Exercises — Wave equation (hyperbolic) 1D — derivation

• $u_{tt}$ = string par ek point ka vertical acceleration (doosra time-derivative).
• $u_{xx}$ = us point par string ki curvature (doosra space-derivative): slope $x$ ke saath chalte hue kaise badalta hai.
• $c$ = wave speed, aur $c=\sqrt{T/\rho }$. Equation yeh kehti hai: ek point ka acceleration is baat ke proportional hai ki string wahan kitni curved hai.

(b).

• (a) mein ek time-derivative hai → yeh heat equation hai (parabolic).
• (c) mein time bilkul nahi → Laplace equation (elliptic).
• (b) mein do time-derivatives hain jo do space-derivatives se match karte hain → wave equation (hyperbolic). Dekho Classification of second-order PDEs.

$A$, $B$, $C$ kya hain? Do variables mein har second-order PDE ko standard form mein likha ja sakta hai: $A{u}_{tt}+B{u}_{tx}+C{u}_{xx}+(\text{lower-order terms})=0,$ jahan $A$ woh number hai jo pure second time-derivative $u_{tt}$ ke saath multiply hota hai, $C$ woh number jo pure second space-derivative $u_{xx}$ ke saath multiply hota hai, aur $B$ woh number jo mixed derivative $u_{tx}$ ke saath multiply hota hai. Wave equation $u_{tt}-c^2{u}_{xx}=0$ ke liye hum padhte hain $A=1$, $B=0$, $C=-c^2$. Discriminant $B^2-4AC$ PDE ko classify karta hai bilkul waise jaise quadratic ka discriminant uske roots classify karta hai.

$c=\sqrt{\frac{T}{\rho }}=\sqrt{\frac{120}{0.006}}=\sqrt{20000}\approx 141.4\text{m/s}.$ Yeh formula kyun? Derivation se, $c^2=T/\rho$: tighter string (bada $T$) → tez; bhari string (bada $\rho$) → dhimi. Guitar tuning se match karta hai (tightening karne se pitch upar jaati hai, yani wave tez hoti hai).

Kyunki $c=\sqrt{T/\rho }$ hai, $c\propto \sqrt{T}$ hoga. $c$ double karne ke liye: $2c=\sqrt{\frac{T^{\prime }}{\rho }}\Rightarrow 4c^2=\frac{T^{\prime }}{\rho }=4\cdot \frac{T}{\rho }\Rightarrow T^{\prime }=4T.$ Tension chaar guna karni padegi. Speed tension ke square root ke saath hi badhti hai, isliye speed double karne ke liye tension mein $4$ ka factor lagta hai.

Differentiate karo. $2x-6t$ ko inside maano; chain rule har baar inside ke derivative se multiply karta hai. $u_t=6\sin (2x-6t),u_{tt}=-36\cos (2x-6t).$ $u_x=-2\sin (2x-6t),u_{xx}=-4\cos (2x-6t).$ $u_{tt}=c^2{u}_{xx}$ ke liye hume chahiye $-36\cos (\cdot )=c^2(-4)\cos (\cdot )$, toh $c^2=36/4=9$, jisse milta hai $c=3\text{m/s}.$ Sanity check: argument $2x-6t=2(x-3t)$ hai, jo $x-3t$ ka function hai, toh yeh speed $3$ par right travel karta hai. Consistent.

d'Alembert ka formula hai: $u(x,t)=\frac{1}{2}[f(x-ct)+f(x+ct)]+\frac{1}{2c}{\int }_{x-ct}^{x+ct}g(s)ds.$ $g\equiv 0$ hone se integral zero ho jaata hai. $c=2$ rakhne par: $u(x,t)=\frac{1}{2}\left[\frac{1}{1+(x-2t{)}^2}+\frac{1}{1+(x+2t{)}^2}\right].$ Physically: ek single bump do half-height copies mein split ho jaata hai, ek right jaata hai, ek left, dono speed $2$ par. Dekho d'Alembert solution.

$u_{tt}=c^2{u}_{xx}$ use karo jahan $c^2>0$ hai, toh $u_{tt}$ ka sign = curvature $u_{xx}$ ka sign.

• $P$ $\cup$-shaped dip mein baitha hai: concave up, $u_{xx}>0$, toh $u_{tt}>0$ → upar accelerate karta hai. Dip bhar jaata hai.
• $Q$ $\cap$-shaped hump par baitha hai: concave down, $u_{xx}<0$, toh $u_{tt}<0$ → neeche accelerate karta hai. Hump flat ho jaata hai. Har point straight ki taraf khicha jaata hai; woh perpetual chasing hi shape ko propagate karti hai.

Acceleration $u_{tt}=c^2{u}_{xx}$ hai, jo curvature $u_{xx}$ par depend karta hai, slope $u_x$ par nahi. Ek smooth bump ke steepest point par graph ek inflection-jaisi region se guzar raha hota hai jahan curvature small ya zero bhi ho sakti hai, jabki hump ka flat top (zero slope) badi curvature rakh sakta hai. Toh steepness (slope) aur acceleration (curvature) alag cheezein hain. Yeh parent note ka classic slope/curvature confusion hai.

Ek newton $\text{kg}\cdot \text{m}/{\text{s}}^2$ hota hai. Toh $\frac{T}{\rho }=\frac{\text{kg}\cdot \text{m}/{\text{s}}^2}{\text{kg}/\text{m}}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}\cdot \frac{\text{m}}{\text{kg}}=\frac{{\text{m}}^2}{{\text{s}}^2}.$ Square root lene par $\text{m}/\text{s}$ milta hai. Ek genuine speed — formula dimensionally honest hai.

Pure travelling wave ka form hota hai ($x-ct$ ka function) ya ($x+ct$ ka function).

• (a) argument $x-5t$ → speed $5$ par right-moving (shape jo $x$ par hai, baad mein bade $x$ par milti hai).
• (b) argument $x+5t$ → speed $5$ par left-moving.
• (c) $\cos x\cos 5t$ ek standing wave hai — na purely left na right. Lekin phir bhi ek solution hai: yeh left- aur right-mover ka sum hai, kyunki $\cos x\cos 5t=\frac{1}{2}[\cos (x-5t)+\cos (x+5t)]$. Yeh Separation of variables for the wave equation se jodta hai.

$f\equiv 0$ hone par sirf integral term bachta hai: $u(x,t)=\frac{1}{2c}{\int }_{x-ct}^{x+ct}\cos sds=\frac{1}{2c}[\sin s{]}_{x-ct}^{x+ct}=\frac{1}{2c}[\sin (x+ct)-\sin (x-ct)].$ $\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$ use karo jahan $A=x+ct$, $B=x-ct$: $u(x,t)=\frac{1}{2c}\cdot 2\cos x\sin (ct)=\frac{\cos x\sin (ct)}{c}.$ Check $t=0$: $u=0$ ✓. Aur $u_t=\cos x\cos (ct)$, $t=0$ par $\cos x=g(x)$ deta hai ✓.

Idea. Hum $u$ ko do naye variables $\xi =x-ct$ (right-movers ke saath constant) aur $\eta =x+ct$ (left-movers ke saath constant) ka function likhte hain. $x$- aur $t$-derivatives ko $\xi$- aur $\eta$-derivatives mein convert karne ke liye hum chain rule use karte hain. Kyunki $\xi$ aur $\eta$ dono $x$ aur $t$ par depend karte hain, har derivative operator split hota hai: $\frac{\partial }{\partial x}=\frac{\partial \xi }{\partial x}\frac{\partial }{\partial \xi }+\frac{\partial \eta }{\partial x}\frac{\partial }{\partial \eta }=\frac{\partial }{\partial \xi }+\frac{\partial }{\partial \eta },\frac{\partial \xi }{\partial x}=1,\frac{\partial \eta }{\partial x}=1.$ $\frac{\partial }{\partial t}=\frac{\partial \xi }{\partial t}\frac{\partial }{\partial \xi }+\frac{\partial \eta }{\partial t}\frac{\partial }{\partial \eta }=-c\frac{\partial }{\partial \xi }+c\frac{\partial }{\partial \eta },\frac{\partial \xi }{\partial t}=-c,\frac{\partial \eta }{\partial t}=+c.$ Signs kahan se aate hain: $\xi =x-ct$ toh $\partial \xi /\partial t=-c$; $\eta =x+ct$ toh $\partial \eta /\partial t=+c$. Woh opposite sign hi poora reason hai ki dono families ki waves alag ho jaati hain.

Second $x$-derivative. ${\partial }_x=({\partial }_{\xi }+{\partial }_{\eta })$ do baar apply karo: $u_{xx}=({\partial }_{\xi }+{\partial }_{\eta })({\partial }_{\xi }+{\partial }_{\eta })u=u_{\xi \xi }+u_{\xi \eta }+u_{\eta \xi }+u_{\eta \eta }.$ Do cross terms $u_{\xi \eta }$ aur $u_{\eta \xi }$ equal hain (mixed partials smooth $u$ ke liye commute karte hain), toh woh $2u_{\xi \eta }$ mein combine ho jaate hain — yahi wajah hai ki cross coefficient $2$ hai: $u_{xx}=u_{\xi \xi }+2u_{\xi \eta }+u_{\eta \eta }.$ Second $t$-derivative. ${\partial }_t=(-c{\partial }_{\xi }+c{\partial }_{\eta })$ do baar apply karo. $c$ factor out karo: ${\partial }_t=c(-{\partial }_{\xi }+{\partial }_{\eta })$, toh $u_{tt}=c^2(-{\partial }_{\xi }+{\partial }_{\eta })(-{\partial }_{\xi }+{\partial }_{\eta })u=c^2(u_{\xi \xi }-u_{\xi \eta }-u_{\eta \xi }+u_{\eta \eta })=c^2(u_{\xi \xi }-2u_{\xi \eta }+u_{\eta \eta }).$ Yahan cross terms minus sign carry karte hain kyunki har factor ek $-{\partial }_{\xi }$ contribute karta hai; $(-{\partial }_{\xi })({\partial }_{\eta })$ aur $({\partial }_{\eta })(-{\partial }_{\xi })$ dono $-{\partial }_{\xi }{\partial }_{\eta }$ dete hain.

Subtract karo. $u_{tt}-c^2{u}_{xx}=c^2(u_{\xi \xi }-2u_{\xi \eta }+u_{\eta \eta })-c^2(u_{\xi \xi }+2u_{\xi \eta }+u_{\eta \eta }).$ $u_{\xi \xi }$ aur $u_{\eta \eta }$ terms cancel ho jaate hain; cross terms $c^2(-2-2)u_{\xi \eta }$ dete hain: $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }.$ Ise zero set karne par $u_{\xi \eta }=0$ force hota hai. $\eta$ mein ek baar integrate karne par $u_{\xi }=(\text{sirf }\xi \text{ ka function})$ milta hai; $\xi$ mein phir se integrate karne par milta hai $u=F(\xi )+G(\eta ),$ jahan $F$ aur $G$ arbitrary twice-differentiable functions hain — "arbitrary" isliye kyunki integration doosre variable ke free functions introduce karta hai, aur "twice-differentiable" isliye taaki $u_{tt}$ aur $u_{xx}$ actually exist karein. Ye exactly right- aur left-moving shapes hain. Dekho Method of characteristics.

Differentiate karo: $u_{tt}=-{\omega }^2\sin (kx)\cos (\omega t),u_{xx}=-k^2\sin (kx)\cos (\omega t).$ $u_{tt}=c^2{u}_{xx}$ mein substitute karo: $-{\omega }^2\sin (kx)\cos (\omega t)=c^2(-k^2)\sin (kx)\cos (\omega t)\Rightarrow {\omega }^2=c^2{k}^2\Rightarrow \boxed{\omega =ck}.$ Yeh string ka dispersion relation hai. $k=3$, $c=2$ ke saath: $\omega =ck=6$. Toh $u=\sin (3x)\cos (6t)$ ise solve karta hai. (Standing waves Separation of variables for the wave equation ke building blocks hain.)

Integral ke andar differentiate karo (interval fixed hai): $\frac{dE}{dt}={\int }_0^L[\rho u_t{u}_{tt}+T{u}_x{u}_{xt}]dx.$ Pehle term mein $\rho u_{tt}=T{u}_{xx}$ replace karo: $={\int }_0^L[T{u}_t{u}_{xx}+T{u}_x{u}_{xt}]dx=T{\int }_0^L\frac{\partial }{\partial x}\left(u_t{u}_x\right)dx,$ kyunki ${\partial }_x(u_t{u}_x)=u_{tx}{u}_x+u_t{u}_{xx}$ (product rule, aur $u_{tx}=u_{xt}$). Integrate karo: $\frac{dE}{dt}=T[u_t{u}_x{]}_0^L=T(u_t(L,t)u_x(L,t)-u_t(0,t)u_x(0,t)).$ Fixed ends matlab $u(0,t)=u(L,t)=0$ sab $t$ ke liye, toh $u_t=0$ dono ends par. Dono boundary terms vanish ho jaate hain: $\frac{dE}{dt}=0.$ Vibrating string ki kinetic + potential energy constant hai — equation energy na banata hai na khatam karta hai. Yeh ki $\rho u_{tt}=T{u}_{xx}$ exactly Newton's second law hai element par apply kiya gaya, yahi ise kaam karwata hai.

$w=u_1-u_2$ lo. Kyunki wave equation linear hai, $w$ bhi $w_{tt}=c^2{w}_{xx}$ satisfy karta hai, fixed ends $w(0,t)=w(L,t)=0$ ke saath, aur zero initial data ke saath: $w(x,0)=f-f=0$ aur $w_t(x,0)=g-g=0$. $w$ par L5.1 ka energy result apply karo: $E_w(t)$ constant hai. $t=0$ par, $w_t=0$ har jagah toh kinetic part $0$ hai; aur $w(x,0)=0$ har jagah matlab $w_x(x,0)=0$ toh potential part $0$ hai. Isliye $E_w(0)=0$, toh sab $t$ ke liye $E_w(t)=0$ hai. Lekin $E_w={\int }_0^L[\frac{1}{2}\rho w_t^2+\frac{1}{2}T{w}_x^2]dx$ squares ka sum hai — yeh zero tabhi ho sakta hai jab $w_t\equiv 0$ aur $w_x\equiv 0$ har jagah ho. Tab $w$ $x$ aur $t$ dono mein constant hai; kyunki boundary par $w=0$ hai, $w\equiv 0$ hoga. Isliye $u_1=u_2$: solution unique hai.

(i) $c$ nikalo. Ek right-mover apni shape rakhta hai aur bina form badle right slide karta hai. Uska peak wahan hota hai jahan argument $x-ct$ woh fixed number $s^{\star }$ ke equal ho jo $F$ ko maximise karta hai. Toh peak position satisfy karta hai: $x_{\text{peak}}-ct=s^{\star }\Rightarrow x_{\text{peak}}=s^{\star }+ct,$ yaani peak speed $c$ par seedhi line par chalta hai. Do snapshots use karke: $c=\frac{\Delta x_{\text{peak}}}{\Delta t}=\frac{3-1}{0.4-0}=\frac{2}{0.4}=5\text{m/s}.$

(ii) Koi bhi smooth $F$ kaam karta hai. $s=x-ct$ inside maano. Chain rule se (derivative of $F$ times derivative of inside): $u_t=\frac{\partial s}{\partial t}{F}^{\prime }(s)=-c{F}^{\prime }(s),u_{tt}=(-c{)}^2{F}^{\prime \prime }(s)=c^2{F}^{\prime \prime }(s),$ $u_x=\frac{\partial s}{\partial x}{F}^{\prime }(s)=F^{\prime }(s),u_{xx}=F^{\prime \prime }(s).$ Isliye $u_{tt}=c^2{F}^{\prime \prime }(s)=c^2{u}_{xx}.✓$ Yeh har twice-differentiable $F$ ke liye hold karta hai — sirf sines ke liye nahi. Yahi wajah hai ki equation arbitrary travelling shapes support karta hai (ek plucked corner, ek Gaussian bump, kuch bhi), jo d'Alembert solution ka content hai.

Equation ko pehchano → $c=\sqrt{T/\rho }$ mein plug karo → motion predict karne ke liye curvature padho → d'Alembert / standing-wave solutions banao → energy se conservation aur uniqueness prove karo.