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2.7 HW 6

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2.7.1 Problem 5.3.2

2.7.1.1 Part (a)

$$\rho \frac{\partial ^{2}u}{\partial t^{2}}=T_{0}\frac{\partial ^{2}u}{\partial t^{2}}+\alpha u+\beta \frac{\partial u}{\partial t}$$ 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.

2.7.1.2 Part (b)

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

$$\begin{align*} T^{\prime \prime }-cT^{\prime }+\lambda T & =0\\ X^{\prime \prime }+X\left ( \frac{\alpha }{T_{0}}+\lambda \frac{\rho }{T_{0}}\right ) & =0 \end{align*}$$

2.7.1.3 Part (c)

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.

2.7.2 Problem 5.3.3

$$\frac{d^{2}\phi }{dx^{2}}+\alpha \left ( x\right ) \frac{d\phi }{dx}+\left ( \lambda \beta \left ( x\right ) +\gamma \left ( x\right ) \right ) \phi =0$$ 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

2.7.3 Problem 5.3.9

$$\begin{align} x^{2}\phi ^{\prime \prime }+x\phi ^{\prime }+\lambda \phi & =0\tag{1}\\ \phi \left ( 1\right ) & =0\nonumber \\ \phi \left ( b\right ) & =0\nonumber \end{align}$$

2.7.3.1 Part (a)

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.

2.7.3.2 Part(b)

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.

2.7.3.3 Part(c)

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

$$\lambda _{1}=\left ( \frac{\pi }{\ln b}\right ) ^{2}$$

2.7.3.4 Part (d)

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.

2.7.3.5 Part(e)

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.

2.7.4 Problem 5.5.1 (b,d,g)

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.

2.7.4.1 Part(b)

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

$$\begin{align*} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b} & =\left [ p\left ( L\right ) \left ( 0\times \frac{dv}{dx}\left ( L\right ) -0\times \frac{du}{dx}\left ( L\right ) \right ) -p\left ( 0\right ) \left ( u\left ( 0\right ) \times 0-v\left ( 0\right ) \times 0\right ) \right ] \\ & =\left [ 0-0\right ] \\ & =0 \end{align*}$$

2.7.4.2 Part (d)

$$\begin{align} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b} & =\left . p\left ( u\frac{dv}{dx}-v\frac{du}{dx}\right ) \right \vert _{b}^{a}\nonumber \\ & =\left [ p\left ( a\right ) \left ( u\left ( a\right ) v^{\prime }\left ( a\right ) -v\left ( a\right ) u^{\prime }\left ( a\right ) \right ) -p\left ( b\right ) \left ( u\left ( b\right ) v^{\prime }\left ( b\right ) -v\left ( b\right ) u^{\prime }\left ( b\right ) \right ) \right ] \nonumber \\ & =p\left ( a\right ) u\left ( a\right ) v^{\prime }\left ( a\right ) -p\left ( a\right ) v\left ( a\right ) u^{\prime }\left ( a\right ) -p\left ( b\right ) u\left ( b\right ) v^{\prime }\left ( b\right ) +p\left ( b\right ) v\left ( b\right ) u^{\prime }\left ( b\right ) \tag{1} \end{align}$$

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

$$\begin{align*} \left . p\left ( uv^{\prime }-vu^{\prime }\right ) \right \vert _{a}^{b} & =u\left ( b\right ) \left ( p\left ( b\right ) v^{\prime }\left ( b\right ) -p\left ( b\right ) v^{\prime }\left ( b\right ) \right ) +v\left ( b\right ) \left ( p\left ( b\right ) u^{\prime }\left ( b\right ) -p\left ( b\right ) u^{\prime }\left ( b\right ) \right ) \\ & =u\left ( b\right ) \left ( 0\right ) +v\left ( b\right ) \left ( 0\right ) \\ & =0-0\\ & =0 \end{align*}$$

2.7.4.3 Part (g)

$p$ 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

$$\alpha \delta -\beta \gamma -1=0$$

2.7.5 Problem 5.5.3

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 $\sigma \left ( x\right )$.

2.7.6 Problem 5.5.8

$$L=\frac{d^{4}}{dx^{4}}$$

2.7.6.1 Part (a)

$$\begin{align*} uL\left [ v\right ] -vL\left [ u\right ] & =u\frac{d^{4}v}{dx^{4}}-v\frac{d^{4}u}{dx^{4}}\\ & =uv^{\left ( 4\right ) }-vu^{\left ( 4\right ) } \end{align*}$$

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.

2.7.6.2 Part (b)

$$\begin{align*} I & =\int _{a}^{b}uL\left [ v\right ] -vL\left [ u\right ] dx\\ & =\int _{a}^{b}\frac{d}{dx}\left ( uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }-vu^{\prime \prime \prime }+v^{\prime }u^{\prime \prime }\right ) dx\\ & =\left . uv^{\prime \prime \prime }-u^{\prime }v^{\prime \prime }-vu^{\prime \prime \prime }+v^{\prime }u^{\prime \prime }\right \vert _{a}^{b}\\ & =u\left ( b\right ) v^{\prime \prime \prime }\left ( b\right ) -u^{\prime }\left ( b\right ) v^{\prime \prime }\left ( b\right ) -v\left ( b\right ) u^{\prime \prime \prime }\left ( b\right ) +v^{\prime }\left ( b\right ) u^{\prime \prime }\left ( b\right ) \\ & -\left ( u\left ( a\right ) v^{\prime \prime \prime }\left ( a\right ) -u^{\prime }\left ( a\right ) v^{\prime \prime }\left ( a\right ) -v\left ( a\right ) u^{\prime \prime \prime }\left ( a\right ) +v^{\prime }\left ( a\right ) u^{\prime \prime }\left ( a\right ) \right ) \end{align*}$$

Or

$$I=u\left ( b\right ) v^{\prime \prime \prime }\left ( b\right ) -u^{\prime }\left ( b\right ) v^{\prime \prime }\left ( b\right ) -v\left ( b\right ) u^{\prime \prime \prime }\left ( b\right ) +v^{\prime }\left ( b\right ) u^{\prime \prime }\left ( b\right ) -u\left ( a\right ) v^{\prime \prime \prime }\left ( a\right ) +u^{\prime }\left ( a\right ) v^{\prime \prime }\left ( a\right ) +v\left ( a\right ) u^{\prime \prime \prime }\left ( a\right ) -v^{\prime }\left ( a\right ) u^{\prime \prime }\left ( a\right )$$

2.7.6.3 Part (c)

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

$$\begin{align*} \int _{0}^{1}uL\left [ v\right ] -vL\left [ u\right ] dx & =\left ( 0\times v^{\prime \prime \prime }\left ( 1\right ) \right ) -0-\left ( 0\times u^{\prime \prime \prime }\left ( 1\right ) \right ) +0-\left ( 0\times v^{\prime \prime \prime }\left ( 0\right ) \right ) +0+\left ( 0\times u^{\prime \prime \prime }\left ( 0\right ) \right ) -0\\ & =0 \end{align*}$$

2.7.6.4 Part (d)

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

$$\begin{align*} \int _{0}^{1}uL\left [ v\right ] -vL\left [ u\right ] dx & =u\left ( 1\right ) v^{\prime \prime \prime }\left ( 1\right ) -u^{\prime }\left ( 1\right ) v^{\prime \prime }\left ( 1\right ) -v\left ( 1\right ) u^{\prime \prime \prime }\left ( 1\right ) +v^{\prime }\left ( 1\right ) u^{\prime \prime }\left ( 1\right ) \\ & -u\left ( 0\right ) v^{\prime \prime \prime }\left ( 0\right ) +u^{\prime }\left ( 0\right ) v^{\prime \prime }\left ( 0\right ) +v\left ( 0\right ) u^{\prime \prime \prime }\left ( 0\right ) -v^{\prime }\left ( 0\right ) u^{\prime \prime }\left ( 0\right ) \\ & =\left ( 0\times v^{\prime \prime \prime }\left ( 1\right ) \right ) -\left ( 0\times v^{\prime \prime }\left ( 1\right ) \right ) -\left ( 0\times u^{\prime \prime \prime }\left ( 1\right ) \right ) +\left ( 0\times u^{\prime \prime }\left ( 1\right ) \right ) \\ & -\left ( 0\times v^{\prime \prime \prime }\left ( 0\right ) \right ) +\left ( 0\times v^{\prime \prime }\left ( 0\right ) \right ) +\left ( 0\times u^{\prime \prime \prime }\left ( 0\right ) \right ) -\left ( 0\times u^{\prime \prime }\left ( 0\right ) \right ) \\ & =0 \end{align*}$$

2.7.6.5 Part (e)

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 $e^{x}.$

2.7.7 Problem 5.5.10

2.7.7.1 Part(a)

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.

2.7.7.2 Part(b)

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.