Applied Numerical Methods for Partial Differential Equations
MATH 783

Summer 2013
Exercise 6.2
With = C (x2 + y 2 ) we get
Lh
C
ar+1/2 [(xr + x)2 x2 ] ar1/2 [x2 (xr x)2 ]
r
r
(x)2
C
2
ar (ys + y )2 2ys + (ys y )2
+
(y )2
ar+1/2 ar1/2
= C ar+1/2 + ar1/2 + 2ar + 2xr
x
4Ca0 ,
=
since ar is monotonic increasing.
Hence Lh 1 if we choose
C
Applied Numerical Methods for Partial Differential Equations
MATH 783

Summer 2013
Exercise 5.1
Substituting the usual Fourier mode n eijkx into the scheme gives a solution provided that
=
or
1
2 (i sin k x) + 2(2 cos k x 1/),
(1 + 2)2 + 2[i sin k x 2 cos k x] + (1 + 2) = 0.
The necessary and sucient conditions for the roots of this qu
Applied Numerical Methods for Partial Differential Equations
MATH 783

Summer 2013
Exercise 3.1
The usual expansion of the left hand side gives
1
2 tutt
+ O(t)2 .
On the right hand side the second xdierence gives
2
x u = 2[ 1 (x)2 uxx +
2
4
1
24 (x) uxxxx
+ . . .]
and similarly for the y dierence. Collecting the terms in the dierence
Applied Numerical Methods for Partial Differential Equations
MATH 783

Summer 2013
Exercise 2.7
Expansion of left hand side:
Ut + 1 t Utt + 1 (t)2 Uttt + . . . .
2
6
For the right hand side
n
Uj +1 = [u + 1 x ux + 1 ( 1 x)2 uxx + 1 ( 1 x)3 uxxx +
2
22
62
4
n
11
24 ( 2 x) uxxxx . . .]j + 1 .
2
n
Uj is the same, with alternating signs, so
Applied Numerical Methods for Partial Differential Equations
MATH 783

Summer 2013
Exercise 2.2
Using Taylor series with remainder,
e K
2 t
= 1 (K 2 t) + 1 (K 2 t)2 e , 0 < < K 2 t
2
= 1 K 2 t + 1 K 4 (t)2
2
where 0 < < 1
and
sin2 ( 1 K x) = 1 (1 cos K x)
2
2
= 0 + 1 (K x)2 + 0 + 1
4
2
= 1 K 2 (x)2
4
4
1
24 (K x) cos ,
4
4
48 K (x)
0
Applied Numerical Methods for Partial Differential Equations
MATH 783

Summer 2013
Exercise 2.1
(i) The Fourier sine series expansion
u0 (x) =
am sin m x
has coecients
am = 2
1
u0 (x) sin m xdx
0
1
2
=2
2x sin m xdx + 2
0
1
1
2
(2 2x) sin m xdx
1
2
cos m x
sin m x
= 2 2x
2
m
m2 2
0
cos m x
sin m x
+ 2 (2 2x)
(2)
m
m2 2
=
8
m2 2
2p+