Sturm-Liouville theory — eigenvalue problems, orthogonality of eigenfunctions

1. What is a Sturm–Liouville problem?

WHAT each piece means:

$p(x)$ — the "stiffness"/diffusion coefficient.
$q(x)$ — a potential/source-like term.
$w(x)$ — the weight; it defines the inner product (CRUCIAL for orthogonality).
$\lambda$ — the eigenvalue (e.g. $\lambda_n=(n\pi/L)^2$ in the heat equation).

Why ANY linear 2nd-order ODE can be put in SL form (derivation)

Start from the general form $a_2(x)y'' + a_1(x)y' + a_0(x)y + \lambda y = 0.$ We want $(p y')' = p y'' + p' y'$ . Divide by $a_2$ and multiply by an integrating factor $\mu(x)$ : $\mu y'' + \mu\frac{a_1}{a_2}y' + \cdots$ For this to equal $(\mu y')' = \mu y'' + \mu' y'$ we need $\mu' = \mu\,\dfrac{a_1}{a_2}$ , so $\boxed{\;\mu(x)=\exp\!\left(\int \frac{a_1}{a_2}\,dx\right)\;}$ Then $p=\mu$ , $q=\mu\,a_0/a_2$ , $w=\mu/a_2$ . Why this step? The integrating factor is exactly what turns two separate terms into one perfect derivative — the same trick as first-order linear ODEs.

2. Boundary conditions that make it "regular"

3. The Main Theorem (the 80/20 core)

3.1 Derive orthogonality FROM SCRATCH (the key result)

Let $y_m,y_n$ solve the SL equation with eigenvalues $\lambda_m,\lambda_n$ : $(p y_m')' + (q+\lambda_m w)y_m = 0,\qquad (p y_n')' + (q+\lambda_n w)y_n=0.$ Multiply the first by $y_n$ , the second by $y_m$ , subtract: $y_n(p y_m')' - y_m(p y_n')' + (\lambda_m-\lambda_n)\,w\,y_m y_n = 0.$ Why subtract? The $q\,y_my_n$ terms cancel, isolating the eigenvalue difference.

Now note Lagrange's identity: $y_n(p y_m')' - y_m(p y_n')' = \frac{d}{dx}\big[p(y_n y_m' - y_m y_n')\big].$ Why this step? Expand the RHS: $p'(y_ny_m'-y_my_n') + p(y_ny_m''-y_my_n'')$ — exactly the LHS. It packages everything into a total derivative so we can integrate cleanly.

Integrate from $a$ to $b$ : $\big[p(y_n y_m' - y_m y_n')\big]_a^b + (\lambda_m-\lambda_n)\int_a^b w\,y_m y_n\,dx = 0.$ The boundary term is the Wronskian-like bracket $\big[p\,W(y_m,y_n)\big]_a^b$ .

Why the boundary term vanishes: At $x=a$ , the separated BC $\alpha_1 y_m(a)+\alpha_2 y_m'(a)=0$ and $\alpha_1 y_n(a)+\alpha_2 y_n'(a)=0$ are two equations for $(\alpha_1,\alpha_2)\ne(0,0)$ . A nonzero solution forces the determinant $y_m(a)y_n'(a)-y_n(a)y_m'(a)=0$ , i.e. $W(y_m,y_n)|_a=0$ . Same at $b$ . Hence the bracket is $0$ .

Therefore $(\lambda_m-\lambda_n)\int_a^b w\,y_m y_n\,dx = 0.$ Since $\lambda_m\ne\lambda_n$ , the integral is zero. $\blacksquare$ Orthogonality with weight $w$ .

3.2 Why eigenvalues are real (same machinery)

Take complex $y$ with eigenvalue $\lambda$ , and its conjugate $\bar y$ with $\bar\lambda$ (since $p,q,w$ real). The same subtraction gives $(\lambda-\bar\lambda)\int w|y|^2dx=0$ . As $\int w|y|^2>0$ , we get $\lambda=\bar\lambda$ , i.e. real.

4. Eigenfunction expansion — getting the coefficients

Why this works: orthogonality kills every term except $m=n$ — exactly like projecting a vector onto an orthogonal axis: $c_m = \dfrac{\langle f,y_m\rangle_w}{\langle y_m,y_m\rangle_w}$ .

Figure — Sturm-Liouville theory — eigenvalue problems, orthogonality of eigenfunctions

5. Worked Examples

Worked example Example 1 — The canonical

$y''+\lambda y=0$ on $[0,L]$ , $y(0)=y(L)=0$ Here $p=1,\,q=0,\,w=1$ . Step 1: Try $\lambda>0$ , write $\lambda=k^2$ . General solution $y=A\cos kx+B\sin kx$ . Why? For $\lambda>0$ solutions oscillate; negative/zero $\lambda$ give exponentials/lines that can't satisfy two zero BCs nontrivially (check!). Step 2: $y(0)=0\Rightarrow A=0$ . Why? $\cos 0=1$ . Step 3: $y(L)=B\sin kL=0$ , $B\ne0\Rightarrow \sin kL=0\Rightarrow kL=n\pi$ . Result: $\lambda_n=\left(\dfrac{n\pi}{L}\right)^2,\quad y_n=\sin\dfrac{n\pi x}{L},\ n=1,2,\dots$ Orthogonality check: $\int_0^L \sin\frac{m\pi x}{L}\sin\frac{n\pi x}{L}dx = 0$ for $m\ne n$ (with $w=1$ ) — that's the ordinary Fourier sine series!

Worked example Example 2 — Put a non-SL ODE into SL form

ODE: $x^2 y'' + x y' + \lambda y = 0$ (an Euler-type). Step 1: $a_2=x^2,\ a_1=x,\ a_0=0$ . Integrating factor $\mu=\exp\int\frac{a_1}{a_2}dx=\exp\int\frac1x dx=x$ . Why? From boxed formula. Wait — $p=\mu=x$ ? Let's recompute: $p=\mu=\exp\int (1/x)dx = x$ . Hmm but standard normalization: multiply original by $\mu/a_2 = x/x^2=1/x$ : $x y'' + y' + \frac{\lambda}{x}y=0 \Rightarrow (x y')' + \frac{\lambda}{x}y=0.$ Result: $p=x,\ q=0,\ w=1/x$ . Eigenfunctions are orthogonal with weight $1/x$ , not $1$ ! Why this matters: forgetting $w$ is the #1 error (see below).

6. Common Mistakes (Steel-man + Fix)

Recall Feynman: explain to a 12-year-old

Imagine a guitar string. When you pluck it, it only vibrates in special shapes — one bump, two bumps, three bumps. You can't get "two and a half bumps." Those allowed shapes are the eigenfunctions, and how fast each one wiggles is the eigenvalue. The cool trick: these shapes are so "independent" that any wiggle of the string is just a recipe — some of shape-1, some of shape-2, etc. To find how much of shape-2 you have, you "compare" your wiggle only against shape-2; all the other shapes politely give zero. That polite-zero property is orthogonality. Sturm–Liouville theory is the rulebook that promises these special shapes always exist and always behave this nicely.

7. Flashcards

What is the standard Sturm–Liouville form of a 2nd-order ODE?

(p(x)y')' + (q(x)+\lambda w(x))y = 0

with

p,w>0

and homogeneous BCs.

What does the weight function

w(x)

determine?

The inner product for orthogonality:

\int_a^b y_m y_n\,w\,dx

State the orthogonality relation of SL eigenfunctions.

\int_a^b y_m(x)y_n(x)w(x)\,dx = 0

for

m\ne n

Integrating factor to convert

a_2y''+a_1y'+a_0y+\lambda y=0

to SL form?

\mu(x)=\exp\left(\int \frac{a_1}{a_2}dx\right)

; then

p=\mu

q=\mu a_0/a_2

w=\mu/a_2

Why are SL eigenvalues real?

Subtracting the equation for

y

and

\bar y

gives

(\lambda-\bar\lambda)\int w|y|^2dx=0

; since the integral

>0

\lambda=\bar\lambda

What identity packages

y_n(py_m')'-y_m(py_n')'

into a derivative?

Lagrange's identity:

\frac{d}{dx}[p(y_ny_m'-y_my_n')]

Why does the boundary term vanish in the orthogonality proof?

Separated BCs force the Wronskian

y_my_n'-y_ny_m'=0

at each endpoint (or

p=0

there).

Formula for expansion coefficient

c_m

f=\sum c_n y_n

c_m=\dfrac{\int_a^b f y_m w\,dx}{\int_a^b y_m^2 w\,dx}

Eigenvalues/eigenfunctions of

y''+\lambda y=0

y(0)=y(L)=0

\lambda_n=(n\pi/L)^2

y_n=\sin(n\pi x/L)

n=1,2,\dots

, weight

w=1

Why can periodic SL problems have multiplicity-2 eigenvalues?

They are not regular SL problems; simplicity is only guaranteed for regular separated-BC problems.

8. Connections

Separation of Variables — produces the SL problem from a PDE.
Fourier Series — the special case $w=1$ , $\sin/\cos$ basis.
Bessel Functions — singular SL with weight $w=x$ .
Legendre Polynomials — singular SL on $[-1,1]$ , $w=1$ .
Hilbert Space and Inner Products — orthogonality as projection.
Heat Equation / Wave Equation — eigenfunction expansions of solutions.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab bhi hum koi PDE — heat equation, wave equation, ya Laplace — ko separation of variables se solve karte hain, to space-wala part hamesha ek special ODE ban jaata hai: $(p y')' + (q + \lambda w)y = 0$ . Isko Sturm–Liouville problem kehte hain. Yahan $\lambda$ wo special numbers hain (eigenvalues) jinke liye non-zero solution milta hai, aur wo solutions $y_n$ eigenfunctions hain — bilkul guitar string ke alag-alag vibration shapes jaise (ek bump, do bump, teen bump).

Sabse bada point yeh hai: yeh eigenfunctions orthogonal hote hain, lekin ek weight function $w(x)$ ke saath: $\int_a^b y_m y_n\,w\,dx = 0$ jab $m \ne n$ . Yeh exactly vectors ke perpendicular hone jaisa hai. Isi wajah se hum kisi bhi function $f$ ko in eigenfunctions ke combination mein likh sakte hain, aur har coefficient $c_m$ ko alag se nikaal sakte hain — kyunki baaki saare terms orthogonality se zero ho jaate hain. Yahi Fourier series, Bessel series, Legendre series sab ke peeche ka asli raaz hai.

Orthogonality ka proof yaad rakhna easy hai: dono equations ko cross-multiply karke subtract karo, Lagrange identity se total derivative banao, integrate karo — to ek boundary term aata hai jo BCs ki wajah se zero ho jaata hai, aur bachta hai $(\lambda_m - \lambda_n)\int w y_m y_n = 0$ . Eigenvalues alag hain to integral zero. Bas yahi pura khel hai.

Sabse common galti: students $w(x)$ ko bhool jaate hain aur weight $1$ maan lete hain. Lekin Euler-type ya Bessel mein weight $1/x$ ya $x$ hota hai. Isliye hamesha pehle ODE ko proper SL form mein likho (integrating factor $\mu = \exp\int (a_1/a_2)dx$ se), tabhi sahi $w$ milega. Yeh chhoti si baat exam mein bohot marks bachati hai!

Test yourself — Partial Differential Equations

Connections