Midterm Solutions
Part 1
a) Analyze wellposedness of the following problem,
ut = (i − 1)uxxxx .
Solution: Using Fourier transform, we have
P̂ (iω) = (i − 1)(iω)4 = (i − 1)ω 4 .
(1)
Then,
P̂ (iω) + P̂ ∗ (iω) = −2ω 4 ≤ 0.
(2)
Therefore, the problem is well posed.
(b) Analyze wellposedness of the following problem
ut = −uxxxx − uxx + ux + u.
(3)
Solution:
P̂ (iω) = −ω 4 + ω 2 + iω + 1.
(4)
Then, ∃α > 0 such that
P̂ (iω) + P̂ ∗ (iω) = −2ω 4 + 2ω 2 + 2 < α.
Therefore, the problem is well posed.
1
(5)
(c) Analyze wellposedness of the following system
 

 

 
u
1 0
u
0 1
u
  =
  + 
  .
v
0 1
v
1 0
v
t
Solution:
x

P̂ (iω) = 
(6)
y

iω1 iω2
iω2 iω1
 = −P̂ ∗ (iω).
(7)
Then,
P̂ (iω) + P̂ ∗ (iω) = 0.
(8)
Thus, the system is well posed.
(d) Analyze wellposedness of the following system
 

 

 

 
u
1 0
u
0 1
u
1 0
u
  =
  + 
  + 
  .
v
0 1
v
1 0
v
0 1
v
t
Solution:
xx

xy
(9)
yy





1 0
0 1
1 0
 (iω1 )2 + 
 (iω1 )(iω2 ) + 
 (iω2 )2
P̂ (iω) = 
0 1
1 0
0 1


−ω12 − ω22
−ω1 ω2
 = P̂ ∗ (iω)
= 
2
2
−ω1 ω2
−ω1 − ω2
(10)
Since P̂ (iω) is normal, ∃ T s.t. Λ = T P̂ T T and ||T || = 1.
1
λ(P̂ ) = −ω12 − ω22 ± ω1 ω2 ≤ − (ω12 + ω22 ) ≤ 0.
2
Therefore, this system is wellposed.
2
(11)
Part 2
Consider the following problem
ut = ux ,
u(x, 0) = f (x).
Assume that the problem will be solved using the downwind scheme
£
¤
n+1
n+1
n
n
n
un+1
=
u
+
λ
α(u
−
u
)
+
(1
−
α)(u
−
u
)
,
j
j
j−1
j
j
j−1
(12)
where λ = k/h, unj = u(jh, nk) represents the grid function, and α ∈ [0, 1].
Discuss the accuracy and stability (which may depend on the parameter
α) of the scheme.
Solution:
Accuracy
Assume u(x, t) ∈ C ∞ is a solution. Then,
un+1
− unj
k
j
= ut + utt + O(k 2 )
k
2
unj − unj−1
h
= ux − uxx + O(h2 )
h
2
n+1
un+1
−
u
h
j
j−1
= ux − uxx + kuxt + O(h2 + kh)
h
2
k
h
h
− α(ux − uxx + kuxt ) − (1 − α)(ux − uxx )
2
2
2
2
2
+ O(h + k + kh)
µ
¶
1
h
= utt k
− α + uxt + O(h2 + k 2 + kh).
2
2
(13)
(14)
(15)
Tjn = ut + utt
(16)
(17)
Therefore, the accuracy is O(h + k) if α 6= 1/2 and O(h + k 2 ) if α = 1/2.
Stability
3
Assume unj = exp(iωxj )ûn . Then,
ûn+1 = Q̂ûn ,
(18)
in which
1 + λ(1 − α)(1 − e−iξ )
1 − λα(1 − e−iξ )
Q̂ =
|Q̂|2 =
(19)
(1 + (1 − α)A)2 + (1 − α)2 B 2
,
(1 − αA)2 + α2 B 2
(20)
where A = λ(1 − cosξ) and B = λsinξ. For stability, |Q̂|2 ≤ 1.
|Q̂|2 ≤ 1 ⇒ (1 + (1 − α)A)2 + (1 − α)2 B 2 ≤ (1 − αA)2 + α2 B 2
⇒ 2A + (1 − 2α)(A2 + B 2 ) ≤ 0 ← A2 + B 2 = 2λA
⇒ A(1 + (1 − 2α)λ) ≤ 0.
(21)
As A ≥ 0, we require α > 1/2, λ ≥ 1/(2α − 1).
Part 3
Consider the following problem
 

 
u
0 1
u
  =
  .
v
1 0
v
t
(22)
x
Remark - taking u = Ez and v = Hz , this system of equations is nothing else
than a normalized form of the one-dimensional Maxwell equations, describing
the propagation of electromagnetic waves in a homogeneous lossless media.
a) Characterize the problem, i.e. , is it hyperbolic (strongly, strictly, symmetric, weakly), parabolic etc., and discuss its wellposedness.
Solution:
4
Let

A=

0 1
1 0
.
(23)
Then,
λ(A) = ±1.
(24)
Since A= AT . this system is symmetric hyperbolic. Thus, this system is
wellposed.
b) now introduce the two ”staggered” grids
xj =
2π
2π
1
j, xj+1/2 =
j + , j ∈ [0, N ],
N +1
N +1
2
i.e. , they represent two equidistant grids separated by a shift by h/2.
Assume that u is given at xj and v at xj+1/2 through the gridfunctions
unj = u(xj , nk) = u(jh, nk),
n
vj+1/2
= v(xj+1/2 , nk) = v((j + 1/2)h, nk).
Consider the following staggered central difference - Leapfrog scheme for
solving Eq. (22)
n
n
vj+1/2
− vj−1/2
un+1
− un−1
j
j
=
,
2k
h
n+1
n−1
vj+1/2
− vj+1/2
unj+1 − unj
=
.
2k
h
Analyze the accuracy and stability of this scheme.
Solution:
5
Accuracy
Assume (u, v) ∈ C ∞ are solutions.
un+1
− un−1
k2
j
j
= ut + uttt + O(k 4 ),
2k
2
unj+1 − unj
h
h2
= ux + uxx + uxxx + O(h3 ),
h
2
6
n+1
n−1
vj+1/2 − vj+1/2
k2
h
= vt + vttt + vxt + O(hk 2 ),
2k
6
2
n
n
vj+1/2
− vj−1/2
h2
= vx + vxxx + O(h4 ).
h
24
(25)
(26)
(27)
(28)
For u,
Tjn = ut +
k2
h2
uttt − vx − vxxx = O(h2 + k 2 ).
6
24
(29)
For v, using vt = ux ⇒ vxt = uxx ,
Tjn = vt +
k2
h
h
h2
vttt + vxt − ux − uxx −
= O(h2 + k 2 ).
6
2
2
6
(30)
Therefore, the scheme is O(h2 + k 2 ).
Stability
n
Assume unj = exp(iωxj )ûn and vj+1/2
= exp(iωxj )eiξ/2 v̂ n . Then,
ûn+1 = ûn−1 + av̂ n ,
(31)
v̂ n+1 = v̂ n−1 + aûn ,
(32)
in which a = 4iλsin(ξ/2).
 

0 1 a
ûn+1
 

 n+1  
v̂  1 0 0
 

 n =
 û  a 0 0
 

0 0 1
v̂ n


ûn
0


 n 


v̂ 
0


  n−1  .
1 û 


v̂ n−1
0
6
(33)
The eigenvalues of the amplification matrix:
µ4 − (2 + a2 )µ2 + 1 = 0.
√
2 + a2 ± a4 + 4a2
2
µ =
.
2
(34)
(35)
For µ distinct, it is sufficient to show |µ| ≤ 1. Take a = i4λ. Then,
µ2 = (1 − 8λ2 ) ±
p
(1 − 8λ2 )2 − 1.
Necessary: |1 − 8λ2 | < 1 ⇒ λ < 12 .
7
(36)