Class 3363. Final Exam.
Problem 1 (70 points) Solve the Laplace equation inside a rectangle
0 ≤ x ≤ L,
0 ≤ y ≤ H,
with the following boundary conditions
u(0, y) = 0,
∂u
(L, y) = f (y),
∂x
and
f (y) :=
∂u
(x, 0) = 0,
∂y
0
1
∂u
(x, H) = 0.
∂y
y < H/2
y > H/2
Solution:
We apply the separation of variables and first look for all possible solutions to the
homogeneous equation of the form u = h(x)φ(y). Substituting it in the ∆u = 0
equation we get two equations
d2 h
= λh, h(0) = 0
dx2
and
d2 φ
dφ
= −λφ,
(0) = 0,
2
dy
dy
dφ
(H) = 0.
dy
nπ 2
All possible solutions to the second equation are φ(y) = c cos( nπy
,
H ), λ =
H
n = 0, 1, 2, . . . . Note that λ = 0 for n = 0. Solutions to the first equations are
h = c x, for λ = 0 and h = c sinh nπx
for n = 1, 2, . . . .
H
Hence applying the principle of decomposition we find that
u(x, y) = B0 x +
∞
X
Bn sinh
n=1
nπx H
cos(
nπy
)
H
(1)
solves the heat equation. It remains to fit the non-homogeneous boundary condition
on the side of the rectangle x = L.
We find for the general solution in (1):
∞
nπ X
∂u
nπy
nπ
(L, y) = B0 +
Bn
cosh
L cos(
)
∂x
H
H
H
n=1
Fitting the boundary condition
B0 +
∞
X
n=1
Bn
∂u
∂x (L, y)
= f (x), we find
nπ nπ
nπy
cosh
L cos(
) = f (y)
H
H
H
(2)
To find Bn we need to find the coefficients of the Fourier’s cosine series for f (y):
f (y) = A0 +
∞
X
n=1
An cos(
nπy
)
H
(3)
Using the formulas for the coefficients, we compute:
A0 =
1
H
Z
H
f (y) dx =
0
1
,
2
Z
Z
nπy
nπy
2 H
2 H
2 H
nπy H
f (y) cos(
cos(
An =
) dy =
) dy =
sin(
)
H 0
H
H H/2
H
H nπ
H H/2
2
nπ
=
sin( ).
nπ
2
Comparing the series in (2) and (3), we obtain
B 0 = A0 ,
Bn =
nπ H
An cosh−1
L
nπ
H
Thus the final solution is
∞
nπ nπx x X
nπy
nπ 2H
u(x, y) = +
L sinh
cos(
)
sin( ) 2 2 cosh−1
2 n=1
2 n π
H
H
H
Problem 2 (70 points) Solve the heat equation
∂u
∂2u
=
+ 3 sin x + et sin 2x,
∂t
∂x2
0 < x < π, t > 0
subject to boundary and initial conditions
u(0, t) = 0,
u(π, t) = 0,
u(x, 0) = 0.
Solution:
Since boundary conditions are Dirichlet and L = π, we look for the solution in the
form
∞
X
u(x, t) =
an (t) sin(nx).
n=1
Note that u(x, t) satisfies boundary conditions and we need to find coefficients
an (t). Substituting such u(x, t) in the given heat equation, we get
∞ X
d an (t)
n=1
dt
+ n2 an (t) sin(nx) = 3 sin x + et sin 2x.
Multiplying this equality by sin(mx) with m = 1, 2, 3, . . . , integrating over x ∈
(0, π), and using the orthogonality property of sinus functions, we get the set of
equations for am :

m = 3, 4, 5 . . .
 0,
d am (t)
3
m=1
+ m2 am (t) =
 t
dt
e
m=2
Taking initial boundary condition into consideration, we conclude am (0) = 0 and
solve the above equations for am :


m = 3, 4, 5 . . .
m = 3, 4, 5 . . .
 0,
 0, R
t
3e−t eτ |t0
m=1
3e−t 0 eτ dt
m=1
=
am (t) =
 −4t 1 5τ t
 −4t R t τ 4τ
e 5 e |0 dt
m=2
e
e e dt
m=2
0


m = 3, 4, 5 . . .
m = 3, 4, 5 . . .
 0,
 0,
3(1 − e−t )
m=1
3e−t (et − 1)
m=1
=
.
=
 1 t
 1 −4t 5t
−4t
e
(e
−
1)
m
=
2
e
−
e
m
=
2
5
5
Thus the final answer is
1
u(x, t) = 3(1 − e−t ) sin(x) + ( et − e−4t ) sin(2x).
5