Visual walkthrough — Boundary value problems — shooting method, finite difference

4.8.28 · D2 · Maths › Numerical Methods › Boundary value problems — shooting method, finite difference

Hum ek running example use karenge taaki numbers concrete rahein: $y^{\prime \prime }=y,y(0)=0,y(1)=\sinh 1\approx 1.1752,$ jiska sach-much answer hai smooth curve $y=\sinh x$.

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Step 1 — Ek aisi curve jo poori ek baar nahi dikh sakti

KYA. Hamare paas ek unknown curve $y(x)$ hai jo interval $[a,b]$ par jee rahi hai. Hum sirf uske do endpoints jaante hain: $y(a)=\alpha$ (left height) aur $y(b)=\beta$ (right height). Beech mein sab kuch mystery hai.

KYUN. Computer ek poori continuous curve hold nahi kar sakta — infinitely many points hote hain. Toh hum ek deal karte hain: poori curve ki jagah, hum sirf kuch finite evenly spaced jagahon par curve ki height track karte hain. Yahi jagahein haara grid hain.

PICTURE. Blue curve sach-much ka answer hai. Orange dots wahi cheezein hain jo hum kabhi compute karenge — heights $y_0,y_1,y_2,\dots$ jo vertical grid lines par baithi hain, ek doosre se $h$ door.

Hamare example mein $a=0,b=1$. Agar hum $N=2$ choose karein toh $h=0.5$ aur interior mystery hai ek single dot $y_1$ at $x_1=0.5$.

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Step 2 — Ek derivative "rise over a tiny run" kyun hoti hai

KYA. Equation $y^{\prime \prime }$ ki baat karta hai, aur $y^{\prime \prime }$ bana hai $y^{\prime }$ se, yaani slope se. Toh second derivatives ko handle karne se pehle hum poochte hain: sirf dots se slope kaise nikaalein?

KYUN tangent-lines, kuch fancier nahi. Slope $y^{\prime }(x_i)$ true curve ki steepness hai us dot par. Hum ise directly measure nahi kar sakte — haare paas sirf neighbouring dots hain. Toh hum steepness ko nearby dots ke beech khiinchi straight line ki tilt se approximate karte hain: rise over run, $\frac{\Delta y}{\Delta x}$. Jab aapke paas sirf heights hoon toh yahi ek tool available hota hai.

PICTURE. Node $i$ se teen candidate lines: ek right neighbour tak (forward), ek left neighbour tak (backward), aur ek dono neighbours ko seedha join karti line (central). Notice karein ki central green line sab se zyada true tangent ke saath milti hai — yeh dono taraf se balanced hai.

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Step 3 — Taylor series: ek wobble ka exact bookkeeping

KYA. Taylor series ek rule hai jo nearby height ko ek anchor point par height, slope, curvature… ke terms mein likhta hai. Hum ise node $i$ ke dono neighbours par apply karte hain.

KYUN yeh tool aur koi nahi. Hum jaanna chahte hain ki har straight-line guess mein exactly kitna error hai. Taylor series woh ek machine hai jo $y(x_i\pm h)$ ko infinite sum ke roop mein expand karta hai jiske terms labeled hain ki kaunsa derivative carry kar rahe hain — toh hum precisely dekh sakte hain kaunse terms bachte hain aur kaunse cancel hote hain.

PICTURE. Anchor dot se $+h$ (right neighbour) aur $-h$ (left neighbour) ki taraf step out karein. Har colored bar ek Taylor term hai jo neighbour ki height mein contribute karta hai.

$y_{i+1}=\underset{\text{anchor height}}{\underbrace{y_i}}+\underset{\text{slope}\times \text{step}}{\underbrace{h{y}_i^{\prime }}}+\underset{\text{curvature term}}{\underbrace{\frac{h^2}{2}{y}_i^{\prime \prime }}}+\underset{\text{next wobble}}{\underbrace{\frac{h^3}{6}{y}_i^{\prime \prime \prime }}}+\dots$

$y_{i-1}=\underset{\text{same anchor}}{\underbrace{y_i}}\underset{\text{step is }-h}{\underbrace{-h{y}_i^{\prime }}}+\underset{h^2\text{ so no sign flip}}{\underbrace{\frac{h^2}{2}{y}_i^{\prime \prime }}}\underset{h^3\text{ flips sign}}{\underbrace{-\frac{h^3}{6}{y}_i^{\prime \prime \prime }}}+\dots$

Term-by-term. Har term hai (anchor par $y$ ka koi derivative) $\times$ (step $\pm h$ kisi power par) $÷$ (ek factorial). $h$ ki odd powers left step par sign flip karti hain; even powers nahi karte. Yeh ek fact hi aage sab kuch drive karta hai.

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Step 4 — Slope $y_i^{\prime }$ paane ke liye subtract karein

KYA. Right expansion mein se left expansion subtract karein.

KYUN. Dono $y_i$ terms equal hain, aur har even term ($\frac{h^2}{2}{y}_i^{\prime \prime }$, …) dono mein identical hai — toh subtract karne se woh kill ho jaate hain. Sirf odd terms bachte hain, aur pehla survivor exactly slope $y_i^{\prime }$ carry karta hai jo hum chahte hain.

PICTURE. Identical even bars annihilate ho jaate hain (grey, crossed out); odd bars double up ho jaate hain. Jo bachta hai woh slope ka ek clean multiple hai plus ek tiny leftover.

$y_{i+1}-y_{i-1}=\underset{\text{bachta hai, yahi chahiye}}{\underbrace{2h{y}_i^{\prime }}}+\underset{\text{leftover error}}{\underbrace{\frac{h^3}{3}{y}_i^{\prime \prime \prime }}}+\dots$

$2h$ se divide karein aur leftover drop karein:

$\boxed{y_i^{\prime }\approx \frac{y_{i+1}-y_{i-1}}{2h}}\text{error shrinks like }{h}^2.$

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Step 5 — Curvature $y_i^{\prime \prime }$ paane ke liye add karein

KYA. Ab dono expansions ko subtract karne ki jagah add karein.

KYUN. Add karne se har odd term kill hota hai (unke opposite signs the), even wale bachte hain. Pehla even survivor hai $h^2{y}_i^{\prime \prime }$ — exactly woh second derivative jo BVP ko chahiye.

PICTURE. Odd slope bars cancel ho jaate hain (opposite arrows), even bars double height par stack ho jaate hain, aur $y_i^{\prime \prime }$ ek $h^2$ coat pehne appear hota hai.

$y_{i+1}+y_{i-1}=\underset{\text{two anchors}}{\underbrace{2y_i}}+\underset{\text{yahi chahiye}}{\underbrace{h^2{y}_i^{\prime \prime }}}+\underset{\text{leftover}}{\underbrace{\frac{h^4}{12}{y}_i^{\prime \prime \prime \prime }}}+\dots$

$2y_i$ move karein aur $h^2$ se divide karein:

$\boxed{y_i^{\prime \prime }\approx \frac{y_{i+1}-2y_i+y_{i-1}}{h^2}}\text{bhi }O(h^2).$

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Step 6 — Stencils ko BVP mein plug karein aur group karein

KYA. General linear BVP $y^{\prime \prime }=p(x)y^{\prime }+q(x)y+r(x)$ lo aur node $i$ par $y^{\prime \prime }$ aur $y^{\prime }$ ko boxed stencils se replace karo.

KYUN. Yeh ek calculus equation (ek curve ke baare mein) ko ek algebra equation (teen neighbouring dots ke baare mein) mein badal deta hai. Hum yeh har interior node par karte hain, toh dots ek doosre se tied ho jaate hain.

PICTURE. Substitution, phir collect-terms step, teen neighbouring dots $y_{i-1},y_i,y_{i+1}$ par landing karte teen "weights" ki tarah dikhaya gaya.

Substitute karein ($p_i=p(x_i)$, etc.): $\frac{y_{i+1}-2y_i+y_{i-1}}{h^2}=p_i\frac{y_{i+1}-y_{i-1}}{2h}+q_i{y}_i+r_i.$

$h^2$ se multiply karein aur har dot gather karein:

$\underset{\text{left dot ka weight}}{\underbrace{\left(1+\frac{h}{2}{p}_i\right)}}{y}_{i-1}\underset{\text{centre dot ka weight}}{\underbrace{-\left(2+h^2{q}_i\right)}}{y}_i+\underset{\text{right dot ka weight}}{\underbrace{\left(1-\frac{h}{2}{p}_i\right)}}{y}_{i+1}=\underset{\text{known number}}{\underbrace{h^2{r}_i}}.$

Term-by-term.

• Left par $+\frac{h}{2}{p}_i$ aur right par $-\frac{h}{2}{p}_i$ sirf slope stencil se aate hain — yeh "flow"/convection term hai. Agar $p_i\ne 0$ toh left aur right weights differ karte hain: stencil asymmetric ho jaata hai.
• Middle mein bare $-2$ curvature ka apna signature hai — famous "minus two in the middle".
• $-h^2{q}_i$ centre weight ko $qy$ term ke liye adjust karta hai.
• $h^2{r}_i$ ek pure known number (forcing) hai, isliye yeh right side par rehta hai.

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Step 7 — Tridiagonal system assemble karein

KYA. Step-6 ki equation $i=1,2,\dots ,N-1$ ke liye likho. Har row sirf teen consecutive unknowns ko touch karti hai, isliye rows stack karne se ek matrix banta hai jo sirf teen diagonals par non-zero hai.

KYUN teen diagonals. Ek node kabhi sirf apne immediate neighbours se baat karta hai (stencil $i-1,i,i+1$ span karta hai). Toh row $i$ mein entries columns $i-1,i,i+1$ mein hain aur kahin nahi — ek tridiagonal matrix, $O(N)$ mein solvable.

PICTURE. Band matrix $A$, unknown column $\mathbf{y}$, aur right-hand side $\mathbf{d}$ — boundary rows highlighted jahan known $y_0=\alpha$ aur $y_N=\beta$ ko $\mathbf{d}$ mein subtract kiya jaata hai.

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Step 8 — Numbers run karo (degenerate one-node case)

KYA. Haare example mein $p=0,q=1,r=0$ hai. Sabse chhota non-trivial grid $N=2$ lo, toh ek single interior node $y_1$ hai $x_1=0.5$ par aur $h=0.5$.

KYUN sabse chhota case dikhao. Ek interior node ke saath tridiagonal system ek single equation mein collapse ho jaata hai — poori machine ek line mein, kuch hidden nahi.

PICTURE. Teen dots $y_0=0$, unknown $y_1$, $y_2=\sinh 1$, aur $y_1$ par balance karti single stencil equation.

Centre weight hai $-(2+h^2{q}_1)=-(2+0.25)=-2.25$; dono side weights $1$ hain (kyunki $p=0$): $\underset{=0}{\underbrace{1\cdot y_0}}\underset{\text{centre}}{\underbrace{-2.25y_1}}+\underset{=\sinh 1}{\underbrace{1\cdot y_2}}=\underset{=0}{\underbrace{h^2{r}_1}}.$ Known values move karein: $-2.25y_1+1.1752=0\Rightarrow y_1=\frac{1.1752}{2.25}=\mathrm{0.5223.}$

Check. True $\sinh (0.5)=0.5211$. Error $\approx 0.0012$ sirf do intervals se — aur Step 4/5 ke mutabik yeh $h^2$ ki tarah girti hai: $N=4$ tak refine karo aur error roughly chaara guna gir jaana chahiye.

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Ek-picture summary

Ek canvas par poora safar: ek smooth curve → dots mein sample kiya → Taylor series subtract/add se dono stencils janam lete hain → BVP mein substitute kiya → har node par teen weights → ek tridiagonal band ek saath solve hota hai.

Recall Feynman retelling — walkthrough plain words mein

Hamare paas ek smooth invisible curve thi jo sirf apne dono ends par pin down thi. Pehle hum raazi ho gaye ki ise sirf evenly spaced dots par dekhenge. Dots se steepness measure karne ke liye, humne ek line across dot ke dono neighbours se khinchi (balanced, taaki errors cancel ho). "Cancel" ko exact banane ke liye humne Taylor series use ki — ek recipe jo har neighbour ki height likhti hai anchor + slope·step + curvature·step² + … ki tarah. Dono neighbours subtract karne se even terms wipe out ho gaye aur humein slope mila; unhe add karne se odd terms wipe out ho gaye aur humein curvature mila (jo bas measure karta hai ki ek dot apne neighbours ko join karne wali line se kitna neeche dip karta hai). Humne woh dono recipes physics equation mein daal diye, calculus ko teen chhote weights mein badal diya jo teen neighbouring dots par baithe hain. Kyunki har dot sirf apne immediate neighbours se baat karta hai, saari equations stack karne se ek slim three-stripe matrix bani jise hum ek sweep mein solve kar sakte hain. Humne ise sabse chhote possible grid par test kiya — ek mystery dot — aur $0.5223$ vs true $0.5211$ mila: close hai, aur har baar dots double karne par chaar guna close hota jaata hai.

Recall

Dono Taylor expansions subtract karne se kya isolate hota hai, aur kyun? ::: Slope $y_i^{\prime }$ — kyunki even-power terms dono mein identical hain aur cancel ho jaate hain, odd terms bachte hain, jinmein pehla $2h{y}_i^{\prime }$ hai. Unhe add karne se kya isolate hota hai? ::: Curvature $y_i^{\prime \prime }$ — odd-power terms ke opposite signs hain aur cancel ho jaate hain, $2y_i+h^2{y}_i^{\prime \prime }$ bachta hai. Resulting matrix tridiagonal kyun hai? ::: Har node ka stencil sirf $y_{i-1},y_i,y_{i+1}$ involve karta hai, isliye har row mein sirf teen diagonals par non-zeros hain. Known boundary values $\alpha ,\beta$ kahan jaate hain? ::: Pehle aur last interior rows mein right-hand side $\mathbf{d}$ par subtract kiye jaate hain. $N=2$ example mein $y_1$ kya hai? ::: $y_1=1.1752/2.25=0.5223$, true $\sinh (0.5)=0.5211$ ke versus.