research32 min studyLesson 87 in Mathematics

Sturm–Liouville problems

Connect weighted inner products and boundary forms to real eigenvalues and normal modes.

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What you’ll understand

  • Identify the weight and boundary data in a regular Sturm–Liouville problem.
  • Use the boundary form to motivate self-adjoint boundary conditions.
  • Compute the complete Dirichlet spectrum of a constant-coefficient interval model.

A vibrating string, a diffusion problem, and a one-dimensional quantum well all lead to second-order eigenvalue equations. Sturm–Liouville theory organizes a broad family of these problems using a weighted inner product and carefully chosen endpoint conditions. The eigenvalues come from the entire boundary-value problem, not from the differential expression alone. We begin with smooth coefficients on a finite regular interval, where the main geometry is visible without the additional subtleties of singular endpoints or infinite spatial extent.

The weight belongs to the geometry

Consider −(p y′)′+q y=λw y on a finite interval [a,b]. For this lesson take real p continuously differentiable and strictly positive, and real continuous q and w with w strictly positive. Dividing by w defines an operator in L2([a,b],w dx), where the weight appears in the inner product. An eigenvalue problem written with λw y on the right is therefore naturally symmetric in the weighted space, rather than automatically in the unweighted one.

The two endpoint conditions complete the problem. Dirichlet conditions fix y to zero, Neumann conditions fix the endpoint flux p y′ to zero, and real separated Robin conditions prescribe a linear combination at each endpoint. These are examples, not the full classification of self-adjoint boundary conditions; coupled periodic conditions also exist. Coefficients that vanish or blow up, an infinite interval, or a nonintegrable endpoint require singular Sturm–Liouville theory with additional domain analysis. The regular results should not be transported there unchanged.

Ly=1w(−(py′)′+qy),⟨f,g⟩w=∫abf(x)‾g(x)w(x) dxLy=\frac1w\left(-(py')'+qy\right),\qquad \langle f,g\rangle_w=\int_a^b\overline{f(x)}g(x)w(x)\,dx
The regular model assumes a finite interval, p∈C1 with p>0, continuous real q, and continuous w>0. With these coefficients, the operator domain consists of H²(a,b) functions satisfying the selected endpoint conditions.↔ Scroll wide equations horizontally.

Integration by parts reveals the endpoint obligation

For sufficiently regular f and g, integration by parts cancels the interior terms in inner-product(f,Lg)−inner-product(Lf,g), leaving an endpoint expression. Dirichlet conditions make both function values vanish, so the boundary form is zero. Real separated Robin conditions also cancel it because both functions obey the same real relation between value and flux. Vanishing of the form on a proposed domain establishes symmetry. Self-adjointness additionally requires that the adjoint domain impose exactly the same endpoint restrictions.

For regular coefficients and self-adjoint endpoint conditions, this domain analysis gives real eigenvalues, orthogonality of eigenfunctions at distinct eigenvalues, and a discrete spectrum with eigenvalues tending upward without bound. The corresponding inverse after a suitable spectral shift is compact. These statements support expansion in normal modes. Orthogonality is with the weight w, and normalization must use that same weight. Changing the boundary conditions changes the eigenfunctions and can change whether a zero-frequency mode is present.

⟨f,Lg⟩w−⟨Lf,g⟩w=[p(f′‾g−f‾ g′)]ab\langle f,Lg\rangle_w-\langle Lf,g\rangle_w=\left[p\left(\overline{f'}g-\overline f\,g'\right)\right]_a^b
The stated inner product is conjugate linear in the first slot. The boundary form, together with domain regularity, determines which realizations are self-adjoint.↔ Scroll wide equations horizontally.

The interval sets the allowed wavelengths

With p=w=1 and q=0, the Dirichlet problem on [0,L] is −y″=λy with y(0)=y(L)=0. A nontrivial solution has positive λ, since integrating against y gives a nonnegative derivative energy and a zero eigenvalue would force a constant that vanishes at both ends. Writing λ=k² leaves a sine after imposing the left endpoint, and the right endpoint requires kL to be a positive integer multiple of π. Boundary conditions quantize the allowed spatial scales.

For Neumann endpoints, the constant function is allowed and gives eigenvalue zero. This difference matters in diffusion: a constant temperature survives when the boundaries prevent flux, while zero Dirichlet boundaries remove it. The mode expansion still needs a declared convergence sense, such as L2 convergence for square-integrable initial data. Termwise differentiation or pointwise boundary recovery may require stronger assumptions. A finite sum can demonstrate shapes without proving those convergence and regularity claims for arbitrary data.

λn=(nπL)2,en(x)=2Lsin⁡ ⁣(nπxL),n=1,2,…\lambda_n=\left(\frac{n\pi}{L}\right)^2,\qquad e_n(x)=\sqrt{\frac2L}\sin\!\left(\frac{n\pi x}{L}\right),\qquad n=1,2,\ldots
These are the normalized eigenfunctions and eigenvalues for the constant-coefficient Dirichlet operator on (0,L), with domain H2∩H1_0.↔ Scroll wide equations horizontally.

Two modes in a well of length π

On (0,π), solve −y″=λy with Dirichlet endpoints. Then compute the squared L2 norm and derivative energy of f(x)=2sin x+sin(2x).

  1. The allowed wave numbers are positive integers, giving eigenvalues n² and sine eigenfunctions. The first two eigenvalues are 1 and 4.

  2. Orthogonality makes the cross integral vanish. Each unnormalized sine has squared L2 norm π/2, so the norm of f squared is (4+1)π/2=5π/2.

  3. Differentiate f to obtain 2cos x+2cos(2x). The cosine cross integral also vanishes, giving derivative energy (4+4)π/2=4π. The Rayleigh quotient is therefore 4π/(5π/2)=8/5.

The result

The mixed profile has Rayleigh quotient 1.6, between its component eigenvalues 1 and 4. The weights are squared L2 amplitudes.

Explore the model

Interactive lab

Interactive lab · spectral

See the spatial scale of sine modes

Pause playback at its initial time and raise N. At that time the sum Σ sin(nx)/nᵖ is a finite Dirichlet sine combination on [0,π]. Changing p shifts relative mode amplitudes. Playback changes the phase to nx−t and does not preserve these endpoint conditions, so it is not the evolution of this boundary-value problem.

See the spatial scale of sine modesA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe initial finite profile illustrates sine scales; the operator and its boundary conditions must be checked separately.

Check your understanding

Why is the weight w included in the natural inner product for Ly=w^{-1}(−(py′)′+qy)?

Try it yourself

On (0,π), replace both Dirichlet conditions by Neumann conditions for −y″. Give the lowest eigenvalue and an associated normalized real eigenfunction.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Test a constant before using oscillating solutions.

Compare with the solution

The lowest eigenvalue is zero, with eigenfunction 1/√π. Its derivative is zero at both endpoints and its squared integral is one. Integration by parts gives nonnegative derivative energy, excluding negative eigenvalues. Higher modes are cos(nx) with eigenvalues n² for positive integers n; the constant mode is the change introduced by the endpoint conditions.

Take it with you

  • A Sturm–Liouville problem specifies a weight, coefficients, interval, and endpoint domain.
  • Boundary forms explain symmetry and guide the full self-adjoint domain check.
  • Regular self-adjoint interval problems have discrete real spectra; singular problems need extra analysis.

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