Boundary value problems — shooting method, finite difference: Exercises

4.8.28 · D4 · Maths › Numerical Methods › Boundary value problems — shooting method, finite difference

Do symbols baar baar aate hain; inhe ek baar pin kar lo taaki koi line undefined mark use na kare:

Level 1 — Recognition

L1.1 — IVP ya BVP?

Har ek ko classify karo. Batao ki tum march karoge (IVP) ya aim / puzzle-solve karoge (BVP), aur kyun.

(a) $y^{''} = - y, y (0) = 1, y^{'} (0) = 0.$
(b) $y^{''} = - y, y (0) = 1, y (π /2) = 0.$

Recall Solution

Kya decide karta hai: dekho conditions kahan hain. Dono ek hi point $x = 0$ par ⇒ IVP (wahan $y$ aur $y^{'}$ dono pata hain, toh aage step kar sakte ho). Ek-ek dono ends par ⇒ BVP.

(a) Dono conditions $x = 0$ par ⇒ IVP. Start par poori state pata hai, toh e.g. RK4 se aage march karo.
(b) $y (0)$ left end par, $y (π /2)$ right end par ⇒ two-point BVP. March nahi kar sakte (starting slope missing hai) — shooting ya finite difference use karo.

L1.2 — Stencil coefficients padh ke nikalo

Linear BVP ko standard form $y^{''} = p (x) y^{'} + q (x) y + r (x)$ mein likhne par, interior node $i$ par finite-difference row hai $(1 + \frac{h}{2} p_{i}) y_{i - 1} - (2 + h^{2} q_{i}) y_{i} + (1 - \frac{h}{2} p_{i}) y_{i + 1} = h^{2} r_{i} .$ $y^{''} = 3 y^{'} - 2 y + x$ ke liye $h = 0.1$ ke saath, teen coefficients (lower, diagonal, upper) aur right-hand side likho jab $x_{i} = 0.4$ ho.

Recall Solution

Identify karo $p = 3, q = - 2, r = x$ toh $r_{i} = x_{i} = 0.4$ .

lower $= 1 + \frac{h}{2} p = 1 + \frac{0.1}{2} (3) = 1 + 0.15 = 1.15 .$
diagonal $= - (2 + h^{2} q) = - (2 + 0.01 (- 2)) = - (2 - 0.02) = - 1.98 .$
upper $= 1 - \frac{h}{2} p = 1 - 0.15 = 0.85 .$
RHS $= h^{2} r_{i} = 0.01 \times 0.4 = 0.004 .$

Notice karo lower $\neq =$ upper — yeh asymmetry exactly first-derivative (convection) term $p$ ki wajah se hai.

Level 2 — Application

L2.1 — Ek secant shooting step

Ek BVP mein landing-miss function $ϕ (s) = y (b; s) - β$ hai. Do shots ne diya $ϕ (s_{0}) = ϕ (1) = - 0.4, ϕ (s_{1}) = ϕ (2) = 0.6.$ $s_{2}$ nikalne ke liye ek secant update karo.

Recall Solution

Kyun secant: $ϕ$ ko saste mein differentiate nahi kar sakte (har evaluation ek poora IVP solve hai), toh hum $ϕ$ ko last do points se ek straight line se approximate karte hain aur wahan zero dhundhhte hain. $s_{2} = s_{1} - ϕ (s_{1}) \frac{s _{1} - s _{0}}{ϕ ( s _{1} ) - ϕ ( s _{0} )} = 2 - 0.6 \cdot \frac{2 - 1}{0.6 - ( - 0.4 )} .$ $s_{2} = 2 - 0.6 \cdot \frac{1}{1.0} = 2 - 0.6 = 1.4 .$ Sign check: $ϕ$ ne $s_{0}$ aur $s_{1}$ ke beech sign badla, toh root $1$ aur $2$ ke beech hai — aur $1.4$ wahan hai. Sahi hai.

L2.2 — Linear shooting exactly do shots mein

Linear BVP $y^{''} = y, y (0) = 0, y (2) = sinh 2$ ko linear-interpolation shortcut se solve karo. Shots $s_{0} = 0$ aur $s_{1} = 1$ use karo. Target $β = sinh 2 \approx 3.6269$ .

Recall Solution

Kyun do shots kaafi hain: linear ODE ke liye landing value $y (b; s)$ , $s$ ka straight-line function hai, toh $ϕ (s)$ ek line hai — do points use exactly fix kar dete hain, iteration nahi chahiye.

Shot $s_{0} = 0$ : IVP $y^{''} = y, y (0) = 0, y^{'} (0) = 0 \Rightarrow y \equiv 0$ , toh $y (2; 0) = 0$ .
Shot $s_{1} = 1$ : linearity se $y = sinh x$ , toh $y (2; 1) = sinh 2 \approx 3.6269$ . $β$ hit karne ke liye interpolate karo: $s^{⋆} = s_{0} + (s_{1} - s_{0}) \frac{β - y ( 2 ; s _{0} )}{y ( 2 ; s _{1} ) - y ( 2 ; s _{0} )} = 0 + (1 - 0) \frac{3.6269 - 0}{3.6269 - 0} = 1.0 .$ $sinh x$ ka exact starting slope $cosh 0 = 1$ hai. ✓ Do shots mein exactly recover hua.

L2.3 — FDM with ek interior node

$y^{''} = y, y (0) = 0, y (1) = sinh 1$ ko $N = 2$ ke saath solve karo (toh $h = 0.5$ , ek interior node $x_{1} = 0.5$ ).

Recall Solution

Yahan $p = 0, q = 1, r = 0$ . Single interior row ( $i = 1$ ): $(1 + 0) y_{0} - (2 + h^{2} \cdot 1) y_{1} + (1 - 0) y_{2} = 0.$ $h^{2} = 0.25$ ke saath: $y_{0} - 2.25 y_{1} + y_{2} = 0$ . Boundary values known hain: $y_{0} = 0, y_{2} = sinh 1 \approx 1.1752$ . Unhe right side move karo: $- 2.25 y_{1} = - 1.1752 \Rightarrow y_{1} = \frac{1.1752}{2.25} = 0.5223 .$ True value $sinh (0.5) = 0.5211$ ; error $\approx 0.0012$ sirf do intervals ke saath.

Level 3 — Analysis

L3.1 — Tridiagonal system build karo aur solve karo ( $N = 4$ )

$y^{''} = y, y (0) = 0, y (1) = sinh 1$ ko $N = 4$ ( $h = 0.25$ ) ke saath solve karo. $y_{1}, y_{2}, y_{3}$ ke liye $3 \times 3$ tridiagonal system set up karo aur solve karo. $y_{2}$ ko truth se compare karo.

Recall Solution

$p = 0, q = 1, r = 0$ , $h^{2} = 0.0625$ . Har interior row: $y_{i - 1} - (2 + 0.0625) y_{i} + y_{i + 1} = 0$ , yaani diagonal $= - 2.0625$ , off-diagonals $= 1$ . Nodes $x_{1} = 0.25, x_{2} = 0.5, x_{3} = 0.75$ . Boundaries $y_{0} = 0$ , $y_{4} = sinh 1 \approx 1.1752$ .

i=1:&\quad -2.0625\,y_1 + y_2 = -y_0 = 0,\\ i=2:&\quad y_1 -2.0625\,y_2 + y_3 = 0,\\ i=3:&\quad y_2 -2.0625\,y_3 = -y_4 = -1.1752. \end{aligned}$$ **Thomas / elimination** (forward sweep): - Row 1: $y_1=y_2/2.0625$. - Row 2 mein substitute karo: $\dfrac{y_2}{2.0625}-2.0625\,y_2+y_3=0\Rightarrow(-2.0625+0.4848)y_2+y_3=0\Rightarrow -1.5777\,y_2+y_3=0$, toh $y_3=1.5777\,y_2$. - Row 3 mein substitute karo: $y_2-2.0625(1.5777\,y_2)=-1.1752\Rightarrow(1-3.2540)y_2=-1.1752\Rightarrow -2.2540\,y_2=-1.1752$. $$y_2=\frac{1.1752}{2.2540}=\mathbf{0.5214}.$$ Back-substitute: $y_1=0.5214/2.0625=\mathbf{0.2528}$, $y_3=1.5777(0.5214)=\mathbf{0.8226}$. **Compare:** $\sinh(0.5)=0.5211$. Midpoint par error ab $\approx0.0003$ hai — versus $N=2$ par $0.0012$. $h$ adha karne par error lagbhag $4\times$ gira, jo $O(h^2)$ confirm karta hai.

Figure — Boundary value problems — shooting method, finite difference

L3.2 — Naive secant bracket miss kyun kar sakta hai

$ϕ (1) = - 0.4$ aur $ϕ (2) = 0.6$ diye hain (same as L2.1), suppose karo ki true $ϕ$ curved hai: $ϕ (1.4) = + 0.05$ actually. Ek line mein explain karo ki ek linear secant step exact answer kyun nahi hai yahan, aur kya karna chahiye.

Recall Solution

Secant step ne assume kiya tha ki $ϕ$ ek straight line hai, toh usne predict kiya ki root $s_{2} = 1.4$ par hai. Lekin real $ϕ (1.4) = + 0.05 \neq = 0$ : function curved hai (ODE nonlinear hai), toh do points se ek line root ek shot mein nahi pakad sakti. Fix: iterate karo — nayi pair $(s_{1}, s_{2}) = (2, 1.4)$ use karo… actually sign-changing pair $(1, 1.4)$ kyunki $ϕ (1) < 0 < ϕ (1.4)$ — aur ek aur secant step lo, tab tak repeat karo jab tak $∣ ϕ ∣ < tol$ . Do shots mein exactness sirf linear ka privilege hai.

Level 4 — Synthesis

L4.1 — FDM with convection term $p \neq = 0$

$y^{''} = y^{'} - y, y (0) = 0, y (1) = 1$ ko $N = 2$ ( $h = 0.5$ , ek interior node $x_{1} = 0.5$ ) ke saath set up karo aur solve karo. Note karo ki standard form $y^{''} = p y^{'} + q y + r$ hai jisme $p = 1, q = - 1, r = 0$ .

Recall Solution

Single interior row par coefficients:

lower $= 1 + \frac{h}{2} p = 1 + \frac{0.5}{2} (1) = 1 + 0.25 = 1.25$ ,
diagonal $= - (2 + h^{2} q) = - (2 + 0.25 (- 1)) = - (2 - 0.25) = - 1.75$ ,
upper $= 1 - \frac{h}{2} p = 1 - 0.25 = 0.75$ ,
RHS $= h^{2} r = 0$ .

Row: $1.25 y_{0} - 1.75 y_{1} + 0.75 y_{2} = 0$ . Boundaries $y_{0} = 0, y_{2} = 1$ : $- 1.75 y_{1} + 0.75 (1) = 0 \Rightarrow y_{1} = \frac{0.75}{1.75} = 0.4286 .$ Asymmetry lower $= 1.25 \neq =$ upper $= 0.75$ convection term $p = 1$ ka fingerprint hai: information flow direction ki taraf jhukti hai.

L4.2 — Order-of-accuracy audit

Usi BVP ke do FDM runs ne midpoint errors diye $E (h = 0.25) = 0.0012$ aur $E (h = 0.125) = 0.0003$ . Observed order $p$ estimate karo $p \approx lo g_{2} \frac{E ( h )}{E ( h /2 )} .$ Kya scheme designed as perform kar rahi hai?

Recall Solution

Kyun $lo g_{2}$ : agar $E \propto h^{p}$ , toh $h$ adha karne par error $2^{- p}$ se multiply hota hai; ratio $E (h) / E (h /2) = 2^{p}$ , aur uska $lo g_{2}$ lene par $p$ milta hai. $p \approx lo g_{2} \frac{0.0012}{0.0003} = lo g_{2} 4 = 2.0 .$ Central-difference scheme $O (h^{2})$ ke liye designed hai, aur measured order $2$ hai — exactly as designed perform kar rahi hai. (Dekho Truncation error and order of accuracy.)

Level 5 — Mastery

L5.1 — Method choose karo aur justify karo

Tumhe ek stiff, nonlinear BVP solve karni hai long domain $[0, 50]$ par jahan $x = 0$ par chhoti si slope error exponentially badhti hai. Kaun sa method — shooting ya FDM — choose karoge, aur do concrete reasons do.

Recall Solution

Finite difference choose karo. Reasons:

Long/stiff domains par stability: shooting ek IVP ko aage march karti hai, toh guessed slope $s$ mein chhoti si error $[0, 50]$ par exponentially amplify hoti hai — yahi toh yahan difficulty ki definition hai. FDM saare nodes simultaneously solve karta hai, errors ko local rakhta hai propagate hone ki bajaye.
Sensitive root-find nahi: stiff problem ke liye $ϕ (s)$ bahut steeply curved hoti hai, toh $s$ par secant/Newton ill-conditioned hota hai. FDM problem ko ek (banded) algebraic system mein convert karta hai; nonlinear part ke liye poore system par Newton apply karo (Jacobian tridiagonal-ish hai, phir bhi Tridiagonal systems — Thomas algorithm se sasta).

Trade-off acknowledged: FDM ko har Newton step mein nonlinear solve chahiye, lekin jab shooting blow up ho jaati hai tab uski stability decisively jeet jaati hai.

L5.2 — Full pipeline, ek clean pass

$y^{''} = y, y (0) = 0, y (1) = sinh 1$ ko RK-free reasoning ke saath shooting se solve karo: kyunki ODE linear hai tum exact IVP solution $y (x; s) = s sinh x$ use kar sakte ho. $s_{0} = 0.5$ aur $s_{1} = 1.5$ shoot karo, phir exact slope par linearly interpolate karo, aur finally recovered solution se $y (0.5)$ state karo.

Recall Solution

Kyun $y = s sinh x$ : $y^{''} = y, y (0) = 0, y^{'} (0) = s$ ke liye, unique IVP solution $s sinh x$ hai (yeh ODE satisfy karta hai, $y (0) = 0$ , aur $y^{'} (0) = s cosh 0 = s$ ).

Shot $s_{0} = 0.5$ : $y (1; 0.5) = 0.5 sinh 1 = 0.5 (1.1752) = 0.5876$ .
Shot $s_{1} = 1.5$ : $y (1; 1.5) = 1.5 sinh 1 = 1.5 (1.1752) = 1.7628$ .
Target $β = sinh 1 = 1.1752$ . Interpolate: $s^{⋆} = 0.5 + (1.5 - 0.5) \frac{1.1752 - 0.5876}{1.7628 - 0.5876} = 0.5 + 1 \cdot \frac{0.5876}{1.1752} = 0.5 + 0.5 = 1.0 .$ $s^{⋆} = 1$ ke saath: $y (x) = sinh x$ , toh $y (0.5) = sinh (0.5) = 0.5211$ — exact midpoint value (shooting ne true solution recover ki kyunki ODE linear hai).

Recall One-line self-test

Central second-difference stencil for $y^{''}$ ::: $\frac{y _{i + 1} - 2 y _{i} + y _{i - 1}}{h ^{2}}$ , accurate to $O (h^{2})$ . Landing-miss function in shooting ::: $ϕ (s) = y (b; s) - β$ ; solve $ϕ (s) = 0$ . Why linear BVP shoots exact in 2 tries ::: $ϕ (s)$ ek straight line hai, aur 2 points ek line fix kar dete hain. Observed order from two errors ::: $p = lo g_{2} (E (h) / E (h /2))$ .