Systems of first-order linear ODEs — matrix method: Visual walkthrough

4.6.21 · D2 · Maths › Ordinary Differential Equations › Systems of first-order linear ODEs — matrix method

Step 0 — Page par objects kya hain?

KYA. Kisi bhi equation se pehle, do characters se milo.

Ek vector $x = (x_{1} x_{2})$ bas ek arrow hai jo origin se coordinates $(x_{1}, x_{2})$ wale dot tak point karta hai. Do numbers = ek arrow.
Ek matrix $A$ ek machine hai: tum usse ek arrow feed karo, woh ek alag arrow bahar nikalti hai. Yeh stretch, shrink, aur rotate kar sakti hai.

KYUN. Poora subject hai "ek arrow jo time ke saath move karta hai." Hume arrow aur machine dono dekhni chahiye inke interact karne se pehle.

PICTURE. Left side par: input arrow (black). Right side par: machine $A$ ne usse kya kiya (red arrow kisi naye jagah point karta hai — rotate bhi hua aur lamba bhi).

Figure — Systems of first-order linear ODEs — matrix method

Step 1 — $x^{'} = A x$ asliyat mein kya kehta hai?

KYA. Equation ko left se right, symbol by symbol padho:

$tip par velocity arrow x^{'} = machine A tip abhi kahan hai x$

Matlab: "tip ki velocity woh machine apply hoti hai current position par." Mujhe batao main kahan hun, aur $A$ mujhe batayega agli baar kis taraf move karna hai.

KYUN. Yahi motion ka rule hai. Yeh ek flow field hai: plane mein har point par, $A$ ek chhoti velocity arrow pin karta hai. Ek solution ek aisi path hai jo hamesha local arrow follow karti hai.

PICTURE. Plane mein chhoti black velocity arrows bichhi hain (field $A x$ ). Ek red curve unke through nikal raha hai, har us arrow ke tangent jo usse touch karta hai — woh red curve ek solution hai $x (t)$ .

Step 2 — Scalar warm-up jis par hum already trust karte hain

KYA. Ek line ke liye matrices bhool jao. Ek akela equation $x^{'} = a x$ (ek number $a$ ) ka solution hai $x (t) = C e^{a t}$ .

$x^{'} = a x ⟹ x (t) = C e^{a t}$

$a$ growth rate hai,
$C$ starting value $x (0)$ hai,
$e^{a t}$ self-reproducing function hai: uski slope khud apne aap ke $a$ times ke barabar hai.

KYUN. Hum exponential ki taraf isliye jaate hain kyunki yahi ek aisi shape hai jiski steepness har instant par uski apni height ke proportional hoti hai — exactly wahi jo $x^{'} = a x$ maangta hai. (Dekho Second-Order Linear ODEs jahan yahi $e^{λ t}$ trick aati hai.) Hum is success ko vector world mein laana chahte hain.

PICTURE. Teen curves: $a > 0$ upar jaata hai, $a = 0$ flat rehta hai, $a < 0$ zero par decay karta hai. Red wala decaying case hai — baad mein yeh ek stable direction ban jaata hai.

Step 3 — Ek aisa guess jo possibly kaam kar sakta hai

KYA. Scalar shape copy karo lekin ek direction fix rakho. Guess karo

$x (t) = scalar clock: grow/shrink karta hai e^{λ t} fixed direction, kabhi rotate nahi hoti v$

Arrow hamesha origin se hoti ek usi line par rehta hai; sirf uski length $e^{λ t}$ ke saath saans leti hai.

KYUN. Agar hum aisi directions dhundh sakein jisme machine $A$ arrow ko rotate nahi karti — sirf scale karti hai — toh un directions mein vector problem collapse ho jaata hai Step 2 ke scalar problem par, jo hum already solve kar chuke hain.

PICTURE. Origin se ek dashed line. Red arrow us par bahar aur andar slide karta hai jab $t$ run karta hai, lekin uski direction kabhi nahi badlti. Woh fixedness hi poora gamble hai.

Step 4 — Guess plug in karo aur dekho yeh simplify hota hai

KYA. Guess differentiate karo (direction $v$ constant hai, toh sirf clock move karta hai):

$x^{'} = λ e^{λ t} v, A x = e^{λ t} A v .$

Dono ko equal karo (ODE yahi demand karta hai) aur clock $e^{λ t}$ cancel karo, jo kabhi zero nahi hota:

\boxed{\,A\mathbf v=\lambda\mathbf v\,}$$ Term by term: $A\mathbf v$ = machine direction par act kar rahi hai; $\lambda\mathbf v$ = same direction, bas scale ki gayi. **Dono equal hone chahiye.** **KYUN.** $e^{\lambda t}$ se divide karna legal hai kyunki ek exponential kabhi $0$ nahi hota. Jo bachta hai woh *ab time ke baare mein nahi hai* — yeh arrow $\mathbf v$ par ek pure geometry condition hai. **PICTURE.** Left: $\mathbf v$ (black) aur $A\mathbf v$ (red) *ek hi taraf* point kar rahe hain — yeh $\mathbf v$ kaam karta hai. Right: ek generic arrow jahan $A\mathbf v$ side mein swing karta hai — woh reject ho jaata hai. ![[deepdives/dd-maths-4.6.21-d2-s05.png]] > [!definition] Eigenvector / eigenvalue > Ek aisi direction $\mathbf v\ne\mathbf 0$ jisko machine sirf ==scale== kare (kabhi rotate na kare) woh **eigenvector** hai. Scaling factor $\lambda$ uski **eigenvalue** hai. Dekho [[Eigenvalues and Eigenvectors]]. --- ## Step 5 — Woh special directions hum *dhundhen* kaise? **KYA.** $A\mathbf v=\lambda\mathbf v$ ko aise likhte hain: $$(A-\lambda I)\,\mathbf v=\mathbf 0,\qquad I=\text{do-nothing machine.}$$ Hum ek **nonzero** $\mathbf v$ chahte hain jo zero arrow par bheja jaaye. Ek machine ek nonzero arrow ko $0$ par tab hi crush kar sakti hai jab woh *plane ko ek line par flat kare* — yaani uski **zero area-scaling** ho. Woh area factor ==determinant== hai: $$\boxed{\;\det(A-\lambda I)=0\;}$$ **KYUN.** Agar $A-\lambda I$ invertible hota (area factor $\ne 0$) toh *sirf* solution hota $\mathbf v=\mathbf 0$ — bekar. Determinant $=0$ demand karna matlab machine ko itna degenerate demand karna ki uski ek genuine null direction ho. **PICTURE.** Ek unit square (black) $A-\lambda I$ se feed kiya gaya. Ek generic $\lambda$ ke liye yeh ek tilted parallelogram mein map hota hai jiski area hai. Jab $\lambda$ ek eigenvalue hit karta hai, red image **ek segment mein collapse ho jaati hai** — area $0$ — aur crushed direction eigenvector hai. ![[deepdives/dd-maths-4.6.21-d2-s06.png]] --- ## Step 6 — Superpose karo: simple motions mix karo **KYA.** Maano hum do independent eigenpairs $(\lambda_1,\mathbf v_1)$ aur $(\lambda_2,\mathbf v_2)$ dhundh chuke hain. Kyunki rule $\mathbf{x}'=A\mathbf{x}$ **linear** hai (koi squares nahi, koi products nahi), do solutions ko add karne se ek aur solution milta hai: $$\mathbf{x}(t)=c_1\,e^{\lambda_1 t}\,\mathbf v_1+c_2\,e^{\lambda_2 t}\,\mathbf v_2.$$ - $e^{\lambda_1 t}\mathbf v_1$ = line 1 par pure breathing, - $e^{\lambda_2 t}\mathbf v_2$ = line 2 par pure breathing, - $c_1,c_2$ = kitna kitna, start point $\mathbf x(0)$ se fix hota hai. **KYUN.** Do independent directions poora plane span karte hain, isliye *har* start point kuch $c_1\mathbf v_1+c_2\mathbf v_2$ hai. Do free constants exactly ek 2-D system se match karte hain — kuch missing nahi hai. **PICTURE.** Do dashed eigen-lines. Ek generic red trajectory ek moving parallelogram ke diagonal ke tip hai jiske do sides eigen-lines ke along independently saans lete hain. ![[deepdives/dd-maths-4.6.21-d2-s07.png]] > [!recall]- Steps 3–6 par quick self-check > Hum $e^{\lambda t}$ cancel kyun kar sakte hain? ::: Kyunki exponential kabhi zero nahi hota, isliye usse divide karna legal hai. > Kaunsa geometric fact $\det(A-\lambda I)=0$ force karta hai? ::: Ek nonzero arrow $\mathbf 0$ par tab hi map ho sakta hai jab machine area ko zero par crush kare. > 2-D mein exactly do free constants kyun? ::: Do independent eigenvectors plane span karte hain, 2-parameter solution family se match karte hue. --- ## Step 7 — Edge case: complex $\lambda$ (koi real fixed line nahi → spiral) **KYA.** Kabhi kabhi $\det(A-\lambda I)=0$ se $\lambda=\alpha\pm i\beta$ milta hai. Koi aise **real direction** nahi hai jise machine sirf scale kare — har real arrow rotate ho jaata hai. [[Euler's Formula]] use karte hue, $e^{(\alpha+i\beta)t}=e^{\alpha t}(\cos\beta t+i\sin\beta t)$, real motions mein split karo: $$\mathbf{x}_1=e^{\alpha t}\big(\mathbf p\cos\beta t-\mathbf q\sin\beta t\big),\qquad \mathbf{x}_2=e^{\alpha t}\big(\mathbf p\sin\beta t+\mathbf q\cos\beta t\big).$$ - $\alpha$ = ==breathing== rate (grow kare agar $>0$, decay kare agar $<0$), - $\beta$ = ==spin== rate (kitni tezi se circle karta hai). **KYUN.** Koi fixed line nahi matlab Step 4 ka pure "andar bahar slide" picture real arrows ke liye exist nahi kar sakta — arrow ko ghoomna hi padega. $\cos,\sin$ exactly ek rotating arrow hain; $e^{\alpha t}$ usse resize karta hai. **PICTURE.** Ek red trajectory ke teen panels: $\alpha<0$ inward spiral (**stable**), $\alpha=0$ closed circle (**center**), $\alpha>0$ outward spiral (**unstable**). Dekho [[Phase Portraits and Stability]]. ![[deepdives/dd-maths-4.6.21-d2-s08.png]] --- ## Step 8 — Degenerate case: repeated $\lambda$, sirf ek eigen-line **KYA.** Ek double root $\lambda$ tumhe **sirf ek** eigenvector $\mathbf v$ de sakta hai — plane span karne ke liye bahut kam. Hum ek doosra solution ek hidden partner $\mathbf w$ (**generalized eigenvector**) se banate hain jo $(A-\lambda I)\mathbf w=\mathbf v$ solve karta hai: $$\mathbf{x}_2=e^{\lambda t}\big(t\,\mathbf v+\mathbf w\big).$$ - extra factor $t$ = woh ==missing direction== jo plain exponential supply nahi kar sakta tha. **KYUN.** Ek eigen-line ke saath tum sirf ek line tak pahunch sakte ho, plane tak nahi. Linear-in-$t$ term arrow ko us line *se door* drift karta hai, doosri degree of freedom restore karta hai. (Yeh [[Second-Order Linear ODEs]] mein repeated-root $t\,e^{\lambda t}$ trick se mirror karta hai.) **PICTURE.** Ek dashed eigen-line. Plain solution us par chipka rehta hai; red generalized solution $t$ badhne ke saath arrow $\mathbf v$ ke along us se peel away karta hai — ek shear-like flow. ![[deepdives/dd-maths-4.6.21-d2-s09.png]] --- ## Ek-picture summary Sab kuch ek saath: guess $e^{\lambda t}\mathbf v$ → clock cancel karo → land on $A\mathbf v=\lambda\mathbf v$ → $\det(A-\lambda I)=0$ force karo → superpose karo. Ek 2-D system ki teen possible personalities (do real lines = node, complex = spiral, repeated = degenerate shear) teen portraits ke roop mein dikhayi gayi hain jo ek hi seed equation se grow karti hain. ![[deepdives/dd-maths-4.6.21-d2-s10.png]] > [!recall]- Pure walkthrough ki Feynman-style retelling > Ek matrix ek chhoti machine hai jo ek arrow pakadti hai aur usse kahi aur phenk deti hai. Hamara equation kehta hai: *jahan bhi tumhare arrow ki tip ho, machine batati hai use agli baar kis taraf slither karna hai.* Humne sabse aasaan possible motion guess ki — ek arrow jo ek line par raha aur bas balloon ki tarah grow ya shrink hua. Woh guess plug in karne par, time-part ($e^{\lambda t}$) cancel ho gaya kyunki woh kabhi zero nahi hota, aur hum bacha rehta hai demand "machine sirf is arrow ko stretch kare, twist na kare." Woh non-twisting arrows eigenvectors hain; stretch factor eigenvalue hai. Unhe dhundhe ke liye hum poochte hain kab machine area ko kuch nahi karta (determinant zero). Jab do aisi lines mil jaati hain toh hum unke along motions mix karte hain har starting point tak pahunchne ke liye. Agar machine *har* real arrow ko twist karne par majboor kare, toh spirals milte hain (complex case). Agar yeh humein sirf ek non-twisting line deti hai, toh hum ek $t$ glue karte hain us par se drift karne aur plane fill karne ke liye (repeated case). Same seed equation, flow ke teen flavours. > [!mnemonic] Guess · Cancel · Crush · Combine > **Guess** $e^{\lambda t}\mathbf v$ → **Cancel** nonzero clock → **Crush** area: > $\det(A-\lambda I)=0$ → **Combine** solutions by superposition.