Boundary value problems — shooting method, finite difference

4.8.28 · Maths › Numerical Methods

BVP KYA hai?

YEH KYUN aate hain? Steady-state physics: ek rod ke saath temperature jab dono ends fixed hon, beam ka deflection jab dono ends clamped hon, do plates ke beech electrostatic potential. Kuch bhi time mein evolve nahi hota — answer ko har jagah ek saath constraints satisfy karne padte hain.

METHOD 1 — Shooting Method

KAISE (algorithm):

Slope $y^{'} (a)$ ke liye $s_{0}$ guess karo. IVP solve karo $\to$ $y (b; s_{0})$ milega.
$s_{1}$ guess karo. Phir solve karo $\to y (b; s_{1})$ .
Secant method use karo (hum $ϕ$ ko cheaply differentiate nahi kar sakte): $s_{n + 1} = s_{n} - ϕ (s_{n}) \frac{s _{n} - s _{n - 1}}{ϕ ( s _{n} ) - ϕ ( s _{n - 1} )} .$
Tab tak repeat karo jab tak $∣ ϕ (s_{n}) ∣ < tol$ .

METHOD 2 — Finite Difference Method (FDM)

KAISE — difference formulas Taylor series se derive karo (first principles):

Grid: $x_{i} = a + ih$ , $h = \frac{b - a}{N}$ , $i = 0, \dots, N$ . Maano $y_{i} \approx y (x_{i})$ .

Expand: $y_{i + 1} = y_{i} + h y_{i}^{'} + \frac{h ^{2}}{2} y_{i}^{''} + \frac{h ^{3}}{6} y_{i}^{'''} + \dots$ $y_{i - 1} = y_{i} - h y_{i}^{'} + \frac{h ^{2}}{2} y_{i}^{''} - \frac{h ^{3}}{6} y_{i}^{'''} + \dots$

Subtract karo → odd terms bachte hain, $y_{i}^{''}$ cancel hota hai: $y_{i + 1} - y_{i - 1} = 2 h y_{i}^{'} + \frac{h ^{3}}{3} y_{i}^{'''} + \dots \Rightarrow y_{i}^{'} \approx \frac{y _{i + 1} - y _{i - 1}}{2 h} (O (h^{2})) .$

Add karo → odd terms cancel hote hain, $y_{i}^{''}$ isolate karo: $y_{i + 1} + y_{i - 1} = 2 y_{i} + h^{2} y_{i}^{''} + \frac{h ^{4}}{12} y_{i}^{''''} + \dots \Rightarrow y_{i}^{''} \approx \frac{y _{i + 1} - 2 y _{i} + y _{i - 1}}{h ^{2}} (O (h^{2})) .$

Ek linear BVP pe apply karo $y^{''} = p (x) y^{'} + q (x) y + r (x)$ . Har interior $x_{i}$ pe stencils substitute karo: $\frac{y _{i + 1} - 2 y _{i} + y _{i - 1}}{h ^{2}} = p_{i} \frac{y _{i + 1} - y _{i - 1}}{2 h} + q_{i} y_{i} + r_{i} .$ $h^{2}$ se multiply karo aur $y_{i - 1}, y_{i}, y_{i + 1}$ group karo: $lower (1 + \frac{h}{2} p_{i}) y_{i - 1} diag - (2 + h^{2} q_{i}) y_{i} + upper (1 - \frac{h}{2} p_{i}) y_{i + 1} = h^{2} r_{i} .$

Yeh ek tridiagonal linear system $A y = d$ hai jisme $y_{0} = α, y_{N} = β$ ko right side mein move kiya gaya hai. Thomas algorithm (tridiagonal Gaussian elimination) se $O (N)$ mein solve karo.

Figure — Boundary value problems — shooting method, finite difference

Shooting vs Finite Difference

	Shooting	Finite Difference
Idea	Slope guess karo, root-find karo	Saare nodes ek saath solve karo
Reuses	IVP solver (RK4)	Linear algebra (Thomas)
Nonlinear BVP	Easy (secant bahar)	System pe Newton chahiye
Stiff/long domain	Blow up ho sakta hai	Stable
Accuracy	RK4 order	$O (h^{2})$ (central)

Common mistake Steel-manned common errors

1. "Shooting hamesha 2 steps mein converge hota hai." Kyun sahi lagta hai: linear shortcut IS exactly 2 shots mein exact hai. Fix: yeh sirf linear ODEs ke liye hai. Nonlinear ones ke liye $ϕ (s)$ curved hoti hai → tumhe secant/Newton ko tab tak iterate karna padega jab tak $∣ ϕ ∣ <$ tol.

2. FDM mein $y^{'}$ ke liye forward difference $y^{'} \approx \frac{y _{i + 1} - y _{i}}{h}$ use karna "simple rakhne ke liye." Kyun sahi lagta hai: yeh ek valid derivative approximation hai. Fix: yeh sirf $O (h)$ hai, jo tumhare poore scheme ko first order tak le aata hai even though $y^{''}$ $O (h^{2})$ tha. Orders match karne ke liye central difference use karo.

3. Boundary values ko RHS pe move karna bhool jaana. Kyun sahi lagta hai: $y_{0}, y_{N}$ "sirf numbers hain." Fix: $i = 1$ pe, term $(1 + \frac{h}{2} p_{1}) y_{0}$ known hai — ise right side $d_{1}$ pe subtract karo. Usi tarah $i = N - 1$ pe. Warna tumhari matrix galat size ki hogi.

4. $N$ bahut chota choose karna aur answer pe trust karna. Kyun sahi lagta hai: tumhe ek number mila. Fix: $h$ halve karo; error $\approx 4 \times$ drop hona chahiye. Agar nahi hota, toh tumhara stencil/setup galat hai (Forecast-then-Verify!).

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho tum ek ball ek door bucket mein exactly land karwani hai. Tumhe sahi throwing angle pata nahi. Shooting: ek baar throw karo (bahut chota), ek baar phir (bahut door), phir beech mein split karo — angle adjust karte raho jab tak bucket mein land na ho jaaye. Finite difference: throw karne ke bajaye, ball ka path bahut saare chhote connected dots se draw karo aur dots ko yeh rule follow karne pe force karo ki "har dot apne neighbours ka roughly average hai," phir dot-puzzle solve karo taaki pehla aur aakhri dot wahan hon jahan hona chahiye. Dono same curve dhundhhte hain — ek aiming se, ek puzzle-solving se.

Flashcards

BVP ko IVP se kya alag karta hai?

BVP har endpoint pe ek condition deta hai (

y (a), y (b)

); IVP saari conditions (

y, y^{'}

) ek single point pe deta hai, toh IVP march kar sakta hai lekin BVP nahi.

Shooting method ka core idea?

Missing slope

y^{'} (a) = s

guess karo, resulting IVP solve karo, landing error

ϕ (s) = y (b; s) - β

treat karo, aur

ϕ (s) = 0

root-find karo (usually secant se).

Linear shooting shortcut exactly 2 shots mein answer kyun deta hai?

Linear ODE ke liye,

y (b; s)

s

mein linear hai, toh

ϕ (s)

ek straight line hai jo exactly do points se determine hoti hai — linear interpolation bina iteration ke root hit karta hai.

Shooting ke liye secant update formula?

s_{n + 1} = s_{n} - ϕ (s_{n}) \frac{s _{n} - s _{n - 1}}{ϕ ( s _{n} ) - ϕ ( s _{n - 1} )}

y^{'}

ke liye central difference aur uska order?

y_{i}^{'} \approx \frac{y _{i + 1} - y _{i - 1}}{2 h}

, accurate to

O (h^{2})

y^{''}

ke liye central difference aur uska order?

y_{i}^{''} \approx \frac{y _{i + 1} - 2 y _{i} + y _{i - 1}}{h ^{2}}

, accurate to

O (h^{2})

(the

+ 1, - 2, + 1

stencil).

y^{''}

stencil kaise derive karte hain?

y_{i + 1}

aur

y_{i - 1}

ki Taylor expansions add karo; odd-power terms cancel ho jaate hain, leaving

y_{i + 1} + y_{i - 1} = 2 y_{i} + h^{2} y_{i}^{''} + O (h^{4})

, phir

y_{i}^{''}

ke liye solve karo.

y^{''} = p y^{'} + q y + r

ke liye tridiagonal FDM coefficients?

Lower

1 + \frac{h}{2} p_{i}

, diagonal

- (2 + h^{2} q_{i})

, upper

1 - \frac{h}{2} p_{i}

, RHS

h^{2} r_{i}

FDM system ko efficiently solve karne wala algorithm aur uski cost?

Thomas algorithm (tridiagonal Gaussian elimination),

O (N)

Shooting ke upar FDM kab prefer karein?

Stiff problems ya long domains jahan shooting mein chote slope errors blow up ho jaate hain; FDM errors ko local rakhta hai aur zyada stable hai.

Central-difference FDM mein

y^{'}

ke liye forward difference use karna buri idea kyun hai?

Yeh sirf

O (h)

hai, poore scheme ko second order se first order tak degrade kar deta hai.

FDM matrix assemble karte waqt

y_{0}

aur

y_{N}

ke saath kya karna padta hai?

Unka (known) contribution pehle aur aakhri interior equations ke right-hand side vector mein move karo.

Connections

Initial Value Problems — shooting ek BVP ko repeated IVPs mein reduce karta hai.
Runge-Kutta methods — shooting ke liye inner IVP solver.
Root finding — Secant and Newton — shooting ka outer loop.
Taylor series — har finite-difference stencil aur uske error order ka source.
Tridiagonal systems — Thomas algorithm — FDM linear system solve karta hai.
Truncation error and order of accuracy — kyun $O (h^{2})$ matter karta hai.
Partial Differential Equations — finite differences — same stencils 2D mein (Laplace/Poisson).