx}\left ( pX^{\prime }\right ) +qX+\lambda \sigma X & =0\nonumber \\ pX^{\prime \prime }+p^{\prime }X^{\prime }+\left ( q+\lambda \sigma \right ) X & =0 \tag{2} \end{align}
Comparing (1) and (2) we see that\begin{align*} p & =1\\ q & =\frac{\alpha }{T_{0}}\\ \sigma & =\frac{\rho }{T_{0}} \end{align*}
To solve the time ODE \(T^{\prime \prime }-cT^{\prime }+\lambda T=0\), since this is second order linear with constant coefficients, then the characteristic equation is\begin{align*} r^{2}-cr+\lambda & =0\\ r & =\frac{-B}{2A}\pm \frac{\sqrt{B^{2}-4AC}}{2A}\\ & =\frac{c}{2}\pm \frac{\sqrt{c^{2}-4\lambda }}{2} \end{align*}
Hence the two solutions are \begin{align*} T_{1}\left ( t\right ) & =e^{\left ( \frac{c}{2}+\frac{\sqrt{c^{2}-4\lambda }}{2}\right ) t}\\ T_{2}\left ( t\right ) & =e^{\left ( \frac{c}{2}-\frac{\sqrt{c^{2}-4\lambda }}{2}\right ) t} \end{align*}
The general solution is linear combination of the above two solution, therefore final solution is\[ T\left ( t\right ) =c_{1}e^{\left ( \frac{c}{2}+\frac{\sqrt{c^{2}-4\lambda }}{2}\right ) t}+c_{2}e^{\left ( \frac{c}{2}-\frac{\sqrt{c^{2}-4\lambda }}{2}\right ) t}\]
Where \(c_{1},c_{2}\) are arbitrary constants of integration.
\[ \frac{d^{2}\phi }{dx^{2}}+\alpha \left ( x\right ) \frac{d\phi }{dx}+\left ( \lambda \beta \left ( x\right ) +\gamma \left ( x\right ) \right ) \phi =0 \] Multiplying by \(H\left ( x\right ) \) gives\begin{equation} H\left ( x\right ) \phi ^{\prime \prime }\left ( x\right ) +H\left ( x\right ) \alpha \left ( x\right ) \phi ^{\prime }\left ( x\right ) +H\left ( x\right ) \left ( \lambda \beta \left ( x\right ) +\gamma \left ( x\right ) \right ) \phi =0 \tag{1} \end{equation} Comparing (1) to Sturm Liouville form, which is\begin{align} \frac{d}{dx}\left ( p\phi ^{\prime }\right ) +q\phi +\lambda \sigma \phi & =0\nonumber \\ p\left ( x\right ) \phi ^{\prime \prime }\left ( x\right ) +p^{\prime }\left ( x\right ) \phi ^{\prime }\left ( x\right ) +\left ( q+\lambda \sigma \right ) \phi \left ( x\right ) & =0 \tag{2} \end{align}
Then we need to satisfy\begin{align*} H\left ( x\right ) & =P\left ( x\right ) \\ H\left ( x\right ) \alpha \left ( x\right ) & =P^{\prime }\left ( x\right ) \end{align*}
Therefore, by combining the above, we obtain one ODE equation to solve for \(H\left ( x\right ) \) \[ H^{\prime }\left ( x\right ) =H\left ( x\right ) \alpha \left ( x\right ) \] This is first order separable ODE. \(\frac{H^{\prime }}{H}=\alpha \) or \(\ln \left \vert H\right \vert =\int \alpha dx+c\) or \[ H=Ae^{\int \alpha \left ( x\right ) dx}\] Where \(A\) is some constant. By comparing (1),(2) again, we see that\[ q+\lambda \sigma =\lambda \beta \left ( x\right ) H\left ( x\right ) +\gamma \left ( x\right ) H\left ( x\right ) \] Summary of solution\begin{align*} \sigma \left ( x\right ) & =\beta \left ( x\right ) H\left ( x\right ) \\ q\left ( x\right ) & =\gamma \left ( x\right ) H\left ( x\right ) \\ P\left ( x\right ) & =H\left ( x\right ) \\ H\left ( x\right ) & =Ae^{\int \alpha \left ( x\right ) dx} \end{align*}
QED
\begin{align} x^{2}\phi ^{\prime \prime }+x\phi ^{\prime }+\lambda \phi & =0\tag{1}\\ \phi \left ( 1\right ) & =0\nonumber \\ \phi \left ( b\right ) & =0\nonumber \end{align}
Multiplying (1) by \(\frac{1}{x}\) where \(x\neq 0\) gives\begin{equation} x\phi ^{\prime \prime }+\phi ^{\prime }+\frac{\lambda }{x}\phi =0 \tag{2} \end{equation} Comparing (2) to Sturm-Liouville form\begin{equation} p\phi ^{\prime \prime }+p^{\prime }\phi ^{\prime }+\left ( q+\lambda \sigma \right ) \phi =0 \tag{3} \end{equation} Then\begin{align*} p & =x\\ q & =0\\ \sigma & =\frac{1}{x} \end{align*}
And since the given boundary conditions also satisfy the Sturm-Liouville boundary conditions, then (2) is a regular Sturm-Liouville ODE.
Using equation 5.3.8 in page 160 of text (called Raleigh quotient), which applies to regular Sturm-Liouville ODE, which relates the eigenvalues to the eigenfunctions\begin{align} \lambda & =\frac{-\left [ p\phi \phi ^{\prime }\right ] _{x=1}^{x=b}+\int _{1}^{b}p\left ( \phi ^{\prime }\right ) ^{2}-q\phi ^{2}dx}{\int _{1}^{b}\phi ^{2}\sigma dx}\tag{5.3.8}\\ & =\frac{-\left [ p\left ( b\right ) \phi \left ( b\right ) \phi ^{\prime }\left ( b\right ) -p\left ( 1\right ) \phi \left ( 1\right ) \phi ^{\prime }\left ( b\right ) \right ] +\int _{1}^{b}p\left ( \phi ^{\prime }\right ) ^{2}-q\phi ^{2}dx}{\int _{1}^{b}\phi ^{2}\sigma dx}\nonumber \end{align}
Using \(p=x,q=0,\sigma =\frac{1}{x}\) and using \(\phi \left ( 1\right ) =0,\phi \left ( b\right ) =0\), then the above simplifies to\[ \lambda =\frac{-\int _{1}^{b}p\left ( \phi ^{\prime }\right ) ^{2}dx}{\int _{1}^{b}\frac{\phi ^{2}}{x}dx}\] The integrands in the numerator and denominator can not be negative, since they are squared quantities, and also since \(x>0\) as the domain starts from \(x=1\), then RHS above can not be negative. This means the eigenvalue \(\lambda \) can not be negative. It can only be \(\lambda \geq 0\). QED.
The possible values of \(\lambda >0\) are determined by trying to solve the ODE and seeing which \(\lambda \) produces non-trivial solutions given the boundary conditions. The ODE to solve is (1) above. Here it is again\begin{equation} x^{2}\phi ^{\prime \prime }+x\phi ^{\prime }+\lambda \phi =0 \tag{1} \end{equation} We know \(\lambda \geq 0,\) so we do not need to check for negative \(\lambda \).
Case \(\lambda =0\)\(.\)
Equation (1) becomes\begin{align*} x^{2}\phi ^{\prime \prime }+x\phi ^{\prime } & =0\\ x\phi ^{\prime \prime }+\phi ^{\prime } & =0\\ \frac{d}{dx}\left ( x\phi ^{\prime }\right ) & =0 \end{align*}
Hence \(x\phi ^{\prime }=c_{1}\) where \(c_{1}\) is constant. Therefore \(\frac{d}{dx}\phi =\frac{c_{1}}{x}\) or\begin{align*} \phi & =c_{1}\int \frac{1}{x}dx+c_{2}\\ & =c_{1}\ln \left \vert x\right \vert +c_{2} \end{align*}
At \(x=1\), \(\phi \left ( 1\right ) =0\), hence\[ 0=c_{1}\ln \left ( 1\right ) +c_{2}\] But \(\ln \left ( 1\right ) =0\), therefore \(c_{2}=0\)\(.\) The solution now becomes \[ \phi =c_{1}\ln \left \vert x\right \vert \] At the right end, \(x=b,\phi \left ( b\right ) =0\), therefore\[ 0=c_{1}\ln b \] But since \(b>1\) the above implies that \(c_{1}=0\). This gives trivial solution. Therefore \(\lambda =0\) is not an eigenvalue.
Case \(\lambda >0\)
\[ x^{2}\phi ^{\prime \prime }+x\phi ^{\prime }+\lambda \phi =0 \] This is non-constant coefficients, linear, second order ODE. Let \(\phi \left ( x\right ) =x^{p}\). Equation (1) becomes\begin{align*} x^{2}p\left ( p-1\right ) x^{p-2}+xpx^{p-1}+\lambda x^{p} & =0\\ p\left ( p-1\right ) x^{p}+px^{p}+\lambda x^{p} & =0 \end{align*}
Dividing by \(x^{p}\neq 0\) gives the characteristic equation\begin{align*} p\left ( p-1\right ) +p+\lambda & =0\\ p^{2}-p+p+\lambda & =0\\ p^{2} & =-\lambda \end{align*}
Since \(\lambda \geq 0\) then \(p\) is complex. Therefore the roots are\[ p=\pm i\sqrt{\lambda }\] Therefore the two solutions (eigenfunctions) are\begin{align*} \phi _{1}\left ( x\right ) & =x^{i\sqrt{\lambda }}\\ \phi _{2}\left ( x\right ) & =x^{-i\sqrt{\lambda }} \end{align*}
To more easily use standard form of solution, the standard trick is to rewrite these solution in exponential form\begin{align*} \phi _{1}\left ( x\right ) & =e^{i\sqrt{\lambda }\ln x}\\ \phi _{2}\left ( x\right ) & =e^{-i\sqrt{\lambda }\ln x} \end{align*}
The general solution to (1) is linear combination of these two solutions, therefore\begin{equation} \phi \left ( x\right ) =c_{1}e^{i\sqrt{\lambda }\ln x}+c_{2}e^{-i\sqrt{\lambda }\ln x} \tag{2} \end{equation} Since \(\lambda >0\) then the above can be written using trig functions as\[ \phi \left ( x\right ) =c_{1}\cos \left ( \sqrt{\lambda }\ln x\right ) +c_{2}\sin \left ( \sqrt{\lambda }\ln x\right ) \] We are now ready to check for allowed values of \(\lambda \) by applying B.C’s. The first B.C. gives\begin{align*} 0 & =c_{1}\cos \left ( \sqrt{\lambda }\ln 1\right ) +c_{2}\sin \left ( \sqrt{\lambda }\ln 1\right ) \\ & =c_{1}\cos \left ( 0\right ) +c_{2}\sin \left ( 0\right ) \\ & =c_{1} \end{align*}
Hence the solution now simplifies to\[ \phi \left ( x\right ) =c_{2}\sin \left ( \sqrt{\lambda }\ln x\right ) \] Applying the second B.C. gives\[ 0=c_{2}\sin \left ( \sqrt{\lambda }\ln b\right ) \] For non-trivial solution we want \begin{align*} \sqrt{\lambda }\ln b & =n\pi \qquad n=1,2,3,\cdots \\ \sqrt{\lambda } & =\frac{n\pi }{\ln b}\\ \lambda _{n} & =\left ( \frac{n\pi }{\ln b}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}
Therefore, there are infinite numbers of eigenvalues. The smallest is when \(n=1\) given by \[ \lambda _{1}=\left ( \frac{\pi }{\ln b}\right ) ^{2}\]
From Equation 5.3.6, page 159 in textbook, the eigenfunction are orthogonal with weight function \(\sigma \left ( x\right ) \)\[ \int _{a}^{b}\phi _{n}\left ( x\right ) \phi _{m}\left ( x\right ) \sigma \left ( x\right ) dx=0\qquad n\neq m \] In this problem, the weight \(\sigma =\frac{1}{x}\) and the solution (eigenfuctions) were found above to be \[ \phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}\ln x\right ) \] Now we can verify the orthogonality\[ \int _{1}^{b}\phi _{n}\left ( x\right ) \phi _{m}\left ( x\right ) \sigma \left ( x\right ) dx=\int _{x=1}^{x=b}\sin \left ( \frac{n\pi }{\ln b}\ln x\right ) \sin \left ( \frac{m\pi }{\ln b}\ln x\right ) \frac{1}{x}dx \] Using the substitution \(z=\ln x\), then \(\frac{dz}{dx}=\frac{1}{x}\). When \(x=1,z=\ln 1=0\) and when \(x=b,z=\ln b\), then the above integral becomes\begin{align*} I & =\int _{z=0}^{z=\ln b}\sin \left ( \frac{n\pi }{\ln b}z\right ) \sin \left ( \frac{m\pi }{\ln b}z\right ) \frac{dz}{dx}dx\\ & =\int _{0}^{\ln b}\sin \left ( \frac{n\pi }{\ln b}z\right ) \sin \left ( \frac{m\pi }{\ln b}z\right ) dz \end{align*}
But \(\sin \left ( \frac{n\pi }{\ln b}z\right ) \) and \(\sin \left ( \frac{m\pi }{\ln b}z\right ) \) are orthogonal functions (now with weight \(1\)). Hence the above gives \(0\) when \(n\neq m\) using standard orthogonality of the \(\sin \) functions we used before many times. QED.
The \(n^{th}\) eigenfunction is \[ \phi _{n}\left ( x\right ) =\sin \left ( \frac{n\pi }{\ln b}\ln x\right ) \] Here, the zeros are inside the interval, not counting the end points \(x=1\) and \(x=b\).\[ \left . \left ( \frac{n\pi }{\ln b}\ln x\right ) \right \vert _{x=1}=\left ( \frac{n\pi }{\ln b}0\right ) =0 \] And\begin{align*} \left . \left ( \frac{n\pi }{\ln b}\ln x\right ) \right \vert _{x=b} & =\frac{n\pi }{\ln b}\ln b\\ & =n\pi \end{align*}
Hence for \(n=1\), The domain of \(\phi _{1}\left ( x\right ) \) is \(0\cdots \pi \). And there are no zeros inside this for \(\sin \) function not counting the end points. For \(n=2\), the domain is \(0\cdots 2\pi \) and \(\sin \) has one zero inside this (at \(\pi \)), not counting end points. And for \(n=3,\) the domain is \(0\cdots 3\pi \) and \(\sin \) has two zeros inside this (at \(\pi ,2\pi \)), not counting end points. And so on. Hence \(\phi _{n}\left ( x\right ) \) has \(n-1\) zeros not counting the end points.
The Sturm-Liouville ODE is\[ \frac{d}{dx}\left ( p\phi ^{\prime }\right ) +q\phi =-\lambda \sigma \phi \] Or in operator form, defining \(L\equiv \frac{d}{dx}\left ( p\frac{d}{dx}\right ) +q\), becomes \[ L\left [ \phi \right ] =-\lambda \sigma \phi \] The operator \(L\) is self adjoined when\[ \int _{a}^{b}uL\left [ v\right ] dx=\int _{a}^{b}vL\left [ u\right ] dx \] For the above to work out, we need to show that \[ \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b}=0 \] And this is what we will do now.
Here \(a=0\) and \(b=L\). \begin{align*} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b} & =\left . p\left ( u\frac{dv}{dx}-v\frac{du}{dx}\right ) \right \vert _{0}^{L}\\ & =\left [ p\left ( L\right ) \left ( u\left ( L\right ) \frac{dv}{dx}\left ( L\right ) -v\left ( L\right ) \frac{du}{dx}\left ( L\right ) \right ) -p\left ( 0\right ) \left ( u\left ( 0\right ) \frac{dv}{dx}\left ( 0\right ) -v\left ( 0\right ) \frac{du}{dx}\left ( 0\right ) \right ) \right ] \end{align*}
Substituting \(u\left ( L\right ) =v\left ( L\right ) =0\) and \(\frac{dv}{dx}\left ( 0\right ) =\frac{du}{dx}\left ( 0\right ) =0\) into the above (since there are the B.C. given) gives\begin{align*} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b} & =\left [ p\left ( L\right ) \left ( 0\times \frac{dv}{dx}\left ( L\right ) -0\times \frac{du}{dx}\left ( L\right ) \right ) -p\left ( 0\right ) \left ( u\left ( 0\right ) \times 0-v\left ( 0\right ) \times 0\right ) \right ] \\ & =\left [ 0-0\right ] \\ & =0 \end{align*}
\begin{align} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b} & =\left . p\left ( u\frac{dv}{dx}-v\frac{du}{dx}\right ) \right \vert _{b}^{a}\nonumber \\ & =\left [ p\left ( a\right ) \left ( u\left ( a\right ) v^{\prime }\left ( a\right ) -v\left ( a\right ) u^{\prime }\left ( a\right ) \right ) -p\left ( b\right ) \left ( u\left ( b\right ) v^{\prime }\left ( b\right ) -v\left ( b\right ) u^{\prime }\left ( b\right ) \right ) \right ] \nonumber \\ & =p\left ( a\right ) u\left ( a\right ) v^{\prime }\left ( a\right ) -p\left ( a\right ) v\left ( a\right ) u^{\prime }\left ( a\right ) -p\left ( b\right ) u\left ( b\right ) v^{\prime }\left ( b\right ) +p\left ( b\right ) v\left ( b\right ) u^{\prime }\left ( b\right ) \tag{1} \end{align}
We are given that \(u\left ( a\right ) =u\left ( b\right ) \) and \(v\left ( a\right ) =v\left ( b\right ) \) and \(p\left ( a\right ) u^{\prime }\left ( a\right ) =p\left ( b\right ) u^{\prime }\left ( b\right ) \) and \(p\left ( a\right ) v^{\prime }\left ( a\right ) =p\left ( b\right ) v^{\prime }\left ( b\right ) \).
We start by replacing \(u\left ( a\right ) \) by \(u\left ( a\right ) \) and replacing \(v\left ( a\right ) \) by \(v\left ( b\right ) \) in (1), this gives\begin{align*} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b} & =p\left ( a\right ) u\left ( b\right ) v^{\prime }\left ( a\right ) -p\left ( a\right ) v\left ( b\right ) u^{\prime }\left ( a\right ) -p\left ( b\right ) u\left ( b\right ) v^{\prime }\left ( b\right ) +p\left ( b\right ) v\left ( b\right ) u^{\prime }\left ( b\right ) \\ & =u\left ( b\right ) \left ( p\left ( a\right ) v^{\prime }\left ( a\right ) -p\left ( b\right ) v^{\prime }\left ( b\right ) \right ) +v\left ( b\right ) \left ( p\left ( b\right ) u^{\prime }\left ( b\right ) -p\left ( a\right ) u^{\prime }\left ( a\right ) \right ) \end{align*}
Now using \(p\left ( a\right ) u^{\prime }\left ( a\right ) =p\left ( b\right ) u^{\prime }\left ( b\right ) \) and \(p\left ( a\right ) v^{\prime }\left ( a\right ) =p\left ( b\right ) v^{\prime }\left ( b\right ) \) in the above gives\begin{align*} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b} & =u\left ( b\right ) \left ( p\left ( b\right ) v^{\prime }\left ( b\right ) -p\left ( b\right ) v^{\prime }\left ( b\right ) \right ) +v\left ( b\right ) \left ( p\left ( b\right ) u^{\prime }\left ( b\right ) -p\left ( b\right ) u^{\prime }\left ( b\right ) \right ) \\ & =u\left ( b\right ) \left ( 0\right ) +v\left ( b\right ) \left ( 0\right ) \\ & =0-0\\ & =0 \end{align*}
\(p\) is constant. Hence\begin{align} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{0}^{L} & =\left . p\left ( u\frac{dv}{dx}-v\frac{du}{dx}\right ) \right \vert _{0}^{L}\nonumber \\ & =p\left [ \left ( u\left ( L\right ) v^{\prime }\left ( L\right ) -v\left ( L\right ) u^{\prime }\left ( L\right ) \right ) -\left ( u\left ( 0\right ) v^{\prime }\left ( 0\right ) -v\left ( 0\right ) u^{\prime }\left ( 0\right ) \right ) \right ] \tag{1} \end{align}
We are given that \begin{align} u\left ( L\right ) +\alpha u\left ( 0\right ) +\beta u^{\prime }\left ( 0\right ) & =0\tag{2}\\ u^{\prime }\left ( L\right ) +\gamma u\left ( 0\right ) +\delta u^{\prime }\left ( 0\right ) & =0 \tag{3} \end{align}
And \begin{align} v\left ( L\right ) +\alpha v\left ( 0\right ) +\beta v^{\prime }\left ( 0\right ) & =0\tag{4}\\ v^{\prime }\left ( L\right ) +\gamma v\left ( 0\right ) +\delta v^{\prime }\left ( 0\right ) & =0 \tag{5} \end{align}
From (2),\[ u\left ( L\right ) =-\alpha u\left ( 0\right ) -\beta u^{\prime }\left ( 0\right ) \] From (3)\[ u^{\prime }\left ( L\right ) =-\gamma u\left ( 0\right ) -\delta u^{\prime }\left ( 0\right ) \] From (4)\[ v\left ( L\right ) =-\alpha v\left ( 0\right ) -\beta v^{\prime }\left ( 0\right ) \] From (5)\[ v^{\prime }\left ( L\right ) =-\gamma v\left ( 0\right ) -\delta v^{\prime }\left ( 0\right ) \] Using these 4 relations in equation (1) gives (where \(p\) is removed out, since it is constant, to simplify the equations)\begin{align*} \left . \left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{0}^{L} & =u\left ( L\right ) v^{\prime }\left ( L\right ) -v\left ( L\right ) u^{\prime }\left ( L\right ) -u\left ( 0\right ) v^{\prime }\left ( 0\right ) +v\left ( 0\right ) u^{\prime }\left ( 0\right ) \\ & =\left ( -\alpha u\left ( 0\right ) -\beta u^{\prime }\left ( 0\right ) \right ) \left ( -\gamma v\left ( 0\right ) -\delta v^{\prime }\left ( 0\right ) \right ) \\ & -\left ( -\alpha v\left ( 0\right ) -\beta v^{\prime }\left ( 0\right ) \right ) \left ( -\gamma u\left ( 0\right ) -\delta u^{\prime }\left ( 0\right ) \right ) \\ & -u\left ( 0\right ) v^{\prime }\left ( 0\right ) +v\left ( 0\right ) u^{\prime }\left ( 0\right ) \end{align*}
Simplifying\begin{align*} \left . \left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{0}^{L} & =\alpha u\left ( 0\right ) \gamma v\left ( 0\right ) +\alpha u\left ( 0\right ) \delta v^{\prime }\left ( 0\right ) +\beta u^{\prime }\left ( 0\right ) \gamma v\left ( 0\right ) +\beta u^{\prime }\left ( 0\right ) \delta v^{\prime }\left ( 0\right ) \\ & -\left ( \alpha v\left ( 0\right ) \gamma u\left ( 0\right ) +\alpha v\left ( 0\right ) \delta u^{\prime }\left ( 0\right ) +\beta v^{\prime }\left ( 0\right ) \gamma u\left ( 0\right ) +\beta v^{\prime }\left ( 0\right ) \delta u^{\prime }\left ( 0\right ) \right ) \\ & -u\left ( 0\right ) v^{\prime }\left ( 0\right ) +v\left ( 0\right ) u^{\prime }\left ( 0\right ) \\ & =\alpha u\left ( 0\right ) \gamma v\left ( 0\right ) +\alpha u\left ( 0\right ) \delta v^{\prime }\left ( 0\right ) +\beta u^{\prime }\left ( 0\right ) \gamma v\left ( 0\right ) +\beta u^{\prime }\left ( 0\right ) \delta v^{\prime }\left ( 0\right ) \\ & -\alpha v\left ( 0\right ) \gamma u\left ( 0\right ) -\alpha v\left ( 0\right ) \delta u^{\prime }\left ( 0\right ) -\beta v^{\prime }\left ( 0\right ) \gamma u\left ( 0\right ) -\beta v^{\prime }\left ( 0\right ) \delta u^{\prime }\left ( 0\right ) -u\left ( 0\right ) v^{\prime }\left ( 0\right ) +v\left ( 0\right ) u^{\prime }\left ( 0\right ) \end{align*}
Collecting\begin{align*} \left . \left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{0}^{L} & =\alpha \delta \left ( u\left ( 0\right ) v^{\prime }\left ( 0\right ) -v\left ( 0\right ) u^{\prime }\left ( 0\right ) \right ) \\ & +\beta \delta \left ( u^{\prime }\left ( 0\right ) v^{\prime }\left ( 0\right ) -v^{\prime }\left ( 0\right ) u^{\prime }\left ( 0\right ) \right ) \\ & +\alpha \gamma \left ( u\left ( 0\right ) v\left ( 0\right ) -v\left ( 0\right ) u\left ( 0\right ) \right ) \\ & +\beta \gamma \left ( u^{\prime }\left ( 0\right ) v\left ( 0\right ) -v^{\prime }\left ( 0\right ) u\left ( 0\right ) \right ) \\ & -u\left ( 0\right ) v^{\prime }\left ( 0\right ) +v\left ( 0\right ) u^{\prime }\left ( 0\right ) \\ & =\alpha \delta \left ( u\left ( 0\right ) v^{\prime }\left ( 0\right ) -v\left ( 0\right ) u^{\prime }\left ( 0\right ) \right ) +\beta \gamma \left ( u^{\prime }\left ( 0\right ) v\left ( 0\right ) -v^{\prime }\left ( 0\right ) u\left ( 0\right ) \right ) -\left ( u\left ( 0\right ) v^{\prime }\left ( 0\right ) -v\left ( 0\right ) u^{\prime }\left ( 0\right ) \right ) \\ & =\alpha \delta \left ( u\left ( 0\right ) v^{\prime }\left ( 0\right ) -v\left ( 0\right ) u^{\prime }\left ( 0\right ) \right ) -\beta \gamma \left ( v^{\prime }\left ( 0\right ) u\left ( 0\right ) -u^{\prime }\left ( 0\right ) v\left ( 0\right ) \right ) -\left ( u\left ( 0\right ) v^{\prime }\left ( 0\right ) -v\left ( 0\right ) u^{\prime }\left ( 0\right ) \right ) \end{align*}
Let \(u\left ( 0\right ) v^{\prime }\left ( 0\right ) -v\left ( 0\right ) u^{\prime }\left ( 0\right ) =\Delta \) then we see that the above is just\begin{align*} \left . \left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{0}^{L} & =\alpha \delta \left ( \Delta \right ) -\beta \gamma \left ( \Delta \right ) -\left ( \Delta \right ) \\ & =\Delta \left ( \alpha \delta -\beta \gamma -1\right ) \end{align*}
Hence, for \(\left . \left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{0}^{L}=0\), we need \[ \alpha \delta -\beta \gamma -1=0 \]
We are given that \begin{equation} \int _{a}^{b}uL\left [ v\right ] -vL\left [ u\right ] dx=0 \tag{1} \end{equation} But \begin{align} L\left [ v\right ] & =-\lambda _{v}\sigma \left ( x\right ) v\tag{2}\\ L\left [ u\right ] & =-\lambda _{u}\sigma \left ( x\right ) u \tag{3} \end{align}
Where \(\sigma \left ( x\right ) \) is the weight function of the corresponding Sturm-Liouville ODE that \(u,v\) are its solution eigenfunctions. Substituting (2,3) into (1) gives\begin{align*} \int _{a}^{b}u\left ( -\lambda _{v}\sigma \left ( x\right ) v\right ) -v\left ( -\lambda _{u}\sigma \left ( x\right ) u\right ) dx & =0\\ \int _{a}^{b}-\lambda _{v}\sigma \left ( x\right ) uv+\lambda _{u}\sigma \left ( x\right ) uvdx & =0\\ \left ( \lambda _{u}-\lambda _{v}\right ) \int _{a}^{b}\sigma \left ( x\right ) uvdx & =0 \end{align*}
Since \(u,v\) are different eigenfunctions, then the \(\lambda _{u}-\lambda _{v}\neq 0\) as these are different eigenvalues. (There is one eigenfunction corresponding to each eigenvalue). Therefore the above says that\[ \int _{a}^{b}\sigma \left ( x\right ) u\left ( x\right ) v\left ( x\right ) dx=0 \] Hence different eigenfunctions \(u\left ( x\right ) ,v\left ( x\right ) \) are orthogonal to each others. The weight is \(\sigma \left ( x\right ) \).
\[ L=\frac{d^{4}}{dx^{4}}\]
\begin{align*} uL\left [ v\right ] -vL\left [ u\right ] & =u\frac{d^{4}v}{dx^{4}}-v\frac{d^{4}u}{dx^{4}}\\ & =uv^{\left ( 4\right ) }-vu^{\left ( 4\right ) } \end{align*}
We want to obtain expression of form \(\frac{d}{dx}\left ({}\right ) \) such that it comes out to be \(uv^{\left ( 4\right ) }-vu^{\left ( 4\right ) }\). If we can do this, then it is exact differential. Now, since \begin{equation} \frac{d}{dx}\left ( uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }\right ) =u^{\prime }v^{\prime \prime \prime }+uv^{\left ( 4\right ) }-u^{\prime \prime }v^{\prime \prime }-u^{\prime }v^{\prime \prime \prime } \tag{1} \end{equation} And\begin{equation} \frac{d}{dx}\left ( vu^{\prime \prime \prime }-v^{\prime }u^{\prime \prime }\right ) =v^{\prime }u^{\prime \prime \prime }+vu^{\left ( 4\right ) }-v^{\prime \prime }u^{\prime \prime }-v^{\prime }u^{\prime \prime \prime } \tag{2} \end{equation} Then (1)-(2) gives\begin{align*} \frac{d}{dx}\left ( uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }\right ) -\frac{d}{dx}\left ( vu^{\prime \prime \prime }-v^{\prime }u^{\prime \prime }\right ) & =\left ( u^{\prime }v^{\prime \prime \prime }+uv^{\left ( 4\right ) }-u^{\prime \prime }v^{\prime \prime }-u^{\prime }v^{\prime \prime \prime }\right ) -\left ( v^{\prime }u^{\prime \prime \prime }+vu^{\left ( 4\right ) }-v^{\prime \prime }u^{\prime \prime }-v^{\prime }u^{\prime \prime \prime }\right ) \\ & =u^{\prime }v^{\prime \prime \prime }+uv^{\left ( 4\right ) }-u^{\prime \prime }v^{\prime \prime }-u^{\prime }v^{\prime \prime \prime }-v^{\prime }u^{\prime \prime \prime }-vu^{\left ( 4\right ) }+v^{\prime \prime }u^{\prime \prime }+v^{\prime }u^{\prime \prime \prime }\\ & =uv^{\left ( 4\right ) }-vu^{\left ( 4\right ) } \end{align*}
Hence we found that \begin{align*} \frac{d}{dx}\left ( uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }-vu^{\prime \prime \prime }+v^{\prime }u^{\prime \prime }\right ) & =uv^{\left ( 4\right ) }-vu^{\left ( 4\right ) }\\ & =uL\left [ v\right ] -vL\left [ u\right ] \end{align*}
Therefore \(uL\left [ v\right ] -vL\left [ u\right ] \) is exact differential.
\begin{align*} I & =\int _{a}^{b}uL\left [ v\right ] -vL\left [ u\right ] dx\\ & =\int _{a}^{b}\frac{d}{dx}\left ( uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }-vu^{\prime \prime \prime }+v^{\prime }u^{\prime \prime }\right ) dx\\ & =\left . uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }-vu^{\prime \prime \prime }+v^{\prime }u^{\prime \prime }\right \vert _{a}^{b}\\ & =u\left ( b\right ) v^{\prime \prime \prime }\left ( b\right ) -u^{\prime }\left ( b\right ) v^{\prime \prime }\left ( b\right ) -v\left ( b\right ) u^{\prime \prime \prime }\left ( b\right ) +v^{\prime }\left ( b\right ) u^{\prime \prime }\left ( b\right ) \\ & -\left ( u\left ( a\right ) v^{\prime \prime \prime }\left ( a\right ) -u^{\prime }\left ( a\right ) v^{\prime \prime }\left ( a\right ) -v\left ( a\right ) u^{\prime \prime \prime }\left ( a\right ) +v^{\prime }\left ( a\right ) u^{\prime \prime }\left ( a\right ) \right ) \end{align*}
Or\[ I=u\left ( b\right ) v^{\prime \prime \prime }\left ( b\right ) -u^{\prime }\left ( b\right ) v^{\prime \prime }\left ( b\right ) -v\left ( b\right ) u^{\prime \prime \prime }\left ( b\right ) +v^{\prime }\left ( b\right ) u^{\prime \prime }\left ( b\right ) -u\left ( a\right ) v^{\prime \prime \prime }\left ( a\right ) +u^{\prime }\left ( a\right ) v^{\prime \prime }\left ( a\right ) +v\left ( a\right ) u^{\prime \prime \prime }\left ( a\right ) -v^{\prime }\left ( a\right ) u^{\prime \prime }\left ( a\right ) \]
From part(b),
\begin{equation} I=\int _{0}^{1}uL\left [ v\right ] -vL\left [ u\right ] dx=\left . uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }-vu^{\prime \prime \prime }+v^{\prime }u^{\prime \prime }\right \vert _{0}^{1} \tag{1} \end{equation}
Since we are given that
\begin{align*} \phi \left ( 0\right ) & =0\\ \phi ^{\prime }\left ( 0\right ) & =0\\ \phi \left ( 1\right ) & =0\\ \phi ^{\prime \prime }\left ( 1\right ) & =0 \end{align*}
The above will give \begin{align*} u\left ( 0\right ) & =v\left ( 0\right ) =0\\ u^{\prime }\left ( 0\right ) & =v^{\prime }\left ( 0\right ) =0\\ u\left ( 1\right ) & =v\left ( 1\right ) =0\\ u^{\prime \prime }\left ( 1\right ) & =v^{\prime \prime }\left ( 1\right ) =0 \end{align*}
Substituting these into (1) gives
\begin{align*} \int _{0}^{1}uL\left [ v\right ] -vL\left [ u\right ] dx & =u\left ( 1\right ) v^{\prime \prime \prime }\left ( 1\right ) -u^{\prime }\left ( 1\right ) v^{\prime \prime }\left ( 1\right ) -v\left ( 1\right ) u^{\prime \prime \prime }\left ( 1\right ) +v^{\prime }\left ( 1\right ) u^{\prime \prime }\left ( 1\right ) \\ & -u\left ( 0\right ) v^{\prime \prime \prime }\left ( 0\right ) +u^{\prime }\left ( 0\right ) v^{\prime \prime }\left ( 0\right ) +v\left ( 0\right ) u^{\prime \prime \prime }\left ( 0\right ) -v^{\prime }\left ( 0\right ) u^{\prime \prime }\left ( 0\right ) \end{align*}
Therefore
\begin{align*} \int _{0}^{1}uL\left [ v\right ] -vL\left [ u\right ] dx & =\left ( 0\times v^{\prime \prime \prime }\left ( 1\right ) \right ) -0-\left ( 0\times u^{\prime \prime \prime }\left ( 1\right ) \right ) +0-\left ( 0\times v^{\prime \prime \prime }\left ( 0\right ) \right ) +0+\left ( 0\times u^{\prime \prime \prime }\left ( 0\right ) \right ) -0\\ & =0 \end{align*}
Any boundary conditions which makes \(\left . uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }-vu^{\prime \prime \prime }+v^{\prime }u^{\prime \prime }\right \vert _{0}^{1}=0\) will do. For example, \begin{align*} \phi \left ( 0\right ) & =0\\ \phi ^{\prime }\left ( 0\right ) & =0\\ \phi \left ( 1\right ) & =0\\ \phi ^{\prime }\left ( 1\right ) & =0 \end{align*}
The above will give \begin{align*} u\left ( 0\right ) & =v\left ( 0\right ) =0\\ u^{\prime }\left ( 0\right ) & =v^{\prime }\left ( 0\right ) =0\\ u\left ( 1\right ) & =v\left ( 1\right ) =0\\ u^{\prime }\left ( 1\right ) & =v^{\prime }\left ( 1\right ) =0 \end{align*}
Substituting these into (1) gives
\begin{align*} \int _{0}^{1}uL\left [ v\right ] -vL\left [ u\right ] dx & =u\left ( 1\right ) v^{\prime \prime \prime }\left ( 1\right ) -u^{\prime }\left ( 1\right ) v^{\prime \prime }\left ( 1\right ) -v\left ( 1\right ) u^{\prime \prime \prime }\left ( 1\right ) +v^{\prime }\left ( 1\right ) u^{\prime \prime }\left ( 1\right ) \\ & -u\left ( 0\right ) v^{\prime \prime \prime }\left ( 0\right ) +u^{\prime }\left ( 0\right ) v^{\prime \prime }\left ( 0\right ) +v\left ( 0\right ) u^{\prime \prime \prime }\left ( 0\right ) -v^{\prime }\left ( 0\right ) u^{\prime \prime }\left ( 0\right ) \\ & =\left ( 0\times v^{\prime \prime \prime }\left ( 1\right ) \right ) -\left ( 0\times v^{\prime \prime }\left ( 1\right ) \right ) -\left ( 0\times u^{\prime \prime \prime }\left ( 1\right ) \right ) +\left ( 0\times u^{\prime \prime }\left ( 1\right ) \right ) \\ & -\left ( 0\times v^{\prime \prime \prime }\left ( 0\right ) \right ) +\left ( 0\times v^{\prime \prime }\left ( 0\right ) \right ) +\left ( 0\times u^{\prime \prime \prime }\left ( 0\right ) \right ) -\left ( 0\times u^{\prime \prime }\left ( 0\right ) \right ) \\ & =0 \end{align*}
Given \[ \frac{d^{4}}{dx^{4}}\phi +\lambda e^{x}\phi =0 \] Therefore\[ L\left [ \phi \right ] =-\lambda e^{x}\phi \] Therefore, for eigenfunctions \(u,v\) we have\begin{align*} L\left [ u\right ] & =-\lambda _{u}e^{x}u\\ L\left [ v\right ] & =-\lambda _{v}e^{x}v \end{align*}
Where \(\lambda _{u},\lambda _{v}\) are the eigenvalues associated with eigenfunctions \(u,v\) and they are not the same. Hence now we can write \begin{align*} 0 & =\int _{0}^{1}uL\left [ v\right ] -vL\left [ u\right ] dx\\ & =\int _{0}^{1}u\left ( -\lambda _{v}e^{x}v\right ) -v\left ( -\lambda _{u}e^{x}u\right ) dx\\ & =\int _{0}^{1}-\lambda _{v}e^{x}uv+\lambda _{u}e^{x}uvdx\\ & =\int _{0}^{1}\left ( \lambda _{u}-\lambda _{v}\right ) \left ( e^{x}uv\right ) dx\\ & =\left ( \lambda _{u}-\lambda _{v}\right ) \int _{0}^{1}\left ( e^{x}uv\right ) dx \end{align*}
Since \(\lambda _{u}-\lambda _{v}\neq 0\) then \[ \int _{0}^{1}\left ( e^{x}uv\right ) dx=0 \] Hence \(u,v\) are orthogonal to each others with weight function \(e^{x}.\)
Equation 5.5.22 is\begin{equation} p\left ( \phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\right ) =\text{constant} \tag{5.5.22} \end{equation} Looking at boundary conditions at one end, say at \(x=a\) (left end), and let the boundary conditions there be \[ \beta _{1}\phi \left ( a\right ) +\beta _{2}\phi ^{\prime }\left ( a\right ) =0 \] Therefore for eigenfunctions \(\phi _{1},\phi _{2}\) we obtain\begin{align} \beta _{1}\phi _{1}\left ( a\right ) +\beta _{2}\phi _{1}^{\prime }\left ( a\right ) & =0\tag{1}\\ \beta _{1}\phi _{2}\left ( a\right ) +\beta _{2}\phi _{2}^{\prime }\left ( a\right ) & =0 \tag{2} \end{align}
From (1), \begin{equation} \phi _{1}^{\prime }\left ( a\right ) =-\frac{\beta _{1}}{\beta _{2}}\phi _{1}\left ( a\right ) \tag{3} \end{equation} From (2)\begin{equation} \phi _{2}^{\prime }\left ( a\right ) =-\frac{\beta _{1}}{\beta _{2}}\phi _{2}\left ( a\right ) \tag{4} \end{equation} Substituting (3,4) into \(\phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\) gives, at end point \(a\), the following\begin{align*} \phi _{1}\left ( a\right ) \phi _{2}^{\prime }\left ( a\right ) -\phi _{2}\left ( a\right ) \phi _{1}^{\prime }\left ( a\right ) & =\phi _{1}\left ( a\right ) \left ( -\frac{\beta _{1}}{\beta _{2}}\phi _{2}\left ( a\right ) \right ) -\phi _{2}\left ( a\right ) \left ( -\frac{\beta _{1}}{\beta _{2}}\phi _{1}\left ( a\right ) \right ) \\ & =-\frac{\beta _{1}}{\beta _{2}}\phi _{2}\left ( a\right ) \phi _{1}\left ( a\right ) +\frac{\beta _{1}}{\beta _{2}}\phi _{2}\left ( a\right ) \phi _{1}\left ( a\right ) \\ & =0 \end{align*}
In the above, we evaluated \(\phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\) at one end point, and found it to be zero. But \(\phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\) is the Wronskian \(W\left ( x\right ) \). It is known that if \(W\left ( x\right ) =0\) at just one point, then it is zero at all points in the range. Hence we conclude that \[ \phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }=0 \] For all \(x.\) This also means the eigenfunctions \(\phi _{1},\phi _{2}\) are linearly dependent. This gives equation 5.5.23. QED.
Equation 5.5.22 is\begin{equation} p\left ( \phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\right ) =\text{constant} \tag{5.5.22} \end{equation} At one end, say end \(x=a\), is where the singularity exist. This means \(p\left ( a\right ) =0.\) Now to show that \(p\left ( \phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\right ) =0\) at \(x=a\), we just need to show that \(\phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\) is bounded. Since in that case, we will have \(0\times A=0\), where \(A\) is some value which is \(\phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\). But boundary conditions at \(x=1\) must be \(\phi \left ( a\right ) <\infty \) and also \(\phi ^{\prime }\left ( a\right ) <\infty \). This is always the case at the end where \(p=0\).
Then let \(\phi \left ( a\right ) =c_{1}\) and \(\phi ^{\prime }\left ( a\right ) =c_{2}\), where \(c_{1},c_{2}\) are some constants. Then we write\begin{align*} \phi _{1}\left ( a\right ) & =c_{1}\\ \phi _{1}^{\prime }\left ( a\right ) & =c_{2}\\ \phi _{2}\left ( a\right ) & =c_{1}\\ \phi _{2}^{\prime }\left ( a\right ) & =c_{2} \end{align*}
Hence it follows immediately that\begin{align*} \phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime } & =c_{1}c_{2}-c_{2}c_{1}\\ & =0 \end{align*}
Hence we showed that \(\phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\) is bounded. Then \(p\left ( \phi _{1}\phi _{2}^{\prime }-\phi _{2}\phi _{1}^{\prime }\right ) =0.\) QED.