Boundary value problems — shooting method, finite difference: Worked examples

4.8.28 · D3 · Maths › Numerical Methods › Boundary value problems — shooting method, finite difference

Shuru karne se pehle, ek self-contained toolkit — neeche use hone waale har symbol ki definition yahan di gayi hai.

Definition Linear-BVP finite-difference template (yahan defined, neeche har jagah use hoga)

Ek general linear second-order BVP aise likha jaata hai $y^{''} = p (x) y^{'} + q (x) y + r (x), y (a) = α, y (b) = β .$

== $p (x)$ == pehli derivative $y^{'}$ ko multiply karta hai — yeh convection term hai (ek "drift" jo ek direction mein push karta hai).
== $q (x)$ == $y$ ko hi multiply karta hai — yeh reaction/stiffness term hai.
== $r (x)$ == source hai — isme koi $y$ nahi hota (forcing, heating waali cheezein).

$[a, b]$ ko $N$ barabar steps mein chop karo, width $h = \frac{b - a}{N}$ ; interior nodes hain $x_{1}, \dots, x_{N - 1}$ . $y^{'}$ aur $y^{''}$ ko central differences se replace karke $h^{2}$ se multiply karne par, har interior node $i$ par yeh three-coefficient stencil milta hai: $lower (1 + \frac{h}{2} p_{i}) y_{i - 1} diag - (2 + h^{2} q_{i}) y_{i} + upper (1 - \frac{h}{2} p_{i}) y_{i + 1} = RHS h^{2} r_{i} .$ jahan $p_{i} = p (x_{i})$ , etc. Yahi ek FDM formula hai jo hum use karte hain; har example bas isme numbers plug karta hai.

Do aur words jinhe hum baar baar use karte hain:

Interior node: ek grid point jo strictly do ends ke beech ho — jinhe hume solve karna hota hai. Ends pehle se diye hote hain.
Residual: humara current answer rule ko satisfy karne se kitna door hai. Shooting ke liye yeh hai $ϕ (s) = y (b; s) - β$ : humari shot jahan aakar giri aur bucket $β$ ke beech ka gap.

Scenario matrix

Cell	Kya cheez alag hai	Example
A. Linear + shooting	$ϕ (s)$ seedhi line hai → 2 shots mein exact	Ex 1
B. Nonlinear + shooting	$ϕ (s)$ curved hai → secant iterate karna padega	Ex 2
C. Linear FDM, $p = 0$	symmetric tridiagonal, koi convection nahi	Ex 3
D. Linear FDM, $p \neq = 0$	asymmetric off-diagonals (convection)	Ex 4
E. Degenerate grid ( $N = 2$ )	sirf EK interior node → single equation	Ex 3 ke andar
F. Limiting case: refine $h$	check karo ki error $\propto h^{2}$ girti hai ( $h$ halve → $\frac{1}{4}$ error)	Ex 5
G. Sign check: kis taraf aakar gire?	overshoot vs undershoot, negative residual	Ex 6
H. Real-world word problem	heated rod, units, physical read-off	Ex 7
I. Exam twist: derivative (Neumann) boundary	$y^{'} (b)$ diya gaya hai, $y (b)$ nahi → ghost node	Ex 8

Har method-cell (A–I) neeche kam se kam ek baar hit hoti hai.

Ex 1 — Cell A: Linear shooting, do shots mein exact

Worked example

$y^{''} = 4 y, y (0) = 0, y (1) = sinh 2$ ko shooting se solve karo. True solution hai $y = sinh (2 x)$ , toh $y^{'} (0) = 2$ . Pretend karo ki hume yeh pata nahi.

Forecast: ODE linear hai, toh exact answer milne se pehle kitne shots chahiye — 2, 5, ya "infinitely many"?

Shot 1: guess $s_{0} = 0$ . $y^{''} = 4 y, y (0) = 0, y^{'} (0) = 0$ solve karne par $y \equiv 0$ milta hai, toh $y (1; 0) = 0$ . Yeh step kyun? Hume ek data point $(s, landing)$ chahiye. Zero slope sabse sasta probe hai — yeh baseline landing batata hai.
Shot 2: guess $s_{1} = 1$ . Linearity se $y = \frac{1}{2} sinh (2 x)$ (slope-1 solution), toh $y (1; 1) = \frac{1}{2} sinh 2 \approx 1.8134$ . Yeh step kyun? Ek line ko do points chahiye. Ab hume do slopes par $ϕ$ pata hai.
Root tak interpolate karo. Target $β = sinh 2 \approx 3.6269$ : $s^{⋆} = s_{0} + (s_{1} - s_{0}) \frac{β - y ( 1 ; s _{0} )}{y ( 1 ; s _{1} ) - y ( 1 ; s _{0} )} = 0 + 1 \cdot \frac{3.6269 - 0}{1.8134 - 0} = 2.0.$ Yeh step kyun? Linear ODEs ke liye $y (b; s)$ , $s$ mein ek seedhi line hai, toh exact root ek interpolation mein mil jaata hai — koi iteration nahi.

Verify: $s^{⋆} = 2$ true $y^{'} (0) = 2$ se exactly match karta hai. Forecast answer: 2 shots. ✓

Ex 2 — Cell B: Nonlinear shooting, secant ko iterate karna hi padega

Worked example

$y^{''} = y^{2}, y (0) = 0, y (1) = 1$ ko shooting se solve karo. $y^{2}$ ki wajah se ODE nonlinear hai, toh $ϕ (s)$ ek curve hai, seedhi line nahi.

Forecast: kya do shots isko nail kar denge, ya do baar interpolate karne ke baad bhi residual nonzero rahega?

Clean, checkable arithmetic ke liye hum secant formula ko made-up-but-consistent landing data par use karte hain. Maan lo IVP solver yeh return karta hai: $y (1; s_{0} = 0.5) = 0.60, y (1; s_{1} = 1.0) = 1.30.$ Target $β = 1$ , toh residuals $ϕ_{0} = 0.60 - 1 = - 0.40$ , $ϕ_{1} = 1.30 - 1 = + 0.30$ .

Secant step. $s_{2} = s_{1} - ϕ_{1} \frac{s _{1} - s _{0}}{ϕ _{1} - ϕ _{0}} = 1.0 - 0.30 \cdot \frac{1.0 - 0.5}{0.30 - ( - 0.40 )} .$ Yeh step kyun? Hum curve $ϕ (s)$ ko last do points se seedhi line se approximate karte hain aur uska root nikalte hain — secant idea, $ϕ$ ki derivative ki zaroorat nahi.
Arithmetic. $\frac{0.5}{0.70} = 0.7143$ ; $0.30 \times 0.7143 = 0.2143$ ; toh $s_{2} = 1.0 - 0.2143 = 0.7857$ . Yeh step kyun? Yeh agla, better slope guess hai.
Abhi khatam nahi! Hume $s_{2}$ par nayi IVP fire karni hai, naya residual lena hai, aur repeat karna hai — kyunki true $ϕ$ hamaari line se curve ho jaata hai. Yeh step kyun? Curved $ϕ$ par, ek secant line sirf root ko approximate karti hai; iteration $∣ ϕ ∣ \to$ tol tak le jaata hai.

Verify: residual ne zero ko straddle kiya ( $ϕ_{0} < 0 < ϕ_{1}$ ), toh root $0.5$ aur $1.0$ ke beech bracketed hai; humara $s_{2} = 0.7857$ andar land karta hai — ek sane secant step. Forecast answer: iterate karo, do shots nahi. ✓

Ex 3 — Cells C & E: Linear FDM, $p = 0$ , degenerate single node

Worked example

$y^{''} = y, y (0) = 0, y (1) = sinh 1$ , $N = 2$ lo toh $h = 0.5$ . Template se match karo: $p = 0, q = 1, r = 0$ . $N = 2$ ke saath exactly ek interior node hai $x_{1} = 0.5$ — yeh degenerate case hai.

Forecast: sirf ek unknown ke saath, kitni equations solve karni padengi — ek system, ya ek single line?

Grouped stencil likho template se $p = 0, q = 1$ use karke: $(1) y_{0} - (2 + h^{2}) y_{1} + (1) y_{2} = 0.$ Yeh step kyun? $p = 0$ substitute karne se off-diagonal asymmetry khatam ho jaati hai (lower $=$ upper $= 1$ ); $q = 1, h^{2} = 0.25$ middle coefficient fix karta hai.
Boundary values insert karo $y_{0} = 0, y_{2} = sinh 1 = 1.1752$ : $0 - 2.25 y_{1} + 1.1752 = 0.$ Yeh step kyun? $y_{0}, y_{N}$ known numbers hain — inhe RHS par move karo (Trap "RHS").
Ek equation solve karo. $y_{1} = \frac{1.1752}{2.25} = 0.5223.$ Yeh step kyun? Ek interior node ⇒ tridiagonal system ek scalar equation mein collapse ho jaata hai (Cell E).

Verify: true $sinh (0.5) = 0.5211$ ; error $\approx 0.0012$ . 2 intervals ke saath bhi chhota, aur yeh $h^{2}$ ki tarah shrink karega (agle example mein prove hoga). ✓

Ex 4 — Cell D: Convection ke saath Linear FDM $p \neq = 0$

Worked example

$y^{''} = 2 y^{'} - y + e^{x}, y (0) = 1, y (1) = e,$ $N = 4$ lo toh $h = 0.25$ . Node $i = 1$ ( $x_{1} = 0.25$ ) par equation likho. Template se match karo: $p = 2, q = - 1, r = e^{x}$ . Ab off-diagonals alag hain — yeh asymmetry hi convection hai.

Forecast: kya $y_{i - 1}$ (lower) ko multiply karne wala coefficient $y_{i + 1}$ (upper) wale se bada ya chhota hoga?

Teen coefficients compute karo $h = 0.25$ ke saath:
- lower $= 1 + \frac{h}{2} p = 1 + 0.25 = 1.25$ ,
- diag $= - (2 + h^{2} q) = - (2 + 0.0625 \cdot (- 1)) = - (2 - 0.0625) = - 1.9375$ ,
- upper $= 1 - \frac{h}{2} p = 1 - 0.25 = 0.75$ . Yeh step kyun? Template ke lower/diag/upper slots mein seedha substitution.
RHS $= h^{2} e^{x_{1}} = 0.0625 e^{0.25} = 0.0625 \cdot 1.2840 = 0.08025$ . Yeh step kyun? $r (x) = e^{x}$ ko node par evaluate karo, $h^{2}$ se scale karo.
$i = 1$ par equation ( $y_{0} = 1$ right par move karke): $- 1.9375 y_{1} + 0.75 y_{2} = 0.08025 - 1.25 (1) = - 1.16975.$ Yeh step kyun? $y_{0}$ known hai; $1.25 y_{0}$ ko RHS par subtract karo (Trap "RHS").

Verify: $p = 2 > 0$ ke saath, term $\frac{h}{2} p$ lower mein add hoti hai ( $1.25$ ) lekin upper se subtract hoti hai ( $0.75$ ), toh lower $>$ upper — upwind side heavy hai, yahi positive convection produce karta hai. Forecast answer: lower bada hai. ✓

Worked example Ex 3 ka FDM lo (

$y^{''} = y$ , exact $sinh$ ) $x = 0.5$ par $N = 2$ aur $N = 4$ ke liye. Confirm karo ki error quarter ho jaata hai.

Forecast: $h$ ko halve karne par (0.5 se 0.25), error lagbhag... 2, 4, ya 8 ke factor se shrink hogi?

$N = 2$ error (Ex 3): $∣0.5223 - 0.5211∣ = 0.00122$ . Yeh step kyun? Yeh humara coarse-grid benchmark hai.
$N = 4$ system explicitly set up karo. Ab $h = 0.25, h^{2} = 0.0625$ ; interior nodes $x_{1} = 0.25, x_{2} = 0.5, x_{3} = 0.75$ . $p = 0, q = 1$ ke saath har diagonal $- (2 + h^{2}) = - 2.0625$ hai aur har off-diagonal $1$ hai. Boundaries $y_{0} = 0, y_{4} = sinh 1 = 1.1752$ RHS par jaati hain (Trap "RHS"), toh $⎩ ⎨ ⎧ - 2.0625 y_{1} + y_{2} = - y_{0} = 0, y_{1} - 2.0625 y_{2} + y_{3} = 0, y_{2} - 2.0625 y_{3} = - y_{4} = - 1.1752.$ Yeh step kyun? Finer grid ka matlab hai teen coupled interior equations — ek genuine tridiagonal system.
Thomas method se solve karo (forward-eliminate, back-substitute).
- Row 1: $y_{1} = y_{2} /2.0625$ .
- Row 2 mein substitute karo: $(y_{2} /2.0625) - 2.0625 y_{2} + y_{3} = 0 \Rightarrow - 1.5776 y_{2} + y_{3} = 0 \Rightarrow y_{3} = 1.5776 y_{2}$ .
- Row 3 mein substitute karo: $y_{2} - 2.0625 (1.5776 y_{2}) = - 1.1752 \Rightarrow y_{2} (1 - 3.2538) = - 1.1752 \Rightarrow y_{2} = \frac{- 1.1752}{- 2.2538} = 0.52140.$ Yeh step kyun? Thomas basically Gaussian elimination hai jo tridiagonal shape ka fayda uthata hai — diagonal ke neeche eliminate karo, phir jo value chahiye uske liye back-substitute karo, $y_{2} = y (0.5)$ .
$N = 4$ error: $∣0.52140 - 0.52110∣ = 0.00030$ . Ratio $\frac{0.00122}{0.00030} \approx 4.1$ . Yeh step kyun? Yeh observed error-shrink factor hai.

Verify: central stencil $O (h^{2})$ hai, toh $h$ halve karne se $\times \frac{1}{4}$ error predict hoti hai. Humne dekha $\approx 4.1 \times$ chhota — match karta hai (Trap "refine": trust karne ke liye hamesha refine karo). Forecast answer: factor 4. ✓

Ex 6 — Cell G: Residual ka sign (over/undershoot)

Figure — Boundary value problems — shooting method, finite difference

Figure: $y^{''} = y, y (0) = 0$ ke liye teen shooting attempts, sabhi left end se launch ki gayi hain. Magenta curve (slope $s_{0} = 0$ ) flat rehti hai aur navy dashed "bucket" line $β = sinh 1$ se neeche khatam hoti hai — undershoot. Violet curve (slope $s_{1} = 2$ ) steeply upar jaati hai aur bucket se upar khatam hoti hai — overshoot. Orange curve (slope $s^{⋆} = 1$ ) exactly bucket par land karti hai. Residual ka sign seedha picture se padho: dashed line ke neeche ⇒ $ϕ < 0$ , upar ⇒ $ϕ > 0$ , us par ⇒ $ϕ = 0$ .

Worked example

$y^{''} = y, y (0) = 0, y (1) = sinh 1 \approx 1.1752$ ke liye, do shooting shots ko sign se interpret karo.

Forecast: agar hum slope $s_{0} = 0$ se shoot karte hain, toh kya hum bucket $β$ se short land karenge ya aage?

Shot $s_{0} = 0$ : $y (1; 0) = 0$ . Residual $ϕ_{0} = 0 - 1.1752 = - 1.1752 < 0$ . Yeh step kyun? Negative residual ka matlab hai hum target se neeche/short land kiye — undershoot. Figure mein yeh magenta path hai jo navy dashed bucket line ke neeche khatam hoti hai.
Shot $s_{1} = 2$ : $y (1; 2) = 2 sinh 1 \approx 2.3504$ . Residual $ϕ_{1} = 2.3504 - 1.1752 = + 1.1752 > 0$ . Yeh step kyun? Positive residual ka matlab hai humne over shoot kiya — violet path bucket ke upar khatam hoti hai.
Root bracketed hai. Kyunki $ϕ_{0} < 0 < ϕ_{1}$ , true slope $0$ aur $2$ ke beech hai; interpolation $s^{⋆} = 1$ par land karti hai (orange path, jo bucket ko touch karti hai). Yeh step kyun? Opposite signs ek crossing guarantee karte hain — "ek chhote shot aur ek lambe shot ke beech aim karo" ka geometric matlab.

Verify: $s^{⋆} = 0 + 2 \cdot \frac{1.1752 - 0}{2.3504 - 0} = 1.0$ , true slope. Sign logic aur root agree karte hain. ✓

Ex 7 — Cell H: Real-world word problem (heated rod)

Worked example

$L = 1 m$ lambe ek rod mein internal source ke saath heat conduct hoti hai. Steady temperature $T (x)$ yeh obey karta hai: $T^{''} = - 100 (^{\circ} C / m^{2}), T (0) = 2 0^{\circ} C, T (1) = 2 0^{\circ} C .$ FDM se $N = 2$ ( $h = 0.5$ ) leke mid-rod temperature $T (0.5)$ estimate karo.

Forecast: dono ends $2 0^{\circ}$ par hain aur andar heating hai, toh kya middle $2 0^{\circ}$ se zyada garam ya thanda hoga?

Template se match karo. $T^{''} = 0 \cdot T^{'} + 0 \cdot T + (- 100)$ likhte hain, toh $p = 0, q = 0, r = - 100$ . Yeh step kyun? $p, q, r$ padhne se hum ek hi stencil formula reuse kar sakte hain.
$x_{1} = 0.5$ par stencil, $h^{2} = 0.25$ : $y_{0} - 2 y_{1} + y_{2} = h^{2} r = 0.25 (- 100) = - 25.$ Yeh step kyun? $p = q = 0$ pure "1, −2, 1" stencil deta hai; RHS mein source $r$ aata hai.
$T_{0} = T_{2} = 20$ insert karo: $20 - 2 T_{1} + 20 = - 25 \Rightarrow - 2 T_{1} = - 65 \Rightarrow T_{1} = 32. 5^{\circ} C .$ Yeh step kyun? Dono ends known hain; single equation midpoint deta hai (Trap "RHS" applied).

Verify: exact solution $T = 20 + 50 x (1 - x)$ deta hai $T (0.5) = 20 + 50 (0.25) = 32. 5^{\circ} C$ — exactly match karta hai (constant $r$ ke liye central stencil is parabola par exact hai). Units poore mein $^{\circ}$ C hain. Forecast answer: zyada garam, $32. 5^{\circ}$ . ✓

Ex 8 — Cell I: Exam twist, ek derivative (Neumann) boundary

Worked example

$y^{''} = y, y (0) = 0, y^{'} (1) = cosh 1 \approx 1.5431$ (right end par slope diya gaya hai). $N = 2$ , $h = 0.5$ lo; equations set up karo. Right boundary ab ek derivative hai, value nahi — classic exam curveball.

Forecast: $x = 1$ par koi $y_{2}$ value humein di nahi gayi. Missing equation kahan se aayegi?

Ghost node $y_{3}$ introduce karo $x_{3} = 1.5$ par (rod ke thoda bahar) aur central derivative boundary node $x_{2} = 1$ par likho: $y^{'} (1) \approx \frac{y _{3} - y _{1}}{2 h} = cosh 1 \Rightarrow y_{3} = y_{1} + 2 h cosh 1 = y_{1} + 1.5431.$ Yeh step kyun? Boundary par central (one-sided nahi) difference use karne se $O (h^{2})$ accuracy maintain rehti hai (Trap "order"). Ghost node $y_{3}$ woh fictitious value hai jo central formula ko chahiye.
DONO unknown nodes $i = 1, 2$ par $y^{''} = y$ stencil likho (yahan $p = 0, q = 1, h^{2} = 0.25$ , diag $= - 2.25$ ): $i = 1 : y_{0} - 2.25 y_{1} + y_{2} = 0, i = 2 : y_{1} - 2.25 y_{2} + y_{3} = 0.$ Yeh step kyun? Node 2 ab ek unknown hai (sirf uska slope diya gaya tha, value nahi), toh isko apna stencil equation chahiye — aur us equation mein ghost $y_{3}$ hai.
Ghost ko eliminate karo. $y_{3} = y_{1} + 1.5431$ (step 1 se) ko node-2 equation mein substitute karo: $y_{1} - 2.25 y_{2} + (y_{1} + 1.5431) = 0 \Rightarrow 2 y_{1} - 2.25 y_{2} = - 1.5431.$ Yeh step kyun? Ghost value ab ek real unknown aur known data ke terms mein express ho gayi, toh woh gayab ho jaati hai aur hum ek solvable $2 \times 2$ system ke saath bach jaate hain.
$2 \times 2$ solve karo $y_{0} = 0$ ke saath: $i = 1$ se: $- 2.25 y_{1} + y_{2} = 0 \Rightarrow y_{2} = 2.25 y_{1}$ . Step-3 equation mein sub karo: $2 y_{1} - 2.25 (2.25 y_{1}) = - 1.5431 \Rightarrow (2 - 5.0625) y_{1} = - 1.5431 \Rightarrow y_{1} = \frac{- 1.5431}{- 3.0625} = 0.5039.$ Phir $y_{2} = 1.1338.$ Yeh step kyun? Back-substitution — ek chhote system ke liye haath se Thomas idea.

Verify: exact $y = sinh x$ deta hai $sinh (0.5) = 0.5211$ aur $sinh (1) = 1.1752$ . Humara coarse $N = 2$ estimate $0.5039, 1.1338$ sahi ballpark mein hai aur $h \to 0$ ke saath converge karega. Ghost-node trick ne correctly missing equation supply ki. ✓

Recall Rapid self-test

Kaun sa matrix cell iteration maangta hai even though shooting use ho rahi hai? ::: Cell B — nonlinear ODE $ϕ (s)$ ko curved banata hai. Negative residual $ϕ (s_{0}) < 0$ ka matlab hai hum shot... ::: short (target $β$ se undershoot kar gaye). Central-difference FDM mein $h$ halve karne se error roughly... ::: $\frac{1}{4}$ factor se change hoti hai (yeh $O (h^{2})$ hai). Neumann boundary $y^{'} (b)$ diye hone par, hum kya add karte hain? ::: ek ghost node plus slope ke liye central-difference equation. $p > 0$ hone par lower off-diagonal upper se bada kyun hota hai? ::: convection term $\frac{h}{2} p$ lower mein add hoti hai, upper se subtract hoti hai.

Worked examples — Boundary value problems — shooting method, finite difference

Scenario matrix

Ex 1 — Cell A: Linear shooting, do shots mein exact

Ex 2 — Cell B: Nonlinear shooting, secant ko iterate karna hi padega

Ex 3 — Cells C & E: Linear FDM, $p = 0$ , degenerate single node

Ex 4 — Cell D: Convection ke saath Linear FDM $p \neq = 0$

Ex 5 — Cell F: Limiting refinement, error $\propto h^{2}$

Ex 6 — Cell G: Residual ka sign (over/undershoot)

Ex 7 — Cell H: Real-world word problem (heated rod)

Ex 8 — Cell I: Exam twist, ek derivative (Neumann) boundary