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Find the eigenfunctions of an irregular Sturm–Liouville problem and hence solve an inhomogeneous boundary value problem by writing the solution as an eigenfunction expansion.

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Consider the eigenvalue problem

\\[x^2\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}x^2} - \\simplify{{b-1}x}\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} + \\lambda x^{\\var{2b}}y = 0 \\qquad \\mbox{for $x \\in [0,\\var{a}]$}\\]

with the boundary conditions $y(0) = 0$ and $y(\\var{a}) = 0$.

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This equation can be rewritten as

\\[\\dfrac{\\mathrm{d}}{\\mathrm{d}x}x^{-\\var{b-1}}\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} + \\lambda \\simplify{x^{{b-1}}}y = 0,\\]

so we have $P(x) = x^{-\\var{b-1}}$, $Q(x) = 0$, and $\\omega(x) = \\simplify{x^{{b-1}}}$. This means that, at $x=0$, $\\omega(x)$ is not strictly positive and $P(x)$ is not defined. Therefore this is an irregular Sturm–Liouville problem.

To put the equation into canonical form, we define a new independent variable $t = \\displaystyle\\int\\dfrac{1}{P(x)}\\,\\mathrm{d}x = \\int \\simplify{x^{{b-1}}}\\,\\mathrm{d}x = \\dfrac{1}{\\var{b}}x^{\\var{b}}$, so that the equation becomes

\\[\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}t^2} + \\lambda y = 0.\\]

When we impose the boundary conditions, we find that the eigenfunctions are $y_n = \\sin(\\sqrt{\\lambda_n}\\,t)$, where $\\lambda_n = \\left(\\simplify{{b}/{a}^{b}}\\pi n\\right)^2$.

The eigenfunctions $y_n(x)$ are orthogonal with respect to the inner product on $[0,\\var{a}]$ with weight fuction $\\omega(x) = \\simplify{x^{{b-1}}}$. Therefore

$\\qquad\\qquad\\displaystyle\\sum_{n=1}^{\\infty}a_ny_n(x) = 1$

$\\qquad\\Rightarrow \\displaystyle\\sum_{n=1}^{\\infty}a_n\\langle y_m,y_n\\rangle = \\langle y_m,1\\rangle$

$\\qquad\\Rightarrow a_m\\langle y_m,y_m\\rangle = \\langle y_m,1\\rangle$

$\\qquad\\Rightarrow a_n = \\dfrac{\\langle y_n,1\\rangle}{\\|y_n\\|^2}$

as indicated in the question. Expressing this result as a ratio of two integrals, we have

$\\qquad\\qquad a_n = \\dfrac{\\int_0^{\\var{a}}\\sin\\bigl(\\pi n\\simplify{(x/{a})^{b}}\\bigr) \\simplify{x^{{b-1}}} \\, \\mathrm{d}x}{\\int_0^{\\var{a}}\\sin\\bigl(\\pi n\\simplify{(x/{a})^{b}}\\bigr)^2 \\simplify{x^{{b-1}}} \\, \\mathrm{d}x}$

$\\qquad\\qquad = \\dfrac{\\int_0^{\\simplify{{a}^{b}/{b}}}\\sin\\bigl(\\simplify{{b}pi n t/{a}^{b}}\\bigr) \\, \\mathrm{d}t}{\\int_0^{\\simplify{{a}^{b}/{b}}}\\sin\\bigl(\\simplify{{b}pi n t/{a}^{b}}\\bigr)^2 \\, \\mathrm{d}t} \\qquad$where $t = \\dfrac{1}{\\var{b}}x^{\\var{b}}$, as in part (b)

$\\qquad\\qquad = \\dfrac{- \\simplify{{a}^{b}/({b}pi n)}\\left[\\cos\\bigl(\\simplify{{b}pi n t/{a}^{b}}\\bigr)\\right]_0^{\\simplify{{a}^{b}/{b}}}}{\\tfrac{1}{2}\\left[t + \\simplify{{a}^{b}/(2{b}pi n)}\\sin\\bigl(2\\simplify{{b}pi n t/{a}^{b}}\\bigr)\\right]_0^{\\simplify{{a}^{b}/{b}}}}$

$\\qquad\\qquad = - \\dfrac{2}{\\pi n}\\bigl[\\cos(\\pi n) - 1\\bigr]$

$\\qquad\\qquad = \\begin{cases}\\dfrac{4}{\\pi n}, \\qquad &\\mbox{if $n$ is odd,} \\\\ 0, \\qquad &\\mbox{if $n$ is even.}\\end{cases}$

We can solve this inhomogeneous problem using the method of eigenfunction expansion, i.e. we can look for a solution in the form

\\[y = \\sum_{n=1}^\\infty b_ny_n(x)\\]

where $y_n$ are the eigenfunctions we found previously. These eigenfunctions have the property that $L[y_n] = - \\lambda_n y$, where $L$ is the operator

\\[L = \\dfrac{1}{\\simplify{x^{b-1}}}\\dfrac{\\mathrm{d}}{\\mathrm{d}x}\\dfrac{1}{\\simplify{x^{b-1}}}\\dfrac{\\mathrm{d}}{\\mathrm{d}x}.\\]

The inhomogeneous equation can be written as

\\[L[y] + \\lambda y = 1\\]

and when we substitute our eigenfunction expansion into this, we get

$\\qquad\\qquad\\sum_{n=1}^\\infty b_n\\bigl(L[y_n] + \\lambda y_n\\bigr) = 1$

$\\qquad\\Rightarrow\\sum_{n=1}^\\infty b_n(\\lambda - \\lambda_n) y_n = 1$

Comparing this with the answer to part (c), we must have

\\[b_n(\\lambda - \\lambda_n) = a_n = \\begin{cases}\\dfrac{4}{\\pi n}, \\qquad &\\mbox{if $n$ is odd,} \\\\ 0, \\qquad &\\mbox{if $n$ is even.}\\end{cases}\\]

So the number of solutions of this inhomogeneous problem depends on the value of $\\lambda$:

If $\\lambda = \\lambda_n$ for some odd $n$, then the coefficient $b_n$ is undefined, and the problem has no solution.
If $\\lambda = \\lambda_n$ for some even $n$, then the coefficient $b_n$ is arbitrary, and the other coefficients are uniquely defined.
If $\\lambda \\neq \\lambda_n$ for any $n$, then the coefficients are all uniquely defined, and the problem has a single solution.

In particular, the problem cannot be solved if $\\lambda = \\lambda_1,\\lambda_3,\\lambda_5,\\ldots = \\left(\\simplify{{b}/{a}^{b}}\\pi\\right)^2,\\left(\\simplify{3{b}/{a}^{b}}\\pi\\right)^2,\\left(\\simplify{5{b}/{a}^{b}}\\pi\\right)^2,\\ldots$

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The end point of the domain in $x$.

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The exponent of $x$.

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Write this differential equation in the Sturm–Liouville form,

\\[\\dfrac{\\mathrm{d}}{\\mathrm{d}x}\\left(P(x)\\dfrac{\\mathrm{d}y}{\\mathrm{d}x}\\right) + Q(x)y + \\lambda\\omega(x)y = 0.\\]

This is an example of an irregular Sturm–Liouville problem, because (tick all that apply):

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By writing the differential equation in canonical form, show that the eigenfunctions are

$y_n(x) = \\sin\\bigl(\\pi n\\simplify{(x/{a})^{b}}\\bigr)$, and find the corresponding eigenvalues $\\lambda_n$.

$\\lambda_n = $[[0]]$(\\pi n)^2.$

Even though this is an irregular Sturm–Liouville problem, you may assume from here on that it has all of the properties of a regular Sturm–Liouville problem.

Find the values of the coefficients $a_n$ such that

$\\displaystyle\\sum_{n=1}^{\\infty}a_ny_n(x) = 1 \\qquad \\mbox{for $\\ 0 \\leqslant x \\leqslant \\var{a}$,}$

where $y_n(x)$ are the eigenfunctions found in part (b).

Hint: The coefficients can be found using the formula $a_n = \\dfrac{\\langle y_n, 1\\rangle}{\\|y_n\\|^2}$, for an appropriate choice of inner product. The integrals can be evaluated using the same change of variable that you made in part (b).

$a_n = $[[0]] if $n$ is odd

$a_n = $[[1]] if $n$ is even

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Now consider the inhomogeneous Boundary Value Problem

\\[x^2\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}x^2} - \\simplify{{b-1}x}\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} + \\lambda x^{\\var{2b}}y = x^{\\var{2b}} \\qquad \\mbox{for $x \\in [0,\\var{a}]$}\\]

with the boundary conditions $y(0) = 0$ and $y(\\var{a}) = 0$.

For which of the following values of $\\lambda$ can this problem be solved?

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