Additional problems


1. Solve
L u = − ∂u
∂t
where
L ≡ − ∂2
∂x2
and

  1. u − 2 ∂u
    ∂x
    		 = 0
    at x=0 (B.C.)
  2. u = 0 at x=1 (B.C.)
  3. u = f(x) at t=0 (I.C.)
2. Solve
c2 ⎛
⎝ 		∂2 w
∂x2
 + ∂2 w
∂y2
⎞
⎠ 		= ∂2 w
∂t2

0 ≤ x ≤ 1,     0 ≤ y ≤ 1
Boundary condition:
w(0,y,t)=w(1,y,t)=w(x,0,t)=w(x,1,t)=0
Initial Condition:
w(x,y,0)=f(x,y),     ∂w
∂t
 (x,y,0)=0.
3. Approximate ex by a linear combination of e1(x) = x and e2(x) = x2−[1/2] in the sense of least square over [0,1] 1.


4.

    (a) A sequence of e1(x)=1, e2(x)=√3(2x−1), e3(x) = √5(1−6x+6x2) forms an ON set in [0,1] (you don't have to prove it). Expand f(x)=x3 by e1(x), e2(x) and e3(x).
    (b) Determine c so that the distance between sinx and c x(π− x) is minimized over [0,π]. See hints below 2.
5.

    (a) Is x2 closer to x or x3 over [0,1] ? Why ?
    (b) What are the advantages of the Sturm-Liouville theory ?
Solution
1. The corresponding eigenvalue problem is

L en = λn en
with
en − 2 ∂en
∂x
=
0    x=0
(1)
en
=
0     x=1.
(2)
Substituting
en = A sin
√
 
λn
 
 x + B cos
√
 
λn
 
 x
(3)
into the first boundary condition at x=0 yields

B − 2 A
√
 
λn
 
 = 0
(4)
and substituting eq.(3) into the second boundary condition at x=1 yields

A sin
√
 
λn
 
 + B cos
√
 
λn
 
 = 0.
(5)
Equations (4-5) are written as

⎛
⎜
⎜
⎜
⎝
−2
√
 
λn
 
1
sin
√
 
λn
 
cos
√
 
λn
 
⎞
⎟
⎟
⎟
⎠ 		⎛
⎜
⎜
⎝
A
B
⎞
⎟
⎟
⎠ 		=0
(6)
For non-trivial solutions, the determinant of the matrix of Eq. (6) must vanish, i.e.

− 2
√
 
λn
 
 cos
√
 
λn
 
 − sin
√
 
λn
 
 = 0
(7)
or

tan
√
 
λn
 
 + 2
√
 
λn
 
 = 0.
(8)
so

B
=
−tan
√
 
λn
 
 A
(9)
=
2
√
 
λn
 
 A
(10)
Thus,
en (x) = An (sin
√
 
λn
 
 x + 2
√
 
λn
 
 cos
√
 
λn
 
 x)
(11)
where

An = 1
  ⎛
√
⌠
⌡ 		1

0 
 (sin
√
 
 [ˉ(λn)] x + 2
√
 
 [ˉ(λn)] cos
√
 
 [ˉ(λn)]x)2 dx
 
(12)
Thus the solution to the original problem can be expressed as

u = ∞
∑
n=1 
 cn en(x)
(13)
Plugging this into the diffusion equation yields

L ∞
∑
n=1 
 cn en(x)
=
− ∂
∂t
 ∞
∑
n=1 
 cn en(x)
(14)
∞
∑
n=1 
 cn Len(x)
=
∞
∑
n=1 
 ⎛
⎝ 		−d cn
d t
⎞
⎠ 		en(x)
(15)
∞
∑
n=1 
 cn λn en(x)
=
∞
∑
n=1 
 ⎛
⎝ 		−d cn
d t
⎞
⎠ 		en(x)
(16)
which yields

dcn
dt
 = − λn cn
(17)
which can be solved as
cn = dn e−λn t
(18)
so eq.(13) is expressed as

u(x,t) = ∞
∑
n=1 
 dn e−λn t en(x)
(19)
From the initial condition at t=0,

f(x) = ∞
∑
n=1 
 dn en(x)
(20)
so
dn = (f, en)
(21)


2. The corresponding eigenvalue problem (in 2-D) is

L emn = λmn emn,
(22)
where
L ≡ −c2 ⎛
⎝ 		∂2
∂x2
 + ∂2
∂y2
⎞
⎠ 		.
(23)
The eigenfunctions and the corresponding eigenvalues are

emn(x,y)
=
2 sinmπx sinnπy
(24)
λmn
=
c2 (m2+n2)π2.
(25)
Thus, the solution to the wave equation is assumed to be expanded as

w = ∞
∑
m,n=1 
 cmn emn(x,y).
(26)
Substituting this into the given wave equation yields

L ∞
∑
m,n=1 
 cmn emn(x,y)
=
− ∂2
∂t2
 ∞
∑
m,n=1 
 cmn emn(x,y)
(27)
∞
∑
m,n=1 
 cmn λmnemn(x,y)
=
∞
∑
m,n=1 
 ⎛
⎝ 		− d2 cmn
dt
⎞
⎠ 		emn(x,y)
(28)
or

d2 cmn
dt2
 + λmn cmn=0
(29)
which can be solved as

cmn = Amn sin
√
 
λmn
 
 t + Bmn cos
√
 
λmn
 
 t.
(30)
Therefore, the solution is expressed as

w(x,y,t) = ∞
∑
m,n=1 
 ⎛
⎝ 		Amn sin
√
 
λmn
 
 t + Bmn cos
√
 
λmn
 
 t ⎞
⎠ 		emn(x,y).
(31)
The second initial condition, ∂w/∂t = 0 at t=0 yields

0 = ∞
∑
m,n=1 
 Amn
√
 
λmn
 
 emn(x,y)
(32)
which results in
Amn=0.
(33)
The first initial condition, w=0 at t=0, yields

f(x,y) = ∞
∑
m,n=1 
 Bmn emn(x,y)
(34)
which results in

Bmn = (f, emn).
(35)
Thus, the solution is

w(x,y,t) = ∞
∑
m,n=1 
 (f, emn)cos
√
 
λmn
 
 t emn(x,y).
(36)


3. Note that (e1, e2)=0 (but not normalized). ex  ∼ c1 e1(x) +c2 e2(x). In the best approximation, it follows (ex − c1 e1(x) −c2 e2(x), e1(x))=0 and (ex − c1 e1(x) −c2 e2(x), e2(x))=0. So (e1,e1)c1 = (ex, e1) and (e2,e2)c2 = (ex, e2) which can be solved as

c1=3, c2= 30
7
 (e−3)
.

4. (a) x3  ∼ c1 e1 +c2 e2+c3 e+3.
c1 = (x3, e1 ) = 1
4

c2 = (x3, e2 ) = 3√3
20

c3 = (x3, e3 ) = 1
4√5
.
(b) When the distance of two vectors, a and cb, is minimized, a−cb must be perpendicular to b. i.e. ( a−cb, b)=0. Therefore,
c= (a,b)
(b,b)
 =
⌠
⌡ 		π

0 
 sinx x (π−x) dx
⌠
⌡ 		π

0 
 x2(π−x)2dx
 = 120
π5
 .
5. (a)
||x2−x||=   ⎛
√
⌠
⌡ 		1

0 
 (x2−x)2 dx
 
 = 1
30
 = 0.1825,

||x2−x3||=   ⎛
√
⌠
⌡ 		1

0 
 (x2−x3)2 dx
 
 = 1
105
 = 0.09759
so x2 is closer to x3.
(b) Discussed in class.


Footnotes:

1Hint: ∫01 x ex dx = 1, ∫01 x2 ex dx = e−2
2∫0π x sinx dx = π, ∫0π x2sinx dx = −4 + π2.

File translated from TEX by TTH, version 4.03.
On 19 Apr 2023, 19:13.