MIT OpenCourseWare
http://ocw.mit.edu
18.085 Computational Science and Engineering I
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
18.085 - Mathematical Methods for Engineers I
Prof. Gilbert Strang
Solutions - Problem Set 7
3.4, Problem 4. Show that u = r cos ∂ + r −1 cos ∂ solves Laplace’s equation (13), and express u in terms
of x and y. Find v = (ux , uy ) and verify that v · n = 0 on the circle x2 + y 2 = 1. This is the velocity of flow
past a circle in Figure 3.18.
Show u = r cos ∂ + r −1 cos ∂ solves Laplace’s equation:
Laplace’s equation in r, ∂ is:
� 2 u 1 �u
1 �2u
+
+ 2 2 = 0.
2
�r
r �r
r �∂
(13)
Now, if u = r cos ∂ + r −1 cos ∂, then
�u
= cos ∂ − r−2 cos ∂
�r
�2u
= 2r−3 cos ∂
�r2
�u
= −r sin ∂ − r−1 sin ∂
�∂
�2u
= −r cos ∂ − r −1 cos ∂.
�∂2
Plugging this into the left-hand side of (13), we get
1
1
1
2r−3 cos ∂ + (cos ∂ − r−2 cos ∂) + 2 (−r cos ∂ − r −1 cos ∂) = (2 − 1 − 1)r −3 cos ∂ + (1 − 1) cos ∂ = 0,
r
r
r
as desired.
U in (x, y):
Now, since x = r cos ∂, y = r sin ∂ � r =
⎛
x2 + y 2 , ∂ = tan−1
u = r cos ∂ + r−1 cos ∂ = x +
So
�y�
x
. So
x
x(1 + x2 + y 2 )
=
.
x2 + y 2
x2 + y 2
�u
(x2 + y 2 ) · 1 − x(2x)
y 2 − x2
=1+
=
1
+
.
�x
(x2 + y 2 )2
(x2 + y 2 )2
�u
−2xy
= 0 + x(−1 · (x2 + y 2 )−2 · 2y) = 2
.
�y
(x + y 2 )2
So
v=
�
1+
y 2 − x2
−2xy
, 2
2
2
2
(x + y ) (x + y 2 )2
�
.
v · n = 0:
n = (x, y) is normal to the circle
v = (1 + y 2 − x2 , −2xy) on the circle
v · n = x − x3 − xy 2 = x − x(x2 + y 2 ) = 0 on the circle
1
sinh(�ky)
3.4, Problem 17. Verify that uk (x, y) = sin(�kx)
solves Laplace’s equation for k = 1, 2, . . .. Its
sinh(�k)
boundary values on the unit square are u0 = sin(�kx) along y = 1 and ub = 0 on the three lower edges.
z
−z
Recall that sinh (x) = e −e
.
2
u = uk (x, y) =
sin(�kx) 12 (e�ky − e−�ky )
1 �k
2 (e
�u
e�ky − e−�ky
= �k cos(�kx) · �k
�x
e − e−�k
2
� u
e�ky − e−�ky
2 2
=
−�
k
sin(�kx)
·
�x2
e�k − e−�k
So
− e−�k )
.
�u
�ke�ky + �ke−�ky
= sin(�kx) ·
�y
e�k − e−�k
2
2 2 �ky
� u
� k (e
− e−�ky )
=
sin(�kx)
·
�y 2
e�k − e−�k
�2u �2u
+ 2 = 0.
�x2
�y
3.6, Problem 6. Solve Poisson’s equation −uxx − uyy = 1 with u = 0 on the standard unit triangle T using
Persson’s code with h = 0.25. The mesh information is in the lists p, t, and b (boundary nodes) : 15 nodes,
16 triangles, 12 boundary nodes (nodes numbered by rows).
1
15 (0, 1)
16
.75
13 (0, .75)
14 (.25, .75)
14
.5
.25
13
15
10 (0, .5)
11 (.25, .5)
9
11
8
10
12
6 (0, .25)
7 (.25, .25)
8 (.5, .25)
2
4
1
1 (0, 0)
0
12 (.5, .5)
6
3
5
2 (.25, 0)
.25
9 (.75, .25)
3 (.5, 0)
.5
2
7
4 (.75, 0)
.75
5 (1, 0)
1
⎤
�
�
�
�
�
�
�
�
�
�
�
�
�
p=�
�
�
�
�
�
�
�
�
�
�
�
0
.25
.5
.75
1
0
.25
.5
.75
0
.25
.5
0
.25
0
0
0
0
0
0
.25
.25
.25
.25
.5
.5
.5
.75
.75
1
⎜
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎢
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
⎤
�
�
�
�
�
�
�
�
�
�
�
�
�
�
t=�
�
�
�
�
�
�
�
�
�
�
�
�
1
2
2
3
3
4
4
6
7
7
8
8
10
11
11
13
2
7
3
8
4
9
5
7
11
8
12
9
11
14
12
14
contourf(xx,yy,zz);
3
6
6
7
7
8
8
9
10
10
11
11
12
13
13
14
15
⎜
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎢
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
⎤
�
�
�
�
�
�
�
�
�
�
b=�
�
�
�
�
�
�
�
�
1
2
3
4
5
6
9
10
12
13
14
15
⎜
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎢
3.6, Problem 9. The square drawn below has 8 nodes rather than the usual 9 for biquadratic Q 2 . Therefore
we remove the x2 y 2 term and keep
U = a1 + a2 x + a3 y + a4 x2 + a5 xy + a6 y 2 + a7 x2 y + a8 xy 2 .
Find the λ(x, y) which equals 1 at x = y = 0 and zero at all other nodes.
�
1
⎝ 0
⎝
⎝ 0
⎝
⎝ 0
⎝
⎝ 0
⎝
⎝ 0
⎝
⎞ 0
0
⎟
⎣
⎣
⎣
⎣
⎣
⎣
⎣
⎣
⎣
⎣
⎠
λ
We solve for a to obtain
�x=0, y=0
�x=1, y=0
�x=0, y=1
�x=.5, y=0
�x=0, y=.5
�x=.5, y=1
�x=1, y=.5
�x=1, y=1
�
1 0 0
0
⎝ 1 1 0
1
⎝
⎝
0
⎝ 1 0 1
⎝ 1 .5 0 .25
= ⎝
⎝
⎝ 1 0 .5 0
⎝
⎝ 1 .5 1 .25
⎞ 1 1 .5 1
1
1 1 1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 .25 0
0
.5 1 .25 .5
.5 .25 .5 .25
1
1
1
1
=
xy
�
1
⎝ −3
⎝
⎝ −3
⎝
⎝ 2
a = xy\λ = ⎝
⎝ 5
⎝
⎝ −2
⎝
⎞ −2
−2
⎟�
⎣⎝
⎣⎝
⎣⎝
⎣⎝
⎣⎝
⎣⎝
⎣⎝
⎣⎝
⎣⎝
⎣⎝
⎠⎞
a1
a2
a3
a4
a5
a6
a7
a8
a
⎟
⎣
⎣
⎣
⎣
⎣
⎣.
⎣
⎣
⎣
⎣
⎠
Alternatively, we can use Method B and set
λ(x, y) = a1 + a2 x + a3 y + a4 x2 + a5 xy + a6 y 2 + a7 x2 y + a8 xy 2 .
Then 1 − 3x + 2x2 = λ(x, 0) = a1 + a2 x + a4 x2 , so that a1 = 1, a2 = −3, a4 = 2.
Likewise, 1 − 3y + 2y 2 = λ(0, y) = a1 + a3 y + a6 y 2 , so that a1 = 1, a3 = −3, a6 = 2.
But 0 = λ(x, 1) = a1 + a2 x + a3 + a4 x2 + a5 x + a6 + a7 x2 + a8 x, so that
a1 + a 3 + a 6
= 0
a2 + a 5 + a 8
a4 + a 7
= 0
= 0.
From this, we can conclude that a7 = −2 and a5 + a8 = 3.
Finally, 0 = λ(1, y) = a1 + a2 + a3 y + a4 + a5 y + a6 y 2 + a7 y + a8 y 2 , so that
a1 + a 2 + a 4
= 0
a3 + a 5 + a 7
a6 + a 8
= 0
= 0.
4
⎟
⎣
⎣
⎣
⎣
⎣
⎣.
⎣
⎣
⎣
⎣
⎠
From this, we know that a8 = −2, a5 = 5. So again,
�
⎝
⎝
⎝
⎝
⎝
a=⎝
⎝
⎝
⎝
⎝
⎞
1
−3
−3
2
5
−2
−2
−2
⎟
⎣
⎣
⎣
⎣
⎣
⎣.
⎣
⎣
⎣
⎣
⎠
⎥
2 −1
−1 2
gives the oscillation frequencies for two unequal masses in a line of springs.
3.6, Problem 16. Find the eigenvalues of KU = �M U if K =
M
−1
KU
KU
det(M −1 K − �I)
M
M
−1
K=
1 0
0 .5
⎦
2
−1
−.5 1
⎦
=
⎥
−1
⎥
⎦
and M =
= �M U
= �U
= 0
⎦
� 0
�I =
0 �
⎥
2−�
−1
M K=
−.5
⎥
(2 − �)(1 − �) − (−1)(−.5) = 0 3 + 3
or −
�2 − 3� + 1.5 = 0 2
�
�
=
5
−1
1−�
�
⎦
3−3
.
2
⎥
1 0
0 2
⎦
. This