SEOUL NATIONAL UNIVERSITY
School of Mechanical & Aerospace Engineering
400.002
Eng Math II
18 2D Wave equation
18.1 Rectangular Membrane. Use of Double Fourier Series
• Two - dimensional wave equation - 2D version of the vibrating string : stretch an elastic membrane, fixed along its bdry in the xy-plane (think of a drumhead)
- deflection u(x, y, t) ?
∂2u
∂t2 =c2 µ∂2u
∂x2 + ∂2u
∂ty2
¶
c2 = T
ρ (1)
T: tension per unit length, ρ: mass per unit area - Boundary condition
u= 0 on the boundary of the membrane for all t≥0 (2) - Two initial conditions
u(x, y,0) = f(x, y) [given initial displacementf(x, y)] (3)
∂u
∂t
¯¯
¯¯
t=0
= g(x, y) [given initial velocityg(x, y)] (4)
• Step I: Three ODEs
u(x, y, t) =F(x, y)G(t) - By substituting this into the wave equation (1),
FG¨ =c2G(Fxx+Fyy) G¨
c2G = 1
F(Fxx+Fyy) =−ν2 - For the time function G(t),
G¨+λ2G= 0 whereλ=cν, - For the amplitude function F(x, y),
Fxx+Fyy+ν2F = 0 two - dimensional Helmholtz equation - Separation of the Helmholtz equation
F(x, y) =H(x)·Q(y)
- Substitution of this into (7) gives d2H
dx2 Q=− µ
Hd2Q
dy2 +ν2HQ
¶ . 1
H ·d2H dx2 =−1
Q µd2Q
dy2 +ν2Q
¶
=−k2
d2H
dx2 +k2H = 0 (5)
d2Q
dy2 +p2Q = 0 wherep2 =ν2−k2. (6)
• Step II: Satisfying the Boundary Conditions - The general solution of (5) and (6) are
H(x) =Acoskx+Bsinkx and Q(y) =Ccospy+Dsinpy - From BC
u(0, y, t) = H(0)Q(y)G(t) = 0 u(a, y, t) = H(a)Q(y)G(t) = 0 u(x,0, t) = H(x)Q(0)G(t) = 0 u(x,0, t) = H(x)Q(b)G(t) = 0
∴H(0) =H(a) =Q(0) =Q(b) = 0 H(0) =A= 0 H(a) =Bsinka= 0 sinka= 0 ⇒ k= mπ
a (m: integer) Q(0) =C= 0 Q(b) =Dsinpb= 0 sinpb= 0 ⇒ p= nπ
b (n: integer) Hm(x) = sinmπx
a and Qn(y) = sinnπy
b , m= 1,2,· · · , n= 1,2,· · · . Fmn(x, y) =Hm(x)Qn(y) = sinmπx
a sinnπy
b , m= 1,2,· · ·, n= 1,2,· · · . -Eigenfunctions and Eigenvalues
p2 =ν2−k2 and λ=cν λ=cp
k2+p2, k= mπa and p= nπb
- Eigenvalues or characteristic values λ=λmn =cπ
rm2 a2 +n2
b2, m= 1,2,· · ·, n= 1,2,· · · . - The corresponding general solution of (6) is
Gmn(t) =Bmncosλmnt+Bmn∗ sinλmnt umn(x, y, t) =Fmn(x, y)Gmn(t) - Eigenfunctions or characteristic functions
umn(x, y, t) = (Bmncosλmnt+Bmn∗ sinλmnt) sinmπx
a sinnπy b
• Step III: Solution of the Entire Problem
u(x, y, t) = X∞ m=1
X∞ n=1
umn(x, y, t)
= X∞ m=1
X∞ n=1
(Bmncosλmnt+Bmn∗ sinλmnt) sinmπx
a sinnπy b - Double Fourier series:
u(x, y,0) = X∞ m=1
X∞ n=1
Bmnsinmπx
a sinnπy
b =f(x, y) (7)
- Setting,
Km(y) = X∞ n=1
Bmnsinnπy
b (8)
- We can write (7) in the form f(x, y) =
X∞ m=1
Km(y) sinmπx a . Km(y) = 2
a Z a
0
f(x, y) sinmπx
a dx (9)
- (8) is the Fourier sine series ofKm(y).
Bmn= 2 b
Z b
0
Km(y) sinnπy
b dy (10)
- (9) and (10):
Bmn = 4 ab
Z b
0
Z a
0
f(x, y) sinmπx
a sinnπy
b dxdy m= 1,2,· · · ,
n= 1,2,· · ·. (11)
∂u
∂t
¯¯
¯¯
t=0
= X∞ m=1
X∞ n=1
Bmn∗ λmnsinmπx
a sinnπy
b =g(x, y) Bmn∗ = 4
abλmn Z b
0
Z a
0
g(x, y) sinmπx
a sinnπy
b dxdy m= 1,2,· · · ,
n= 1,2,· · ·. (12) Example . Vibrations of a rectangular membrane
a= 4 ft, b= 2 ft, T = 12.5 lb/ft,ρ = 2.5 slugs/ft2
f(x, y) = 0.1(4x−x2)(2y−y2) tf and g(x, y) = 0 solution)
c2 = T
ρ = 12.5
2.5 = 5 (ft2/sec2) - From (12) Bmn∗ = 0.
Bmn = 4 4·2
Z 2
0
Z 4
0
0.1(4x−x2)(2y−y2) sinmπx
4 sinnπy 2 dx dy
= 1
20 Z 4
0
(4x−x2) sinmπx 4 dx
Z 2
0
(2y−y2) sinnπy 2 dy - The 1st integral:
Z 4
0
(4x−x2) sinmπx
4 dx = − 4
mπ(4x−x2) cosmπx 4
¯¯
¯¯
4 0
+ 4 mπ
Z 4
0
(4−2x) cosmπx 4 dx
= 4·4
m2π2(4−2x) sinmπx 4
¯¯
¯¯
4 0
+4·4·2 m2π2
Z 4
0
sinmπx 4 dx
= − 128
(mπ)3 cosmπx 4
¯¯
¯¯
4 0
= 128
(πm)3(1−cosmπ)
= 128
(πm)3[1−(−1)m] - The 2nd integral:
Z 2
0
(2y−y2) sinnπy
2 dy = − 2
nπ(2y−y2) cosnπy 2
¯¯
¯¯
2 0
+ 2 nπ
Z 2
0
(2−2y) cosnπy 2 dy
= µ 2
nπ
¶2
·2·(1−y) sinnπy 2
¯¯
¯¯
¯
2
0
+ µ 2
nπ
¶2
·2 Z 2
0
sinnπy 2 dy
= −2 µ 2
nπ
¶3
cosnπy 2
¯¯
¯¯
¯
2
0
= µ 16
n3π3
¶
(1−cosnπ)
= 16
(nπ)3[1−(−1)n]
- For even m orn, we get 0.
- For odd m andn,
128
m3π3[1−(−1)m] = 256
(πm)3 (m= odd) 16
(nπ)3[1−(−1)n] = 32
(πn)3 (n= odd) Bmn= 256·32
20m3n3π6 ≈ 0.426050
m3n3 (m and nboth odd) u(x, y, t) = 0.426050X
m,
X
nodd
1 m3n3 cos
µ5π 4
pm2+ 4n2
¶
tsinmπx
4 sinnπy
2 ].
