𝜕2 𝑦 𝜕 𝑦2

Wave equation in one dimension: = 𝑐 2 𝜕𝑥 2

𝜕2 𝑢 2
2 𝜕 𝑢 𝜕2 𝑢
Wave equation in two-dimension: = 𝑐 (𝜕𝑥 2 + 𝜕𝑦 2 )
𝜕𝑡 2

𝜕2 𝑦 𝜕 𝑦2
Solution of the wave equation: = 𝑐 2 𝜕𝑥 2 -------------------------------------- (1)
𝜕𝑡 2

Assume that a solution of (1) is of the form 𝑦 = 𝑋(𝑥 )𝑇(𝑡) --------------------- (2)
where 𝑋 is a function of 𝑥 only and 𝑇 is a function of 𝑡 only.
Substituting in Eq. (1), we have
′′ 2 ′′ 𝑋 ′′ 1 𝑇 ′′
𝑋𝑇 = 𝑐 𝑋 𝑇 ⇒ = 𝑐2 = 𝑘(say)
𝑋 𝑇

𝑋′′ 𝑑2 𝑋
Then =𝑘⇒ − 𝑘𝑋 = 0 ----------------------------------------------------------- (3)
𝑋 𝑑𝑥 2

Auxiliary equation is 𝑚2 − 𝑘 = 0 ⇒ 𝑚2 = 𝑘
 1 𝑇 ′′ ′′ 2 𝑑2 𝑇
And = 𝑘 ⇒ 𝑇 − 𝑘𝑐 𝑇 = 0 ⇒ − 𝑘𝑐 2 𝑇 = 0 ----------------------------- (4)
𝑐2 𝑇 𝑑𝑡 2

Auxiliary equation is 𝑚2 − 𝑘𝑐 2 = 0 ⇒ 𝑚2 = 𝑘𝑐 2
Solving (3) and (4), we get
1) When 𝑘 is positive and 𝑘 = 𝑝2 , say 𝑋 = 𝑐1 𝑒 𝑝𝑥 + 𝑐2 𝑒 −𝑝𝑥 ; 𝑇 = 𝑐3 𝑒 𝑐𝑝𝑡 + 𝑐4 𝑒 −𝑐𝑝𝑡 .
2) When 𝑘 is negative and 𝑘 = −𝑝2 , 𝑋 = 𝑐5 cos 𝑝𝑥 + 𝑐6 sin 𝑝𝑥 ; 𝑇 = 𝑐7 cos 𝑐𝑝𝑡 + 𝑐8 sin 𝑐𝑝𝑡.
3) When 𝑘 is zero, 𝑋 = 𝑐9 𝑥 + 𝑐10 ; 𝑇 = 𝑐11 𝑡 + 𝑐12 .
Thus, the various possible solutions of wave equation (1) are
𝑦(𝑥, 𝑡) = (𝑐1 𝑒 𝑝𝑥 + 𝑐2 𝑒 −𝑝𝑥 )(𝑐3 𝑒 𝑐𝑝𝑡 + 𝑐4 𝑒 −𝑐𝑝𝑡 ) --------------------------------- (5)
𝑦(𝑥, 𝑡) = (𝑐5 cos 𝑝𝑥 + 𝑐6 sin 𝑝𝑥)(𝑐7 cos 𝑐𝑝𝑡 + 𝑐8 sin 𝑐𝑝𝑡) ------------------------- (6)
𝑦(𝑥, 𝑡) = (𝑐9 𝑥 + 𝑐10 )(𝑐11 𝑡 + 𝑐12 ) --------------------------------------------------- (7)
As we will be dealing with problems on vibrations, 𝑦 must be a periodic function of 𝑥
and 𝑡. Hence their solution must involve trigonometric terms. Accordingly, the solution
given by Eq. (6) is the only suitable solution of the wave equation.
Q.1 A string is stretched and fastened to two points 𝑙 apart. Motion is started by
𝜋𝑥
displacing the string in the form 𝑦 = 𝑎 sin ( 𝑙 ) from which it is released at time 𝑡 = 0.
Show that the displacement of any point at a distance 𝑥 from one end at time 𝑡 is given
𝜋𝑥 𝜋𝑐𝑡
by 𝑦(𝑥, 𝑡) = 𝑎 sin ( 𝑙 ) cos ( ).
𝑙

𝜕2 𝑦 2
𝜕 𝑦
Ans: The vibration of the string is given by = 𝑐 2 𝜕𝑥 2 ------------------------ (1)
𝜕𝑡 2

As the end points of the string are fixed, for all time
𝑦(0, 𝑡) = 0 ---------------------------------------------------------- (2)
And 𝑦(𝑙, 𝑡) = 0 ---------------------------------------------------------- (3)
Since the initial transverse velocity of any point of the string is zero, therefore,
𝜕𝑦
( 𝜕𝑡 ) = 0 ------------------------------------- (4)
𝑡=0
𝜋𝑥
Also 𝑦(𝑥, 0) = 𝑎 sin ( 𝑙 ) -------------------------- (5)
Now we have to solve (1) subject to the boundary conditions (2) and (3) and initial
conditions (4) and (5). Since the vibration of the string is periodic, therefore, the solution
of (1) is of the form
𝑦(𝑥, 𝑡) = (𝐶1 cos 𝑝𝑥 + 𝐶2 sin 𝑝𝑥 )(𝐶3 cos 𝑐𝑝𝑡 + 𝐶4 sin 𝑐𝑝𝑡) ---------------- (6)
By (2), 𝑦(0, 𝑡) = 0 ⇒ 𝐶1 (𝐶3 cos 𝑐𝑝𝑡 + 𝐶4 sin 𝑐𝑝𝑡) = 0 ⇒ 𝐶1 = 0
Hence 𝑦(𝑥, 𝑡) = 𝐶2 sin 𝑝𝑥 (𝐶3 cos 𝑐𝑝𝑡 + 𝐶4 sin 𝑐𝑝𝑡) ------------------------ (7)
𝜕𝑦
And = 𝐶2 sin 𝑝𝑥 {𝐶3 (−𝑐𝑝 sin(𝑐𝑝𝑡)) + 𝐶4 (𝑐𝑝. cos(𝑐𝑝𝑡))}
𝜕𝑡
𝜕𝑦
By Eq. (4), ( 𝜕𝑡 )𝑡=0 = 𝐶2 sin 𝑝𝑥 . (𝐶4 . 𝑐𝑝) = 0 ⇒ 𝐶2 𝐶4 𝑐𝑝 = 0

If 𝐶2 = 0, Eq. (7) will lead to the trivial solution 𝑦(𝑥, 𝑡) = 0. Therefore, the only
possibility is 𝐶4 = 0.
Thus Eq. (7) becomes 𝑦(𝑥, 𝑡) = 𝐶2 𝐶3 sin 𝑝𝑥 cos 𝑐𝑝𝑡 ------------------------- (8)
By Eq. (3), 𝑦(𝑙, 𝑡) = 𝐶2 𝐶3 sin 𝑝𝑙 cos 𝑐𝑝𝑡 = 0 for all 𝑡.
𝑛𝜋
Since 𝐶2 and 𝐶3 are non-zero, we have sin 𝑝𝑙 = 0 ⇒ 𝑝𝑙 = 𝑛𝜋 ⇒ 𝑝 = 𝑙
, 𝑛 ∈ 𝑍.
 𝑛𝜋𝑥 𝑛𝜋𝑐𝑡
Hence, Eq. (8) reduces to 𝑦(𝑥, 𝑡) = 𝐶2 𝐶3 sin 𝑙
cos 𝑙
.
𝑛𝜋𝑥 𝜋𝑥
Finally, imposing the last condition (5), we have 𝑦(𝑥, 0) = 𝐶2 𝐶3 sin = 𝑎 sin which
𝑙 𝑙
will be satisfied by taking 𝐶2 𝐶3 = 𝑎 and 𝑛 = 1.
𝜋𝑥 𝜋𝑐𝑡
Hence, the required solution is 𝑦(𝑥, 𝑡) = 𝑎 sin cos .
𝑙 𝑙

𝜕2 𝑦 2
𝜕 𝑦
Q.1 Solve completely the equation 𝜕𝑡 2 = 𝑐 2 𝜕𝑥 2 , representing the vibrations of a string
of length 𝑙, fixed at both ends, given that 𝑦(0, 𝑡) = 0; 𝑦(𝑙, 𝑡) = 0; 𝑦(𝑥, 0) = 𝑓(𝑥) and
𝜕𝑦
( 𝜕𝑡 ) = 0, 0 < 𝑥 < 𝑙.
𝑡=0