Eigenfunction expansion

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Toby Wood

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Toby Wood 3 years, 5 months ago

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Toby Wood 3 years, 6 months ago

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Question statement

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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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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$P(x)$ is not defined over the whole domain

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$Q(x)$ is not strictly positive over the whole domain

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$\omega(x)$ is not strictly positive over the whole domain

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the boundary conditions do not have the right format

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$\omega(x)$ is not defined over the whole domain

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Advice

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)

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

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

$\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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This question is used in the following exams:

MAS3810 Assessment 3 by Toby Wood in NCL MSP3801.
Semester 1 assignment 3 by Toby Wood in NCL MSP3801.
Semester 1 assignment 3 by Toby Wood in NCL MSP3801.
Assignment 3 by Toby Wood in NCL MSP3801.
Assignment 2 by Toby Wood in NCL MSP3801.

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Eigenfunction expansion

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Toby Wood 3 years, 5 months ago

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