Question

Solve the following eigenvalue problem (see end of Section 2 and problem 11): Given the differential equation ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\left(\frac{\lambda }{x}-\frac{1}{4}-\frac{l(l+1)}{x^2}\right)\mathit{y}\mathbf{=}\mathbf{0}$where $l$ is an integerlocalid="1654860659044" $\mathbf{\ge }\mathbf{0}$ , find values of localid="1654860714122" $\mathit{\lambda }$such that localid="1654860676211" $\mathit{y}\mathbf{\to }\mathbf{0}$ aslocalid="1654860742759" role="math" $\mathit{x}\mathbf{\to }\mathit{\infty }$ , and find the corresponding eigenfunctions. Hint: letlocalid="1654860764612" $\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{l}\mathbf{+}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{x}\mathbf{/}\mathbf{2}}\mathit{v}\mathbf{\left(}\mathit{x}\mathbf{\right)}$, and show that localid="1654860784518" $\mathit{v}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ satisfies the differential equationlocalid="1654860800910" $\mathit{x}{\mathit{v}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{(}\mathbf{2}\mathit{l}\mathbf{+}\mathbf{2}\mathbf{-}\mathit{x}\mathbf{)}{\mathit{v}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{(}\mathit{\lambda }\mathbf{-}\mathit{l}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{v}\mathbf{=}\mathbf{0}$.Comparelocalid="1654860829619" $\mathbf{(}\mathbf{22}\mathbf{.26}\mathbf{)}$ to show that if localid="1654860854431" $\mathit{\lambda }$ is an integerlocalid="1654860871428" $\mathbf{>}\mathit{l}$, there is a polynomial solution localid="1654860888067" $\mathit{v}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{L}}_{\mathbf{\lambda }\mathbf{-}\mathbf{t}\mathbf{-}\mathbf{1}}^{\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$.Solve the eigenvalue problem localid="1654860910472" ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\left(\frac{\lambda }{x}-\frac{1}{4}-\frac{l(l+1)}{x^2}\right)\mathit{y}\mathbf{=}\mathbf{0}$.

Short answer

The solution of the equation$y^{\text{'}\text{'}}+\left(\frac{\lambda }{x}-\frac{1}{4}-\frac{l(l+1)}{x^2}\right)y=0$is$y\left(x\right)=x^{l+1}{e}^{-x/2}{L}_{\lambda -l-1}^{2l+1}\left(x\right)$.

Step-by-step solution

Concept of eigenvalue

Consider the equation is$x{y}^{\text{'}\text{'}}+(k+1-x)y^{\text{'}}+ny=0$ .

Find the solutions of equation   y''+λx−14−l(l+1)x2y=0

As the given equation is,.$y^{\text{'}\text{'}}+\left(\frac{\lambda }{x}-\frac{1}{4}-\frac{l(l+1)}{x^2}\right)y=0$

Let,$y=x^{l+1}{e}^{-x/2}v\left(x\right)$.

So, calculate:

$\begin{array}{c}{y}^{\text{'}}=(l+1)x^l{e}^{-x/2}v\left(x\right)++x^{l+1}\left(-\frac{1}{2}\right)e^{-x/2}v\left(x\right)+x^{l+1}{e}^{-x/2}{v}^{\text{'}}\left(x\right)\\ y^{\text{'}}=\left[\begin{array}{l}\left\{(l+1)l{x}^{l-1}{e}^{-x/2}v\left(x\right)+(l+1)x^l\left(-\frac{1}{2}\right)e^{-x/2}v\left(x\right)+(l+1)x^l{e}^{-x/2}\right.\\ +\left\{(l+1)x^l\left(-\frac{1}{2}\right)e^{-x/2}v\left(x\right)+x^{l+1}{\left(-\frac{1}{2}\right)}^2{e}^{-x/2}v\left(x\right)+x^{l+1}\left(-\frac{1}{2}\right)\right.\\ +\left\{(l+1)x^l{e}^{-x/2}{v}^{\text{'}}\left(x\right)+x^{l+1}\left(-\frac{1}{2}\right)e^{-x/2}{v}^{\text{'}}\left(x\right)+x^{l+1}{e}^{-x/2}{v}^{\text{'}\text{'}}\left(x\right)\right\}\end{array}\right.\end{array}$

Now, substitute the values ofandin the given equation and simplify to obtain the equation as follows:

$x{v}^{\text{'}\text{'}}+(2l+2-x)v^{\text{'}}+(\lambda -l-1)v=0$

Compare this equation with equation$x{y}^{\text{'}\text{'}}+(k+1-x)y^{\text{'}}+ny=0$that has solution.

$y=L_n^k\left(x\right)$

To get the solution as follows;

$v\left(x\right)=L_{\lambda -l-1}^{2l+1}\left(x\right)$

Therefore, the solution of the given equation is$y\left(x\right)=x^{l+1}{e}^{-x/2}{L}_{\lambda -l-1}^{2l+1}\left(x\right)$.