Math 201 Assignment #11
Problem 1 (10.5 2) Find a formal solution to the given initial-boundary value problem.
∂u
∂t
u(0, t)
u(x, 0)
=
∂2u
, 0 < x < π, t > 0
∂x2
u(π, t) = 0, t > 0
=
x2 , 0 < x < π
=
Problem 2 (10.5 5) Find a formal solution to the given initial-boundary value problem.
∂u
∂t
∂u
(0, t)
∂x
u(x, 0)
=
=
=
∂2u
, 0 < x < π, t > 0
∂x2
∂u
(π, t) = 0, t > 0
∂x
ex , 0 < x < 1
Problem 3 (10.5 8) Find a formal solution to the given initial-boundary value problem.
∂u
∂t
u(0, t)
u(x, 0)
=
∂2u
, 0 < x < π, t > 0
∂x2
0, u(π, t) = 3π, t > 0
=
0, 0 < x < π
=
Problem 4 (10.5 10) Find a formal solution to the given initial-boundary value problem.
∂u
∂t
u(0, t)
u(x, 0)
=
∂2u
+ x, 0 < x < π, t > 0
∂x2
u(π, t) = 0, t > 0
=
sin(x), 0 < x < π
=
3
Problem 5 (10.5 14) Find a formal solution to the given initial-boundary value problem.
∂u
∂t
u(0, t)
u(x, 0)
=
∂2u
+ 5, 0 < x < π, t > 0
∂x2
u(π, t) = 1, t > 0
=
1, 0 < x < π
=
3
1
Problem 6 (10.6 2) Find a formal solution.
∂2u
∂t2
u (0, t) = u (π, t)
u (x, 0)
∂u
(x, 0)
∂t
(Ans.
P∞
n=1
=
∂2u
,
0 < x < π, t > 0
∂x2
0,
t>0
=
sin2 x,
=
1 − cos x,
=
16
[an cos (4nt) + bn sin (4nt)] sin (nx),
(
− π8 n n21−4
n odd
(
)
an =
,
0
n even
0<x<1
bn =
0 < x < π.
(
1
πn2
−π
1
(n2 −1)
n odd
n even .
Problem 7 (10.6 8) Find a formal solution.
∂2u
∂t2
u (0, t) = u (π, t)
u (x, 0)
∂u
(x, 0)
∂t
(Ans. [sin t − t cos t] sin x +
2(−1)n
n=2 n2 (n2 −1)
P∞
=
∂2u
+ x sin t,
∂x2
0,
t>0
=
0,
0 < x < π,
=
0,
0 < x < π.
=
0 < x < π, t > 0.
[sin (nt) − n sin t] sin (nx) .)
Problem 8 (10.6 10) Derive a formal formula for the solution.
∂2u
∂t2
u (0, t) = U1
u (x, 0)
∂u
(x, 0)
∂t
=
∂2u
,
0 < x < L,
∂x2
u (L, t) = U2 ,
t>0
α2
=
f (x) ,
0 < x < L,
=
g (x) ,
0 < x < L,
t>0
where U1 , U2 are constants. (Ans.
„
«
„
«–
∞ »
“ nπx ”
X
nπαt
nπαt
U2 − U1
an cos
+ bn sin
sin
x+
L
L
L
L
n=1
»
–ff
Z L
“
”
2
nπx
U2 − U1
an =
f (x) − U1 +
x sin
dx,
L 0
L
L
Z L
“ nπx ”
2
bn =
g (x) sin
dx.
nπα 0
L
u (x, t) = U1 +
2
Problem 9 (10.6 14) Consider
∂2u
∂t2
u (x, 0)
∂u
(x, 0)
∂t
=
∂2u
,
∂x2
f (x) ,
−∞ < x < ∞
=
g (x) ,
−∞ < x < ∞
=
α2
−∞ < x < ∞,
t>0
with f (x) = x2 , g (x) = 0. (Ans. x2 + α2 t2 )
Problem 10 (10.6 16) Same as above, but with f (x) = sin 3x, g (x) = 1.
(Ans. sin (3x) cos (3αt) + t.)
3
Solution 1. (10.5 2) Using separation of variables, assume
u(x, t) = X(x)T (t)
then
∂u
∂t
∂2u
∂x2
= X(x)T ′ (t)
= X ′′ (x)T (t)
Substituting back into the PDE we have
X(x)T ′ (t) = X ′′ (x)T (t)
X ′′ (x)
T ′ (t)
=
= K, K constant
⇒
T (t)
X(x)
½ ′′
X (x) − KX(x) = 0
⇒
T ′ (t) − KT (t)
= 0
Consider the boundary conditions
u(0, t) = u(π, t) = 0
then
X(0)T (t) = X(π)T (t) = 0
so either T (t) = 0, the trivial solution, or
X(0) = X(π) = 0
Solving the boundary value problem
X ′′ (x) − KX(x) = 0, X(0) = X(π) = 0
this is second order linear homogeneous equation with auxiliary equation r2 − K = 0 We investigate
three possible cases for K.
√
Case 1: K > 0 ⇒ r = ± K, real and distinct roots.
This has the general solution
√
√
X(x) = c1 e Kx + c2 e− Kx
Using the boundary conditions:
X(0) = c1 + c2 = 0 ⇒ c2 = −c1
³ √
´
√
√
√
X(π) = c1 e Kπ − c1 e− Kπ = c1 e− Kπ e2 Kπ − 1 = 0
Since K > 0, we must have c1 = 0, which is the trivial solution.
4
Case 2: K = 0 ⇒ r = 0, double root.
This has the general solution
X(x) = c1 + c2 x
Using the boundary conditions:
X(0) = c1 = 0
X(π) = c2 π = 0 ⇒ c2 = 0
This case also gives us the trivial solution.
√
Case 3: K < 0 ⇒ r = ± −Ki, complex roots.
This has the general solution
√
√
X(x) = c1 cos( −Kx) + c2 sin( −Kx)
Using the boundary conditions:
X(0) = c1 = 0
√
X(π) = c2 sin( −Kπ) = 0
Either c2 = 0 (the trivial solution) or
√
sin( −Kπ) = 0
√
⇒ −Kπ = nπ
⇒K
= −n2 , n = 1, 2, . . .
These eigenvalues have the corresponding eigenfunctions
Xn (x) = cn sin(nx)
Using the eigenvalues in the equation for T (t):
T ′ (t) − KT (t) =
⇒ T ′ (t) + n2 T (t) =
0
0
⇒ Tn (t) = an e−n
2
t
Putting the equations for X(x) and T (t) together and combining the constants, we have
un (x, t) = Xn (x)Tn (t) = bn sin(nx)e−n
2
t
Since un (x, t) is a solution to the PDE for any value of n, we take the infinite series
u(x, t) =
∞
X
bn sin(nx)e−n
2
n=1
From the general solution and initial condition, we have
u(x, 0)
=
∞
X
n=1
2
= x
5
bn sin(nx)
t
which means we can choose the bn ’s as the coefficients in the Fourier sine series for f (x) = x2 :
Z
2 π 2
bn =
x sin(nx)dx
π 0
·
¸π
2 −n2 x2 cos(nx) + 2 cos(nx) + 2nx sin(nx)
=
π
n3
0
¢
2 ¡
2 2
n
=
(2 − n π )(−1) − 2
πn3
Thus the formal solution to the given initial-boundary value problem is
∞
¢
2
2X 1 ¡
(2 − n2 π 2 )(−1)n − 2 sin(nx)e−n t
u(x, t) =
π n=1 n3
6
Solution 2. (10.5 5) Following a similar argument as for 10.5 2, we use separation of variables
and the boundary conditions to arrive at the system of ODEs:
½ ′′
X (x) − KX(x) = 0, X ′ (0) = X ′ (π) = 0
T ′ (t) − KT (t)
= 0
Consider the boundary value problem:
X ′′ (x) − KX(x) = 0, X ′ (0) = X ′ (π) = 0
This has the auxiliary equation
Again we consider three cases as above.
r2 − K = 0
Case 1: K > 0
Using a similar argument as in 10.5 2, the only solution in this case is the trivial solution.
Case 2: K = 0
Here
X(x) = c0 + c1 x ⇒ X ′ (x) = c1
Using the boundary conditions:
X ′ (0) = c1 = 0
and c0 is arbitrary. So here we have the nontrivial solution
X(x) = c0
Case 3: K < 0
In this case the general solution is
X(x)
⇒ X ′ (x)
√
√
= c1 cos( −Kx) + c2 sin( −Kx)
√
√
√
√
= −c1 −K sin( −Kx) + c2 −K cos( −Kx)
Using the boundary conditions:
X ′ (0) = c2 = 0
√
√
X ′ (π) = −c1 −K sin( −Kπ) = 0
Either c1 = 0 (trivial solution) or
√
sin( −Kπ) = 0
⇒ K = −n2
These eigenvalues have the corresponding eigenfunctions
Xn (x) = cn cos(nx)
Combining cases 2 and 3, we have the eigenvalues and eigenfunctions
K
Xn (x)
= −n2
= cn cos(nx)
7
for n = 0, 1, 2, . . .
Returning to the equation for T (t):
T ′ (t) − KT (t) =
⇒ T ′ (t) + n2 T (t) =
0
0
⇒ Tn (t) = an e−n
2
t
Putting the equations for X(x) and T (t) together and merging the constants, we have
2
un (x, t) = an cos(nx)e−n t , n = 0, 1, 2, . . .
Taking the infinite series of these solutions, we have
u(x, t) =
∞
2
a0 X
an cos(nx)e−n t
+
2
n=1
From the general solution and the initial condition, we have
u(x, 0)
=
∞
a0 X
an cos(nx)
+
2
n=1
= ex
which means we can choose the an ’s as the coefficients in the Fourier cosine series for f (x) = ex :
Z
2 π x
2
a0 =
e dx = (eπ − 1)
π 0
π
an
=
=
Z
2 π x
e cos(nx)dx
π 0
µ
¶
2 eπ (−1)n − 1
π
n2 + 1
Thus the formal solution to the given initial-boundary value problem is
u(x, t) =
∞
2
1 π
2 X eπ (−1)n − 1
(e − 1) +
cos(nx)e−n t
2
π
π n=1
n +1
8
Solution 3. (10.5 8) We assume the solution consists of a steady-state solution v(x) and a
transient solution w(x, t) so that
u(x, t) = v(x) + w(x, t)
where w(x, t) and its derivatives tend to zero as t → ∞. Then
∂u
∂t
∂2u
∂x2
=
∂w
∂t
= v ′′ (x) +
∂2w
∂x2
Substituting these back into the original PDE, we obtain the problem
∂w
∂t
v(0) + w(0, t)
=
∂2w
, 0 < x < π, t > 0
∂x2
0, v(π) + w(π, t) = 3π, t > 0
(2)
=
0, 0 < x < π
(3)
= v ′′ (x) +
v(x) + w(x, 0)
Letting t → ∞ we obtain the steady-state boundary value problem
v ′′ (x)
=
0, 0 < x < π
v(0)
=
0, v(π) = 3π
Integrating twice, we find
v(x) = c1 x + c2
Using the boundary conditions, we find
v(0) = c2 = 0
Thus
v(π) = c1 π = 3π ⇒ c1 = 3
v(x) = 3x
Substituting this back into equations (1)-(3) we have
∂w
∂t
w(0, t)
w(x, 0)
∂2w
, 0 < x < π, t > 0
∂x2
= w(π, t) = 0, t > 0
= −3x, 0 < x < π
=
Following a similar argument as in 10.5 2, we have the general solution
w(x, t) =
∞
X
bn sin(nx)e−n
2
t
n=1
From the general solution and the initial condition w(x, 0) we have
w(x, 0)
=
∞
X
bn sin(nx)
n=1
= −3x
9
(1)
which means we can choose the bn ’s as the coefficients in the Fourier sine series for g(x) = −3x:
Z
2 π
bn =
−3x sin(nx)dx
π 0
6
(−1)n
=
n
Thus,
w(x, t) = 6
∞
X
2
(−1)n
sin(nx)e−n t
n
n=1
and the solution to the initial-boundary value problem is
u(x, t)
= v(x) + w(x, t)
∞
X
2
(−1)n
= 3x + 6
sin(nx)e−n t
n
n=1
10
Solution 4. (10.5 10) As in question 10.5 8, assume the solution consists of a steady-state
solution and a transient solution:
u(x, t) = v(x) + w(x, t)
where w(x, t) and its derivatives tend to zero as t → ∞. Substituting into the original PDE we
have
¶
µ
∂w
∂2w
+ x, 0 < x < π, t > 0
= 3 v ′′ (x) +
∂t
∂x2
v(0) + w(0, t) = v(π) + w(π, t) = 0, t > 0
v(x) + w(x, 0)
=
sin(x), 0 < x < π
Letting t → ∞, we obtain the steady-state boundary value problem
x
v ′′ (x) = −
3
v(0) = v(π) = 0
Integrating twice to find v(x):
v(x) = −
1 3
x + c1 x + c2
18
Using the boundary conditions to find the constants:
v(0) = c2 = 0
v(π) = −
π2
1 3
π + c1 π = 0 ⇒ c1 =
18
18
Thus
1 3 π2
x +
x
18
18
We then have the following problem for w(x, t):
v(x) = −
∂w
∂t
w(0, t)
w(x, 0)
∂2w
, 0 < x < π, t > 0
∂x2
= w(π, t) = 0, t > 0
1
π2
= sin(x) + x3 −
x, 0 < x < π
18
18
= 3
Following a similar argument as in question 10.5 2, we have the general solution
w(x, t) =
∞
X
bn sin(nx)e−3n
2
t
n=1
From the general solution and the initial condition for w(x, 0), we have
w(x, 0)
=
∞
X
bn sin(nx)
n=1
=
sin(x) +
11
1 3 π2
x −
x
18
18
which means we can choose the bn ’s as the coefficients in the Fourier sine series for g(x) = sin(x) +
1 3
π2
18 x − 18 x:
bn
=
=
=
b1
=
=
¶
Z µ
1 3 π2
2 π
sin(x) + x −
x sin(nx)dx
π 0
18
18
µ
¶
2 −3π cos(nπ) + 3n2 π cos(nπ)
Note: this is not valid for n = 1
π
9n3 (n2 − 1)
2
(−1)n
3n3 µ
¶
Z
1
π2
2 π
sin(x) + x3 −
x sin(x)dx
π 0
18
18
1
3
Thus,
w(x, t) =
∞
X
2
1
2
sin(x)e−3t +
(−1)n sin(nx)e−3n t
3
3
3n
n=2
and the formal solution to the initial-boundary value problem is
u(x, t)
= v(x) + w(x, t)
=
∞
2
π2
1
1
2 X (−1)n
x − x3 + sin(x)e−3t +
sin(nx)e−3n t
18
18
3
3 n=2 n3
12
Solution 5. (10.5 14) As in question 10.5 8, assume the solution consists of a steady state
solution and a transient solution:
u(x, t) = v(x) + w(x, t)
where w(x, t) and its derivatives tend to zero as t → ∞. Substituting into the original PDE we
have
¶
µ
∂w
∂2w
′′
+ 5, 0 < x < π, t > 0
(4)
= 3 v (x) +
∂t
∂x2
v(0) + w(0, t) = v(π) + w(π, t) = 1, t > 0
(5)
v(x) + w(x, 0)
=
1, 0 < x < π
(6)
Letting t → ∞ we obtain the steady-state boundary value problem
5
3
= v(π) = 1
v ′′ (x)
= −
v(0)
Integrating twice, we find
5
v(x) = − x2 + c1 x + c2
6
Using the boundary conditions:
v(0) = c2 = 1
5
5
v(π) = − π 2 + c1 π + 1 = 1 ⇒ c1 = π
6
6
Thus
5π
5
x − x2
6
6
Substituting this back into equations (4)-(6), we have
v(x) = 1 +
∂w
∂t
w(0, t)
w(x, 0)
∂2w
, 0 < x < π, t > 0
∂x2
= w(/pi, t) = 0, t > 0
5
5π
= − x + x2 , 0 < x < π
6
6
=
3
Following a similar argument as in question 10.5 2, we have the general solution
w(x, t) =
∞
X
bn sin(nx)e−3n
2
t
n=1
From the general solution and the initial condition for w(x, 0), we have
w(x, 0)
=
∞
X
bn sin(nx)
n=1
= −
13
5π
5
x + x2
6
6
which means we can choose the bn ’s as the coefficients in the Fourier since series for g(x) = − 5π
6 x+
5 2
x
:
6
¶
Z µ
2 π
5π
5
bn =
− x + x2 sin(nx)dx
π 0
6
6
10
((−1)n − 1)
=
3n3 π
Here, cn = 0 when n is even, and for n odd
c2k+1 = −
20
, k = 0, 1, 2, . . .
3(2k + 1)3 π
Thus,
w(x, t) = −
∞
2
20 X
1
sin(nx)e−3n t
3π
(2k + 1)3
k=0
and the formal solution to the initial-boundary value problem is
u(x, t)
= v(x) + w(x, t)
=
1+
∞
2
5π
5
20 X
1
sin ((2k + 1)x) e−3(2k+1) t
x − x2 −
6
6
3π
(2k + 1)3
k=0
14
Solution 6. (10.6 2) We give a complete solution here, with every step included. Some steps will
be omitted in the following problems.
1. Separate variables
Plug in u = X (x) T (t) gives
T ′′ X = 16X ′′ T =⇒
X ′′
T ′′
= 16
.
T
X
As the left is a function of t alone and the right is a function of x alone, that they are equal
for all x, t means both are constants. Call it λ.
We get
T ′′ − 16λT = 0,
X ′′ − λX = 0.
2. Solve the eigenvalue problem.
The X equation combined with boundary conditions
u (0, t) = 0 =⇒ X (0) T (t) = 0 =⇒ X (0) = 0;
u (π, t) = 0 =⇒ X (π) T (t) = 0 =⇒ X (π) = 0.
gives the eigenvalue problem
X ′′ − λX = 0,
X (0) = X (π) = 0.
Recall that an “eigenvalue” is a specific value of λ such that the above problem has nonzero
solutions, and these nonzero solutions (clearly dependent on λ!) are the corresponding eigenfunctions.
i. Write down the general solutions to the equation: The formulas for the general solutions
depend on the sign of λ.
• λ > 0,
√
X = C1 e
• λ = 0,
• λ < 0,
λx
+ C2 e−
√
λx
;
X = C1 + C2 x;
X = C1 cos
√
√
−λx + C2 sin −λx.
ii. Check whether there are any λ’s such that the corresponding solutions can satisfy the
boundary conditions while being nonzero (that is at least one of C1 , C2 is nonzero.
• Any λ > 0?
X (0) = 0
=⇒ C1 + C2 = 0;
X (π) = 0
=⇒ C1 eπ
√
λ
+ C2 e−π
√
λ
= 0.
These two requirements combined =⇒ C1 = C2 = 0. So there is no positive eigenvalue.
15
• Is λ = 0 an eigenvalue?
X (0) = 0
X (π) = 0
=⇒ C1 = 0
=⇒ C1 + C2 π = 0
Combined we have C1 = C2 = 0. So 0 is not an eigenvalue.
• Any λ < 0?
X (0) = 0
=⇒ C1 = 0;
√
√
=⇒ C1 cos −λπ + C2 sin −λπ = 0.
X (π) = 0
Combine these two we have
√
C2 sin −λπ = 0.
C1 = 0;
Now “at least one of C1 , C2 is nonzero” is equivalent to
√
√
sin −λπ = 0 ⇐⇒ −λπ = nπ
for some integer n. This then becomes
λ = −n2
for integer n. Notice that 1. n and −n give the same λ; 2. we are discussing the case
λ < 0. We finally conclude that
λn = −n2 ,
n = 1, 2, 3, . . .
√
√
The corresponding Xn are C1 cos −λx + C2 sin −λx with C1 = 0 and λ = λn :
Xn = An sin nx,
n = 1, 2, 3, . . .
Summary: The eigenvalues are λn = −n2 , eigenfunctions are Xn = An sin nx, and the range
of n is 1, 2, 3, . . ..
3. Solve for Tn . Recall that T ′′ − 16λT = 0. With the particular λn ’s, we have
Tn′′ + 16n2 Tn = 0
which gives
Tn = Dn cos (4nt) + En sin (4nt)
with Dn , En arbitrary constants and n ranging 1 to ∞.
P
4. Write down u = cn Xn Tn :
u (x, t) =
∞
X
cn An sin (nx) [Dn cos (4nt) + En sin (4nt)]
n=1
which simplifies to
u (x, t) =
∞
X
[an cos (4nt) + bn sin (4nt)] sin (nx) .
n=1
16
5. Determine an , bn through initial conditions.
The above formula gives
u (x, 0) =
∞
X
an sin (nx)
n=1
and
∞
X
∂u
4nbn sin (nx) .
(x, 0) =
∂t
n=1
Comparing with initial conditions
∂u
(x, 0) = 1 − cos x,
∂t
u (x, 0) = sin2 x,
we have
sin2 x =
1 − cos x =
∞
X
n=1
∞
X
an sin (nx)
4nbn sin (nx) .
n=1
In other words an and 4nbn are coefficients for the Fourier Sine expansion of sin2 x and
1 − cos x, respectively.
• Fourier Sine expansion of sin2 x. We have L = π. So
Z
2 π 2
sin x sin (nx) dx
an =
π 0
Z
2 π 1 − cos 2x
sin (nx) dx
=
π 0
2
¸
·Z π
Z π
1
=
cos 2x sin (nx) dx .
sin nxdx −
π 0
0
We evaluate the two integrals.
Z π
sin nxdx =
0
¯π
¯
1
− cos nx¯¯
n
0
1
= − [cos (nπ) − 1]
n
½
n
1 − (−1)
0
=
=
2
n
n
17
n even
.
n odd
Z
π
cos 2x sin (nx) dx =
0
=
=
=
=
=
=
π
sin [(n + 2) x] + sin [(n − 2) x]
dx
2
0
·
¸¯π
1 cos (n + 2) x cos (n − 2) x ¯¯
+
−
¯
2
n+2
n−2
"
#0
n+2
n−2
− 1 (−1)
−1
1 (−1)
+
−
2
n+2
n−2
·
¸
n
n
1 (−1) − 1 (−1) − 1
+
−
2
n+2
n−2
n
1 − (−1)
2n
2
n2 − 4
n
[1 − (−1) ] n
n2 − 4
½
0
n even
.
2n
n odd
n2 −4
Z
Note that, the above calculation is only correct when n 6= 2, as when n = 2 dividing by
n − 2 becomes meaningless. However checking the n = 2 case separately, we see that the
result is 0 so the above is in fact true for all n.
Thus we have
½ 8
− π n(n21−4) n odd
.
an =
0
n even
• Fourier Sine expansion of 1 − cos x. We have
Z
2 π
(1 − cos x) sin (nx) dx
4nbn =
π 0
Z π
Z
2 π
2
sin (nx) dx −
cos x sin (nx) dx.
=
π 0
π 0
We have
Z
π
sin nxdx =
0
18
½
0
2
n
n even
.
n odd
and
Z
π
cos x sin (nx) dx =
0
=
=
=
=
=
1
2
Z
π
[sin (n + 1) x + sin (n − 1) x] dx
¸¯π
1 cos (n + 1) x cos (n − 1) x ¯¯
+
−
¯
2
n+1
n−1
"
#0
n+1
n−1
− 1 (−1)
−1
1 (−1)
+
−
2
n+1
n−1
³
´
n−1
1 − (−1)
2n
2
2
n −1
´ n
³
n−1
1 − (−1)
n2 − 1
½
0
n odd
2n
n even.
2
n −1
0
·
Again we need to discuss the n = 1 case separately. In the case
Z π
Z π
cos x sin x = 0
cos x sin (nx) dx =
0
0
so the general formula still holds.
Thus we have
½
bn =
1
πn2
− π(n21−1)
19
n odd
.
n even
Solution 7. (10.6 8)
1. Separate variables. Writing u = X (x) T (t) and plug into equation, neglecting x sin t for now:
T ′′ − λT = 0;
X ′′ − λX = 0.
2. Solve eigenvalue problem. The eigenvalue problem is
X ′′ − λX = 0,
X (0) = X (π) = 0
which leads to
λn = −n2 ;
3. Write down u:
u (x, t) =
Xn = An sin (nx) ;
∞
X
cn Tn sin (nx) =
∞
X
n = 1, 2, 3, . . .
Tn (t) sin (nx) .
n=1
n=1
Note that we have integrated cn into Tn .
4. Apply the initial conditions and integrate the forcing:
u (x, 0) = 0 =⇒
∂u
(x, 0) = 0 =⇒
∂t
∂2u
∂2u
=
+ x sin t =⇒
∂t2
∂x2
Thus we must have
∞
X
n=1
∞
X
n=1
∞
X
n=1
Tn (0) sin (nx) = 0 =⇒ Tn (0) = 0;
Tn′ (0) sin (nx) = 0 =⇒ Tn′ (0) = 0;
Tn′′ sin (nx) = −
∞
X
n2 Tn sin (nx)
n=1
∞
X
£ ′′
¤
Tn + n2 Tn sin (nx) = x sin t.
+x sin t.
n=1
To determine Tn , we need to expand x into sin(nx)’s.
P∞
5. Expand x. x = n=1 an sin (nx) with
Z
2 π
x sin (nx) dx
an =
π 0
Z π
2
= −
xd [cos (nx)]
nπ 0
¸
·
Z π
2
π
cos (nx) dx
cos (nx) x|0 −
= −
nπ
0
2
n
= −
[π (−1) ]
nπ
n+1
2 (−1)
=
.
n
20
6. Solve Tn . We have
∞
∞
X
X
£ ′′
¤
Tn + n2 Tn sin (nx) =
n=1
n=1
Ã
2 (−1)
n
!
n+1
sin t sin (nx)
which gives
n+1
2 (−1)
n
Together with initial conditions Tn (0) = Tn′ (0) = 0.
Tn′′ + n2 Tn =
sin t.
First solve the homogeneous equation:
Tn′′ + n2 Tn = 0 =⇒ Tn = C1 cos (nt) + C2 sin (nt) .
Next we try to find a particular solution. We use undetermined coefficients. The form of the
particular solution is
Tp = ts [A sin t + B cos t] .
Where s is the number of times sin t appears in the general solution of the homogeneous
problem. There are two cases:
• n = 1. In this case s = 1. Substitute Tp = t [A sin t + B cos t] into the equation we find
out
A = 0,
B = −1.
So the particular solution is
Tp = −t cos t.
The general solution to the non-homogeneous problem is then
Tn = C1 cos t + C2 sin t − t cos t.
Applying T (0) = T ′ (0) = 0 we reach
C1 = 0,
C2 = 1.
So T1 = sin t − t cos t.
• n 6= 1. In this case s = 0 and
Tp = A sin t + B cos t.
Substituting into equation, we have
n+1
A=
2 (−1)
,
n (n2 − 1)
B = 0.
So
n+1
Tn = C1 cos (nt) + C2 sin (nt) +
21
2 (−1)
sin t.
n (n2 − 1)
Applying T (0) = T ′ (0) = 0 we reach
n
C1 = 0,
Thus
C2 =
2 (−1)
.
n2 (n2 − 1)
n
Tn =
2 (−1)
[sin (nt) − n sin t] .
2
n (n2 − 1)
7. Write down solution. Finally we have
u (x, t) =
∞
X
Tn Xn
n=1
= T1 X1 +
∞
X
T n Xn
n=2
=
[sin t − t cos t] sin x
∞
n
X
2 (−1)
[sin (nt) − n sin t] sin (nx) .
+
n2 (n2 − 1)
n=2
22
Solution 8. (10.6 10)
1. Take care of the boundary conditions. We need to find an appropriate function w such that
it satisfies both the equation and the boundary conditions, and then set v = u − w. The
boundary conditions for v would be v (0, t) = v (L, t) = 0 and separation of variables can then
be applied.
The first try is usually w = w (x) independent of t. Then w must satisfy
0 = α2 w′′ ,
w (0) = U1 , w (L) = U2 .
Such w exists. We have
w (x) = U1 +
U2 − U1
x.
L
2. Now set v = u − w. This gives u = v + w and the equation becomes
2
2
∂ 2 (v + w)
∂2v
2 ∂ (v + w)
2∂ v
=
α
⇐⇒
=
α
,
∂t2
∂x2
∂t2
∂x2
the boundary conditions are
v (0, t) = v (L, t) = 0,
and the initial condition becomes
¸
·
U2 − U1
x ,
v (x, 0) = f (x) − w = f (x) − U1 +
L
∂v
∂w
(x, 0) = g (x) −
(x, 0) = g (x) .
∂t
∂t
So we need to solve
∂2v
∂t2
v (0, t) = v (L, t)
v (x, 0)
∂v
(x, 0)
∂t
= α2
=
∂2v
∂x2
0,
·
¸
U2 − U1
= f (x) − U1 +
x
L
= g (x) .
3. Solve v. The solution is given by
µ
¶
µ
¶¸
∞ ·
³ nπx ´
X
nπαt
nπαt
an cos
v (x, t) =
+ bn sin
sin
L
L
L
n=1
with
an =
2
L
L
·
¸¾
³ nπx ´
U2 − U1
dx,
f (x) − U1 +
x sin
L
L
0
Z L
³ nπx ´
2
bn =
g (x) sin
dx.
nπα 0
L
Z
½
23
4. Write down u. Recalling u = v + w we have
¶
µ
¶¸
µ
∞ ·
³ nπx ´
X
U2 − U1
nπαt
nπαt
u (x, t) = U1 +
+ bn sin
sin
an cos
x+
L
L
L
L
n=1
with
an =
2
L
L
·
¸¾
³ nπx ´
U2 − U1
dx,
f (x) − U1 +
x sin
L
L
0
Z L
³ nπx ´
2
bn =
g (x) sin
dx.
nπα 0
L
Z
½
24
Solution 9. (10.6 14) As −∞ < x < ∞ we need to apply d’Alembert formula:
u (x, t) =
f (x − αt) + f (x + αt)
1
+
2
2α
Z
x+αt
g (y) dy.
x−αt
Substituting f = x2 , g = 0 we have
2
u (x, t)
2
(x − αt) + (x + αt)
2
= x2 + α2 t2 .
=
25
Solution 10. (10.6 16) As f = sin 3x, g = 1 we have
u (x, t)
=
=
sin [3 (x − αt)] + sin [3 (x + αt)]
1
+
2
2α
sin (3x) cos (3αt) + t.
26
Z
x+αt
x−αt
1dx