General form se shuru karo:
a2(x)y′′+a1(x)y′+a0(x)y+λy=0.
Hum chahte hain (py′)′=py′′+p′y′. a2 se divide karo aur ek integrating factorμ(x) se multiply karo:
μy′′+μa2a1y′+⋯
Yeh (μy′)′=μy′′+μ′y′ ke barabar ho iske liye humein chahiye μ′=μa2a1, toh
μ(x)=exp(∫a2a1dx)
Phir p=μ, q=μa0/a2, w=μ/a2. Yeh step kyun? Integrating factor exactly wahi cheez hai jo do alag terms ko ek perfect derivative mein convert karta hai — same trick jaise first-order linear ODEs mein hota hai.
Maano ym,yn SL equation solve karte hain eigenvalues λm,λn ke saath:
(pym′)′+(q+λmw)ym=0,(pyn′)′+(q+λnw)yn=0.
Pehle ko yn se multiply karo, doosre ko ym se multiply karo, subtract karo:
yn(pym′)′−ym(pyn′)′+(λm−λn)wymyn=0.Subtract kyun?qymyn wale terms cancel ho jaate hain, eigenvalue difference isolate ho jaata hai.
Ab note karo Lagrange's identity:
yn(pym′)′−ym(pyn′)′=dxd[p(ynym′−ymyn′)].Yeh step kyun? RHS expand karo: p′(ynym′−ymyn′)+p(ynym′′−ymyn′′) — exactly LHS hi hai. Yeh sab kuch ek total derivative mein pack kar deta hai taaki hum cleanly integrate kar sakein.
a se b tak integrate karo:
[p(ynym′−ymyn′)]ab+(λm−λn)∫abwymyndx=0.
Boundary term Wronskian-jaisa bracket hai: [pW(ym,yn)]ab.
Boundary term zero kyun hota hai:x=a par, separated BC α1ym(a)+α2ym′(a)=0 aur α1yn(a)+α2yn′(a)=0 do equations hain (α1,α2)=(0,0) ke liye. Nonzero solution ka matlab determinant ym(a)yn′(a)−yn(a)ym′(a)=0 hona chahiye, yani W(ym,yn)∣a=0. Same b par. Isliye bracket 0 hai.
Therefore
(λm−λn)∫abwymyndx=0.
Kyunki λm=λn, integral zero hai. ■Weight w ke saath Orthogonality.
Complex y lo eigenvalue λ ke saath, aur uska conjugate yˉλˉ ke saath (kyunki p,q,w real hain). Same subtraction se milta hai (λ−λˉ)∫w∣y∣2dx=0. Kyunki ∫w∣y∣2>0, hume milta hai λ=λˉ, yani real.
Yeh kyun kaam karta hai: orthogonality har term ko kill kar deta hai sivaaye m=n ke — exactly jaise ek vector ko orthogonal axis par project karna: cm=⟨ym,ym⟩w⟨f,ym⟩w.
Ek guitar ki string imagine karo. Jab tum use pluck karte ho, toh woh sirf special shapes mein vibrate karti hai — ek bump, do bumps, teen bumps. "Dhai bumps" nahi mil sakta. Woh allowed shapes eigenfunctions hain, aur har ek kitni tezi se hilti hai woh eigenvalue hai. Cool trick yeh hai: yeh shapes itni "independent" hain ki string ki koi bhi movement bas ek recipe hai — shape-1 ka kuch, shape-2 ka kuch, etc. Yeh pata karne ke liye ki shape-2 ka kitna hai, tum apni movement ko sirf shape-2 se "compare" karte ho; baaki saari shapes politely zero deti hain. Woh polite-zero property orthogonality hai. Sturm–Liouville theory woh rulebook hai jo promise karta hai ki yeh special shapes hamesha exist karengi aur hamesha itni acchi tarah behave karengi.