Initial value problems (IVP) vs boundary value problems (BVP): Worked examples

4.7.2 · D3 · Maths › Partial Differential Equations › Initial value problems (IVP) vs boundary value problems (BVP

Yeh page parent topic on IVPs and BVPs ka exhaustive case-book hai. Parent ne bataya tha ki distinction kya hai. Yahan hum har tarah ke outcome ko walk through karte hain jo ek second-order problem produce kar sakta hai, taaki koi bhi scenario aisa na ho jise tumne pehle solve hote nahi dekha ho.

The scenario matrix

Kuch bhi naya define karne se pehle, yeh hai har cell ka map jo hume cover karna hai. Har row ek class of situation hai; aakhri column us example ka naam hai jo use hit karta hai.

#	Cell (scenario class)	Key question	Outcome	Example
A	IVP, dono data ek hi point par	Wronskian $\neq = 0$ ?	Hamesha unique	Ex 1
B	IVP with $x_{0} \neq = 0$ (the " $t = 0$ trap")	Same point, zero nahi?	Phir bhi unique	Ex 2
C	BVP, $det M \neq = 0$	Matrix invertible?	Unique	Ex 3
D	BVP, $det M = 0$ , target reachable	Compatible RHS?	Infinitely many	Ex 4
E	BVP, $det M = 0$ , target unreachable	Incompatible RHS?	No solution	Ex 5
F	Eigenvalue sweep (kaun sa $λ$ todta hai?)	Kis parameter ke liye $det M = 0$ ?	Discrete "resonant" values	Ex 6
G	Degenerate ODE ( $y^{''} = 0$ ) — real-world rod	Straight-line family	Unique linear profile	Ex 7
H	Mixed / Robin BVP (ek end par value, doosre par slope)	Different points phir bhi?	BVP, check $det M$	Ex 8
I	Exam twist: same numbers, IVP vs BVP swap	Conditions kahan hain?	Ek unique, ek singular	Ex 9

Do symbols baar baar aate hain, dono parent mein banaye gaye hain — aao unhe ek baar mein re-anchor karte hain taaki yahan kuch bhi unexplained na rahe.

Group 1 — IVPs hamesha kaam karte hain (cells A, B)

Group 2 — BVPs: teen fates (cells C, D, E)

Yahan hum same ODE $y^{''} + y = 0$ ko teen baar lete hain, sirf endpoints badal ke. Yahi topic ka core hai: conditions count karna kaafi nahi. Figure dekho — yeh dikhata hai kyun $[0, π]$ "dangerous" interval hai.

Figure — Initial value problems (IVP) vs boundary value problems (BVP)

Recall

$det M = 0$ par ka fork Jab $det M = 0$ ho, toh jawab "no solution" hai ya "infinitely many"? ::: Target par depend karta hai: agar boundary data singular system ke saath compatible hai (RHS column space mein) → infinitely many; agar incompatible → none.

Group 3 — Eigenvalue sweep (cell F)

Ab ek parameter vary karo aur pucho: kin values ke liye BVP singular ho jaata hai? Yeh special values eigenvalues hain; Sturm-Liouville Theory ka poora subject yahi sawaal systematically poochta hai.

Worked example Example 6 — Cell F: kaun sa

$λ$ nontrivial solutions deta hai? ODE: $y^{''} + λ y = 0$ on $[0, π]$ with $y (0) = 0, y (π) = 0$ , aur $λ > 0$ . Likho $λ = μ^{2}$ .

Forecast: "Zyaadatar" $λ$ ke liye sirf $y \equiv 0$ kaam karta hai. Kaun se special $λ$ ke liye suddenly real curves milti hain?

General solution $y = c_{1} cos μx + c_{2} sin μx$ . Kyun? $λ = μ^{2} > 0$ ke liye characteristic roots $\pm i μ$ hain, jo cos/sin dete hain.
$y (0) = c_{1} = 0$ , toh $y = c_{2} sin μx$ . Kyun? Left endpoint cosine part ko kill karta hai.
$y (π) = c_{2} sin (μ π) = 0$ . Nontrivial ( $c_{2} \neq = 0$ ) solution ke liye humein $sin (μ π) = 0$ chahiye. Kyun? Tabhi $c_{2}$ free reh sakta hai. Matlab $μ π = nπ \Rightarrow μ = n$ , toh $λ_{n} = n^{2}$ , $n = 1, 2, 3, \dots$
Eigenfunctions: $y_{n} = sin (n x)$ .

Verify: $λ = 4$ ( $n = 2$ ) ke liye: $y = sin 2 x$ , $y (0) = 0$ ✓, $y (π) = sin 2 π = 0$ ✓ — nontrivial. $λ = 2$ ke liye: $μ = 2$ , $sin (2 π) \approx sin (4.4429) \approx - 0.9633 \neq = 0$ → sirf $c_{2} = 0$ → trivial solution $y \equiv 0$ . Toh discrete list ${1, 4, 9, \dots}$ exactly resonant set hai. (Parent Ex 4 se compare karo, jo $λ = 4$ ko length sweep ki tarah read karta tha.)

Group 4 — Degenerate ODE, real-world word problem (cell G)

Worked example Example 7 — Cell G: rod mein steady heat (units check!)

Ek $L = 2 m$ lambi metal rod steady state mein hai. Temperature $y^{''} = 0$ follow karta hai (koi time change nahi, koi source nahi). Ends $y (0) = T_{0} = 10 0^{\circ} C$ aur $y (L) = T_{1} = 2 0^{\circ} C$ par hold ki gayi hain.

Forecast: Straight line, curve, ya kuch aur? Aur midpoint $x = 1$ par temperature kya hai?

Twice integrate karo: $y = c_{1} x + c_{2}$ . Kyun? $y^{''} = 0$ matlab slope kabhi nahi badlta → ek straight line; do integrations do constants dete hain.
$y (0) = c_{2} = 100$ . Kyun? Left end intercept fix karta hai.
$y (2) = 2 c_{1} + 100 = 20 \Rightarrow c_{1} = \frac{20 - 100}{2} = - 40$ . Kyun? Right end slope fix karta hai.
Profile: $y (x) = 100 - 40 x$ (units: slope ke liye $^{\circ}$ C per metre, yaani $- 40$ ).
Midpoint $x = 1$ : $y (1) = 100 - 40 = 6 0^{\circ} C$ .

Verify: $det M = y_{1} (0) y_{2} (L) - y_{2} (0) y_{1} (L)$ with $y_{1} = x, y_{2} = 1$ : $= 0 \cdot 1 - 1 \cdot 2 = - 2 \neq = 0$ → unique ✓. Endpoints: $y (0) = 100$ ✓, $y (2) = 100 - 80 = 20$ ✓. Units: slope $\times$ length $= - 40 \frac{^{\circ} C}{m} \times 2 m = - 8 0^{\circ} C$ , toh $100 - 80 = 20$ ✓. Yeh Heat Equation ki steady-state limit hai.

Group 5 — Mixed / Robin BVP (cell H)

Ek condition ek value aur ek slope ko mix kar sakti hai — lekin jo cheez ise BVP banati hai woh yeh hai ki yeh alag-alag points par hote hain. Ise parent ki warning se compare karo: $y (a)$ aur $y^{'} (a)$ (same point) ek IVP hota.

Worked example Example 8 — Cell H: ek end par value, doosre par slope

ODE: $y^{''} + y = 0$ , conditions $y (0) = 0, y^{'} (\frac{π}{2}) = 4$ . Alag points → genuine BVP.

Forecast: Value aur derivative mix karna — kya $det M$ phir bhi checkable hoga? Unique hai ya nahi?

$y = c_{1} cos x + c_{2} sin x$ ; $y (0) = c_{1} = 0$ . Kyun? Left endpoint value condition.
$y^{'} (x) = - c_{1} sin x + c_{2} cos x$ ; $x = \frac{π}{2}$ par: $- c_{1} \cdot 1 + c_{2} \cdot 0 = - c_{1}$ . Ruko — kya yeh $- c_{1} = 4$ hai? Dhyan se kyun check karein? $cos \frac{π}{2} = 0$ se $c_{2}$ kill ho jaata hai. Toh slope condition padh ke aati hai $- c_{1} = 4$ .
Lekin step 1 ne already $c_{1} = 0$ diya tha. Toh $- 0 = 4$ → $0 = 4$ , contradiction → no solution yahan.

Verify: Mixed BVP matrix mein rows hain "value at $0$ " aur "slope at $π /2$ ": $M = (cos 0 - sin \frac{π}{2} sin 0 cos \frac{π}{2}) = (1 - 1 00)$ , $det M = 1 \cdot 0 - 0 \cdot (- 1) = 0$ → singular, aur target $(0, 4)$ incompatible hai (doosri row force karti hai $- c_{1} = 4$ jabki pehli force karti hai $c_{1} = 0$ ) → no solution ✓. Lesson: alag-alag points par mixed conditions phir bhi ek BVP hain aur exactly Ex 5 ki tarah singular ho sakti hain.

Group 6 — Exam twist (cell I)

Worked example Example 9 — Cell I: identical numbers, IVP vs BVP swap

Same ODE $y^{''} + y = 0$ aur same do numbers " $0$ aur $0$ ", do tarike se padhe. (a) IVP: $y (0) = 0, y^{'} (0) = 0$ (dono $x = 0$ par). (b) BVP: $y (0) = 0, y (π) = 0$ ( $x = 0$ aur $x = π$ par).

Forecast: Kya dono same answer denge? (Trap: yeh almost identical lagte hain.)

(a) $y (0) = c_{1} = 0$ ; $y^{'} (0) = c_{2} = 0$ . Dono dials zero par force ho gaye → unique $y \equiv 0$ . Unique kyun? Wronskian $= 1 \neq = 0$ , IVPs hamesha well-posed hote hain.
(b) Ex 4 se: $c_{1} = 0$ aur $c_{2}$ free → $y = c_{2} sin x$ , infinitely many. Alag kyun? Doosri condition ab $x = π$ par hai, jahan $sin π = 0$ se $c_{2}$ pin hone ki jagah erase ho jaata hai.
Toh sirf location se "unique trivial" palatke "infinitely many" ho jaata hai.

Verify: (a) $W = cos^{2} x + sin^{2} x = 1 \neq = 0$ → unique, aur $y \equiv 0$ , $y (0) = y^{'} (0) = 0$ satisfy karta hai ✓. (b) $det M = cos 0 sin π - sin 0 cos π = 0$ , RHS $(0, 0)$ consistent → infinitely many, har $c_{2} sin x$ ke liye $y (0) = 0, y (π) = 0$ ✓.

Recall check

Recall Poori matrix par rapid-fire

Cell C vs D: "unique" aur "infinitely many" ko kaun si ek quantity alag karti hai? ::: $det M$ — nonzero se unique milta hai, zero se none/∞ ke fork par pahunchte hain. Cell F: $y^{''} + λ y = 0$ , $y (0) = y (π) = 0$ ke liye eigenvalues list karo. ::: $λ_{n} = n^{2}, n = 1, 2, 3, \dots$ with eigenfunctions $sin (n x)$ . Cell H: kya $y (0)$ aur $y^{'} (π /2)$ saath mein IVP hain ya BVP? ::: BVP — different points, chahe ek slope hi kyun na ho. Cell G: $y^{''} = 0$ , $y (0) = 100, y (2) = 20$ ka solution? ::: $y = 100 - 40 x$ , midpoint $6 0^{\circ}$ C, unique kyunki $det M = - 2 \neq = 0$ . Cell I: same $(0, 0)$ data — IVP ek solution kyun deta hai lekin BVP infinitely many kyun deta hai? ::: IVP mein dono conditions ek hi point par hain (Wronskian $\neq = 0$ ); BVP doosri condition $x = π$ par padh ta hai jahan $sin π = 0$ se $c_{2}$ free ho jaata hai.

Connections

Sturm-Liouville Theory — cell F entry point hai: $det M = 0$ eigenvalues generate karta hai.
Wronskian and Linear Independence — yahi reason hai ki cells A, B kabhi fail nahi karte.
Picard-Lindelöf Existence and Uniqueness Theorem — IVP uniqueness ke peeche ka deeper guarantee.
Heat Equation — cell G iski steady-state limit hai.
Wave Equation — same $y^{''} + λ y = 0$ eigenvalues standing waves drive karte hain.
Separation of Variables (PDE) — PDEs ko exactly inhi BVP eigenvalue problems mein turn karta hai.
Green's Functions — ek invertible BVP (cell C) ka systematic inverse.