← 2023 Paper 2
UPSC 2023 Maths Optional Paper 2 Q7c — Step-by-Step Solution 20 marks · Section B
Wave equation · PDEs · asked 7× in 14 yrs · Read the full method →
Question
Solve the partial differential equation
a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 , 0 < x < L , t > 0 , a^2\dfrac{\partial^2 u}{\partial x^2}=\dfrac{\partial^2 u}{\partial t^2},\;0<x<L,\;t>0, a 2 ∂ x 2 ∂ 2 u = ∂ t 2 ∂ 2 u , 0 < x < L , t > 0 , subject to the conditions
u ( 0 , t ) = 0 , u ( L , t ) = 0 , t > 0 ; u ( x , 0 ) = x , ( ∂ u ∂ t ) t = 0 = 1 , 0 < x < L . u(0,t)=0,\;u(L,t)=0,\;t>0;\;u(x,0)=x,\;\Bigl(\dfrac{\partial u}{\partial t}\Bigr)_{t=0}=1,\;0<x<L. u ( 0 , t ) = 0 , u ( L , t ) = 0 , t > 0 ; u ( x , 0 ) = x , ( ∂ t ∂ u ) t = 0 = 1 , 0 < x < L . Technique
Separation of variables; Dirichlet ends force a sine basis; expand both initial profiles in that basis.
Solution
Strategy. Separation of variables on ( 0 , L ) (0,L) ( 0 , L ) { sin ( n π x / L ) } n ≥ 1 \{\sin(n\pi x/L)\}_{n\ge 1} { sin ( nπ x / L ) } n ≥ 1
Step 1 — Separated solutions
Set u = X ( x ) T ( t ) u=X(x)T(t) u = X ( x ) T ( t ) a 2 X ′ ′ T = X T ′ ′ a^{2}X''T=XT'' a 2 X ′′ T = X T ′′
X ′ ′ X = T ′ ′ a 2 T = − λ . \frac{X''}{X}=\frac{T''}{a^{2}T}=-\lambda. X X ′′ = a 2 T T ′′ = − λ . Spatial problem X ′ ′ + λ X = 0 X''+\lambda X=0 X ′′ + λ X = 0 X ( 0 ) = X ( L ) = 0 X(0)=X(L)=0 X ( 0 ) = X ( L ) = 0
X n ( x ) = sin n π x L , λ n = n 2 π 2 L 2 , n = 1 , 2 , 3 , … . X_n(x)=\sin\frac{n\pi x}{L},\qquad\lambda_n=\frac{n^{2}\pi^{2}}{L^{2}},\quad n=1,2,3,\ldots. X n ( x ) = sin L nπ x , λ n = L 2 n 2 π 2 , n = 1 , 2 , 3 , … . Temporal problem T ′ ′ + a 2 λ n T = 0 T''+a^{2}\lambda_n T=0 T ′′ + a 2 λ n T = 0
T n ( t ) = A n cos n π a t L + B n sin n π a t L . T_n(t)=A_n\cos\frac{n\pi at}{L}+B_n\sin\frac{n\pi at}{L}. T n ( t ) = A n cos L nπ a t + B n sin L nπ a t . General solution:
u ( x , t ) = ∑ n = 1 ∞ sin n π x L [ A n cos n π a t L + B n sin n π a t L ] . ( ∗ ) u(x,t)=\sum_{n=1}^{\infty}\sin\frac{n\pi x}{L}\!\left[A_n\cos\frac{n\pi at}{L}+B_n\sin\frac{n\pi at}{L}\right]. \tag{$\ast$} u ( x , t ) = n = 1 ∑ ∞ sin L nπ x [ A n cos L nπ a t + B n sin L nπ a t ] . ( ∗ ) Step 2 — Use u ( x , 0 ) = x u(x,0)=x u ( x , 0 ) = x A n A_n A n
From ( ∗ ) (\ast) ( ∗ ) t = 0 t=0 t = 0 ∑ A n sin ( n π x / L ) = x \sum A_n\sin(n\pi x/L)=x ∑ A n sin ( nπ x / L ) = x A n A_n A n x x x ( 0 , L ) (0,L) ( 0 , L )
A n = 2 L ∫ 0 L x sin n π x L d x . A_n=\frac{2}{L}\int_0^L x\sin\frac{n\pi x}{L}\,dx. A n = L 2 ∫ 0 L x sin L nπ x d x . Integration by parts (let u = x , d v = sin ( n π x / L ) d x u=x,\,dv=\sin(n\pi x/L)dx u = x , d v = sin ( nπ x / L ) d x d u = d x , v = − ( L / n π ) cos ( n π x / L ) du=dx,\,v=-(L/n\pi)\cos(n\pi x/L) d u = d x , v = − ( L / nπ ) cos ( nπ x / L )
∫ 0 L x sin n π x L d x = [ − L x n π cos n π x L ] 0 L + L n π ∫ 0 L cos n π x L d x . \int_0^L x\sin\frac{n\pi x}{L}\,dx=\left[-\frac{Lx}{n\pi}\cos\frac{n\pi x}{L}\right]_0^L+\frac{L}{n\pi}\int_0^L\cos\frac{n\pi x}{L}\,dx. ∫ 0 L x sin L nπ x d x = [ − nπ Lx cos L nπ x ] 0 L + nπ L ∫ 0 L cos L nπ x d x . The bracket evaluates to − L 2 n π cos ( n π ) = L 2 ( − 1 ) n + 1 n π -\dfrac{L^{2}}{n\pi}\cos(n\pi)=\dfrac{L^{2}(-1)^{n+1}}{n\pi} − nπ L 2 cos ( nπ ) = nπ L 2 ( − 1 ) n + 1 sin ( n π ) = sin 0 = 0 \sin(n\pi)=\sin 0=0 sin ( nπ ) = sin 0 = 0
A n = 2 L ⋅ L 2 ( − 1 ) n + 1 n π = 2 L ( − 1 ) n + 1 n π . A_n=\frac{2}{L}\cdot\frac{L^{2}(-1)^{n+1}}{n\pi}=\frac{2L\,(-1)^{n+1}}{n\pi}. A n = L 2 ⋅ nπ L 2 ( − 1 ) n + 1 = nπ 2 L ( − 1 ) n + 1 . Step 3 — Use u t ( x , 0 ) = 1 u_t(x,0)=1 u t ( x , 0 ) = 1 B n B_n B n
Differentiating ( ∗ ) (\ast) ( ∗ ) t t t t = 0 t=0 t = 0
u t ( x , 0 ) = ∑ n = 1 ∞ B n ⋅ n π a L sin n π x L = 1. u_t(x,0)=\sum_{n=1}^{\infty}B_n\cdot\frac{n\pi a}{L}\sin\frac{n\pi x}{L}=1. u t ( x , 0 ) = n = 1 ∑ ∞ B n ⋅ L nπ a sin L nπ x = 1. So n π a L B n \dfrac{n\pi a}{L}B_n L nπ a B n 1 1 1
n π a L B n = 2 L ∫ 0 L sin n π x L d x = 2 L ⋅ L n π [ 1 − ( − 1 ) n ] = 2 [ 1 − ( − 1 ) n ] n π . \frac{n\pi a}{L}B_n=\frac{2}{L}\int_0^L\sin\frac{n\pi x}{L}\,dx=\frac{2}{L}\cdot\frac{L}{n\pi}\bigl[1-(-1)^{n}\bigr]=\frac{2\bigl[1-(-1)^{n}\bigr]}{n\pi}. L nπ a B n = L 2 ∫ 0 L sin L nπ x d x = L 2 ⋅ nπ L [ 1 − ( − 1 ) n ] = nπ 2 [ 1 − ( − 1 ) n ] . Therefore
B n = 2 L [ 1 − ( − 1 ) n ] n 2 π 2 a = { 4 L n 2 π 2 a , n odd , 0 , n even . B_n=\frac{2L\bigl[1-(-1)^{n}\bigr]}{n^{2}\pi^{2}\,a}=\begin{cases}\dfrac{4L}{n^{2}\pi^{2}\,a},&n\text{ odd},\\[4pt]0,&n\text{ even}.\end{cases} B n = n 2 π 2 a 2 L [ 1 − ( − 1 ) n ] = ⎩ ⎨ ⎧ n 2 π 2 a 4 L , 0 , n odd , n even . Step 4 — Assemble the solution
Answer
u ( x , t ) = ∑ n = 1 ∞ 2 L ( − 1 ) n + 1 n π sin n π x L cos n π a t L + ∑ n odd 4 L n 2 π 2 a sin n π x L sin n π a t L . \boxed{\;u(x,t)=\sum_{n=1}^{\infty}\frac{2L(-1)^{n+1}}{n\pi}\sin\frac{n\pi x}{L}\cos\frac{n\pi at}{L}\;+\;\sum_{n\text{ odd}}\frac{4L}{n^{2}\pi^{2}\,a}\sin\frac{n\pi x}{L}\sin\frac{n\pi at}{L}.\;} u ( x , t ) = n = 1 ∑ ∞ nπ 2 L ( − 1 ) n + 1 sin L nπ x cos L nπ a t + n odd ∑ n 2 π 2 a 4 L sin L nπ x sin L nπ a t .