Processing math: 100%

2.8 HW 7

PDF (letter size)

PDF (legal size)

2.8.1 Problem 5.6.1 (a)

2.8.1.1 part (a)

$$\begin{align*} \frac{d^{2}\phi }{dx^{2}}+\left ( \lambda -x^{2}\right ) \phi & =0\\ \phi ^{\prime }\left ( 0\right ) & =0\\ \phi \left ( 1\right ) & =0 \end{align*}$$

Putting the equation in the form

$$\frac{d^{2}\phi }{dx^{2}}-x^{2}\phi =-\lambda \phi$$ And comparing it to the standard Sturm-Liouville form $$p\frac{d^{2}\phi }{dx^{2}}+p^{\prime }\frac{d\phi }{dx}+q\phi =-\lambda \sigma \phi$$ Shows that $$\begin{align*} p & =1\\ q & =-x^{2}\\ \sigma & =1 \end{align*}$$

Now the Rayleigh quotient is

$$\lambda =\frac{-\left ( p\phi \phi ^{\prime }\right ) _{0}^{1}+\int _{0}^{1}p\left ( \phi ^{\prime }\right ) ^{2}-q\phi ^{2}dx}{\int _{0}^{1}\sigma \phi ^{2}dx}$$ Substituting known values, and since $\phi ^{\prime }\left ( 0\right ) =0,\phi \left ( 1\right ) =0$ the above simplifies to $$\lambda =\frac{\int _{0}^{1}\left ( \phi ^{\prime }\right ) ^{2}+x^{2}\phi ^{2}dx}{\int _{0}^{1}\phi ^{2}dx}$$ Now we can say that $$\begin{equation} \lambda _{\min }=\lambda _{1}\leq \frac{\int _{0}^{1}\left ( \phi ^{\prime }\right ) ^{2}+x^{2}\phi ^{2}dx}{\int _{0}^{1}\phi ^{2}dx} \tag{1} \end{equation}$$ We now need a trial solution $\phi _{trial}$ to use in the above, which needs only to satisfy boundary conditions to use to estimate lowest $\lambda _{\min }$. The simplest such function will do. The boundary conditions are $\phi ^{\prime }\left ( 0\right ) =0$,$\phi \left ( 1\right ) =0.$ We see for example that $\phi _{trial}\left ( x\right ) =x^{2}-1$ works, since $\phi _{trial}^{\prime }\left ( x\right ) =2x,$ and $\phi _{trial}^{\prime }\left ( 0\right ) =0$ and $\phi _{trial}\left ( 1\right ) =1-1=0$. So will use this in (1) $$\begin{align*} \lambda _{\min } & =\lambda _{1}\leq \frac{\int _{0}^{1}\left ( 2x\right ) ^{2}+x^{2}\left ( x^{2}-1\right ) ^{2}dx}{\int _{0}^{1}\left ( x^{2}-1\right ) ^{2}dx}\\ & =\frac{\int _{0}^{1}\left ( 2x\right ) ^{2}+x^{2}\left ( x^{4}-2x^{2}+1\right ) dx}{\int _{0}^{1}\left ( x^{4}-2x^{2}+1\right ) dx}\\ & =\frac{\int _{0}^{1}4x^{2}+x^{6}-2x^{4}+x^{2}dx}{\int _{0}^{1}\left ( x^{4}-2x^{2}+1\right ) dx}\\ & =\frac{\int _{0}^{1}3x^{2}+x^{6}dx}{\int _{0}^{1}x^{4}-2x^{2}+1dx}\\ & =\frac{\left ( x^{3}+\frac{1}{7}x^{7}\right ) _{0}^{1}}{\left ( \frac{1}{5}x^{5}-\frac{2}{3}x^{3}+x\right ) _{0}^{1}}=\frac{\left ( 1+\frac{1}{7}\right ) }{\left ( \frac{1}{5}-\frac{2}{3}+1\right ) }\\ & =\frac{15}{7}\\ & =2.1429 \end{align*}$$

Hence

$$\lambda _{1}\leq 2.1429$$

2.8.2 Problem 5.6.2

$$\begin{align*} \frac{d^{2}\phi }{dx^{2}}+\left ( \lambda -x^{2}\right ) \phi & =0\\ \phi ^{\prime }\left ( 0\right ) & =0\\ \phi ^{\prime }\left ( 1\right ) & =0 \end{align*}$$

Putting the equation in the form

$$\frac{d^{2}\phi }{dx^{2}}-x^{2}\phi =-\lambda \phi$$ And comparing it to the standard Sturm-Liouville form $$p\frac{d^{2}\phi }{dx^{2}}+p^{\prime }\frac{d\phi }{dx}+q\phi =-\lambda \sigma \phi$$ Shows that $$\begin{align*} p & =1\\ q & =-x^{2}\\ \sigma & =1 \end{align*}$$

Now the Rayleigh quotient is

$$\lambda =\frac{-\left ( p\phi \phi ^{\prime }\right ) _{0}^{1}+\int _{0}^{1}p\left ( \phi ^{\prime }\right ) ^{2}-q\phi ^{2}dx}{\int _{0}^{1}\sigma \phi ^{2}dx}$$ Substituting known values, and since $\phi ^{\prime }\left ( 0\right ) =0,\phi \left ( 1\right ) =0$ the above simplifies to $$\lambda =\frac{\int _{0}^{1}\left ( \phi ^{\prime }\right ) ^{2}+x^{2}\phi ^{2}dx}{\int _{0}^{1}\phi ^{2}dx}$$ Since eigenfunction $\phi$ can not be identically zero, the denominator in the above expression can only be positive, since the integrand is positive. So we need now to consider the numerator term only: $$\int _{0}^{1}\left ( \phi ^{\prime }\right ) ^{2}dx+\int _{0}^{1}x^{2}\phi ^{2}dx$$ For the second term, again, this can only be positive since $\phi$ can not be zero. For the first term, there are two cases. If $\phi ^{\prime }$ zero or not. If it is not zero, then the term is positive and we are done. This means $\lambda >0$. if $\phi ^{\prime }=0$ then $\int _{0}^{1}\left ( \phi ^{\prime }\right ) ^{2}dx=0$ and also conclude $\lambda >0$ thanks to the second term $\int _{0}^{1}x^{2}\phi ^{2}$ being positive. So we conclude that $\lambda$ can only be positive.

2.8.3 Problem 5.6.4

Problem

Consider eigenvalue problem $\frac{d}{dr}\left ( r\frac{d\phi }{dr}\right ) =-\lambda r\phi ,0<r<1$ subject to B.C. $\left \vert \phi \left ( 0\right ) \right \vert <\infty$ (you may also assume $\frac{d\phi }{dr}$ bounded). And $\frac{d\phi }{dr}\left ( 1\right ) =0$. (a) prove that $\lambda \geq 0$. (b) Solve the boundary value problem. You may assume eigenfunctions are known. Derive coefficients using orthogonality.

Notice: Correction was made to problem per class email. Book said to show that $\lambda >0$ which is error changed to $\lambda \geq 0$.

2.8.3.1 Part (a)

From the problem we see that $p=r,q=0,\sigma =r$. The Rayleigh quotient is

$$\begin{align} \lambda & =\frac{-\left ( p\phi \phi ^{\prime }\right ) _{0}^{1}+\int _{0}^{1}p\left ( \phi ^{\prime }\right ) ^{2}-q\phi ^{2}dr}{\int _{0}^{1}\sigma \phi ^{2}dr}\nonumber \\ & =\frac{-\left ( r\phi \phi ^{\prime }\right ) _{0}^{1}+\int _{0}^{1}r\left ( \phi ^{\prime }\right ) ^{2}dr}{\int _{0}^{1}r\phi ^{2}dr} \tag{1} \end{align}$$

The term $-\left ( r\phi \phi ^{\prime }\right ) _{0}^{1}$ expands to

$$-\left ( \left ( 1\right ) \phi \left ( 1\right ) \phi ^{\prime }\left ( 1\right ) -\left ( 0\right ) \phi \left ( 0\right ) \phi ^{\prime }\left ( 0\right ) \right )$$ Since $\phi ^{\prime }\left ( 1\right ) =0$, the above is zero and Equation(1) reduces to $$\lambda =\frac{\int _{0}^{1}r\left ( \phi ^{\prime }\right ) ^{2}dr}{\int _{0}^{1}r\phi ^{2}dr}$$ The denominator above can only be positive, as an eigenfunction $\phi$ can not be identically zero. For the numerator, we have to consider two cases.

case 1 If $\phi ^{\prime }\neq 0$ then we are done. The numerator is positive and we conclude that $\lambda >0$.  

case 2 If $\phi ^{\prime }=0$ then $\phi$ is constant and this means $\lambda =0$ is possible hence $\lambda \geq 0$. Now we need to show $\phi$ being constant is also possible. Since $\phi ^{\prime }\left ( 1\right ) =0$, then for $\phi ^{\prime }=0$ to be true everywhere, it should also be $\phi ^{\prime }\left ( 0\right ) =0$ which means $\phi \left ( 0\right )$ is some constant. We are told that $\left \vert \phi \left ( 0\right ) \right \vert <\infty$. Hence means $\phi \left ( 0\right )$ is constant is possible value (since bounded). Hence $\phi ^{\prime }=0$ is possible.

Therefore $\lambda \geq 0.$ QED.

2.8.3.2 Part (b)

The ODE is

$$\begin{align} r\phi ^{\prime \prime }+\phi ^{\prime }+\lambda r\phi & =0\qquad 0<r<1\tag{1}\\ \left \vert \phi \left ( 0\right ) \right \vert & <\infty \nonumber \\ \phi ^{\prime }\left ( 1\right ) & =0\nonumber \end{align}$$

In standard form the ODE is $\phi ^{\prime \prime }+\frac{1}{r}\phi ^{\prime }+\lambda \phi =0$. This shows that $r=0$ is a regular singular point. Therefore we try

$$\phi \left ( r\right ) =\sum _{n=0}^{\infty }a_{n}r^{n+\alpha }$$ Hence $$\begin{align*} \phi ^{\prime }\left ( r\right ) & =\sum _{n=0}^{\infty }\left ( n+\alpha \right ) a_{n}r^{n+\alpha -1}\\ \phi ^{\prime \prime }\left ( r\right ) & =\sum _{n=0}^{\infty }\left ( n+\alpha \right ) \left ( n+\alpha -1\right ) a_{n}r^{n+\alpha -2} \end{align*}$$

Substituting back into the ODE gives

$$\begin{align*} r\sum _{n=0}^{\infty }\left ( n+\alpha \right ) \left ( n+\alpha -1\right ) a_{n}r^{n+\alpha -2}+\sum _{n=0}^{\infty }\left ( n+\alpha \right ) a_{n}r^{n+\alpha -1}+\lambda r\sum _{n=0}^{\infty }a_{n}r^{n+\alpha } & =0\\ \sum _{n=0}^{\infty }\left ( n+\alpha \right ) \left ( n+\alpha -1\right ) a_{n}r^{n+\alpha -1}+\sum _{n=0}^{\infty }\left ( n+\alpha \right ) a_{n}r^{n+\alpha -1}+\lambda \sum _{n=0}^{\infty }a_{n}r^{n+\alpha +1} & =0 \end{align*}$$

To make all powers of $r$ the same, we subtract $2$ from the power of last term, and add $2$ to the index, resulting in

$$\sum _{n=0}^{\infty }\left ( n+\alpha \right ) \left ( n+\alpha -1\right ) a_{n}r^{n+\alpha -1}+\sum _{n=0}^{\infty }\left ( n+\alpha \right ) a_{n}r^{n+\alpha -1}+\lambda \sum _{n=2}^{\infty }a_{n-2}r^{n+\alpha -1}=0$$ For $n=0$ we obtain $$\begin{align*} \left ( \alpha \right ) \left ( \alpha -1\right ) a_{0}r^{\alpha -1}+\left ( \alpha \right ) a_{0}r^{\alpha -1} & =0\\ \left ( \alpha \right ) \left ( \alpha -1\right ) a_{0}+\left ( \alpha \right ) a_{0} & =0\\ a_{0}\left ( \alpha ^{2}-\alpha +\alpha \right ) & =0\\ a_{0}\alpha ^{2} & =0 \end{align*}$$

But $a_{0}\neq 0$ (we always enforce this condition in power series solution), which implies

$$\fbox{$\alpha =0$}$$ Now we look at $n=1$, which gives $$\begin{align*} \left ( 1\right ) \left ( 1-1\right ) a_{1}r^{n+\alpha -1}+\left ( 1\right ) a_{1}r^{n+\alpha -1} & =0\\ a_{1} & =0 \end{align*}$$

For $n\geq 2$, now all terms join in, and we get a recursive relation

$$\begin{align*} \left ( n\right ) \left ( n-1\right ) a_{n}r^{n-1}+\left ( n\right ) a_{n}r^{n-1}+\lambda a_{n-2}r^{n-1} & =0\\ \left ( n\right ) \left ( n-1\right ) a_{n}+\left ( n\right ) a_{n}+\lambda a_{n-2} & =0\\ a_{n} & =\frac{-\lambda a_{n-2}}{\left ( n\right ) \left ( n-1\right ) +n}\\ & =\frac{-\lambda }{n^{2}}a_{n-2} \end{align*}$$

For example, for $n=2$, we get

$$a_{2}=\frac{-\lambda }{2^{2}}a_{0}$$ All odd powers of $n$ result in $a_{n}=0$. For $n=4$ $$a_{4}=\frac{-\lambda }{4^{2}}a_{2}=\frac{-\lambda }{4^{2}}\left ( \frac{-\lambda }{2^{2}}a_{0}\right ) =\frac{\lambda ^{2}}{\left ( 2^{2}\right ) \left ( 4^{2}\right ) }a_{0}$$ And for $n=6$ $$a_{6}=\frac{-\lambda }{6^{2}}a_{4}=\frac{-\lambda }{6^{2}}\frac{\lambda ^{2}}{\left ( 2^{2}\right ) \left ( 4^{2}\right ) }a_{0}=\frac{-\lambda ^{3}}{\left ( 2^{2}\right ) \left ( 4^{2}\right ) \left ( 6^{2}\right ) }a_{0}$$ And so on. The series is $$\begin{align} \phi \left ( r\right ) & =\sum _{n=0}^{\infty }a_{n}r^{n}\nonumber \\ & =a_{0}+0+a_{2}r^{2}+0+a_{4}r^{4}+0+a_{6}r^{6}+\cdots \nonumber \\ & =a_{0}-\frac{\lambda r^{2}}{2^{2}}a_{0}+\frac{\lambda ^{2}r^{4}}{\left ( 2^{2}\right ) \left ( 4^{2}\right ) }a_{0}-\frac{\lambda ^{3}r^{6}}{\left ( 2^{2}\right ) \left ( 4^{2}\right ) \left ( 6^{2}\right ) }a_{0}+\cdots \nonumber \\ & =a_{0}\left ( 1-\frac{\left ( \sqrt{\lambda }r\right ) ^{2}}{2^{2}}+\frac{\left ( \sqrt{\lambda }r\right ) ^{4}}{\left ( 2^{2}\right ) \left ( 4^{2}\right ) }-\frac{\left ( \sqrt{\lambda }r\right ) ^{6}}{\left ( 2^{2}\right ) \left ( 4^{2}\right ) \left ( 6^{2}\right ) }+\cdots \right ) \tag{2} \end{align}$$

From tables, Bessel function of first kind of order zero, has series expansion given by

$$\begin{align} J_{o}\left ( z\right ) & =\sum _{n=0}^{\infty }\frac{\left ( -1\right ) ^{n}}{\left ( n!\right ) ^{2}}\left ( \frac{z}{2}\right ) ^{2n}\nonumber \\ & =1-\left ( \frac{z}{2}\right ) ^{2}+\frac{1}{\left ( 2\right ) ^{2}}\left ( \frac{z}{2}\right ) ^{4}-\frac{1}{\left ( \left ( 2\right ) \left ( 3\right ) \right ) ^{2}}\left ( \frac{z}{2}\right ) ^{6}+\cdots \nonumber \\ & =1-\frac{z^{2}}{2^{2}}+\frac{1}{2^{2}4^{2}}z^{4}-\frac{1}{2^{2}3^{2}2^{6}}z^{6}+\cdots \nonumber \\ & =1-\frac{z^{2}}{2^{2}}+\frac{z^{4}}{2^{2}4^{2}}-\frac{z^{6}}{2^{2}4^{2}6^{2}}+\cdots \tag{3} \end{align}$$

By comparing (2),(3) we see a match between $J_{o}\left ( z\right )$ and $\phi \left ( r\right )$, if we let $z=\sqrt{\lambda }r$ we conclude that

$$\fbox{$\phi _1\left ( r\right ) =a_0J_0\left ( \sqrt{\lambda }r\right ) $}$$

We can now normalized the above eigenfunction so that $a_{0}=1$ as mentioned in class. But it is not needed. The above is the first solution. We now need second solution. For repeated roots, the second solution will be

$$\phi _{2}\left ( r\right ) =\phi _{1}\left ( r\right ) \ln \left ( r\right ) +r^{\alpha }\sum _{n=0}^{\infty }b_{n}r^{n}$$ But $\alpha =0$, hence $$\phi _{2}\left ( r\right ) =\phi _{1}\left ( r\right ) \ln \left ( r\right ) +\sum _{n=0}^{\infty }b_{n}r^{n}$$ Hence the solution is $$\phi \left ( r\right ) =c_{1}\phi _{1}\left ( r\right ) +c_{2}\phi _{2}\left ( r\right )$$ Since $\phi \left ( 0\right )$ is bounded, then $c_{2}=0$ (since $\ln \left ( 0\right )$ not bounded at zero), and the solution becomes (where $a_{0}$ is now absorbed with the constant $c_{1}$) $$\begin{align*} \phi \left ( r\right ) & =c_{1}\phi _{1}\left ( r\right ) \\ & =cJ_{0}\left ( \sqrt{\lambda }r\right ) \end{align*}$$

The boundedness condition has eliminated the second solution altogether. Now we apply the second boundary conditions $\phi ^{\prime }\left ( 1\right ) =0$ to find allowed eigenvalues. Since

$$\phi ^{\prime }\left ( r\right ) =-cJ_{1}\left ( \sqrt{\lambda }r\right )$$ Then $\phi ^{\prime }\left ( 1\right ) =0$ implies $$0=-cJ_{1}\left ( \sqrt{\lambda }\right )$$ The zeros of this are the values of $\sqrt{\lambda }$. Using the computer, these are the first few such values. $$\begin{align*} \sqrt{\lambda _{1}} & =3.8317\\ \sqrt{\lambda _{2}} & =7.01559\\ \sqrt{\lambda _{3}} & =10.1735\\ & \vdots \end{align*}$$

or

$$\begin{align*} \lambda _{1} & =14.682\\ \lambda _{2} & =49.219\\ \lambda _{3} & =103.5\\ & \vdots \end{align*}$$

Hence

$$\begin{align} \phi _{n}\left ( r\right ) & =c_{n}J_{0}\left ( \sqrt{\lambda _{n}}r\right ) \tag{4}\\ \phi \left ( r\right ) & =\sum _{n=1}^{\infty }\phi _{n}\left ( r\right ) \nonumber \\ & =\sum _{n=1}^{\infty }c_{n}J_{0}\left ( \sqrt{\lambda _{n}}r\right ) \nonumber \end{align}$$

To find $c_{n}$, we use orthogonality. Per class discussion, we can now assume this problem was part of initial value problem, and that at $t=0$ we had initial condition of $f\left ( r\right )$, therefore, we now write

$$f\left ( r\right ) =\sum _{n=1}^{\infty }c_{n}J_{0}\left ( \sqrt{\lambda _{n}}r\right )$$ Multiplying both sides by $J_{0}\left ( \sqrt{\lambda _{m}}r\right ) \sigma$ and integrating gives (where $\sigma =r$) $$\begin{align*} \int _{0}^{1}\sum _{n=1}^{\infty }f\left ( r\right ) J_{0}\left ( \sqrt{\lambda _{m}}r\right ) rdr & =\sum _{n=1}^{\infty }\int _{0}^{1}c_{n}J_{0}\left ( \sqrt{\lambda _{m}}r\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr\\ & =\int _{0}^{1}c_{m}J_{0}^{2}\left ( \sqrt{\lambda _{m}}r\right ) rdr\\ & =c_{m}\int _{0}^{1}J_{0}^{2}\left ( \sqrt{\lambda _{m}}r\right ) rdr\\ & =c_{m}\Omega \end{align*}$$

Where $\Omega$ is some constant. Therefore

$$c_{n}=\frac{\int _{0}^{1}\sum _{n=1}^{\infty }f\left ( r\right ) J_{0}\left ( \sqrt{\lambda _{m}}r\right ) rdr}{\Omega }$$ And $\phi \left ( r\right )$ $$\sum _{n=1}^{\infty }c_{n}J_{0}\left ( \sqrt{\lambda _{n}}r\right )$$ With the eigenvalues given as above, which have to be computed for each $n$ using the computer.

2.8.4 Problem 5.9.1 (b)

From textbook, equation 5.9.8, we are given that for large $\lambda$

$$\begin{align} \phi \left ( x\right ) & \approx \left ( \sigma p\right ) ^{\frac{-1}{4}}\exp \left ( \pm i\sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \nonumber \\ & =\left ( \sigma p\right ) ^{\frac{-1}{4}}\left ( c_{1}\cos \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) +c_{2}\sin \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \right ) \tag{1} \end{align}$$

Where $c_{1},c_{2}$ are the two constants of integration since this is second order ODE. For $\phi \left ( 0\right ) =0$, the integral $\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt=\int _{0}^{0}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt=0$ and the above becomes

$$\begin{align*} 0 & =\phi \left ( 0\right ) \\ & =\left ( \sigma p\right ) ^{\frac{-1}{4}}\left ( c_{1}\cos \left ( 0\right ) +c_{2}\sin \left ( 0\right ) \right ) \\ & =c_{1}\left ( \sigma p\right ) ^{\frac{-1}{4}} \end{align*}$$

Hence $c_{1}=0$ and (1) reduces to

$$\phi \left ( x\right ) =c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\sin \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right )$$ Hence $$\begin{align*} \phi ^{\prime }\left ( x\right ) & =c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\cos \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \left ( \frac{d}{dx}\sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \\ & =c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\cos \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \left ( \sqrt{\lambda }\sqrt{\frac{\sigma \left ( x\right ) }{p\left ( x\right ) }}\right ) \\ & =c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\sqrt{\frac{\lambda \sigma }{p}}\cos \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \end{align*}$$

Since $\phi ^{\prime }\left ( L\right ) =0$ then

$$0=c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\sqrt{\frac{\lambda \sigma }{p}}\cos \left ( \sqrt{\lambda }\int _{0}^{L}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right )$$ Which means, for non-trivial solution, that $$\sqrt{\lambda _{n}}\int _{0}^{L}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt=\left ( n-\frac{1}{2}\right ) \pi$$ Therefore, for large $\lambda$, (i.e. large $n$) the estimate is $$\begin{align*} \sqrt{\lambda _{n}} & =\frac{\left ( n-\frac{1}{2}\right ) \pi }{\int _{0}^{L}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt}\\ \lambda _{n} & =\left ( \frac{\left ( n-\frac{1}{2}\right ) \pi }{\int _{0}^{L}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt}\right ) ^{2} \end{align*}$$

2.8.5 Problem 5.9.2

$$\phi ^{\prime \prime }+\lambda \left ( 1+x\right ) \phi =0$$ Comparing the above to Sturm-Liouville form $$\left ( p\phi ^{\prime }\right ) ^{\prime }+q\phi =-\lambda \sigma \phi$$ Shows that $$\begin{align*} p & =1\\ q & =0\\ \sigma & =1+x \end{align*}$$

Now, from textbook, equation 5.9.8, we are given that for large $\lambda$

$$\begin{align} \phi \left ( x\right ) & \approx \left ( \sigma p\right ) ^{\frac{-1}{4}}\exp \left ( \pm i\sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \nonumber \\ & =\left ( \sigma p\right ) ^{\frac{-1}{4}}\left ( c_{1}\cos \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) +c_{2}\sin \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \right ) \tag{1} \end{align}$$

Where $c_{1},c_{2}$ are the two constants of integration since this is second order ODE. For $\phi \left ( 0\right ) =0$, the integral $\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt=\int _{0}^{0}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt=0$ and the above becomes

$$\begin{align*} 0 & =\phi \left ( 0\right ) \\ & =\left ( \sigma p\right ) ^{\frac{-1}{4}}\left ( c_{1}\cos \left ( 0\right ) +c_{2}\sin \left ( 0\right ) \right ) \\ & =c_{1}\left ( \sigma p\right ) ^{\frac{-1}{4}} \end{align*}$$

Hence $c_{1}=0$ and (1) reduces to

$$\phi \left ( x\right ) =c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\sin \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right )$$ Applying the second boundary condition $\phi \left ( 1\right ) =0$ on the above gives $$\begin{align*} 0 & =\phi \left ( 1\right ) \\ & =c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\sin \left ( \sqrt{\lambda }\int _{0}^{1}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \end{align*}$$

Hence for non-trivial solution we want, for large positive integer $n$

$$\begin{align*} \sqrt{\lambda }\int _{0}^{1}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt & =n\pi \\ \sqrt{\lambda } & =\frac{n\pi }{\int _{0}^{1}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt}\\ & =\frac{n\pi }{\int _{0}^{1}\sqrt{1+t}dt} \end{align*}$$

But $\int _{0}^{1}\sqrt{1+t}dt=1.21895$, hence

$$\begin{align*} \sqrt{\lambda } & =\frac{n\pi }{1.21895}=2.5773n\\ \lambda & =6.6424n^{2} \end{align*}$$

Therefore, solution for large $\lambda$ is

$$\begin{align*} \phi \left ( x\right ) & =c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\sin \left ( \sqrt{\lambda }\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \\ & =c_{2}\left ( \sigma p\right ) ^{\frac{-1}{4}}\sin \left ( 2.5773n\int _{0}^{x}\sqrt{\frac{\sigma \left ( t\right ) }{p\left ( t\right ) }}dt\right ) \\ & =c_{2}\left ( 1+x\right ) ^{\frac{-1}{4}}\sin \left ( 2.5773n\int _{0}^{x}\sqrt{1+t}dt\right ) \\ & =c_{2}\left ( 1+x\right ) ^{\frac{-1}{4}}\sin \left ( 2.5773n\left ( -\frac{2}{3}+\frac{2}{3}\left ( 1+x\right ) ^{\frac{3}{2}}\right ) \right ) \end{align*}$$

To plot this, let us assume $c_{2}=1$ (we have no information given to find $c_{2}$). What value of $n$ to use? Will use different values of $n$ in increasing order. So the following is plot of

$$\phi \left ( x\right ) =\left ( 1+x\right ) ^{\frac{-1}{4}}\sin \left ( 2.5773n\left ( -\frac{2}{3}+\frac{2}{3}\left ( 1+x\right ) ^{\frac{3}{2}}\right ) \right )$$ For $x=0\cdots 1$ and for $n=10,20,30,\cdots ,80$.