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AB2.13: Separation of Variables: Use of Fourier Series
Consider the boundary value problem for theone-dimensional wave
equationdescribingvibrations of an elastic string fixed at its
ends:
2
ut2 =c2
2
ux2 , u= u(x, t), t >0, 0< x < L,
u(x, 0) =f(x),ut(x, 0) =g(x), (1)u(0, t) = 0, u(L, t) = 0, t
0.
The solution is determined by the separation of variables (the
Fourier method):
u(x, t) =F(x)G(t).
Then2u
t2
=F G, 2u
x2
=FG
Substituting this into one-dimensional wave equation and
separating variables,
F G =c2FG
G
c2G=
F
F =const= k
we obtain the differential equations forG(t) andF(x)
G c2kG = 0,
F kF = 0.
Satisfy the boundary conditions:
u(0, t) =F(0)G(t) = 0, u(L, t) =F(L)G(t) = 0, t 0.
Thus,F(0) = 0, F(L) = 0.
Fork 0, the solution of
F kF = 0, F(0) = 0, F(L) = 0
is identically zero.Fork = p2
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F =Fn= sinpnx= sinn
Lx (n= 1, 2, . . .).
Fork = kn= p2n
, the equation forG becomes
G +2n
G= 0, n=cn
L .
The general solution of this equation is
G(t) =Gn(t) =Bncos nt+B
nsin nt.
Hence the solutions of2u
t2 =c2
2u
x2, 0< x < L
satisfyingu(0, t) = 0, u(L, t) = 0, t 0.
are
un(x, t) =Fn(x)Gn(t) = (Bncos nt+B
nsin nt)sinn
Lx (n= 1, 2, . . .).
These functions are called eigenfunctionsand
n=cn
L
are called eigenvalues.Now we can solve the entire problem by
setting
u(x, t) =n=1
un(x, t) =n=1
(Bncos nt+B
nsin nt)sin
n
Lx.
Satisfy the initial conditions:
u(x, 0) =n=1
Bnsinn
Lx= f(x).
Thus,
Bn= 2
L
L
0f(x)sin
n
Lxdx, n= 1, 2, . . . .
u
t
t=0
=
n=1(Bnnsin nt+B
nncos nt)sin
n
Lx
t=0
=
n=1
Bn
nsinn
Lx= g(x).
Thus,
Bn
n= 2
L
L
0g(x)sin
n
Lxdx,
so that
Bn
= 2
cn
L
0g(x)sin
n
Lxdx, n= 1, 2, . . . .
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Analyze the solution in the case when the initial velocity g(x)
0. Then allBn
= 0 and
u(x, t) =n=1
Bncos nt sinn
Lx, n=
cn
L .
We have
coscnL
t sinnL
x= 12
sinn
L(x ct) + sinn
L(x+ct)
and
u(x, t) =1
2
n=1
Bnsinn
L(x ct) +
1
2
n=1
Bnsinn
L(x+ct)
so that
u(x, t) =1
2[f(x ct) +f(x+ct)]
EXAMPLE 1 Vibrating string if the initial deflection is
triangular.
Find the solution of the wave equation corresponding to the
triangular initial deflection
f(x) =
2kL
x if 0< x < L/2,2kL
(L x) ifL/2< x < L.
Solution. For the odd periodic extension off(x) the Fourier
coefficients are
Bn= 2
L
L
0f(x)sin
n
Lxdx=
8k
n22sin
n
2 =
0 ifn = 2l,(1)l+1 8k(2l1)22 ifn = 2l 1.
, l= 1, 2, . . . .
and the desired solution of the wave equation is
u(x, t) =n=1
Bncos nt sinn
Lx=
8k
2
l=1
(1)l+1
(2l 1)2cos 2l1t sin
(2l 1)
L x=
8k
2
1
1sin
Lx cos
c
Lt
1
9sin
3
Lx cos
3c
L t+. . .
.
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PROBLEM 11.3.1
Findu(x, t) of the string of length L = whenc2 = 1, the initial
velocity is zero and the initialdeflection is
f(x) =k(sin x1
2sin 2x)
Solution. f(x) is given in the form of a finite trigonometric
series consisting of two sineterms with n = 1, 2 and corresponding
coefficients
B1= k, B2= k 12
.Also, n= cnL
=n,
so that
u(x, t) =n=1
Bncos nt sinn
Lx= B1cos 1t sin x+B2cos 2t sin2x= k(cos t sin x
1
2cos 2t sin2x)
PROBLEM 11.3.7
Find u(x, t) of the string of length L= when c2 = 1, the initial
velocity is zero and theinitial deflection is triangular (k= 0.5,
EX. 1),
f(x) =
1
x if 0< x < /2,1
(/2 x) if/2< x < .
Solution. For the odd periodic extension off(x) the Fourier
coefficients are
Bn= 2
0f(x)sin nxdx=
4
n22sin
n
2 ) =
0 ifn= 2l,(1)l+1 4(2l1)22 ifn= 2l 1.
, l= 1, 2, . . . .
Also,n=
cn
L =n,
so that the desired solution of the wave equation is
u(x, t) =n=1
Bncos nt sinn
Lx=
4
2
l=1
(1)l+1
(2l 1)2cos 2l1t sin
(2l 1)
L x=
4
2
sin x cos t
1
9sin 3x cos3t+
1
25sin 5x cos5t . . .
.
