SRI / PDE / Wave Equations 1

WAVE EQUATIONS

ƒ Consider a perfectly flexible elastic string stretched between two points with
uniform tension T.

ƒ The string is displaced slightly from its initial position of rest and released, with
the end points remaining fixed. Then the string will vibrate.

ƒ The position of any point P in the string will depend on its distance from one end
and on the instant in time.

ƒ Its displacement u at any time t. Thus be expressed as u = f ( x, t ) where x is its

distance from the left-hand end.

y
P u(x, t)
u = f (x, t)

x
x L

WAVE EQUATION
∂ 2u 1 ∂ 2u
= 2 ⋅ 2 where c 2 = T / ρ
∂x 2
c ∂t
T - tension in the string
ρ - the mass per unit length of the string

The displacement of the string is regarded as small so that T and ρ remain constant.

PROBLEM OF THE WAVE EQUATION

ƒ Boundary conditions :
– The string is fixed at both ends i.e. at x = 0 and x = L for all values of time t.
– Therefore u ( x, t ) becomes
u (0, t ) = 0 ⎫
⎬ for all values of t ≥ 0
u ( L, t ) = 0 ⎭
ƒ Initial conditions :
– If the initial deflection of P at t = 0 is denoted by f (x), then
u ( x , 0) = f ( x )
– Let the initial velocity of P be g(x), then
⎡ ∂u ⎤
⎢ ∂t ⎥ = u t = 0 = g ( x)
⎣ ⎦ t =0
SRI / PDE / Wave Equations 2

SOLUTION BY SEPARATING THE VARIABLES

1st Step
Assume u ( x, t ) = X ( x)T (t ) where
ƒ X (x) : function of x only
ƒ T (t) : function of t only

Express the equation in terms of X and T and their derivatives

∂u ∂u
u = XT = X 'T = XT '
∂x ∂t
∂ 2u ∂ 2u
= X "T = XT "
∂x 2 ∂t 2

2nd Step
Substitute into the given PDE
1
X "T = XT ′
c2

Then by separation of the variables,

X"
ƒ =k : X "−kX = 0
X
1 T"
ƒ =k : T "−c 2 kT = 0
c2 T

3rd Step
Choose k = − p 2 to give an oscillatory solution.
ƒ X "+ p 2 X = 0
m2 + p2 = 0
m2 = − p2
m = ±i p

So the solution is X = A cos px + B sin px

ƒ T "+c 2 p 2 T = 0
m2 + c2 p2 = 0
m 2 = −c 2 p 2
m = ± i cp

So the solution is T = C cos cpt + D sin cpt

SRI / PDE / Wave Equations 3

4th Step
u ( x, t ) = XT
= ( A cos px + B sin px)(C cos cpt + D sin cpt )

5th Step
λ
Let pc = λ , i.e. p = , then
c
⎛ λ λ ⎞
u ( x, t ) = ⎜ A cos x + B sin x ⎟(C cos λt + D sin λt )
⎝ c c ⎠

6th Step
Apply boundary conditions to determine A and B.

7th Step
Apply the remaining initial conditions to get the eigen values and eigen functions for
n = 1, 2, 3, ... and determine the general solution in t.
ncπ
Eigen Value : λ=
L
λx
Eigen Function : u ( x, t ) = B sin (C cos λt + D sin λt )
c

8th Step
Apply the remaining initial or boundary conditions.

9th Step
Determine the coefficients Cn and Dn by Fourier series technique.
SRI / PDE / Wave Equations 4

Example 1
A stretched string of length 20cm is set oscillating by displacing its mid-point a distance
1cm from its rest position and releasing it with zero initial velocity. Solve the wave
equation
∂ 2u 1 ∂ 2u
= ⋅
∂ x2 c2 ∂ t 2
where c 2 = 1 to determine the resulting motion, u(x, t).

Solution
y
1
u = f (x, t)

x
10 20

Boundary Conditions : u (0, t ) = 0

u ( 20, t ) = 0

⎧x
⎪10 , 0 ≤ x ≤ 10,
Initial Conditions : u ( x, 0) = ⎨
20 − x
⎪ , 10 ≤ x ≤ 20.
⎩ 10
ut ( x, 0) = 0

1st Step
∂u ∂u
u = XT = X 'T = XT '
∂x ∂t
∂ 2u ∂ 2u
= X "T = XT "
∂x 2 ∂t 2

2nd Step
Since c 2 = 1
∂ 2u ∂ 2u
=
∂x 2 ∂t 2
X "T = XT "

Then by separation of the variables,

X"
=k : X "−kX = 0
X
T"
=k : T "−kT = 0
T
SRI / PDE / Wave Equations 5

3rd Step
Choose k = − p 2
X "+ p 2 X = 0
m2 + p 2 = 0
m = ±i p

So the solution is X = A cos px + B sin px

T "+ p 2T = 0
m2 + p 2 = 0
m = ±i p

So the solution is T = C cos pt + D sin pt

4th Step
u ( x, t ) = XT
= ( A cos px + B sin px)(C cos pt + D sin pt )

5th Step
Let p = λ , then
u ( x, t ) = ( A cos λx + B sin λx ) (C cos λt + D sin λt )

6th Step
Determine A and B from boundary condition
(a) u (0, t ) = 0
0 = A(C cos λt + D sin λt )
A=0

∴ u ( x, t ) = ( B sin λx ) (C cos λt + D sin λt )

(b) u ( 20, t ) = 0
0 = ( B sin 20λ )(C cos λt + D sin λt )
Since B ≠ 0
∴ sin 20λ = 0 = sin nπ
20λ = nπ
nπ
λ= ; n = 1, 2, 3, ...
20
SRI / PDE / Wave Equations 6

Then,
nπ ⎛ nπ nπ ⎞
u ( x, t ) = sin x ⎜ P cos t + Q sin t ⎟ ; n = 1, 2, 3, ...
20 ⎝ 20 20 ⎠
where P = B × C and Q = B × D

7th Step

Eigen Values Eigen Functions

π π ⎛ π π ⎞
1 λ1 = u1 = sin x ⎜ P1 cos t + Q1 sin t ⎟
20 20 ⎝ 20 20 ⎠

π π ⎛ π π ⎞
2 λ2 = u 2 = sin x ⎜ P2 cos t + Q2 sin t ⎟
10 10 ⎝ 10 10 ⎠

M M M
nπ nπ ⎛ nπ nπ ⎞
n λn = u n = sin x ⎜ Pn cos t + Qn sin t⎟
20 20 ⎝ 20 20 ⎠

∞
nπ ⎛ nπ nπ ⎞
∴ u ( x, t ) = ∑ sin x ⎜ Pn cos t + Qn sin t⎟
n =1 20 ⎝ 20 20 ⎠

8th Step
Apply the remaining initial condition.
⎧x
⎪ , 0 ≤ x ≤ 10,
(a) u ( x, 0) = f ( x) = ⎨10
20 − x
⎪ , 10 ≤ x ≤ 20.
⎩ 10
∞
nπ
u ( x, 0) = ∑ sin x [ Pn cos(0) + Qn sin(0)]
n =1 20
∞
nπ
= ∑ Pn sin x
n =1 20

Then, find Pn .

(b) u t ( x , 0) = 0

– apply after find Pn .

SRI / PDE / Wave Equations 7

9th Step
Determine the coefficients Pn and Qn by Fourier series technique.
2 L nπ
Pn = ∫ f ( x) sin x dx
L 0 L
2 20 nπ
= ∫
20 0
f ( x ) sin
20
x dx

1 ⎡ 10 x nπ 20 ⎛ 20 − x ⎞ nπ ⎤
= ⎢∫ sin x dx + ∫ ⎜ ⎟ sin x dx ⎥
10 ⎣ 0 10 20 10 ⎝ 10 ⎠ 20 ⎦
1 ⎡ 10 nπ 20 nπ ⎤
= ⎢ ∫
100 ⎣ 0
x sin
20
x dx + ∫ 10
( 20 − x) sin
20
x dx ⎥
⎦
1 ⎛⎜ ⎡ − 20 x nπ nπ ⎡ 20 x − 400 nπ nπ ⎤ ⎞⎟
10 20
400 ⎤ 400
= ⎢ cos x + 2 2 sin x⎥ + ⎢ cos x − 2 2 sin x⎥
100 ⎜⎝ ⎣ nπ 20 nπ 20 ⎦ 0 ⎣ nπ 20 nπ 20 ⎦ 10 ⎟⎠
1 ⎛ − 200 nπ 400 nπ 400 200 nπ 400 nπ ⎞
= ⎜ cos + 2 2 sin + 0 − 0 + 0− 2 2 sin nπ + cos + 2 2 sin ⎟
100 ⎝ nπ 2 nπ 2 nπ nπ 2 nπ 2 ⎠
1 ⎛ 800 nπ 400 ⎞
= ⎜ 2 2 sin − 2 2 sin nπ ⎟
100 ⎝ n π 2 nπ ⎠
8 nπ 4
= 2 2 sin − sin nπ
nπ 2 n 2π 2

8 nπ
n = 1, 2, 3, K : Pn = sin
nπ2 2
2

Thus,
∞
⎛nπ nπ nπ ⎞
u ( x, t ) = ∑ sinx ⎜ Pn cos t + Qn sin t⎟
n =1 ⎝20 20 20 ⎠
∞
nπ ⎡ 8 ⎛ nπ ⎞⎛ nπ ⎞ nπ ⎤
= ∑ sin x ⎢ 2 2 ⎜ sin ⎟⎜ cos t ⎟ + Qn sin t
n =1 20 ⎣ n π ⎝ 2 ⎠⎝ 20 ⎠ 20 ⎥⎦

(b) ut ( x, 0) = 0

∂u ∞ nπ ⎡ 8 ⎛ nπ ⎞⎛ nπ nπ ⎞ nπ nπ ⎤
= ∑ sin x ⎢ 2 2 ⎜ sin ⎟⎜ − sin t ⎟ + Qn cos t
∂t n =1 20 ⎣n π ⎝ 2 ⎠⎝ 20 20 ⎠ 20 20 ⎥⎦

∞
nπ ⎡ nπ ⎤
ut ( x, 0) = 0 = ∑ sin x ⎢0 + Qn
n =1 20 ⎣ 20 ⎥⎦
∴ Qn = 0

So
8 ∞
1⎛ nπ ⎞⎛ nπ ⎞⎛ nπ ⎞
u ( x, t ) =
π 2 ∑n
n =1
2 ⎜
⎝
sin x ⎟⎜ sin
20 ⎠⎝
⎟⎜ cos
2 ⎠⎝
t⎟
20 ⎠
SRI / PDE / Wave Equations 8

Example 2
A perfectly elastic string is stretched between two points 10cm apart. Its center point is
displaced 2cm from its position of rest at right angles to the original direction of the string
and then released with zero initial velocity. By applying the equation
∂ 2u 1 ∂ 2u
= ⋅ ,
∂x 2 c 2 ∂t 2
determine the subsequent motion, u ( x, t ) .

Solution
y
2
u = f (x, t)

x
5 10

Boundary Conditions : u (0, t ) = 0

u (10, t ) = 0

⎧ 2x
⎪5 , 0 ≤ x ≤ 5,
Initial Conditions : u ( x, 0) = ⎨
20 − 2 x
⎪ , 5 ≤ x ≤ 10.
⎩ 5
ut ( x, 0) = 0

1st Step
∂u ∂u
u = XT = X 'T = XT '
∂x ∂t
∂ 2u ∂ 2u
= X "T = XT "
∂x 2 ∂t 2

2nd Step
Since c 2 = 1
∂ 2u 1 ∂ 2u
= ⋅
∂x 2 c 2 ∂t 2
1
X "T = 2 XT "
c

Then by separation of the variables,

X"
=k : X "−kX = 0
X
T"
=k : T "−c 2 kT = 0
c 2T
SRI / PDE / Wave Equations 9

3rd Step
Choose k = − p 2
X "+ p 2 X = 0
m2 + p 2 = 0
m = ±i p
So the solution is X = A cos px + B sin px

T "+c 2 p 2T = 0
m2 + c2 p 2 = 0
m = ± i cp
So the solution is T = C cos cpt + D sin cpt

4th Step
u ( x, t ) = XT
= ( A cos px + B sin px )(C cos cpt + D sin cpt )

5th Step
Let pc = λ , then
⎛ λ λ ⎞
u ( x, t ) = ⎜ A cos x + B sin x ⎟ (C cos λt + D sin λt )
⎝ c c ⎠

6th Step
Determine A and B from boundary condition
(a) u (0, t ) = 0
0 = A(C cos λt + D sin λt )
A=0

⎛ λ ⎞
∴ u ( x, t ) = ⎜ B sin x ⎟ (C cos λt + D sin λt )
⎝ c ⎠

(b) u (10, t ) = 0
⎛ 10λ ⎞
0 = ⎜ B sin ⎟ (C cos λt + D sin λt )
⎝ c ⎠
Since B ≠ 0
10λ
∴ sin = 0 = sin nπ
c
λ nπ
= ; n = 1, 2, 3, ...
c 10
SRI / PDE / Wave Equations 10

Then,
nπ ⎛ ncπ ncπ ⎞
u ( x, t ) = sin x ⎜ P cos t + Q sin t ⎟ ; n = 1, 2, 3, ...
10 ⎝ 10 10 ⎠
where P = B × C and Q = B × D

7th Step

Eigen Values Eigen Functions

cπ π ⎛ cπ cπ ⎞
1 λ1 = u1 = sin x ⎜ P1 cos t + Q1 sin t⎟
10 10 ⎝ 10 10 ⎠

cπ π ⎛ cπ cπ ⎞
2 λ2 = u 2 = sin x ⎜ P2 cos t + Q 2 sin t⎟
5 5 ⎝ 5 5 ⎠

M M M
ncπ nπ ⎛ ncπ ncπ ⎞
n λn = u n = sin x ⎜ Pn cos t + Qn sin t⎟
10 10 ⎝ 10 10 ⎠

∞
nπ ⎛ ncπ ncπ ⎞
∴ u ( x, t ) = ∑ sin x ⎜ Pn cos t + Qn sin t⎟
n =1 10 ⎝ 10 10 ⎠

8th Step
Apply the remaining initial condition.
⎧ 2x
⎪ , 0 ≤ x ≤ 5,
(a) u ( x, 0) = f ( x) = ⎨ 5
20 − 2 x
⎪ , 5 ≤ x ≤ 10.
⎩ 5
∞
nπ
u ( x, 0) = ∑ sin x [ Pn cos(0) + Qn sin(0)]
n =1 10
∞
nπ
= ∑ Pn sin x
n =1 10

Then, find Pn .

(b) u t ( x, 0) = 0

– apply after find Pn .

SRI / PDE / Wave Equations 11

9th Step
ƒ Determine the coefficients Pn and Qn by Fourier series technique.
2 L nπ
Pn = ∫ f ( x) sin x dx
L 0 L
2 10 nπ
= ∫ f ( x) sin x dx
10 0 10
1 ⎡ 5 2x nπ 10 ⎛ 20 − 2 x ⎞ nπ ⎤
= ⎢∫ sin x dx + ∫ ⎜ ⎟ sin x dx ⎥
5⎣ 0 5 10 5 ⎝ 5 ⎠ 10 ⎦
2 ⎡ 5 nπ 10 nπ ⎤
= ⎢ ∫
25 ⎣ 0
x sin
10
x dx + ∫ (10 − x) sin
5 10
x dx ⎥
⎦
2 ⎛ ⎡ − 10 x nπ nπ ⎤ ⎡10 x − 100 nπ⎞ nπ
5 10
100 100 ⎤
= ⎜⎢ cos x + 2 2 sin x⎥ + ⎢ cos ⎟x − 2 2 sin x⎥
25 ⎜⎝ ⎣ nπ 10 nπ 10 ⎦ 0 ⎣ nπ 10⎟
⎠ nπ 10 ⎦5
2 ⎛ − 50 nπ 100 nπ 100 50 nπ 100 nπ ⎞
= ⎜ cos + 2 2 sin + 0 − 0 + 0 − 2 2 sin nπ + cos + 2 2 sin ⎟
25 ⎝ nπ 2 nπ 2 nπ nπ 2 nπ 2 ⎠
2 ⎛ 200 nπ 100 ⎞
= ⎜ 2 2 sin − 2 2 sin nπ ⎟
25 ⎝ n π 2 n π ⎠
16 nπ 8
= 2 2 sin − sin nπ
nπ 2 n 2π 2

16 nπ
n = 1, 2, 3, K : Pn = sin
nπ
2 2
2

Thus,
∞
nπ ⎛ ncπ ncπ ⎞
u ( x, t ) = ∑ sin x ⎜ Pn cos t + Qn sin t⎟
n =1 10 ⎝ 10 10 ⎠
∞
nπ ⎡ 16 ⎛ nπ ⎞⎛ ncπ ⎞ ncπ ⎤
= ∑ sin x ⎢ 2 2 ⎜ sin ⎟⎜ cos t ⎟ + Qn sin t⎥
n =1 10 ⎣ n π ⎝ 2 ⎠⎝ 10 ⎠ 10 ⎦

(b) ut ( x, 0) = 0
∂u ∞
nπ ⎡ 16 ⎛ nπ ⎞⎛ ncπ ncπ ⎞ ncπ ncπ ⎤
= ∑ sin x ⎢ 2 2 ⎜ sin ⎟⎜ − sin t ⎟ + Qn cos t
∂t n =1 10 ⎣ n π ⎝ 2 ⎠⎝ 10 10 ⎠ 10 10 ⎥⎦

∞
nπ ⎡ ncπ ⎤
u t ( x,0) = 0 = ∑ sin x ⎢0 + Q n
n =1 10 ⎣ 10 ⎥⎦
∴ Qn = 0

So
16 ∞
1 ⎛ nπ ⎞⎛ nπ ⎞⎛ ncπ ⎞
u ( x, t ) =
π 2 ∑n
n =1
2 ⎜
⎝
sin x ⎟⎜ sin
10 ⎠⎝
⎟⎜ cos
2 ⎠⎝
t⎟
10 ⎠
SRI / PDE / Wave Equations 12

Example 3
The center point of a perfectly elastic string stretched between two points A and B, 4m
apart, is deflected a distance 2m from its position of rest perpendicular to AB and released
initially with zero velocity. Solve the wave equation 4u xx = u tt with to determine an
expression for the subsequent motion of any point in the string at time t if
∞
⎛ ncπ ncπ ⎞ nπ
u ( x, t ) = ∑ ⎜ An cos t + Bn sin t ⎟ sin x
n =1 ⎝ L L ⎠ L

Solution
∞
⎛ nπ nπ ⎞ nπ
c = 2, L = 4 : u ( x, t ) = ∑ ⎜ An cos t + Bn sin t ⎟ sin x
n =1 ⎝ 2 2 ⎠ 4
1st Initial Condition :
u (x, 0) = f (x )
∞
nπ
∑
n =1
[ An (1) + Bn (0)] sin
4
x = f ( x)

∞
nπ ⎧ x, 0 ≤ x ≤ 2,
∑ An sin x=⎨
n =1 4 ⎩4 − x, 2 ≤ x ≤ 4.

2 L nπ
An = ∫
L 0
f ( x) sin
L
x dx

2⎛ 2 nπ 4 nπ ⎞
= ⎜ ∫ x sin x dx + ∫ (4 − x) sin x dx ⎟
4⎝ 0 4 2 4 ⎠
1 ⎛⎜ nπ nπ nπ nπ ⎤ ⎞
2
⎡ 4( 4 − x )
4
⎡ 4x 16 ⎤ 16
= −
⎢ nπ cos x + sin x ⎥ + ⎢− cos x − 2 2 sin x⎥ ⎟
2 ⎜⎝ ⎣ nπ ⎦0 ⎣ nπ nπ ⎦ 2 ⎟⎠
2 2
4 4 4 4
1⎛ 8 nπ 16 nπ 16
= ⎜− cos + 2 2 sin + 0 − 2 2 sin(0)
2 ⎝ nπ 2 nπ 2 nπ
16 8 nπ 16 nπ ⎞
+0− sin nπ + cos + 2 2 sin ⎟
nπ
2 2
nπ 2 nπ 2 ⎠
16 nπ
= sin
nπ
2 2
2

Thus,
∞
⎛ nπ nπ ⎞ nπ
u ( x, t ) = ∑ ⎜ An cos t + Bn sin t ⎟ sin x
n =1 ⎝ 2 2 ⎠ 4
∞
nπ ⎡ 16 ⎛ nπ ⎞ ⎛ nπ ⎞ nπ ⎤
= ∑ sin
n =1
x ⎢ 2 2 ⎜ sin
4 ⎣n π ⎝
⎟ ⎜ cos
2 ⎠⎝
t ⎟ + Bn sin
2 ⎠
t
2 ⎥⎦

∂u ∞ nπ ⎡ 16 ⎛ nπ ⎞ ⎛ nπ nπ ⎞ nπ nπ ⎤
= ∑ sin x ⎢ 2 2 ⎜ sin ⎟⎜ − sin t ⎟+ Bn cos t⎥
∂t n=1 4 ⎣n π ⎝ 2 ⎠⎝ 2 2 ⎠ 2 2 ⎦
SRI / PDE / Wave Equations 13

2nd Initial Condition :

u t ( x, 0) = 0
∞
⎛ nπ ⎞⎡ nπ ⎤
∑ ⎜⎝ sin
n =1 4
x ⎟ ⎢0 +
⎠⎣ 2
Bn ⎥ = 0
⎦
Bn = 0

OR
2 L nπ
Bn =
ncπ 0∫ g ( x) sin
L
x dx

2 4 nπ
= ∫
2 nπ 0
(0) sin
4
x dx

=0

∞
16 ⎛ nπ ⎞⎛ nπ ⎞⎛ nπ ⎞
∴ u ( x, t ) = ∑ 2 2 ⎜
sin ⎟⎜ sin x ⎟⎜ cos t⎟
n =1 nπ ⎝ 2 ⎠⎝ 4 ⎠⎝ 2 ⎠
SRI / PDE / Wave Equations 14

Example 4
Given the wave equation c 2 u xx = u tt with c = 4 ,

boundary conditions : u (0, t ) = 0 ,

u (3, t ) = 0 ,
initial conditions : u ( x, 0) = 4 sin 3π x ,
ut ( x, 0) = 0 .
Determine the resulting motion, u(x, t) if
∞
⎛ ncπ ncπ ⎞ nπ
u ( x, t ) = ∑ ⎜ An cos t + Bn sin t ⎟ sin x
n =1 ⎝ L L ⎠ L

Solution
∞
⎛ 4 nπ 4 nπ ⎞ nπ
c = 4, L = 3 : u ( x, t ) = ∑ ⎜ An cos t + Bn sin t ⎟ sin x
n =1 ⎝ 3 3 ⎠ 3

1st Initial Condition : u ( x, 0) = 4 sin 3π x

∞
nπ
∑
n =1
[ An (1) + Bn (0)]sin
3
x = 4 sin 3π x

π 9π
A1 sin x + L + A9 sin x + L = 4 sin 3π x
3 3

By equating the coefficients,

A9 = 4 ; n=9
A1 = L = A8 = A10 = A11 = L = 0 ; n≠9

Thus,
u ( x, t ) = (4 cos12π t + Bn sin 12π t ) sin 3π x
∂u
= (sin 3π x )(−48π sin 12π t + 12π Bn cos 12π t )
∂t

2nd Initial Condition :

ut ( x, 0) = 0
(sin 3π x)(−0 + 12π Bn ] = 0
Bn = 0

OR
2 L nπ 2 4 nπ
Bn =
ncπ ∫ 0
g ( x ) sin
L
x dx =
4 nπ ∫ 0
(0) sin
3
x dx = 0

∴ u ( x, t ) = 4(cos 12π t ) sin 3π x