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3.6 HW 6

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3.6.1 Problem 1

Consider the equation $xy^{\prime \prime }+\left ( c-x\right ) y^{\prime }-ay=0$. Identify a regular singular point and find two series solutions around this point. Test the solutions for convergence.

Solution

Writing the ODE as

$$y^{\prime \prime }+A\left ( x\right ) y^{\prime }+B\left ( x\right ) y=0$$ Where $$\begin{align*} A\left ( x\right ) & =\frac{\left ( c-x\right ) }{x}\\ B\left ( x\right ) & =\frac{-a}{x} \end{align*}$$

The above shows that $x_{0}=0$ is a singularity point for both $A\left ( x\right )$ and $B\left ( x\right )$. Examining $A\left ( x\right )$ and $B\left ( x\right )$ to determine what type of singular point it is

$$\lim _{x\rightarrow x_{0}}\left ( x-x_{0}\right ) A\left ( x\right ) =\lim _{x\rightarrow 0}x\frac{\left ( c-x\right ) }{x}=\lim _{x\rightarrow 0}\left ( c-x\right ) =c$$ Because the limit exists, then $x_{0}=0$ is regular singular point for $A\left ( x\right ) .$ $$\lim _{x\rightarrow x_{0}}\left ( x-x_{0}\right ) ^{2}B\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}\left ( \frac{-a}{x}\right ) =\lim _{x\rightarrow 0}\left ( -ax\right ) =0$$ Because the limit exists, then $x_{0}=0$ is also regular singular point for $B\left ( x\right )$.

Therefore $x_{0}=0$ is a regular singular point for the ODE.

Assuming the solution is Frobenius series  gives

$$\begin{align*} y\left ( x\right ) & =x^{r}\sum _{n=0}^{\infty }C_{n}\left ( x-x_{0}\right ) ^{n}\qquad C_{0}\neq 0\\ & =x^{r}\sum _{n=0}^{\infty }C_{n}x^{n}\\ & =\sum _{n=0}^{\infty }C_{n}x^{n+r} \end{align*}$$

Therefore

$$\begin{align*} y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) C_{n}x^{n+r-1}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) C_{n}x^{n+r-2} \end{align*}$$

Substituting the above in the original ODE $xy^{\prime \prime }+\left ( c-x\right ) y^{\prime }-ay=0$ gives

$$\begin{align*} x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) C_{n}x^{n+r-2}+\left ( c-x\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) C_{n}x^{n+r-1}-a\sum _{n=0}^{\infty }C_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-1}+c\sum _{n=0}^{\infty }\left ( n+r\right ) C_{n}x^{n+r-1}-x\sum _{n=0}^{\infty }\left ( n+r\right ) C_{n}x^{n+r-1}-\sum _{n=0}^{\infty }aC_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) C_{n}x^{n+r-1}+\sum _{n=0}^{\infty }c\left ( n+r\right ) C_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\left ( n+r\right ) C_{n}x^{n+r}-\sum _{n=0}^{\infty }aC_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +c\left ( n+r\right ) \right ) C_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\left ( \left ( n+r\right ) +a\right ) C_{n}x^{n+r} & =0 \end{align*}$$

Since all powers of $x$ have to be the same, adjusting indices and exponents gives (where in the second sum above, the outside index $n$ is increased by $1$ and $n$ inside the sum is decreased by $1$)

$$\begin{equation} \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +c\left ( n+r\right ) \right ) C_{n}x^{n+r-1}-\sum _{n=1}^{\infty }\left ( \left ( n-1+r\right ) +a\right ) C_{n-1}x^{n+r-1}=0 \tag{1} \end{equation}$$ Setting $n=0$ gives the indicial equation, which only comes from the first sum above as the second sum starts from $n=1$. $$\left ( \left ( r\right ) \left ( r-1\right ) +cr\right ) C_{0}=0$$ Since $C_{0}\neq 0$ then $$\begin{align*} \left ( r\right ) \left ( r-1\right ) +cr & =0\\ r^{2}-r+cr & =0\\ r\left ( r+c-1\right ) & =0 \end{align*}$$

The roots are

$$\begin{align*} r_{1} & =1-c\\ r_{2} & =0 \end{align*}$$

Assuming that $r_{2}-r_{1}$ is not an integer, in other words, assuming $1-c$ is not an integer (problem did not say), then In this case, two linearly independent solutions can be constructed directly. The first is associated with $r_{1}=1-c$ and the second is associated with $r_{2}=0$. These solutions are

$$\begin{align*} y_{1}\left ( x\right ) & =\sum _{n=0}^{\infty }C_{n}x^{n+1-c}\qquad C_{0}\neq 0\\ y_{2}\left ( x\right ) & =\sum _{n=0}^{\infty }D_{n}x^{n}\qquad D_{0}\neq 0 \end{align*}$$

The coefficients are not the same in each solution. For the first one $C_{n}$ is used and for the second $D_{n}$ is used.

The solution $y_{1}\left ( x\right )$ associated with $r_{1}=1-c$ is now found. From (1), and replacing $r$ by $1-c$ gives

$$\begin{align*} \sum _{n=0}^{\infty }\left ( \left ( n+1-c\right ) \left ( n+1-c-1\right ) +c\left ( n+1-c\right ) \right ) C_{n}x^{n+1-c-1}-\sum _{n=1}^{\infty }\left ( \left ( n-1+1-c\right ) +a\right ) C_{n-1}x^{n+1-c-1} & =0\\ \sum _{n=0}^{\infty }\left ( \left ( n+1-c\right ) \left ( n-c\right ) +c\left ( n+1-c\right ) \right ) C_{n}x^{n-c}-\sum _{n=1}^{\infty }\left ( \left ( n-c\right ) +a\right ) C_{n-1}x^{n-c} & =0\\ \sum _{n=0}^{\infty }n\left ( n-c+1\right ) C_{n}x^{n-c}-\sum _{n=1}^{\infty }\left ( \left ( n-c\right ) +a\right ) C_{n-1}x^{n-c} & =0 \end{align*}$$

For $n>0$ the above gives the recursive relation ($n=0$ is not used, since it was used to find $r$). For $n>0$ the last equation above gives

$$\begin{align*} n\left ( n-c+1\right ) C_{n}-\left ( \left ( n-c\right ) +a\right ) C_{n-1} & =0\\ C_{n} & =\frac{\left ( \left ( n-c\right ) +a\right ) }{n\left ( n-c+1\right ) }C_{n-1} \end{align*}$$

Few terms are generated to see the pattern. For $n=1$

$$C_{1}=\frac{\left ( 1-c+a\right ) }{1\left ( 1-c+1\right ) }C_{0}=\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }C_{0}$$ For $n=2$ $$\begin{align*} C_{2} & =\frac{\left ( 2-c+a\right ) }{2\left ( 2-c+1\right ) }C_{1}\\ & =\frac{\left ( 2-c+a\right ) }{2\left ( 3-c\right ) }\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }C_{0} \end{align*}$$

For $n=3$

$$\begin{align*} C_{3} & =\frac{\left ( 3-c+a\right ) }{3\left ( 3-c+1\right ) }C_{2}\\ & =\frac{\left ( 3-c+a\right ) }{3\left ( 4-c\right ) }\frac{\left ( 2-c+a\right ) }{2\left ( 3-c\right ) }\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }C_{0} \end{align*}$$

And so on. The pattern for general term is

$$\begin{align*} C_{n} & =\frac{\left ( \left ( n-c\right ) +a\right ) }{n\left ( n-c+1\right ) }\cdot \cdot \cdot \frac{\left ( 3-c+a\right ) }{3\left ( 3-c+1\right ) }\frac{\left ( 2-c+a\right ) }{2\left ( 2-c+1\right ) }\frac{\left ( 1-c+a\right ) }{1\left ( 1-c+1\right ) }C_{0}\\ & ={\displaystyle \prod \limits _{m=1}^{n}} \frac{\left ( \left ( m-c\right ) +a\right ) }{m\left ( n-c+1\right ) } \end{align*}$$

Therefore the solution associated with $r_{1}=1-c$ is

$$\begin{align*} y_{1}\left ( x\right ) & =\sum _{n=0}^{\infty }C_{n}x^{n+r}\\ & =\sum _{n=0}^{\infty }C_{n}x^{n+1-c}\\ & =C_{0}x^{1-c}+C_{1}x^{2-c}+C_{2}x^{3-c}+\cdots \end{align*}$$

Using results found above, and looking at few terms gives the first solution as

$$y_{1}\left ( x\right ) =C_{0}x^{1-c}\left ( 1+\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }x+\frac{1}{2}\frac{\left ( 2-c+a\right ) }{\left ( 3-c\right ) }\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }x^{2}+\frac{1}{6}\frac{\left ( 3-c+a\right ) }{\left ( 4-c\right ) }\frac{\left ( 2-c+a\right ) }{\left ( 3-c\right ) }\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }x^{3}+\cdots \right )$$ The second solution associated with $r_{2}=0$ is now found.  As above, using (1) but with $D_{n}$ instead of $C_{n}$ for coefficients and replacing $r$ by zero gives $$\sum _{n=0}^{\infty }\left ( n\left ( n-1\right ) +cn\right ) D_{n}x^{n-1}-\sum _{n=1}^{\infty }\left ( \left ( n-1\right ) +a\right ) D_{n-1}x^{n-1}=0$$ For $n>0$ the above gives the recursive relation for the second solution $$\begin{align*} \left ( n\left ( n-1\right ) +cn\right ) D_{n}-\left ( \left ( n-1\right ) +a\right ) D_{n-1} & =0\\ D_{n} & =\frac{n-1+a}{n\left ( n-1\right ) +cn}D_{n-1}\\ & =\frac{n-1+a}{cn-n+n^{2}}D_{n-1} \end{align*}$$

Few terms are now generated to see the pattern. For $n=1$

$$D_{1}=\frac{a}{c}D_{0}$$ For $n=2$ $$\begin{align*} D_{2} & =\frac{1+a}{2c-2+4}D_{1}\\ & =\frac{1+a}{2\left ( c+1\right ) }\frac{a}{c}D_{0} \end{align*}$$

For $n=3$

$$\begin{align*} D_{3} & =\frac{3-1+a}{3c-3+9}D_{2}\\ & =\frac{2+a}{3\left ( c+2\right ) }\frac{1+a}{2\left ( c+1\right ) }\frac{a}{c}D_{0} \end{align*}$$

And so on. Hence the solution $y_{2}\left ( x\right )$ is

$$\begin{align*} y_{2}\left ( x\right ) & =\sum _{n=0}^{\infty }D_{n}x^{n}\\ & =D_{0}+D_{1}x+D_{2}x^{2}+\cdots \end{align*}$$

Using result found above gives the second solution as

$$y_{2}\left ( x\right ) =D_{0}\left ( 1+\frac{a}{c}x+\frac{1}{2}\frac{\left ( 1+a\right ) a}{c\left ( c+1\right ) }x^{2}+\frac{1}{6}\frac{a\left ( 1+a\right ) \left ( 2+a\right ) }{c\left ( c+2\right ) \left ( c+1\right ) }x^{3}+\cdots \right )$$ The final solution is therefore the sum of the two solutions $$\begin{align} y\left ( x\right ) & =C_{0}x^{1-c}\left ( 1+\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }x+\frac{1}{2}\frac{\left ( 2-c+a\right ) }{\left ( 3-c\right ) }\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }x^{2}+\frac{1}{6}\frac{\left ( 3-c+a\right ) }{\left ( 4-c\right ) }\frac{\left ( 2-c+a\right ) }{\left ( 3-c\right ) }\frac{\left ( 1-c+a\right ) }{\left ( 2-c\right ) }x^{3}+\cdots \right ) \tag{2}\\ & +D_{0}\left ( 1+\frac{a}{c}x+\frac{1}{2}\frac{\left ( 1+a\right ) a}{c\left ( c+1\right ) }x^{2}+\frac{1}{6}\frac{a\left ( 1+a\right ) \left ( 2+a\right ) }{c\left ( c+2\right ) \left ( c+1\right ) }x^{3}+\cdots \right ) \nonumber \end{align}$$

Where $C_{0},D_{0}$ are the two constant of integration.

Testing for convergence. For $y_{1}\left ( x\right )$ solution, the general term from above was

$$C_{n}x^{n}=\frac{\left ( \left ( n-c\right ) +a\right ) }{n\left ( n-c+1\right ) }C_{n-1}x^{n}$$ Hence by ratio test $$\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{C_{n}x^{n}}{C_{n-1}x^{n-1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{\left ( \left ( n-c\right ) +a\right ) }{n\left ( n-c+1\right ) }C_{n-1}x^{n}}{C_{n-1}x^{n-1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( \left ( n-c\right ) +a\right ) x}{\left ( n\left ( n-c+1\right ) \right ) }\right \vert \\ & =\left \vert x\right \vert \lim _{n\rightarrow \infty }\left \vert \frac{n-c+a}{n^{2}-nc+n}\right \vert \\ & =\left \vert x\right \vert \lim _{n\rightarrow \infty }\left \vert \frac{\frac{1}{n}-\frac{c}{n^{2}}+\frac{a}{n^{2}}}{1-\frac{c}{n}+\frac{1}{n}}\right \vert \\ & =\left \vert x\right \vert \left \vert \frac{0}{1}\right \vert \\ & =0 \end{align*}$$

Therefore the series $y_{1}\left ( x\right )$ converges for all $x$.

Testing for convergence. For $y_{2}\left ( x\right )$ solution, the general term is

$$D_{n}x^{n}=\frac{n-1+a}{cn-n+n^{2}}D_{n-1}x^{n}$$ Hence by ratio test $$\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{D_{n}x^{n}}{D_{n-1}x^{n-1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{n-1+a}{cn-n+n^{2}}D_{n-1}x^{n}}{D_{n-1}x^{n-1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{n-1+a}{cn-n+n^{2}}x\right \vert \\ & =\left \vert x\right \vert \lim _{n\rightarrow \infty }\left \vert \frac{n-1+a}{cn-n+n^{2}}\right \vert \\ & =\left \vert x\right \vert \lim _{n\rightarrow \infty }\left \vert \frac{\frac{1}{n}-\frac{1}{n^{2}}+\frac{a}{n^{2}}}{\frac{c}{n}-\frac{1}{n}+1}\right \vert \\ & =\left \vert x\right \vert \left \vert \frac{0}{1}\right \vert \\ & =0 \end{align*}$$

Therefore the series $y_{2}\left ( x\right )$ also converges for all $x$.  This means the solution $y\left ( x\right ) =y_{1}\left ( x\right ) +y_{2}\left ( x\right )$ found in (2) above also converges for all $x$.

3.6.2 Problem 2

The Sturm Liouville equation can be expressed as

$$L\left [ u\left ( x\right ) \right ] =\lambda \rho \left ( x\right ) u\left ( x\right )$$ Where $L$ is given as in class. Show $L$ is Hermitian on the domain $a\leq x\leq b$ with boundary conditions $u\left ( a\right ) =u\left ( b\right ) =0$. Find the orthogonality condition.

Solution

$$L=-\left ( p\frac{d^{2}}{dx^{2}}+p^{\prime }\frac{d}{dx}-q\right )$$ The operator $L$ is Hermitian if $$\int _{a}^{b}\bar{v}L\left [ u\right ] dx=\overline{\int _{a}^{b}\bar{u}L\left [ v\right ] dx}$$ Where in the above $u,v$ are any two functions defined over the domain that satisfy the boundary conditions given. Starting from the left integral to show it will result in the right integral. Replacing $L\left [ u\right ]$ by $-\left ( p\frac{d^{2}}{dx^{2}}+p^{\prime }\frac{d}{dx}-q\right ) u$ in the LHS of the above gives $$\begin{align} -\int _{a}^{b}\bar{v}\left ( p\frac{d^{2}}{dx^{2}}+p^{\prime }\frac{d}{dx}-q\right ) u\ dx & =-\int _{a}^{b}\bar{v}\left ( p\frac{d^{2}u}{dx^{2}}+p^{\prime }\frac{du}{dx}-qu\right ) \ dx\nonumber \\ & =-\int _{a}^{b}\bar{v}p\frac{d^{2}u}{dx^{2}}+\bar{v}p^{\prime }\frac{du}{dx}-q\bar{v}u\ dx\nonumber \\ & =-\overset{I_{1}}{\overbrace{\int _{a}^{b}p\bar{v}\frac{d^{2}u}{dx^{2}}dx}}-\int _{a}^{b}\bar{v}p^{\prime }\frac{du}{dx}dx+\int _{a}^{b}q\bar{v}u\ dx \tag{1} \end{align}$$

Looking at the first integral above, which is $I_{1}=\int _{a}^{b}\left ( p\bar{v}\right ) \left ( \frac{d^{2}u}{dx^{2}}\right ) dx$.  The idea is to integrate this twice to move the second derivative from $u$ to $\bar{v}$. Applying $\int AdB=AB-\int BdA$, where

$$\begin{align*} A & \equiv p\bar{v}\\ dB & \equiv \frac{d^{2}u}{dx^{2}} \end{align*}$$

Hence

$$\begin{align*} dA & =p\frac{d\bar{v}}{dx}+p^{\prime }\bar{v}\\ B & =\frac{du}{dx} \end{align*}$$

Therefore the integral $I_{1}$ in (1) becomes

$$\begin{align*} I_{1} & =\int _{a}^{b}p\bar{v}\frac{d^{2}}{dx^{2}}u\\ & =\left [ p\bar{v}\frac{du}{dx}\right ] _{a}^{b}-\int _{a}^{b}\frac{du}{dx}\left ( p\frac{d\bar{v}}{dx}+p^{\prime }\bar{v}\right ) dx \end{align*}$$

But $\bar{v}\left ( a\right ) =0$ and $\bar{v}\left ( b\right ) =0$, hence the boundary terms above vanish and simplifies to

$$\begin{align} I_{1} & =-\int _{a}^{b}p\frac{du}{dx}\frac{d\bar{v}}{dx}+p^{\prime }\bar{v}\ \frac{du}{dx}dx\nonumber \\ & =-\int _{a}^{b}p\frac{du}{dx}\frac{d\bar{v}}{dx}dx-\int _{a}^{b}p^{\prime }\bar{v}\ \frac{du}{dx}dx \tag{2} \end{align}$$

Before integrating by parts a second time, putting the result of $I_{1}$ back into (1) first simplifies the result. Substituting (2) into (1) gives

$$\begin{align*} \int _{a}^{b}\bar{v}L\left [ u\right ] dx & =-I_{1}-\int _{a}^{b}\bar{v}p^{\prime }\frac{du}{dx}dx+\int _{a}^{b}q\bar{v}u\ dx\\ & =-\overset{I_{1}}{\overbrace{\left ( -\int _{a}^{b}p\frac{du}{dx}\frac{d\bar{v}}{dx}dx-\int _{a}^{b}p^{\prime }\bar{v}\ \frac{du}{dx}dx\right ) }}-\int _{a}^{b}\bar{v}p^{\prime }\frac{du}{dx}dx+\int _{a}^{b}q\bar{v}u\ dx\\ & =\int _{a}^{b}p\frac{du}{dx}\frac{d\bar{v}}{dx}dx+\int _{a}^{b}p^{\prime }\bar{v}\ \frac{du}{dx}dx-\int _{a}^{b}\bar{v}p^{\prime }\frac{du}{dx}dx+\int _{a}^{b}q\bar{v}u\ dx \end{align*}$$

The second and third terms above cancel and the result becomes

$$\begin{equation} \int _{a}^{b}\bar{v}L\left [ u\right ] dx=\overset{I_{2}}{\overbrace{\int _{a}^{b}p\frac{du}{dx}\frac{d\bar{v}}{dx}dx}}+\int _{a}^{b}q\bar{v}u\ dx \tag{3} \end{equation}$$ Now integration by parts is applied on the first integral above. Let $I_{2}=\int _{a}^{b}\frac{du}{dx}\left ( p\frac{d\bar{v}}{dx}\right ) dx$. Applying $\int AdB=AB-\int BdA$, where $$\begin{align*} A & \equiv p\frac{d\bar{v}}{dx}\\ dB & \equiv \frac{du}{dx} \end{align*}$$

Hence

$$\begin{align*} dA & =p\frac{d^{2}\bar{v}}{dx^{2}}+p^{\prime }\frac{d\bar{v}}{dx}\\ B & =u \end{align*}$$

Therefore the integral $I_{2}$ becomes

$$I_{2}=\left [ p\frac{d\bar{v}}{dx}u\right ] _{a}^{b}-\int _{a}^{b}u\left ( p\frac{d^{2}\bar{v}}{dx^{2}}+p^{\prime }\frac{d\bar{v}}{dx}\right ) dx$$ But $u\left ( a\right ) =0,u\left ( b\right ) =0$, hence the boundary term vanishes and the above simplifies to $$I_{2}=-\int _{a}^{b}u\left ( p\frac{d^{2}\bar{v}}{dx^{2}}+p^{\prime }\frac{d\bar{v}}{dx}\right ) dx$$ Substituting the above back into (3) gives $$\begin{align*} \int _{a}^{b}\bar{v}L\left [ u\right ] dx & =-\int _{a}^{b}u\left ( p\frac{d^{2}\bar{v}}{dx^{2}}+p^{\prime }\frac{d\bar{v}}{dx}\right ) dx+\int _{a}^{b}q\bar{v}u\ dx\\ & =-\int _{a}^{b}u\left ( p\frac{d^{2}\bar{v}}{dx^{2}}+p^{\prime }\frac{d\bar{v}}{dx}-q\bar{v}\right ) dx \end{align*}$$

But $-\left ( p\frac{d^{2}\bar{v}}{dx^{2}}+p^{\prime }\frac{d\bar{v}}{dx}-q\bar{v}\right ) =L\left [ \bar{v}\right ]$ by definition, and the above becomes

$$\int _{a}^{b}\bar{v}L\left [ u\right ] dx=\int _{a}^{b}uL\left [ \bar{v}\right ] dx$$ But $\int _{a}^{b}uL\left [ \bar{v}\right ] dx=\overline{\int _{a}^{b}\bar{u}L\left [ v\right ] dx}$, and the above becomes $$\int _{a}^{b}\bar{v}L\left [ u\right ] dx=\overline{\int _{a}^{b}\bar{u}\left ( L\left [ v\right ] \right ) dx}$$ Therefore $L$ is Hermitian.

3.6.3 Problem 3

1.

For the equation $y^{\prime \prime }+\frac{1-\alpha ^{2}}{4x^{2}}y=0$ show that two solutions are $y_{1}\left ( x\right ) =a_{0}x^{\frac{1+\alpha }{2}}$ and $y_{2}\left ( x\right ) =a_{0}x^{\frac{1-\alpha }{2}}$

2.

For $\alpha =0$, the two solutions are not independent. Find a second solution $y_{20}$ by solving $W^{\prime }=0$ ($W$ is the Wronskian).

3.

Show that the second solution found in (2) is a limiting case of the two solutions from part (1). That is $$y_{20}=\lim _{\alpha \rightarrow 0}\frac{y_{1}-y_{2}}{\alpha }$$

Solution

Part 1

The point $x_{0}=0$ is a regular singular point. This is shown as follows.

$$\begin{align*} \,\lim _{x\rightarrow x_{0}}\left ( x-x_{0}\right ) ^{2}\frac{1-\alpha ^{2}}{4x^{2}} & =\lim _{x\rightarrow 0}x^{2}\frac{1-\alpha ^{2}}{4x^{2}}\\ & =\lim _{x\rightarrow 0}\frac{1-\alpha ^{2}}{4}\\ & =\frac{1-\alpha ^{2}}{4} \end{align*}$$

Since the limit exist, then $x_{0}=0$ is a regular singular point. Assuming the solution is a Frobenius series given by

$$y\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}x^{n+r}\qquad c_{0}\neq 0$$ Therefore $$\begin{align*} y^{\prime }\left ( x\right ) & =\sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n+r-1}\\ y^{\prime \prime }\left ( x\right ) & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-2} \end{align*}$$

Substituting the above 2 expressions back into the original ODE gives

$$\begin{align} 4x^{2}\left ( \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-2}\right ) +\left ( 1-\alpha ^{2}\right ) \left ( \sum _{n=0}^{\infty }c_{n}x^{n+r}\right ) & =0\nonumber \\ \sum _{n=0}^{\infty }4\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r}+\left ( 1-\alpha ^{2}\right ) \left ( \sum _{n=0}^{\infty }c_{n}x^{n+r}\right ) & =0\tag{1} \end{align}$$

Looking at $n=0$ first, in order to obtain the indicial equation gives

$$\begin{align*} 4\left ( r\right ) \left ( r-1\right ) c_{0}+\left ( 1-\alpha ^{2}\right ) c_{0} & =0\\ c_{0}\left ( 4r^{2}-4r+\left ( 1-\alpha ^{2}\right ) \right ) & =0 \end{align*}$$

But $c_{0}\neq 0$, therefore

$$r^{2}-r+\frac{\left ( 1-\alpha ^{2}\right ) }{4}=0$$ The roots are $r=\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}$, but $a=1,b=-1,c=\frac{\left ( 1-\alpha ^{2}\right ) }{4}$, hence the roots are $$\begin{align*} r & =\frac{1}{2}\pm \frac{1}{2}\sqrt{1-\left ( 1-\alpha ^{2}\right ) }\\ & =\frac{1}{2}\pm \frac{1}{2}\sqrt{\alpha ^{2}}\\ & =\frac{1}{2}\pm \frac{1}{2}\alpha \end{align*}$$

Hence $r_{1}=\frac{1}{2}\left ( 1+\alpha \right )$ and $r_{2}=\frac{1}{2}\left ( 1-\alpha \right )$. Each one of these roots gives a solution. The difference is

$$\begin{align*} r_{2}-r_{1} & =\frac{1}{2}\left ( 1+\alpha \right ) -\frac{1}{2}\left ( 1-\alpha \right ) \\ & =\alpha \end{align*}$$

Therefore, to use the same solution form $y_{1}\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}x^{n+r_{1}}$ and $y_{2}\left ( x\right ) =\sum _{n=0}^{\infty }d_{n}x^{n+r_{2}}$ for each, it is assumed that $\alpha$ is not an integer. In this case, the recursive relation for $y_{1}\left ( x\right )$ is found from (1) by using $r=\frac{1}{2}\left ( 1+\alpha \right )$ which results in

$$\sum _{n=0}^{\infty }4\left ( n+\frac{1}{2}\left ( 1+\alpha \right ) \right ) \left ( n+\frac{1}{2}\left ( 1+\alpha \right ) -1\right ) c_{n}x^{n+\frac{1}{2}\left ( 1+\alpha \right ) }+\left ( 1-\alpha ^{2}\right ) \left ( \sum _{n=0}^{\infty }c_{n}x^{n+\frac{1}{2}\left ( 1+\alpha \right ) }\right ) =0$$ For $n>0$ the above becomes $$\begin{align*} 4\left ( n+\frac{1}{2}\left ( 1+\alpha \right ) \right ) \left ( n+\frac{1}{2}\left ( 1+\alpha \right ) -1\right ) c_{n}+\left ( 1-\alpha ^{2}\right ) c_{n} & =0\\ \left ( 4\left ( n+\frac{1}{2}\left ( 1+\alpha \right ) \right ) \left ( n+\frac{1}{2}\left ( 1+\alpha \right ) -1\right ) +\left ( 1-\alpha ^{2}\right ) \right ) c_{n} & =0\\ 4n\left ( n+\alpha \right ) c_{n} & =0 \end{align*}$$

The above can be true for all $n>0$ only when $c_{n}=0$ for $n>0$. Therefore the solution is only the term with $c_{0}$

$$y_{1}\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}x^{n+r_{1}}=c_{0}x^{r_{1}}=c_{0}x^{\frac{1}{2}\left ( 1+\alpha \right ) }$$ To find the second solution $y_{2}\left ( x\right )$, the above is repeated but with $$y_{2}\left ( x\right ) =\sum _{n=0}^{\infty }d_{n}x^{n+r_{2}}$$ Where the constants are not the same and by replacing $r$ in (1) by $r_{2}=\frac{1}{2}\left ( 1-\alpha \right )$. This results in $$\sum _{n=0}^{\infty }4\left ( n+\frac{1}{2}\left ( 1-\alpha \right ) \right ) \left ( n+\frac{1}{2}\left ( 1-\alpha \right ) -1\right ) d_{n}x^{n+\frac{1}{2}\left ( 1-\alpha \right ) }+\left ( 1-\alpha ^{2}\right ) \left ( \sum _{n=0}^{\infty }d_{n}x^{n+\frac{1}{2}\left ( 1-\alpha \right ) }\right ) =0$$ For $\,n>0$ $$\begin{align*} \left ( 4\left ( n+\frac{1}{2}\left ( 1-\alpha \right ) \right ) \left ( n+\frac{1}{2}\left ( 1-\alpha \right ) -1\right ) +\left ( 1-\alpha ^{2}\right ) \right ) d_{n} & =0\\ 4n\left ( n-\alpha \right ) d_{n} & =0 \end{align*}$$

The above is true for all $n>0$ only when $c_{n}=0$ for $n>0$. Therefore the solution is just the term with $d_{0}$

$$y_{2}\left ( x\right ) =\sum _{n=0}^{\infty }d_{n}x^{n+r_{2}}=d_{0}x^{r_{2}}=d_{0}x^{\frac{1}{2}\left ( 1-\alpha \right ) }$$ Therefore the two solutions are $$\begin{align*} y_{1}\left ( x\right ) & =c_{0}x^{\frac{1}{2}\left ( 1+\alpha \right ) }\\ y_{2}\left ( x\right ) & =d_{0}x^{\frac{1}{2}\left ( 1-\alpha \right ) } \end{align*}$$

Part 2

When $\alpha =0$ then the ODE becomes

$$4x^{2}y^{\prime \prime }+y=0$$ And the two solutions found in part (1) simplify to $$\begin{align*} y_{1}\left ( x\right ) & =c_{0}\sqrt{x}\\ y_{2}\left ( x\right ) & =d_{0}\sqrt{x} \end{align*}$$

Therefore the two solutions are not linearly independent. Let $y_{20}\left ( x\right )$ be the second solution. The Wronskian is

$$\begin{equation} W\left ( x\right ) =\begin{vmatrix} y_{1} & y_{20}\\ y_{1}^{\prime } & y_{20}^{\prime }\end{vmatrix} =y_{1}y_{20}^{\prime }-y_{20}y_{1}^{\prime }\tag{1} \end{equation}$$ Using Abel’s theorem which says that for ODE of form $y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0$, the Wronskian is $W\left ( x\right ) =Ce^{-\int p\left ( x\right ) dx}$. Applying this to the given ODE above and since $p\left ( x\right ) =0$ then the above becomes $$W\left ( x\right ) =C$$ Where $C$ is constant. For $y_{20}$ to be linearly independent from $y_{1}$ $W\left ( x\right ) \neq 0$. Using $W\left ( x\right ) =C$ in (1) results in the following equation (here it is also assumed that $y_{1}\neq 0$, or $x\neq 0$, because the equation is divided by $y_{1}$) $$\begin{align*} y_{1}y_{20}^{\prime }-y_{20}y_{1}^{\prime } & =C\\ y_{20}^{\prime }-y_{20}\frac{y_{1}^{\prime }}{y_{1}} & =\frac{C}{y_{1}} \end{align*}$$

Since $y_{1}=\sqrt{x}$ and $y_{1}^{\prime }=\frac{1}{2}\frac{1}{\sqrt{x}}$ the above simplifies to

$$\begin{align} y_{20}^{\prime }-y_{20}\frac{\frac{1}{2}\frac{1}{\sqrt{x}}}{\sqrt{x}} & =\frac{C}{\sqrt{x}}\nonumber \\ y_{20}^{\prime }-y_{20}\frac{1}{2x} & =\frac{C}{\sqrt{x}}\tag{2} \end{align}$$

But the above is linear first order ODE of the form $Y^{\prime }+pY=q$, therefore the standard integrating factor to use is $I=e^{\int p\left ( x\right ) dx}$ which results in

$$\begin{align*} I & =e^{\int \frac{-1}{2x}dx}\\ & =e^{-\frac{1}{2}\int \frac{1}{x}dx}\\ & =e^{-\frac{1}{2}\ln x}\\ & =\frac{1}{\sqrt{x}} \end{align*}$$

Multiplying both sides of (2) by this integrating factor, makes the left side of (2) an exact differential

$$\frac{d}{dx}\left ( y_{20}\frac{1}{\sqrt{x}}\right ) =\frac{C}{x}$$ Integrating both sides gives $$\begin{align*} y_{20}\frac{1}{\sqrt{x}} & =C\int \frac{1}{x}dx+C_{1}\\ y_{20}\frac{1}{\sqrt{x}} & =2C\ln x+C_{1}\\ y_{20} & =2C\ln x\sqrt{x}+C_{1}\sqrt{x} \end{align*}$$

Or

$$\begin{equation} y_{20}=C_{1}\ln x\sqrt{x}+C_{2}\sqrt{x}\tag{3} \end{equation}$$ The above is the second solution. Therefore the final solution is $$y\left ( x\right ) =C_{0}y_{1}\left ( x\right ) +C_{3}y_{20}\left ( x\right )$$ Substituting $y_{1}=\sqrt{x}$ and $y_{20}$ found above and combining the common term $\sqrt{x}$ and renaming constants gives $$y\left ( x\right ) =C_{1}\sqrt{x}+C_{2}\ln x\sqrt{x}$$ Another method to find the second solution

This method is called the reduction of order method. It does not require finding $W\left ( x\right )$ first. Let the second solution be

$$\begin{equation} y_{20}=Y=v\left ( x\right ) y_{1}\left ( x\right ) \tag{4} \end{equation}$$ Where $v\left ( x\right )$ is unknown function to be determined, and $y_{1}\left ( x\right ) =\sqrt{x}$ which is the first solution that is already known. Therefore $$\begin{align*} Y^{\prime } & =v^{\prime }y_{1}+vy_{1}^{\prime }\\ Y^{\prime \prime } & =v^{\prime \prime }y_{1}+v^{\prime }y_{1}^{\prime }+v^{\prime }y_{1}^{\prime }+vy_{1}^{\prime \prime }\\ & =v^{\prime \prime }y_{1}+2v^{\prime }y_{1}^{\prime }+vy_{1}^{\prime \prime } \end{align*}$$

Since $Y$ is a solution to the ODE $4x^{2}y^{\prime \prime }+y=0$, then substituting the above equations back into the ODE $4x^{2}y^{\prime \prime }+y=0$ gives

$$\begin{align*} 4x^{2}\left ( v^{\prime \prime }y_{1}+2v^{\prime }y_{1}^{\prime }+vy_{1}^{\prime \prime }\right ) +vy_{1} & =0\\ v^{\prime \prime }\left ( 4x^{2}y_{1}\right ) +v^{\prime }\left ( 8x^{2}y_{1}^{\prime }\right ) +v\left ( \overset{0}{\overbrace{4x^{2}y_{1}^{\prime \prime }+y_{1}}}\right ) & =0 \end{align*}$$

But $4x^{2}y_{1}^{\prime \prime }+y_{1}=0$ because $y_{1}$ is a solution. The above simplifies to

$$v^{\prime \prime }\left ( 4x^{2}y_{1}\right ) +v^{\prime }\left ( 8x^{2}y_{1}^{\prime }\right ) =0$$ But $y_{1}=x^{\frac{1}{2}}$, hence $y_{1}^{\prime }=\frac{1}{2}x^{\frac{-1}{2}}$ and the above simplifies to $$\begin{align*} v^{\prime \prime }\left ( 4x^{2}x^{\frac{1}{2}}\right ) +v^{\prime }\left ( 4x^{2}x^{\frac{-1}{2}}\right ) & =0\\ x^{\frac{5}{2}}v^{\prime \prime }+v^{\prime }x^{\frac{3}{2}} & =0\\ xv^{\prime \prime }+v^{\prime } & =0\\ v^{\prime \prime }+\frac{1}{x}v^{\prime } & =0 \end{align*}$$

This ODE is now easy to solve because the $v\left ( x\right )$ term is missing. Let $w=v^{\prime }$ and the above first order ODE $w^{\prime }+\frac{1}{x}w=0$. This is linear in $w$. Hence using integrating factor $I=e^{\int \frac{1}{x}dz}=x$, this ODE becomes

$$\begin{align*} \frac{d}{x}\left ( wx\right ) & =0\\ wx & =C\\ w & =\frac{C}{x} \end{align*}$$

Where $C$ is constant of integration. Since $v^{\prime }=w$, then $v^{\prime }=\frac{C_{1}}{x}$. Now $v\left ( x\right )$ is found by integrating both sides

$$v=C_{1}\ln x+C_{2}$$ Therefore the second solution from (4) becomes $$\begin{align} y_{20} & =C_{1}\ln xy_{1}+C_{2}y_{1}\nonumber \\ & =C_{1}\sqrt{x}\ln x+C_{2}\sqrt{x}\tag{5} \end{align}$$

Comparing the above to (3), shows it is the same solution. Both methods can be used, but reduction of order method is a more common method and it does not require finding the Wronskian first, although it is not hard to find by using Abel’s theorem.

Part 3

The solutions we found in part (1) are

$$\begin{align*} y_{1}\left ( x\right ) & =C_{1}x^{\frac{1}{2}\left ( 1+\alpha \right ) }\\ y_{2}\left ( x\right ) & =C_{2}x^{\frac{1}{2}\left ( 1-\alpha \right ) } \end{align*}$$

Therefore

$$\lim _{\alpha \rightarrow 0}\frac{y_{1}-y_{2}}{\alpha }=\lim _{\alpha \rightarrow 0}\frac{C_{1}x^{\frac{1}{2}\left ( 1+\alpha \right ) }-C_{2}x^{\frac{1}{2}\left ( 1-\alpha \right ) }}{\alpha }$$ Applying L’Hopital’s $$\begin{equation} \lim _{\alpha \rightarrow 0}\frac{y_{1}-y_{2}}{\alpha }=\lim _{\alpha \rightarrow 0}\frac{C_{1}\frac{d}{d\alpha }\left ( x^{\frac{1}{2}\left ( 1+\alpha \right ) }\right ) -C_{2}\frac{d}{d\alpha }\left ( x^{\frac{1}{2}\left ( 1-\alpha \right ) }\right ) }{1} \tag{1} \end{equation}$$ But $$\begin{align*} \frac{d}{d\alpha }\left ( x^{\frac{1}{2}\left ( 1+\alpha \right ) }\right ) & =\frac{d}{d\alpha }e^{\frac{1}{2}\left ( 1+\alpha \right ) \ln x}\\ & =\frac{d}{d\alpha }e^{\left ( \frac{1}{2}\ln x+\alpha \ln x\right ) }\\ & =\ln xe^{\left ( \frac{1}{2}\ln x+\alpha \ln x\right ) } \end{align*}$$

And

$$\begin{align*} \frac{d}{d\alpha }\left ( x^{\frac{1}{2}\left ( 1-\alpha \right ) }\right ) & =\frac{d}{d\alpha }e^{\frac{1}{2}\left ( 1-\alpha \right ) \ln x}\\ & =\frac{d}{d\alpha }e^{\left ( \frac{1}{2}\ln x-\alpha \ln x\right ) }\\ & =-\ln xe^{\left ( \frac{1}{2}\ln x-\alpha \ln x\right ) } \end{align*}$$

Therefore (1) becomes

$$\begin{align*} \lim _{\alpha \rightarrow 0}\frac{y_{1}-y_{2}}{\alpha } & =\lim _{\alpha \rightarrow 0}C_{1}\ln xe^{\left ( \frac{1}{2}\ln x+\alpha \ln x\right ) }+C_{2}\ln xe^{\left ( \frac{1}{2}\ln x-\alpha \ln x\right ) }\\ & =\ln x\left ( \lim _{\alpha \rightarrow 0}C_{1}e^{\left ( \frac{1}{2}\ln x+\alpha \ln x\right ) }+C_{2}e^{\left ( \frac{1}{2}\ln x-\alpha \ln x\right ) }\right ) \\ & =\ln x\left ( C_{1}e^{\frac{1}{2}\ln x}+C_{2}e^{\frac{1}{2}\ln x}\right ) \\ & =\ln x\left ( C_{1}\sqrt{x}+C_{2}\sqrt{x}\right ) \\ & =C\sqrt{x}\ln x \end{align*}$$

The above is the same as (3) found in part (2). Hence

$$y_{20}\left ( x\right ) =\lim _{\alpha \rightarrow 0}\frac{y_{1}-y_{2}}{\alpha }$$ Which is what the problem asked to show.

3.6.4 key solution to HW 6

key.pdf