Homework #9 Solutions
Math 182, Spring 2016
Instructor: Dr. Doreen De Leon
Exercises 4.4: 6, 8
In problems 6 and 8, the question asks you to solve the nonhomogeneous heat initial-boundary-value
problem. In each case, rather than following the example in the text, start from scratch by assuming a
solution of the correct form, as in (4.11). Also, calculate the steady state temperature, lim u(x, t); if
t→∞
the limit doesn’t exist, describe the system’s behavior as t → ∞. (Hint: Use l’Hôpital’s rule and the
fundamental theorem of calculus.)
6.
ut = uxx + 10,
u(x, 0) = 50,
u(0, t) = u(π, t) = 0.
Solution: By the boundary conditions, our solution takes the form
u(x, t) =
∞
X
Tn (t) sin nx.
n=1
Next, we find the Fourier sine series of F (x, t) = 10.
Z
2 π
Fn (t) =
F (x, t) sin nx dx
π 0
Z π
2
=
10 sin nx dx
π 0
π 2
10
=
− cos nx
π
n
0
2
10
=
−
(cos nπ − 1)
π
n
20
=
(1 − cos nπ)
nπ
20
=
(1 − (−1)n ).
nπ
1
(1)
Now, substitute (1) into the PDE to obtain
∞
X
Tn0 (t) sin nx = −
n=1
=
∞
X
n2 Tn (t) sin nx +
n=1
∞
X
∞
X
Fn (t) sin nx
n=1
(−n2 Tn (t) + Fn (t)) sin nx.
n=1
Equating coefficients of like terms gives the ODE
Tn0 (t) + n2 Tn (t) = Fn (t),
R
a linear ODE with integrating factor e
n2 dt
= en
2t
i
d h n2 t
2
e Tn (t) = Fn (t)en t
dt
Z t
2
n2 t
=⇒ e Tn (t) − Tn (0) =
en τ Fn (τ ) dτ.
0
Z t
20
2
2
en t Tn (t) =
en τ (1 − (−1)n ) dτ + Tn (0)
nπ 0
20
2 t
= 3 (1 − (−1)n ) en τ + Tn (0)
n π
0
2
20
= 3 (1 − (−1)n ) en t − 1 + Tn (0)
n π
20
2
2
=⇒ Tn (t) = 3 (1 − (−1)n )(1 − e−n t ) + Tn (0)e−n t ,
n π
=⇒
where Tn (0) are determined from the initial condition.
50 = u(x, 0) =
∞
X
Tn (0) sin nx,
n=1
and, so, Tn (0) = bn , the Fourier sine series coefficients of 50 on 0 ≤ x ≤ π.
Z
2 π
Tn (0) =
50 sin nx dx
π 0
100
=−
(cos nπ − 1)
πn
100
=
(1 − (−1)n ),
nπ
∞ 100
X
20
n
−n2 t
n −n2 t
u(x, t) =
(1
−
(−1)
)
1
−
e
+
(1
−
(−1)
)
e
sin nx,
n3 π
nπ
n=1
or
∞
20 X 1
1
n
−n2 t
−n2 t
u(x, t) =
(1 − (−1) ) 2 1 − e
+ 5e
sin nx.
π
n
n
n=1
2
Steady state solution:
∞
1
20 X 1
−n2 t
−n2 t
n
+ 5e
sin nx.
(1 − (−1) ) 2 1 − e
lim u(x, t) = lim
t→∞
t→∞ π
n
n
n=1
∞
20 X 1
1
n
−n2 t
−n2 t
=
(1 − (−1) ) sin nx lim
(1 − e
) + 5e
t→∞ n2
π
n
=
20
π
n=1
∞
X
n=1
1
(1 − (−1)n ) sin nx
n3
8.
ut = uxx + cos 2x − cos 5x,
u(x, 0) = cos 3x,
ux (0, t) = ux (π, t) = 0.
Solution: The boundary conditions suggest that our solution atkes the form
u(x, t) = T0 (t) +
∞
X
Tn (t) cos nx.
(2)
n=1
Next, we find the Fourier cosine series of F (x, t) = cos 2x − cos 5x. We could directly integrate to
determine the coefficients for the Fourier cosine series, but since we see that we have
∞
F0 (t) X
+
Fn (t) cos nx,
cos 2x − cos 5x =
2
n=1
we see that simply equating coefficients of corresponding terms gives us
F0 (t) = 0, F2 (t) = 1, F5 (t) = −1, and Fn (t) = 0 if n 6= 2, 5.
Therefore, the Fourier cosine series of F (x, t) is, in fact, F (x, t).
Now, substitute (2) into the PDE to obtain
T00 (t)
+
∞
X
Tn0 (t) cos nx
=−
n=1
=
∞
X
2
n Tn (t) cos nx +
n=1
∞
X
∞
X
Fn (t) cos nx
n=1
(−n2 Tn (t) + Fn (t)) cos nx.
n=1
Equating coefficients of like terms gives us the following ODEs.
T00 (t) = 0 =⇒ T0 (t) = T0 (0),
Tn0 (t) = −n2 Tn (t) + Fn (t),
=⇒ T20 (t) + 4T2 (t) = 1,
T50 (t) + 25T5 (t) = −1,
Tn0 (t) + n2 Tn (t) = 0, n 6= 2, 5.
3
R
The last three equations can all be viewed as linear ODEs with integrating factor e
So, we have
n2 dt
2
= en t .
d T2 (t)e4t = e4t
dt
Z t
=⇒ T2 (t)e4t =
e4τ dτ + T2 (0)
0
1
=⇒ T2 (t) =
1 − e−4t + T2 (0)e−4t .
4
d T5 (t)e25t = −e25t .
dt
Z t
25t
=⇒ T5 (t)e =
−e25τ dτ + T5 (0)
0
1
−1 + e−25t + T5 (0)e−25t .
=⇒ T5 (t) =
25
i
d h
2
Tn (t)e−n t = 0
dt
Z t
−n2 t
=⇒ Tn (t)e
=
0 dt + Tn (0)
0
2
=⇒ Tn (t) = Tn (0)e−n t .
Applying the initial condition gives
cos 3x = T0 (0) +
∞
X
Tn (0) cos nx,
n=1
and, so, T0 (0) = a0 and Tn (0) = an , the Fourier cosine coefficients of cos 3x on 0 ≤ x ≤ π. But, as
above, we obtain
T0 (0) = a0 = 0,
T3 (0) = 1,
Tn (0) = 0, n 6= 3,
so we obtain
u(x, t) = T0 (t) +
∞
X
Tn (t) cos nx
n=1
= 0 + T2 (t) cos 2x + T5 (t) cos 5x + T3 (t) cos 3x
1
1
1 − e−4t cos 2x +
−1 + e−25t cos 5x + e−9t cos 3x.
=⇒ u(x, t) =
4
25
Steady state solution:
1
1
1 − e−4t cos 2x +
−1 + e−25t cos 5x + e−9t cos 3x
lim u(x, t) = lim
t→∞
t→∞ 4
25
1
1
= cos 2x −
cos 5x.
4
25
4
Exercises 5.1: 2, 6, 8, 17
Exercises 2, 6, 8: Find all possible solutions of the PDE, and check your answer. Then, solve the
initial-value problem, and check your answer, or say why the initial-value problem cannot be solved.
2. 3ux + uy = 0, u(4, y) = e−y
2
Solution: Let ξ = x and η = Ax + By. Then, since ux = uξ + Auη and uy = Buη , the PDE
becomes
3(uξ + Auη ) + Buη = 0 =⇒ 3uξ + (3A + B)uη = 0.
1
If we choose A = 1, B = −3, then ξ = x, η = x − 3y =⇒ x = ξ, y = (ξ − η), and the PDE
3
becomes
3uξ = 0 =⇒ uξ = 0.
The solution to this equation is
u(ξ, η) = g(η) =⇒ u(x, y) = g(x − 3y).
To verify that this solves the PDE, note that ux = g 0 (x−3y) and uy = −3g 0 (x−3y), so 3ux +uy = 0.
Next, we need to determine if we can find a function g so that the initial condition is satisfied.
2
e−y = u(4, y) = g(4 − 3y).
1
Let z = 4 − 3y. Then, y = (4 − z), and
3
1
2
g(z) = e− 9 (4−z)
1
2
=⇒ u(x, y) = e− 9 (4−(x−3y))
1
2
=⇒ u(x, y) = e− 9 (3y−x+4) .
2
It is easy to see that u(4, y) = e−y .
6. 2ux + uy − 5u = 0, u(y 2 , y) = 3(−y 2 + 2y)e
5y 2
2
Solution: Let ξ = x and η = Ax + By. The, since ux = uξ + Auη and uy = Buη , the PDE becomes
2(uξ + Auη ) + Buη − 5u = 0 =⇒ 2uξ + (2A + B)uη − 5u = 0.
1
If we choose A = 1, B = −2, then ξ = x, η = x − 2y =⇒ x = ξ, y = (ξ − η), and the PDE
2
becomes
5
2uξ − 5u = 0 =⇒ uξ − u = 0.
2
5
We can view this either as a separable equation or as a linear equation with integrating factor e− 2 ξ .
Then, we have
5
u(ξ, η) = g(η)e 2 ξ .
This gives
5
u(x, y) = g(x − 2y)e 2 x .
5
5
5
g(x − 2y)e 2 x and uy =
2
5
−2g 0 (x − 2y)e 2 x , so 2ux + uy − 5u = 0. Next, we need to determine if we can find a function g so
that the initial condition is satisfied.
5
To verify that this solves the PDE, note that ux = g 0 (x − 2y)e 2 x +
5 2
5 2
3(−y 2 + 2y)e 2 y = u(y 2 , y) = g(y 2 − 2y)e 2 y .
So, we have
5 2
5 2
3(−y 2 + 2y)e 2 y = g(y 2 − 2y)e 2 y
=⇒ 3(−y 2 + 2y) = g(y 2 − 2y)
=⇒ −3(y 2 − 2y) = g(y 2 − 2y).
Let z = y 2 − 2y. Then, we have
g(z) = −3z
5
=⇒ u(x, y) = −3(x − 2y)e 2 x .
5 2
It is easy to see that u(y 2 , y) = −3(y 2 − 2y)e 2 y .
8. ux + 4uy − 2u = ex+y , u(x, 0) = cos x
Solution: Let ξ = x and η = Ax + By. The, since ux = uξ + Auη and uy = Buη , the PDE becomes
uξ + Auη + 4Buη − 2u = eξ+
η−Aξ
B
.
If we choose A = 4, B = −1, then ξ = x, η = 4x − y =⇒ x = ξ, y = 4ξ − η, and the PDE becomes
uξ − 2u = eξ+(4ξ−η) ,
or
uξ − 2u = e5ξ−η .
We can view this as a linear equation with integrating factor e−2ξ . Then, we have
d −2ξ e u = e5ξ−η e−2ξ
dξ
d −2ξ e u = e3ξ−η
dξ
1
e−2ξ u = e3ξ−η + g(η)
3
1
u(ξ, η) = e5ξ−η + g(η)e2ξ
3
1
=⇒ u(x, y) = ex+y + g(4x − y)e2x .
3
1
To verify that this solves the PDE, note that ux = ex+y + 4g 0 (4x − y)e2x + 2g(4x − y)e2x and
3
1 x+y
0
2x
x+y
uy = e
− g (4x − y)e , so ux + 4uy − 2u = e . Next, we need to determine if we can find a
3
function g so that the initial condition is satisfied.
1
cos x = u(x, 0) = ex + g(4x)e2x .
3
6
So, we have
1
cos x = ex + g(4x)e2x
3
1
=⇒ g(4x) = e−2x cos x − e−x .
3
1
Let z = 4x =⇒ x = z. Then, we have
4
1
g(z) = e− 2 z cos
z 1 −1z
− e 4
4 3
1
1
4x − y 1 − 4x−y
− e 4
=⇒ u(x, y) = ex+y + e− 2 (4x−y) cos
3
4
3
y
1
1
1 x+y
=⇒ u(x, y) = e 2 cos x − y +
e
− ex+ 4 y
4
3
e2x
We can easily verify that u(x, 0) = cos x.
17. Solve the initial-value problem ux − uy = 0, u(x, x3 − x) = x2 . Where is the solution not differentiable? Why does this example not contradict Theorem 5.1?
Solution: Let ξ = x and η = Ax + By. The, since ux = uξ + Auη and uy = Buη , the PDE becomes
uξ + Auη − Buη = 0.
If we choose A = 1, B = 1, then ξ = x, η = x + y =⇒ x = ξ, y = η − ξ, and the PDE becomes
uξ = 0.
Therefore, u(ξ, η) = g(η) =⇒ u(x, y) = g(x + y). Find g(η) using the initial condition.
x2 = u(x, x3 − x) = g(x + x3 − x)
=⇒ x2 = g(x3 ).
1
Let z = x3 =⇒ x = z 3 , so
2
g(z) = z 3
2
=⇒ u(x, y) = (x + y) 3 .
We have that
1
1
2
2
ux (x, y) = (x + y)− 3 and uy (x, y) = (x + y)− 3 .
3
3
Therefore, the solution is not differentiable at any point of the form (x, −x). Why does this not
contradict Theorem 5.1? The characteristics are x + y = c, where c is a constant, and the initial
curve is given by y = x3 − x. Tangents to the initial curve take the form y − y1 = (3x2 − 1)(x − x1 ).
At the point (x, −x), we have
y − (−x) = (3x2 − 1)(x − x) =⇒ y = −x,
or x + y = 0, which is a characteristic. This violates the third condition in the theorem.
7