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6.3: The Eigenfunction Expansion Method

( \newcommand{\kernel}{\mathrm{null}\,}\)

In section 4.3.2 we saw generally how one can use the eigenfunctions of a differential operator to solve a nonhomogeneous boundary value problem. In this chapter we have seen that Sturm-Liouville eigenvalue problems have the requisite set of orthogonal eigenfunctions. In this section we will apply the eigenfunction expansion method to solve a particular nonhomogenous boundary value problem.

Recall that one starts with a nonhomogeneous differential equation

Ly=f

where y(x) is to satisfy given homogeneous boundary conditions. The method makes use of the eigenfunctions satisfying the eigenvalue problem

Lϕn=λnσϕn

subject to the given boundary conditions. Then, one assumes that y(x) can be written as an expansion in the eigenfunctions,

y(x)=n=1cnϕn(x)

and inserts the expansion into the nonhomogeneous equation. This gives

f(x)=L(n=1cnϕn(x))=n=1cnλnσ(x)ϕn(x).

The expansion coefficients are then found by making use of the orthogonality of the eigenfunctions. Namely, we multiply the last equation by ϕm(x) and integrate. We obtain

baf(x)ϕm(x)dx=n=1cnλnbaϕn(x)ϕm(x)σ(x)dx

Orthogonality yields

baf(x)ϕm(x)dx=cmλmbaϕ2m(x)σ(x)dx

Solving for cm, we have

cm=baf(x)ϕm(x)dxλmbaϕ2m(x)σ(x)dx

Example 6.3.1

As an example, we consider the solution of the boundary value problem

(xy)+yx=1x,x[1,e],

y(1)=0=y(e).

Solution

This equation is already in self-adjoint form. So, we know that the associated Sturm-Liouville eigenvalue problem has an orthogonal set of eigenfunctions. We first determine this set. Namely, we need to solve

(xϕ)+ϕx=λσϕ,ϕ(1)=0=ϕ(e).

Rearranging the terms and multiplying by x, we have that

x2ϕ+xϕ+(1+λσx)ϕ=0.

This is almost an equation of Cauchy-Euler type. Picking the weight function σ(x)=1x, we have

x2ϕ+xϕ+(1+λ)ϕ=0.

This is easily solved. The characteristic equation is

r2+(1+λ)=0.

One obtains nontrivial solutions of the eigenvalue problem satisfying the boundary conditions when λ>1. The solutions are

ϕn(x)=Asin(nπlnx),n=1,2,

where λn=n2π21
It is often useful to normalize the eigenfunctions. This means that one chooses A so that the norm of each eigenfunction is one. Thus, we have

1=e1ϕn(x)2σ(x)dx=A2e1sin(nπlnx)1xdx=A210sin(nπy)dy=12A2

Thus, A=2
We now turn towards solving the nonhomogeneous problem, Ly=1x. We first expand the unknown solution in terms of the eigenfunctions,

y(x)=n=1cn2sin(nπlnx)

Inserting this solution into the differential equation, we have

1x=Ly=n=1cnλn2sin(nπlnx)1x

Next, we make use of orthogonality. Multiplying both sides by ϕm(x)=2sin(mπlnx) and integrating, gives

λmcm=e12sin(mπlnx)1xdx=2mπ[(1)m1]

Solving for cm, we have

cm=2mπ[(1)m1]m2π21.

Finally, we insert our coefficients into the expansion for y(x). The solution is then

y(x)=n=12nπ[(1)n1]n2π21sin(nπln(x)).


This page titled 6.3: The Eigenfunction Expansion Method is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform.

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