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Mathematics LibreTexts

10.6: Sturm-Liouville theory

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    In the end we shall want to write a solution to an equation as a series of Bessel functions. In order to do that we shall need to understand about orthogonality of Bessel function – just as sines and cosines were orthogonal. This is most easily done by developing a mathematical tool called Sturm-Liouville theory. It starts from an equation in the so-called self-adjoint form

    \[[r(x) y'(x)]' + [p(x)+\lambda s(x)] y(x) = 0 \label{eq:selfadj}\]

    where \(\lambda\) is a number, and \(r(x)\) and \(s(x)\) are greater than 0 on \([a,b]\). We apply the boundary conditions

    \[\begin{aligned} a_1 y(a)+ a_2 y'(a)&=0,\nonumber\\ b_1 y(b)+ b_2 y'(b)&=0,\end{aligned}\]

    with \(a_1\) and \(a_2\) not both zero, and \(b_1\) and \(b_2\) similar.

    Theorem \(\PageIndex{1}\)

    If there is a solution to (\ref{eq:selfadj}) then \(\lambda\) is real.


    Assume that \(\lambda\) is a complex number (\(\lambda = \alpha + i \beta\)) with solution \(\Phi\). By complex conjugation we find that

    \[\begin{aligned} [r(x) \Phi'(x)]' + [p(x)+\lambda s(x)] \Phi(x) &= 0\nonumber\\ {}[r(x) (\Phi^*)'(x)]' + [p(x)+\lambda^* s(x)] (\Phi^*)(x) &= 0\end{aligned}\]

    where \(*\) note complex conjugation. Multiply the first equation by \(\Phi^*(x)\) and the second by \(\Phi(x)\), and subtract the two equations: \[(\lambda^*-\lambda)s(x) \Phi^*(x)\Phi(x)=\Phi(x)[r(x) (\Phi^*)'(x)]'-\Phi^*(x)[r(x) \Phi'(x)]'.\] Now integrate over \(x\) from \(a\) to \(b\) and find

    \[(\lambda^*-\lambda)\int_a^bs(x) \Phi^*(x)\Phi(x)\,dx = \int_a^b\Phi(x)[r(x) (\Phi^*)'(x)]'-\Phi^*(x)[r(x) \Phi'(x)]'\,dx\]

    The second part can be integrated by parts, and we find

    \[\begin{aligned} (\lambda^*-\lambda)\int_a^b s(x) \Phi^*(x)\Phi(x)\,dx &= \left[\Phi'(x) r(x) (\Phi^*)'(x)-\Phi^*(x)r(x) \Phi'(x)\right|^b_a \nonumber\\ &= r(b)\left[\Phi'(b) (\Phi^*)'(b)-\Phi^*(b)\Phi'(b)\right] \nonumber\\&& -r(a)\left[\Phi'(a) (\Phi^*)'(a)-\Phi^*(a)\Phi'(a)\right] \nonumber\\ &=0,\end{aligned}\]

    where the last step can be done using the boundary conditions. Since both \(\Phi^*(x)\Phi(x)\) and \(s(x)\) are greater than zero we conclude that \(\int_a^bs(x) \Phi^*(x)\Phi(x)\,dx > 0\), which can now be divided out of the equation to lead to \(\lambda=\lambda^*\).

    Theorem \(\PageIndex{2}\)

    Let \(\Phi_n\) and \(\Phi_m\) be two solutions for different values of \(\lambda\), \(\lambda_n\neq \lambda_m\), then \[\int_a^b s(x) \Phi_n(x) \Phi_m(x)\,dx = 0.\]


    The proof is to a large extent identical to the one above: multiply the equation for \(\Phi_n(x)\) by \(\Phi_m(x)\) and vice-versa. Subtract and find \[(\lambda_n-\lambda_m)\int_a^b s(x) \Phi_m(x)\Phi_n(x)\,dx = 0\] which leads us to conclude that \[\int_a^b s(x) \Phi_n(x) \Phi_m(x)\,dx = 0.\]

    Theorem \(\PageIndex{3}\)

    Under the conditions set out above

    1. There exists a real infinite set of eigenvalues \(\lambda_0,\ldots,\lambda_n, \ldots\) with \(\lim_{n\rightarrow \infty} = \infty\).
    2. If \(\Phi_n\) is the eigenfunction corresponding to \(\lambda_n\), it has exactly \(n\) zeroes in \([a,b]\). No proof shall be given.

    No proof shall be given.

    Clearly the Bessel equation is of self-adjoint form: rewrite \[x^2 y'' + xy' + (x^2-\nu^2) y = 0\] as (divide by \(x\)) \[[x y']' + (x-\frac{\nu^2}{x}) y = 0\] We cannot identify \(\nu\) with \(\lambda\), and we do not have positive weight functions. It can be proven from properties of the equation that the Bessel functions have an infinite number of zeroes on the interval \([0,\infty)\). A small list of these:

    \[\begin{array}{lclllll} J_0 &:& 2.42&5.52 &8.65&11.79&\ldots\\ J_{1/2} &:& \pi & 2 \pi & 3\pi & 4\pi & \ldots\\ J_{8} &:& 11.20 & 16.04 & 19.60 & 22.90 & \ldots \end{array}\]