Visual walkthrough — Wave equation (hyperbolic) 1D — derivation

4.7.10 · D2 · Maths › Partial Differential Equations › Wave equation (hyperbolic) 1D — derivation

Neeche sab kuch ek lambi kahani hai: ek tight string, uska ek chota sa tukda, aur do padosiyon ke beech ka tug-of-war jo ek bump ko aage badhata hai.

---

Step 0 — Woh picture jo hum explain kar rahe hain

Koi bhi maths se pehle, cheez ko khud dekho.

• Picture mein horizontal line string ki rest position hai.
• Wavy magenta curve string ka abhi ka haal hai, jab humne use pluck kiya.
• Position string ke saath saath letter $x$ se measure karte hain (kitna daayein).
• String kitna upar ya neeche gayi hai, woh letter $u$ se measure karte hain.

Hamara poora kaam: ek aisa equation dhundhna jo bataye ki yeh height time ke aage badhne par kaise change hoti hai.

---

Step 1 — String ka ek chota sa bead kaato

Hum poori wavy curve ke baare mein ek saath reason nahi kar sakte. Toh hum wahi karte hain jo physicists hamesha karte hain: zoom in karo jab tak curve ek chote, lagbhag seedhe segment jaisi na lage.

KYA: Hum position $x$ aur position $x+\Delta x$ ke beech ka tukda kaatke nikalte hain.

KYU: Newton ka law $F=ma$ (dekho Newton's second law) mass wali objects ke baare mein hai. Ek single point ka koi mass nahi hota. Is chote segment ka mass hota hai, isliye hum ise $F=ma$ apply kar sakte hain.

PICTURE: Figure mein segment woh thick violet chunk hai; uske do ends $x$ (left) aur $x+\Delta x$ (right) par hain.

---

Step 2 — Sirf ek force: tension string ke saath pull kar raha hai

Is bead ko kaun hila raha hai? Uske do padosi. Ek stretched string har tukde ko apni direction mein pull karti hai — jaise do log ek rope ko kheench rahe hon, ek taraf se ek.

KYA: Right end par string bead ko upar-aur-daayein kheenchti hai; left end par woh neeche-aur-bayein kheenchti hai. Dono pulls ka size $T$ hai, lekin woh alag alag directions mein point karte hain kyunki string dono ends par alag tilt hoti hai.

KYU direction matter karta hai: Ek force jo partly sideways aur partly upar point karti hai use do sawaalon mein todna padta hai — kitni sideways hai? aur kitni upar hai? Woh do pieces bilkul alag behave karte hain, isliye hum unhe Steps 3 aur 4 mein alag handle karte hain.

---

Step 3 — Tilt, aur do magic ratios

Har pull ko "sideways part" aur "up part" mein todne ke liye, hume string ka tilt measure karna hoga.

Woh right triangle banao jiska slanted side (hypotenuse) pulling force $T$ hai. Uski horizontal side aur vertical side exactly woh sideways aur upward parts hain jo humhe chahiye.

KYU yeh tools aur koi nahi? Humare paas ek slanted arrow hai jiska known length ($T$) aur known tilt ($\theta$) hai, aur hum chahte hain uski horizontal aur vertical shadows. Yeh exactly woh sawaal hai jiske liye sine aur cosine banaye gaye the: woh "length + angle" ko "kitna across, kitna up" mein convert karte hain.

PICTURE: Figure mein, orange arm $T\cos \theta$ (sideways) hai aur magenta arm $T\sin \theta$ (up) hai. Saath mein woh slanted tension arrow ko rebuild karte hain.

---

Step 4 — Sideways forces cancel ho jaate hain (string bhaagti nahi)

KYA: Do horizontal (sideways) pulls ko jodo: right end daayein kheenchta hai, left end bayein kheenchta hai.

$$\underset{\text{sideways pull, right end}}{\underbrace{T\cos \theta (x+\Delta x)}}-\underset{\text{sideways pull, left end}}{\underbrace{T\cos \theta (x)}}=0.$$

KYU yeh zero hona chahiye: String sirf upar aur neeche chalti hai (assumption 4 — motion purely transverse hai). Agar sideways forces cancel nahi hote, toh bead sideways accelerate karta, jo humne mana kiya hai. Toh woh cancel ho jaate hain.

---

Step 5 — Upward forces cancel NAHI hote — yahi bead ko hilata hai

Ab interesting direction: upar. Yahan motion rehti hai.

KYA: Do upward pulls jodo. Right end $T\sin \theta (x+\Delta x)$ se upar kheenchta hai; left end $T\sin \theta (x)$ se upar kheenchta hai (yeh negative ho sakta hai = neeche kheenchna). Net upward force difference hai:

$$F_{\text{net}}=\underset{\text{up-pull, right end}}{\underbrace{T\sin \theta (x+\Delta x)}}-\underset{\text{up-pull, left end}}{\underbrace{T\sin \theta (x)}}.$$

KYU difference? Agar dono ends same amount se upar kheenchein, toh woh simply balance ho jaate hain aur bead ko koi net vertical force nahi lagti. Dono ends ke beech ka mismatch hi bead ko push karta hai. Figure dekho: valley (concave-up tukde) mein right end left se zyada upar tilt hota hai, isliye net pull upar hoti hai.

Ab Newton apply karo, $F=ma$, vertical direction mein:

$$\underset{F=\text{net upward force}}{\underbrace{T[\sin \theta (x+\Delta x)-\sin \theta (x)]}}=\underset{m=\text{mass}}{\underbrace{\rho \Delta x}}\times \underset{a=\text{vertical acceleration}}{\underbrace{\frac{{\partial }^{2}u}{\partial t^2}}}.$$

---

Step 6 — Angles ko slopes se replace karo

Angles awkward hain; slopes calculus-friendly hain. Convert karte hain.

Ek chote angle ke liye, sine aur tangent almost equal hote hain (figure dekho — dono curves $0$ ke paas ek dusre se chipke hain):

$$\sin \theta \approx \tan \theta =\underset{\text{string ka slope}}{\underbrace{u_x}}.$$

KYU $\sin \theta$ ki jagah $u_x$ rakhein na ki $\theta$? Kyunki slope $u_x$ woh quantity hai jise hum ek baar aur differentiate karke curvature bana sakte hain. $\theta$ ka peecha karne se kuch useful nahi milta; slope ka peecha karne se seedha wave equation milti hai. Step 5 mein substitute karo:

$$T[\underset{\text{right end par slope}}{\underbrace{u_x(x+\Delta x,t)}}-\underset{\text{left end par slope}}{\underbrace{u_x(x,t)}}]=\rho \Delta x{u}_{tt}.$$

Physics ab pure calculus hai: bead ke across slope ka difference uski acceleration drive karta hai.

---

Step 7 — Bead ko ek point tak shrink karo → curvature aata hai

Ab hum $\Delta x$ ko zero jaane dete hain, "slopes ka difference" ko ek sachcha derivative mein badal dete hain.

KYA: Dono sides ko $\Delta x$ se divide karo:

$$T\underset{\text{slope mein change, per unit width}}{\underbrace{\frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x}}}=\rho u_{tt}.$$

KYU divide karo? Taaki left side padhe "slope daayein jaane par kitni tezi se change hota hai." Jab $\Delta x\to 0$ toh woh fraction, derivative ki bilkul definition se, slope ka derivative hai:

$$\underset{\Delta x\to 0}{\lim }\frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x}=\frac{{\partial }^{2}u}{\partial x^2}=u_{xx}.$$

Toh hum yahan pahunchte hain:

$$T{u}_{xx}=\rho u_{tt}.$$

---

Step 8 — Constant ka naam rakho, meaning padho

KYA: $\rho$ se divide karo aur bachi hui ratio ko ek naam do.

$$u_{tt}=\underset{\text{ise }{c}^2\text{ kaho}}{\underbrace{\frac{T}{\rho }}}{u}_{xx}⟹\boxed{u_{tt}=c^2{u}_{xx}},c=\sqrt{\frac{T}{\rho }}.$$

Edge & degenerate cases (koi gap mat chodo)

Jahan naive picture toot jaati — bade slopes — wahan assumption 2 fail ho jaata hai: tab $\sin \theta \ne u_x$, horizontal pull cleanly cancel nahi hoti, aur tumhe ek nonlinear wave milti hai jiska speed amplitude par depend kar sakta hai. Hamari poori derivation strictly small-slope world mein rehti hai; uske bahar, yeh equation sirf approximate hai.

---

Ek-picture summary

Upar ka single diagram sab aath steps compress karta hai: bead kaato → tension dono ends par tangentially kheenchti hai → sideways (cancel) aur upward (mismatched) mein todo → mismatch hai $T\Delta (\text{slope})$ → $\Delta x$ se divide karo aur shrink karo → curvature $u_{xx}$, acceleration $u_{tt}$ drive karta hai → naam rakho $c^2=T/\rho$.

Recall Feynman retelling — poora walkthrough plain words mein

Ek tight string pakdo aur uske ek chote bead ko ghoor ke dekho. Do andeekhe haath us bead ko pakde hain — uski left mein string aur uski right mein string — aur dono exactly string ke slant ke saath kheenchte hain. Har pull ko "sideways" aur "up" mein todo. Sideways pulls draw ho jaate hain (bead kabhi left ya right nahi bhaagta), toh unhe ignore karo. Up pulls hi poora show hain: agar bead ki right side left side se zyada steeply upar tilt ho, toh bead ko net yank upar milta hai. "Right tilt minus left tilt" simply hai slope bead ke across kaise change ho raha hai — aur slope kaise change hota hai exactly curvature hai. Toh: curved-up bead upar kheencha jaata hai, curved-down bead neeche kheencha jaata hai, har bead hamesha "straight" ka peecha karta hai, overshoot karta hai, aur nudge aage bhaijta hai. Newton ka $F=ma$ us sentence ko $T\times (\text{curvature})=(\text{mass})\times (\text{acceleration})$ mein badal deta hai, aur constants saaf karne par padhta hai $u_{tt}=c^2{u}_{xx}$ jahan $c=\sqrt{T/\rho }$. String tight karo, waves udti hain; mass daalo, woh creep karti hain.

Recall

Woh kaun si single physical quantity hai, jo $u_{xx}$ mein aati hai, jo bead ki vertical acceleration drive karti hai? ::: String ki curvature (bead ke across slope kaise change hota hai), na ki uski steepness. Horizontal tension components kyun cancel ho jaate hain? ::: Kyunki string sirf upar aur neeche move karti hai (purely transverse), isliye koi net sideways force nahi ho sakti. Derivation mein derivative actually kahan aata hai? ::: Jab hum slope-difference ko $\Delta x$ se divide karte hain aur $\Delta x\to 0$ jaane dete hain, tab derivative ki definition se $u_{xx}$ milta hai. $\sin \theta$ ko $\theta$ ki jagah $u_x$ se kyun replace karte hain? ::: Kyunki $u_x$ slope hai, jise hum ek baar aur differentiate karke curvature $u_{xx}$ pa sakte hain.

Aage ke related paths: d'Alembert solution · Method of characteristics · Separation of variables for the wave equation.