The Sturm–Liouville Problem

Notes on Chapter 11.4 of my MATH 285 course (see also Boyce and DiPrima [1]). This chapter is quite interesting: a second-order ODE turns out to behave exactly like a real symmetric matrix.

The plan is to first recall two facts from linear algebra, then show that the Sturm–Liouville operator is the infinite-dimensional analogue of a symmetric matrix, and finally see why this matters for solving PDEs.

I. Prerequisite: the spectral theorem

For a real symmetric matrix \(A\) (\(A^\top = A\)) of size \(n \times n\):

1. All eigenvalues \(\lambda\) of \(A\) are real.
2. Eigenvectors \(\mathbf{v}_i, \mathbf{v}_j\) for different eigenvalues \(\lambda_i \neq \lambda_j\) are orthogonal: \(\mathbf{v}_i \perp \mathbf{v}_j\).
3. The eigenvectors of \(A\) span all of \(\mathbb{R}^n\), so \(A\) has an orthonormal basis of eigenvectors.

We prove the first two. The third needs a little more work (usually an induction on \(n\)), and we take it as known.

Proof of 1. Let \(A\mathbf{v} = \lambda\mathbf{v}\) with \(\mathbf{v} \neq \mathbf{0}\), where \(\lambda\) and \(\mathbf{v}\) may be complex. Since \(A\) is real, conjugating gives \(A\bar{\mathbf{v}} = \bar\lambda\,\bar{\mathbf{v}}\). Transposing the original equation and using \(A^\top = A\) gives \(\mathbf{v}^\top A = \lambda\,\mathbf{v}^\top\). Now compute \(\mathbf{v}^\top A \bar{\mathbf{v}}\) in two ways:

\[ \mathbf{v}^\top A\,\bar{\mathbf{v}} = (\mathbf{v}^\top A)\,\bar{\mathbf{v}} = \lambda\,\mathbf{v}^\top\bar{\mathbf{v}}, \qquad \mathbf{v}^\top A\,\bar{\mathbf{v}} = \mathbf{v}^\top (A\bar{\mathbf{v}}) = \bar\lambda\,\mathbf{v}^\top\bar{\mathbf{v}} . \]

Since \(\mathbf{v}^\top\bar{\mathbf{v}} = \sum_k |v_k|^2 = \|\mathbf{v}\|^2 > 0\), we get \(\lambda = \bar\lambda\): \(\lambda\) is real.

Proof of 2. Let \(A\mathbf{v}_1 = \lambda_1\mathbf{v}_1\) and \(A\mathbf{v}_2 = \lambda_2\mathbf{v}_2\) with \(\lambda_1 \neq \lambda_2\). Again compute one number in two ways:

\[ \mathbf{v}_1^\top A\,\mathbf{v}_2 = (A\mathbf{v}_1)^\top\mathbf{v}_2 = \lambda_1\,\mathbf{v}_1^\top\mathbf{v}_2, \qquad \mathbf{v}_1^\top A\,\mathbf{v}_2 = \mathbf{v}_1^\top (A\mathbf{v}_2) = \lambda_2\,\mathbf{v}_1^\top\mathbf{v}_2 . \]

Subtracting, \((\lambda_1 - \lambda_2)\,\mathbf{v}_1^\top\mathbf{v}_2 = 0\). Since \(\lambda_1 \neq \lambda_2\), \(\mathbf{v}_1^\top\mathbf{v}_2 = 0\).

Both proofs use only one thing about \(A\): that it can be moved from one side of a dot product to the other, \((A\mathbf{u})^\top\mathbf{v} = \mathbf{u}^\top(A\mathbf{v})\). That is the property we will look for in differential operators.

II. Prerequisite: the inner product of functions

For complex-valued functions on \([a, b]\), define

\[ \langle u, v\rangle = \int_a^b u^*(t)\,v(t)\,dt , \]

where \(^*\) is complex conjugation. For real functions, \(u^* = u\) and this is just \(\int_a^b u(t)v(t)\,dt\), the continuous version of \(\mathbf{u}^\top\mathbf{v} = \sum_k u_k v_k\). Swapping the arguments conjugates the result:

\[ \overline{\langle v, u\rangle} = \overline{\int_a^b v^*(t)\,u(t)\,dt} = \int_a^b v(t)\,u^*(t)\,dt = \langle u, v\rangle . \]

In the real case, \(\langle u, v\rangle = \langle v, u\rangle\), just like \(\mathbf{u}^\top\mathbf{v} = \mathbf{v}^\top\mathbf{u}\). We will also need a weighted inner product, with a weight function \(p(t) > 0\):

\[ \langle u, v\rangle_p = \int_a^b u^*(t)\,v(t)\,p(t)\,dt . \]

III. Self-adjoint operators

A linear operator \(\mathcal{L}\) is self-adjoint (mathematics) or Hermitian (quantum physics) if

\[ \langle \mathcal{L}u, v\rangle = \langle u, \mathcal{L}v\rangle \]

for all \(u, v\) in its domain. For a differential operator, the domain includes the boundary conditions, and as we will see, the boundary conditions are exactly what make this identity hold. For a real matrix, \(\langle A\mathbf{u}, \mathbf{v}\rangle = \mathbf{u}^\top A^\top\mathbf{v}\) and \(\langle \mathbf{u}, A\mathbf{v}\rangle = \mathbf{u}^\top A\mathbf{v}\), so self-adjoint means \(A^\top = A\): a symmetric matrix.

These prerequisites are enough. Now let us set up the Sturm–Liouville problem.

IV. The regular Sturm–Liouville problem

Let \(r(t)\), \(r'(t)\), \(p(t)\) and \(q(t)\) be real functions, continuous on \([a, b]\), with \(r(t) > 0\) and \(p(t) > 0\) on \([a, b]\). We look for numbers \(\lambda\) and nonzero functions \(y\) that solve

\[ \begin{equation} \frac{d}{dt}\big[r(t)\,y'(t)\big] + \big(q(t) + \lambda\,p(t)\big)\,y(t) = 0 \label{eq:sl} \end{equation} \]

subject to the separated boundary conditions

\[ \begin{equation} A_1\,y(a) + B_1\,y'(a) = 0, \qquad A_2\,y(b) + B_2\,y'(b) = 0, \label{eq:bc} \end{equation} \]

where \(A_1, B_1, A_2, B_2\) are real constants with \((A_1, B_1) \neq (0, 0)\) and \((A_2, B_2) \neq (0, 0)\). Each condition involves only one endpoint. This is a regular Sturm–Liouville problem.

Moving the \(\lambda\) term to the right-hand side,

\[ -\frac{d}{dt}\big[r(t)\,y'(t)\big] - q(t)\,y(t) = \lambda\,p(t)\,y(t) . \]

Define the operator

\[ \mathcal{L} = -\frac{d}{dt}\left(r(t)\,\frac{d}{dt}\right) - q(t) . \]

Then the problem reads

\[ \mathcal{L}\,y = \lambda\,p\,y . \]

This is an eigenvalue problem, the analogue of \(A\mathbf{v} = \lambda\mathbf{v}\). We will show that \(\mathcal{L}\) is self-adjoint, and then every property of the spectral theorem carries over.

V. Two properties of the operator

Linearity. For complex-valued functions \(x_1, x_2\) and complex constants \(\alpha, \beta\),

\[ \begin{aligned} \mathcal{L}\{\alpha x_1 + \beta x_2\} &= -\frac{d}{dt}\Big[r\,(\alpha x_1' + \beta x_2')\Big] - q\,(\alpha x_1 + \beta x_2) \\ &= \alpha\Big(-\frac{d}{dt}\big[r\,x_1'\big] - q\,x_1\Big) + \beta\Big(-\frac{d}{dt}\big[r\,x_2'\big] - q\,x_2\Big) = \alpha\,\mathcal{L}x_1 + \beta\,\mathcal{L}x_2 . \end{aligned} \]

Conjugation. Because \(r\) and \(q\) are real, \(\mathcal{L}\) commutes with complex conjugation: \((\mathcal{L}x)^* = \mathcal{L}\{x^*\}\). Indeed, write \(x = \alpha + i\beta\) with \(\alpha, \beta\) real functions. Then \(\mathcal{L}\alpha\) and \(\mathcal{L}\beta\) are real, and by linearity

\[ (\mathcal{L}x)^* = (\mathcal{L}\alpha + i\,\mathcal{L}\beta)^* = \mathcal{L}\alpha - i\,\mathcal{L}\beta = \mathcal{L}\{\alpha - i\beta\} = \mathcal{L}\{x^*\} . \]

VI. The operator is self-adjoint

We want \(\langle \mathcal{L}u, v\rangle = \langle u, \mathcal{L}v\rangle\) for any two functions \(u, v\) that satisfy the boundary conditions \(\eqref{eq:bc}\). Using the conjugation property and integrating by parts once,

\[ \begin{aligned} \langle \mathcal{L}u, v\rangle &= \int_a^b (\mathcal{L}u)^*\,v\,dt = \int_a^b \mathcal{L}\{u^*\}\,v\,dt = \int_a^b \Big(-\frac{d}{dt}\big[r\,u^{*\prime}\big] - q\,u^*\Big)\,v\,dt \\ &= -\Big[\,r\,u^{*\prime}\,v\,\Big]_a^b + \int_a^b r\,u^{*\prime}\,v'\,dt - \int_a^b q\,u^*\,v\,dt . \end{aligned} \]

In the same way,

\[ \langle u, \mathcal{L}v\rangle = \int_a^b u^*\,\mathcal{L}v\,dt = -\Big[\,r\,u^*\,v'\,\Big]_a^b + \int_a^b r\,u^{*\prime}\,v'\,dt - \int_a^b q\,u^*\,v\,dt . \]

The two integrals are identical, so only the boundary terms survive in the difference:

\[ \begin{equation} \langle \mathcal{L}u, v\rangle - \langle u, \mathcal{L}v\rangle = \Big[\,r(t)\,\big(u^*(t)\,v'(t) - u^{*\prime}(t)\,v(t)\big)\Big]_a^b . \label{eq:boundary} \end{equation} \]

The boundary conditions kill the boundary term

Both \(u\) and \(v\) satisfy the condition at \(t = a\). Since \(A_1\) and \(B_1\) are real, conjugating the condition for \(u\) shows that \(u^*\) satisfies it too (using \((u^*)' = (u')^*\)):

\[ A_1\,u^*(a) + B_1\,u^{*\prime}(a) = 0, \qquad A_1\,v(a) + B_1\,v'(a) = 0 . \]

In matrix form,

\[ \begin{bmatrix} u^*(a) & u^{*\prime}(a) \\ v(a) & v'(a) \end{bmatrix} \begin{bmatrix} A_1 \\ B_1 \end{bmatrix} = \mathbf{0} . \]

Since \((A_1, B_1) \neq (0, 0)\), this homogeneous system has a nonzero solution, so its determinant must vanish:

\[ u^*(a)\,v'(a) - u^{*\prime}(a)\,v(a) = 0 . \]

The same argument at \(t = b\) gives \(u^*(b)\,v'(b) - u^{*\prime}(b)\,v(b) = 0\). Both ends of \(\eqref{eq:boundary}\) are zero, so

\[ \langle \mathcal{L}u, v\rangle = \langle u, \mathcal{L}v\rangle : \]

\(\mathcal{L}\) is self-adjoint. Real-valued functions are a special case of complex-valued ones, so this holds in the real case too. Notice that the proof never used \(r > 0\) or \(p > 0\). Self-adjointness comes from the form of the operator together with the boundary conditions; the positivity of \(r\) and \(p\) matters later.

VII. The spectral theorem for Sturm–Liouville problems

We now get the analogue of the spectral theorem:

1. All eigenvalues are real.
2. Eigenfunctions for different eigenvalues are orthogonal with respect to the weight \(p\): \(\langle y_i, y_j\rangle_p = 0\).
3. The eigenfunctions form a complete orthogonal basis. The eigenvalues form an infinite increasing sequence \(\lambda_1 < \lambda_2 < \cdots \to \infty\), and any reasonable function on \([a, b]\) (for example, any square-integrable one) can be expanded in the eigenfunctions, in the sense of the weighted inner product.

The first two follow in a few lines, exactly as for matrices. The third is a deeper theorem [1, 2], which we take as known.

Proof of property 1: the eigenvalues are real

Let \(\mathcal{L}u = \lambda\,p\,u\) with \(u \neq 0\). Self-adjointness gives \(\langle u, \mathcal{L}u\rangle = \langle \mathcal{L}u, u\rangle\), so

\[ \langle u, \lambda\,p\,u\rangle = \langle \lambda\,p\,u, u\rangle \quad\Longrightarrow\quad \int_a^b u^*\,\lambda\,p\,u\,dt = \int_a^b (\lambda\,p\,u)^*\,u\,dt . \]

Since \(p\) is real, \(p^* = p\), and this becomes

\[ \lambda \int_a^b p(t)\,|u(t)|^2\,dt = \lambda^* \int_a^b p(t)\,|u(t)|^2\,dt . \]

Because \(p > 0\) and \(u \neq 0\), the integral is positive, so \(\lambda = \lambda^*\): \(\lambda\) is real. This is where \(p > 0\) is needed: it makes \(\langle u, u\rangle_p\) a genuine (positive) squared length.

Proof of property 2: orthogonality

Let \(\mathcal{L}u_1 = \lambda_1\,p\,u_1\) and \(\mathcal{L}u_2 = \lambda_2\,p\,u_2\) with \(\lambda_1 \neq \lambda_2\). Self-adjointness gives \(\langle u_1, \mathcal{L}u_2\rangle = \langle \mathcal{L}u_1, u_2\rangle\), so

\[ \int_a^b u_1^*\,\lambda_2\,p\,u_2\,dt = \int_a^b (\lambda_1\,p\,u_1)^*\,u_2\,dt . \]

Since \(\lambda_1\) is real (property 1) and \(p\) is real,

\[ (\lambda_2 - \lambda_1) \int_a^b u_1^*(t)\,u_2(t)\,p(t)\,dt = 0 . \]

Since \(\lambda_1 \neq \lambda_2\),

\[ \langle u_1, u_2\rangle_p = \int_a^b u_1^*(t)\,u_2(t)\,p(t)\,dt = 0 . \]

The eigenfunctions are orthogonal, but with respect to the weight \(p\), not the plain inner product. The weight is the coefficient of \(\lambda\) in \(\eqref{eq:sl}\).

VIII. More on Sturm–Liouville problems

1. Why \(r(t) > 0\)?

• \(r < 0\) everywhere is no real change. Multiplying \(\eqref{eq:sl}\) by \(-1\) gives a problem with \(-r > 0\), \(-q\), the same \(p\), and eigenvalue \(-\lambda\). So we may as well assume \(r > 0\).
• \(r = 0\) somewhere: there the coefficient of \(y''\) vanishes and the equation stops being a proper second-order equation. At an interior point this breaks the theory. At an endpoint it is allowed, and gives a singular problem (Section 4 below).
• \(r\) changes sign: \(\mathcal{L}\) is still self-adjoint (Section VI never used the sign of \(r\)), but the eigenvalues lose their order. For \(r > 0\), integrating by parts gives \(\langle -(r u')', u\rangle = \int_a^b r\,|u'|^2\,dt\) up to boundary terms, which is nonnegative. This is what bounds the eigenvalues from below and makes them increase to \(+\infty\). If \(r\) changes sign, this quadratic form is indefinite, and eigenvalues can run off to both \(+\infty\) and \(-\infty\).

2. From any second-order linear ODE to Sturm–Liouville form

Consider a general second-order linear equation with \(a(t) \neq 0\) on \([a, b]\):

\[ a(t)\,x''(t) + b(t)\,x'(t) + c(t)\,x(t) = 0 . \]

Divide by \(a(t)\) and multiply by an integrating factor \(\mu(t)\) to be chosen:

\[ \mu\,x'' + \mu\,\frac{b}{a}\,x' + \mu\,\frac{c}{a}\,x = 0 . \]

We want the first two terms to be a single derivative, \((\mu\,x')' = \mu\,x'' + \mu'\,x'\). That requires \(\mu' = \mu\,b/a\), so

\[ \mu(t) = \exp\!\left(\int \frac{b(t)}{a(t)}\,dt\right), \qquad\text{and then}\qquad \big(\mu\,x'\big)' + \mu\,\frac{c}{a}\,x = 0 . \]

This is Sturm–Liouville form with \(r = \mu > 0\) and \(q = \mu c/a\). (For an eigenvalue problem \(a x'' + b x' + (c + \lambda d)\,x = 0\), the weight is \(p = \mu d/a\).) Every second-order linear ODE can be put in Sturm–Liouville form by multiplying by an integrating factor.

3. Eigenvalues are simple

In a regular Sturm–Liouville problem, each eigenvalue has only one eigenfunction, up to a constant factor. Suppose \(y_1\) and \(y_2\) are eigenfunctions for the same \(\lambda\). Both satisfy the condition at \(t = a\), so, by the determinant argument of Section VI, their Wronskian vanishes there:

\[ W(a) = \det\begin{bmatrix} y_1(a) & y_1'(a) \\ y_2(a) & y_2'(a) \end{bmatrix} = 0 . \]

For two solutions of \(\eqref{eq:sl}\), \(r(t)\,W(t)\) is constant (Abel's identity: differentiate \(r(y_1 y_2' - y_1' y_2)\) and use the equation). Since \(r > 0\), \(W(a) = 0\) forces \(W \equiv 0\), so \(y_1\) and \(y_2\) are linearly dependent.

With periodic boundary conditions this argument fails, and an eigenvalue can have two independent eigenfunctions. For example, \(y'' + \lambda y = 0\) on \([-\pi, \pi]\) with \(y(-\pi) = y(\pi)\), \(y'(-\pi) = y'(\pi)\) has both \(\cos nt\) and \(\sin nt\) for \(\lambda = n^2\).

4. Regular, periodic and singular problems

• Regular: a finite interval \([a, b]\), \(r > 0\) and \(p > 0\) on the whole closed interval, and separated boundary conditions \(\eqref{eq:bc}\).
• Periodic: \(r(a) = r(b)\), with the conditions \(y(a) = y(b)\) and \(y'(a) = y'(b)\). The boundary term \(\eqref{eq:boundary}\) still cancels, so \(\mathcal{L}\) is self-adjoint, but eigenvalues may be double (as above). Ordinary Fourier series come from this case.
• Singular: \(r\) vanishes at an endpoint (or the interval is infinite). At such an endpoint the usual condition is replaced by requiring \(y\) to stay bounded:
  • \(r(a) = 0\), with a condition \(A_2 y(b) + B_2 y'(b) = 0\) at \(t = b\);
  • \(r(b) = 0\), with a condition \(A_1 y(a) + B_1 y'(a) = 0\) at \(t = a\);
  • \(r(a) = r(b) = 0\), with no condition at either end (only boundedness).
  The boundary term vanishes because \(r = 0\) there, so \(\mathcal{L}\) is still self-adjoint.

For the classical singular problems, such as Legendre's equation on \([-1, 1]\) and Bessel's equation on \([0, R]\), the eigenfunctions are still complete. In general, though, a singular problem can also have a continuous spectrum. For example, on the whole real line the sum over eigenfunctions becomes an integral, and Fourier series become the Fourier transform [1].

5. What the theory is for

• Sturm–Liouville theory does not give a faster way to solve a particular ODE, and it is not needed to show that solutions exist; ordinary ODE theory does that. What it guarantees is the structure of the solutions: real eigenvalues, orthogonal eigenfunctions, and completeness.
• Completeness is the key for PDEs. It means any initial or boundary data \(f\) can be expanded as a generalized Fourier series, \(f = \sum_n c_n y_n\), and orthogonality gives each coefficient directly: \[ c_n = \frac{\langle y_n, f\rangle_p}{\langle y_n, y_n\rangle_p} . \]
• The functions are not arbitrary. \(r\) comes from the integrating factor (Section 2), and \(p\) comes from the physics of the problem; it often represents a non-uniform density or heat capacity. So the weighted inner product \(\langle\cdot,\cdot\rangle_p\), rather than the plain one, is forced on us by the problem.

IX. Examples from PDEs

1. The heat equation

Heat conduction in a rod with non-uniform material is described by

\[ \frac{\partial}{\partial x}\left[r(x)\,\frac{\partial u}{\partial x}\right] = \rho(x)\,\frac{\partial u}{\partial t}, \]

where \(u(x, t)\) is the temperature, \(r(x) > 0\) is the thermal conductivity, and \(\rho(x) > 0\) is the heat capacity per unit length. Separate variables, \(u(x, t) = y(x)\,T(t)\):

\[ T(t)\,\frac{d}{dx}\left[r(x)\,\frac{dy}{dx}\right] = \rho(x)\,y(x)\,\frac{dT}{dt} . \]

Dividing by \(\rho\,y\,T\), the left side depends only on \(x\) and the right side only on \(t\), so both equal a constant, which we call \(-\lambda\):

\[ \frac{1}{\rho(x)\,y(x)}\,\frac{d}{dx}\left[r(x)\,\frac{dy}{dx}\right] = \frac{1}{T(t)}\,\frac{dT}{dt} = -\lambda . \]

• Time equation: \(T'(t) = -\lambda\,T(t)\), so \(T(t) = C\,e^{-\lambda t}\).
• Space equation, in Sturm–Liouville form with \(q = 0\) and weight \(p = \rho\): \[ \frac{d}{dx}\left[r(x)\,\frac{dy}{dx}\right] + \lambda\,\rho(x)\,y(x) = 0 . \]

With separated boundary conditions at the ends of the rod, the eigenfunctions \(y_n\) are orthogonal with weight \(\rho\) and complete. So the solution for an initial temperature \(f(x)\) is

\[ u(x, t) = \sum_{n} c_n\,y_n(x)\,e^{-\lambda_n t}, \qquad c_n = \frac{\langle y_n, f\rangle_\rho}{\langle y_n, y_n\rangle_\rho} . \]

2. The Schrödinger equation

\[ i\hbar\,\frac{\partial \psi(x, t)}{\partial t} = -\frac{\hbar^2}{2m}\,\frac{\partial^2 \psi(x, t)}{\partial x^2} + V(x)\,\psi(x, t) . \]

Separate variables, \(\psi(x, t) = \psi(x)\,T(t)\), and divide by \(\psi(x)\,T(t)\):

\[ \frac{i\hbar}{T(t)}\,\frac{dT}{dt} = \frac{1}{\psi(x)}\left(-\frac{\hbar^2}{2m}\,\frac{d^2\psi}{dx^2} + V(x)\,\psi(x)\right) = E, \]

where the constant \(E\) is the energy.

• Time equation: \(T'(t) = -\dfrac{iE}{\hbar}\,T(t)\), so \(T(t) = C\,e^{-iEt/\hbar}\).
• Space equation, the time-independent Schrödinger equation: \[ -\frac{\hbar^2}{2m}\,\frac{d^2\psi}{dx^2} + V(x)\,\psi = E\,\psi \quad\Longleftrightarrow\quad \frac{d}{dx}\left[\frac{\hbar^2}{2m}\,\frac{d\psi}{dx}\right] + \big(-V(x) + E\big)\,\psi = 0 . \]

This is Sturm–Liouville form with \(r = \hbar^2/2m > 0\), \(q = -V\), \(p = 1\) and eigenvalue \(\lambda = E\). The operator \(\mathcal{L} = -\frac{d}{dx}\big(r\,\frac{d}{dx}\big) - q = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V\) is exactly the Hamiltonian. So Section VII explains two basic facts of quantum mechanics: energies are real, and states with different energies are orthogonal.

References