Massachusetts Institute of Technology
18.085: Computational Science and Engineering
Gilbert Strang
Spring 2016
Solution to Problem Set 1
Section 1.1
Exercise 4


1 −1 0
0
0 1 −1 0 
,
U4 = 
0 0
1 −1
0 0
0
1

S4 = U4−1
1
0
=
0
0
1
1
0
0
1
1
1
0

1
1
.
1
1
We get

4

3
S4 S4T = 
2
1
3
3
2
1
2
2
2
1

1
1
,
1
1
which is indeed equal to T4−1 . We can see it via MATLAB:
>> S_4 = triu(ones(4));
>> S_4*S_4’;
To show SS T is symmetric, we will use the fact that (AB)T = B T AT for any square matrix
A, B. Hence, (SS T )T = (S T )T S T = SS T so SS T is symmetric. In fact it holds for any
matrix S (doesn’t need to be symmetric or upper-triangular or even square-shaped).
Exercise 5
K2−1
1 2 1
=
. We may expect that det(K5 ) = 6.
3 1 2

5 4 3
4 8 6
1
3 6 9
K −1 = 
6
2 4 6
1 2 3
We get det(K) = 6 and
2
4
6
8
4

1
2

3

4
5
MATLAB code:
K = diag(2*ones(5,1))-diag(ones(4,1),1)-diag(ones(4,1),-1);
det(K);
det(K)*inv(K);
Solution to Problem Set 1
Spring 2016
Exercise 20
2 3
4 5
1 2
2 2 4 6 12
8 16
1 2 =
=
=
.
2 4
4
4 8 10 20
14 28
For n by n, this algorithm will operate n3 multiplications.
Section 1.2
Exercise 2
Homogeneous solution would be
uhom (x) = c1 x + c0 ,
and particular solution would be
(
0
if x < 0
upart (x) = −S(x) =
−x if x ≥ 0.
Hence the solution u(x) = uhom (x) + upart (x) is
(
c1 x + c0
u(x) =
(c1 − 1)x + c0
if x < 0
if x ≥ 0.
Letting u(−2) = u(3) = 0, we get
−2c1 + c0 = 0
3c1 + c0 = 3.
So c0 = 6/5 and c1 = 3/5 and
(
u(x) =
Note that u is continuous at x = 0, i.e.,

2 −1
−1 2
KU = 
 0 −1
0
0
3
(x + 2)
5
− 25 (x − 3)
if x < 0
if x ≥ 0.
two lines meet at x = 0. From finite difference,
   
3/5
0
0
0




−1 0  6/5 1
=  = F.
2 −1 4/5 0
2/5
0
−1 2
2
Solution to Problem Set 1
Spring 2016
3
Exercise 6
∆2 u(x) = u(x + ∆x) − 2u(x) + u(x − ∆x)
= (x + ∆x)4 − 2x4 + (x − ∆x)4
= 12x2 (∆x)2 + 2(∆x)4 .
Hence ∆2 u/(∆x)2 = 12x2 + 2(∆x)2 . On the other hand, u00 (x) = 12x2 . From
∆2 u(x) = (∆x)2 u00 (x) + c(∆x)4 u0000 (x),
we have c = 1/12 (which is equal to 2/4!).
Exercise 18
True solution will be the form of u(x) = c3 x3 + c2 x2 + c1 x + c0 , where
6c3 = 1,
c2 = 0,
c0 = 0,
c3 + c2 + c1 + c0 = 0.
First two equalities come from u00 (x) = x, and last two come from boundary condition.
Hence u(x) = 61 x3 − 61 x.
We would like to solve u00 = −f (x) using K with n = 4. Here f (x) = −x and h = 15 , and

 
 


 

u(h)
u(1/5)
f (h)
−h
−1/5
u(2h) u(2/5)

 
 

=
 F = f (2h) = −2h = −2/5 .
U =
u(3h) u(3/5)
f (3h) −3h −3/5
u(4h)
u(4/5)
f (4h)
−4h
−4/5
1
KU = F , or explicitly
h2
 



2 −1 0
0
u(1/5)
−1/5
−1 2 −1 0  u(2/5) −2/5

 

25 
 0 −1 2 −1 u(3/5) = −3/5 .
0
0 −1 2
u(4/5)
−4/5
We approximate d2 /dx2 by
1
K
h2 n
so
Using MATLAB, we get

4
1 1
1 −1
3
· 
U= K F =

25
25 5 2
1
which exactly agrees with true solution.
3
6
4
2
2
4
6
3

 

−4/125
1 −1/5

 

2
 −2/5 = −7/125 ,




3 −3/5
−8/125
−6/125
4 −4/5
Solution to Problem Set 1
Spring 2016
Exercise 21
Pluging in u0 (0) = 0 and −u00 = f to Taylor expansion, we get
1
u(h) ≈ u(0) − h2 f (0).
2
Since u0 = u(0) and u1 = u(h), we get
1
u0 − u1 ≈ h2 f (0).
2
If we use this approximation instead of u0 − u1 ≈ u0 (0) = 0, then


 

1 −1 0
0
u(0)
f (0)/2

 

1 
−1 2 −1 0   u(h)  =  f (h)  .





2
0 −1 2 −1 u(2h)
f (2h) 
h
0
0 −1 2
u(3h)
f (3h)
By solving this equation, we get exact solution for Figure 1.4.
4