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=1cnλn∫baϕn(x)ϕm(x)σ(x)dx
Orthogonality yields
∫baf(x)ϕm(x)dx=−cmλm∫baϕ2m(x)σ(x)dx
Solving for cm, we have
cm=−∫baf(x)ϕm(x)dxλm∫baϕ2m(x)σ(x)dx
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=n2π2−1
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=A2∫e1sin(nπlnx)1xdx=A2∫10sin(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=1cn√2sin(nπlnx)
Inserting this solution into the differential equation, we have
1x=Ly=−∞∑n=1cnλn√2sin(nπlnx)1x
Next, we make use of orthogonality. Multiplying both sides by ϕm(x)=√2sin(mπlnx) and integrating, gives
λmcm=∫e1√2sin(mπlnx)1xdx=√2mπ[(−1)m−1]
Solving for cm, we have
cm=√2mπ[(−1)m−1]m2π2−1.
Finally, we insert our coefficients into the expansion for y(x). The solution is then
y(x)=∞∑n=12nπ[(−1)n−1]n2π2−1sin(nπln(x)).