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Chapter 6: Problem 24

Show that a Sturm-Liouville operator with periodic boundary condion \([a, b]\) is self-adjoint if and only if \(p(a)=p(b)\). [Recall that periodic dary conditions are given as \(u(a)=u(b)\) and \(\left.u^{\prime}(a)=u^{\prime}(b) .\right]\).

Short Answer

Expert verified
A Sturm-Liouville operator with periodic boundary conditions on \([a, b]\) is self-adjoint if and only if \(p(a) = p(b)\).

Step by step solution

01

Recall the definitions

A differential operator \(L\) is said to be self-adjoint if its adjoint \(L^{*}\) is the same as \(L\). For an operator to be a Sturm-Liouville operator, it has to be of the form \(Ly = -(p(x)y')' + q(x)y\) for some functions \(p(x)\) and \(q(x)\). The operator \(L\) is said to satisfy periodic boundary conditions if \(y(a) = y(b)\) and \(y'(a) = y'(b)\).
02

Prove self-adjointness

Consider the inner product \(\langle u, Lv \rangle = \langle Lu, v \rangle\) which is implicit in the definition of a self-adjoint operator. Using integration by parts, the left-hand side can be written as \(\int_{a}^{b}u(x)(-(p(x)v'(x))' + q(x)v(x))dx = -[u(x)p(x)v'(x)]_{a}^{b} + \int_{a}^{b} u'(x)p(x)v'(x) + u(x)q(x)v(x)dx\). The right-hand side can be rewritten as \(\int_{a}^{b}(-u'(x)p(x)v'(x) + u(x)q(x)v(x))dx + [u'(x)p(x)v(x)]_{a}^{b}\). The two sides are equal if and only if the boundary terms cancel each other out, which holds if \(p(a) = p(b)\) and the periodic boundary conditions are satisfied.
03

Conclusion

A Sturm-Liouville operator with periodic boundary conditions on \([a, b]\) is self-adjoint if and only if \(p(a) = p(b)\).

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