sol1.pdf
TRANSCRIPT
MAT 3310 (2009-2010, Second Term)
Suggested Solution for Assignment 1∗
1. The third main submatrix∣
∣
∣
∣
∣
∣
1 5 92 6 103 7 11
∣
∣
∣
∣
∣
∣
= 0
so we use the elementary matrix to interchange the third and the fourth row of A
P =
1 0 0 00 1 0 00 0 0 10 0 1 0
, PA =
1 5 9 72 6 10 94 8 14 143 7 11 12
The linear system now is PA = Pb, then we do LU factorization to the modified matrix
L1 =
1 0 0 0−2 1 0 0−4 0 1 0−3 0 0 1
→ L1A =
1 5 9 70 −4 −8 −50 −12 −22 −140 −8 −16 −9
→ L2 =
1 0 0 00 1 0 00 −3 1 00 −2 0 1
→ L2L1A =
1 5 9 70 −4 −8 −50 0 2 10 0 0 1
= U
L = L−11
L−12
=
1 0 0 02 1 0 04 3 1 03 2 0 1
Solve the lower triangular system Ly = Pb by forward substitution,
~y =
24−1911
∗Please email to the tutors if there are mistakes: [email protected].
1
Then solve the upper triangular system Ux = y by backward substitution,
~x =
−0.53.501
.
2. (a)
A =
1 0 0−1 1 00 −1 10 0 1
C =
c1 0 0 00 c2 0 00 0 c3 00 0 0 c4
~x =
x1
x2
x3
, ~f =
f1
f2
f3
(b) ∀x 6= 0, xT (AT CA)x = (Ax)T C(Ax).Let y = Ax, then since C is diagonal and positive definite, so yT Cy > 0 and yT Cy = 0is equivalent to y = 0.Then we show the equation Ax = y = 0 has only one trivial solution. Solve thesystem
A~x =
1 0 0−1 1 00 −1 10 0 1
x1
x2
x3
=
0000
we get ~x = ~0. So for x 6= 0, xT (AT CA)x > 0. A is positive definite.
(c) With
C =
1 0 0 00 1 0 00 0 1 00 0 0 1
, ~f =
111
we get
AT CA~x =
2 −1 0−1 2 −10 −1 2
= ~f,
Solve the linear system,
~x =
1.52
1.5
2
3. (a) Let the stress function be w(x), the displacement be u(x), the elastic constant bec(x), the external force be f(x), then we have
w(x) = c(x)dudx
−dwdx = f
w(0) = w(1) = 0
(1)
(b) Integrate the above second equation in [0, 1],
−∫
1
0
dw
dx= w(0) − w(1) =
∫
1
0
f(x)dx = 0
In order to have a solution, we need the condition∫
1
0f(x)dx = 0.
(c) If u(x) is a solution of the model in (a), then let v(x) = u(x) + C, C is any constant.Since v′(x) = u′(x),
−d
(
c(x)dvdx
)
dx= −
d(
c(x)dudx
)
dx= f
v(x) satisfies the equation. For the boundary conditions
c(x)dv
dx= c(x)
du
dx
so it also has w(0) = w(1) = 0 for v(x), and v(x) is a solution.
4. (a) Substitute u = d1ex/
√c + d2e
−x/√
c + 1 into the equation,
−cu′′(x) = −c(d1ex/
√c + d2e
−x/√
c + 1)′′ (2)
= −c(d1√
cex/
√c +
−d2√c
e−x/√
c)′ (3)
= −c(d1
cex/
√c +
d2
ce−x/
√c) (4)
= −(d1ex/
√c + d2e
−x/√
c) (5)
= −(u + 1) (6)
So u = d1ex/
√c + d2e
−x/√
c + 1 is an exact solution of the equation −cu” + u = 1.
(b) Set the boundary conditions u(0) = u(1) = 0, we get
{
d1 + d2 + 1 = 0
d1e1/
√c + d2e
−1/√
c + 1 = 0(7)
Solve the system, we get
d1 =e−1/
√c − 1
e1/√
c − e−1/√
c, d2 =
1 − e1/√
c
e1/√
c − e−1/√
c
3
(c) For 0 < x < 1,
limc→0
u(x) = limc→0
ex√c (
e−1/√
c − 1
e1/√
c − e−1/√
c) + e
− x√c
1 − e1/√
c
e1/√
c − e−1/√
c+ 1 (8)
= limc→0
ex−1√
c (e−1/
√c − 1
1 − e−2/√
c) + e
− x√c
e−1/√
c − 1
1 − e−2/√
c+ 1 (9)
= 0 · (−1) + 0 · (−1) + 1 (10)
= 1 (11)
(d) For any c, substitute x = 0 and x = 1 into u(x), u(0) = u(1) = 0 are always satisfied.So both boundary conditions are kept, and both ends have the boundary layer.
5. (a) Solve the equation−u′′ = ex
we get the general solution is u(x) = −ex + c1x + c2, where c1, c2 are constants.Substitute the boundary conditions u(0) = 0 and u(1) = 0,
{
c2 − 1 = 0
−e + c1 + c2 = 0(12)
so c1 = e − 1, c2 = 1, and the solution is u(x) = −ex + (e − 1)x + 1.
(b) The original equation is:
{
−u′′ = δ(x − 1
4), 0 < x < 1
u(0) = 0, u′(1) = 0.(13)
so −u′′ = 0 at x 6= 1
4, solve this equation:
u =
{
c1x + c2, 0 < x < 1
4
c3x + c4,1
4< x < 1
where c1, c2, c3, c4 are constants to be determined from the following conditions:u(0) = 0 =⇒ c2 = 0,u′(1) = 0 =⇒ c3 = 0,u is continuous at x = 1
4=⇒ 1
4c1 = c4,
−∫
1
4+ε
1
4−ε
u′′dx = −∫
1
4+ε
1
4−ε
δ(x− 1
4)dx = −u′(1
4+ ε)+u′(1
4− ε) = 1 =⇒ u′
+(1
4)−u′
−(1
4) =
1,=⇒ 0 + c1 = 1 =⇒ c4 = 1
4. so we have the solution:
u =
{
x, 0 < x < 1
41
4, 1
4< x < 1
4