sin m
m =
cos m
m = m2
m = 0, 1, 2,
r (r R )
T2
r=
T
R
T
=
T
1
r
(r R )
R
2
n
H
n
H
2
r 2 m2
m2
=
r2
T + T = 0
1
r
(r R )
= +
R
n
H
r (r R )
= r 2 +
R
r (r R ) (
n
H
2
+
2
n
H
2
m2
r2
r 2 + m2
) r 2 R m2 R = 0
BC :  R (0)  <
R (a) = 0
Rnm = Im
1c. uxx + uxy = 0
A=1
B=1
+1 1
dy
=
dx
2
C=0
=1
hyperbolic
+1
0
y = + x + K1
y = K2
= y x x = 1 y = 1
=y
x = 0
y = 1
ux = u + u x = u
=0
uy = u + u
uxx = u
uxy = u u
uxx + uxy = u = 0
The solution in the original domain is then:
u = f (y x) + g (y )
22
5
5.1
Separation of VariablesNonhomogeneous Problems
Inhomogeneous Boundary Conditions
Problems
1. For each of the following problems obtain the function w (x, t) that satises the boundary
conditions and obtain the PDE
a.
ut (x, t) = kuxx (x, t) + x,
0<x
2. b. This function is already in a Fourier sine series form and thus we can read the
coecients
an = 0
n = 0, 1, 2, . . .
bn = 0
n=1
b1 = 1
2. c.
8
6
4
2
0
L
L/2
L
L/2
2
4
6
4
2
0
2
4
6
Figure 14: graph of f (x) for problem 2c
Since the function is even,
4. In attempting to solve a simple PDE, a system of nitedierence equations of the form
un+1
j
1+
= 0

1+
0
0
un .
j
1+
Investigate the stability of the scheme.
For the stability, we need A 1 or that eig (A) 1. To compute the eigenvalues we
need to
1d. uxx + 10ux y + 9uyy = y
A=1
B = 10
10 8
dy
=
dx
2
= y 9x
=yx
= 100 36 = 64
C=9
hyperbolic
9
1
x = 9
x = 1
y = 1
y = 1
ux = 9u u
uy = u + u
uxx = 9 (9u u ) (9u u )
= 81u + 18u + u
uxy = 9 (u + u ) (u + uyy )
= 9u 10u u
uyy = u + 2u + u
uxx + 10ux y
1a. ux (0, t) = 1
u(L, t) = t
w (x, t) = A(t)x + B (t)
1 = wx (0, t) = A(t)
A(t) = 1
t = w (L, t) = A(t)L + B (t)
B (t) = t L
w (x, t) = x + t L
wt = 1
wxx = 0
v=uw
u=v+w
vt + 1 = kvxx + x
b. w = Ax + B
1 = w (0, t) = B (t)
1 = wx (L, t) = A(t)
w=x+1
wt =
3.
1
f (x) a0 +
2
An cos
n=1
n
n
x + bn sin
x
L
L
n
n
x + Bn sin
x
L
L
1
g (x) A0 +
2
where
an cos
n=1
1
L
1
bn =
L
1
An =
L
1
Bn =
L
an =
n
x dx
L
L
L
n
x dx
f (x) sin
L
L
L
n
x dx
g (x) cos
L
L
L
n
x dx
g (x) sin
L
L
L
f (x) cos
For f (x) + g (x) we hav
1. Derive the modied equation for the Lax Wendro method.
un+1 un1
un+1 un
c2 t n
j
j
j
j
+c
=
u 2un + un1
(1)
j
j
t
2x
2x2 j +1
Expand each fraction in Taylor series (all terms on the right side of the following equations
are evaluated at the point (j, n)
3.4
Relationship to Least Squares
3.5
Convergence
3.6
Fourier Cosine and Sine Series
Problems
1. For each of the following functions
i. Sketch f (x)
ii. Sketch the Fourier series of f (x)
iii. Sketch the Fourier sine series of f (x)
iv. Sketch the Fourier
c. wx (0, t) = t
wx (L, t) = t2
try w = A(t) x + B
wx = A(t) and we can not satisfy the 2 conditions.
try w = A(t) x2 + B (t) x
wx = 2A(t) x + B (t)
t = wx (0, t) = B (t)
2
t2 = wx (L, t) = 2A(t)L + B (t) A(t) = t 2 t
L
=t
t2 t 2
w=
x + tx
2L
wt =
wxx =
v
2a. uxx uyy + 3ux 2uy + u = 0
U = u e( + )
A=1
C = 1
B=0
=4
hyperbolic
2
dy
=
= 1
dx
2
=yx
=y+x
ux = u + u
uy = u + u
uxx = (u + u ) + (u + u ) = u 2u + u
uyy = u + 2u + u
4u 3u + 3u 2u 2u + u = 0
4u 5u + u + u = 0
U = u e( + ) u = U e( + )
u = U e( + )
Dierentiate (2) with respect to x twice
utxx = cuxxx + linear terms
Linear terms are enough because we have t2 in front. Also get
uttx = cuxxt + linear terms
= c (cuxxx ) + linear terms
= c2 uxxx + linear terms
uttt = cuxtt + linear terms
= c3 uxxx + line
3. Determine the stability requirement to solve the 1D heat equation with a source term
2u
u
= 2 + ku
t
x
Use the centralspace, forwardtime dierence method. Does the von Neumann necessary
condition make physical sense for this type of computational pro
The last term vanishes at both end points L
=
2L
1 2L cos n
= (1)n
n
L
n
L
Thus
bn =
and the Fourier series is
x
2L
n=1
2L
(1)n+1
n
(1)n+1
n
sin
x
n
L
33
3.
The equation becomes
u + cot u +
1
u = 0,
sin2
0 < < , 0 < < 2
Using separation of variables
u(, ) = ()()
+ cot +
1
= 0
sin2
Divide by and multiply by sin2 we have
sin2
+ cos sin =
=
Thus
+ = 0
sin2 + sin cos = 0
Because of periodicity, the equa
7
6
5
4
3
2
1
0
1
10
L
8
6
4
2
L
0
2
4
6
8
10
Figure 6: Graph of f (x) = ex
7
6
5
4
3
2
1
0
1
10
4L
8
3L
6
2L
4
L
2
L
0
2L
3L
4L
2
4
6
8
10
Figure 7: Graph of its periodic extension
1. c. f (x) = ex
Since the periodic extension of f (x) is discontinuous,
Now nd the eigenvalues
50
27
= 2 95 + 1521 = 0
27
45
So
=
1
95 2941
2
The spectral radius is
(AT A) =
1
95 + 2941
2
and the spectral norm is
A2 8.638
319
1. Let the vector U be dened as
u
v
U=
Then the equation can be written as
Ut = AUx + BUy
where
A=
0
1
1 0
B=
1 0
0
1
The eigenvalues of A are
1
1
or
2 + 1 = 0
= i
So the system behaves like elliptic in the x, t space
The eigenvalues of B are
+1 0
0
1
o
4.7
Laplaces equation in a sphere
Problems
1. Solve Laplaces equation on the sphere
2
1
cot
1
urr + ur + 2 u + 2 u + 2 2 u = 0,
r
r
r
r sin
0 r < a, 0 < < , 0 < < 2,
subject to the boundary condition
ur (a, , ) = f ().
2. Solve Laplaces equation on the
8
6
4
2
0
L
L
2
4
10
8
6
4
2
0
2
4
6
8
10
Figure 8: Graph of f (x)
8
6
4
2
0
4L
3L
2L
L
L
2L
3L
4L
2
4
6
8
2
4
10
8
6
4
2
0
10
Figure 9: Graph of its periodic extension
1. d.
Since the periodic extension of f (x) is discontinuous, the Fourier series is id
9.7
Matrix Method for Stability
9.8
Derivative Boundary Conditions
9.9
Hyperbolic Equations
9.9.1
Stability
Problems
1. Use a von Neumann stability analysis to show for the wave equation that a simple explicit
Euler predictor using central dierencing in s
6.6
General Solution
Problems
1. Determine the general solution of
a.
b.
c.
d.
c = constant
uxx c12 uyy = 0
uxx 3uxy + 2uyy = 0
uxx + uxy = 0
uxx + 10uxy + 9uyy = y
2. Transform the following equations to
U = cU
by introducing the new variables
U = ue( +)
1. u (r, , ) =
n=0
An 0 r n Pn (cos )
+
n=0 m=1
m
r n Pn (cos ) (Anm cos m + Bnm sin m )
(7.7.37)
ur (a, , ) = f () =
=
n=0
n An0 an1 Pn (cos )
+
n=0 m=1
n1
An0 n a
=
n an1 Anm =
n an1 Bnm =
0
m
n an1 Pn (cos ) (Anm cos m + Bnm sin m )
f () Pn (cos ) sin
8
6
4
2
0
L
L
2
4
10
8
6
4
2
0
2
4
6
8
10
Figure 10: Graph of f (x)
8
6
4
2
0
4L
3L
2L
L
L
2L
3L
4L
2
4
6
8
2
4
10
8
6
4
2
0
10
Figure 11: Graph of its periodic extension
1. e.
Since the periodic extension of f (x) is discontinuous, the Fourier series is
1. Use a von Neumann stability analysis to show for the wave equation that a simple explicit
Euler predictor using central dierencing in space is unstable. The dierence equation is
un+1 = un c
j
j
Substitute a Fourier mode
un+1 un1
j
j
2
n eikm j x
where
1a.
uxx
A=1
1
uyy = 0
c2
B=0
C=
1
c2
=
4
>0
c2
hyperbolic
2
1
dy
c
=
=
dx
2
c
y=
1
x +K
c
=y+
1
x
c
=y
1
x
c
Canonical form:
u = 0
The solution is:
u = f ( ) + g ( )
Substitute for and to get the solution in the original domain:
1
1
u (x, y ) = f (y + x)