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Unformatted text preview: a) with the trapezoidalrule variant of the weighted average scheme (4.4.20b). Assuming that Dirichlet data is
prescribed at x = 0 and 1, let's calculate the Jacobian of this system. Thus, di erentiating Fj with respect to Uin+1 .
8
f i=j
m t n+1 m;1
@Fj = < ;+ 2mx2 (Uti (U ) +1 )m;1 iif i = j ; 1 j + 1 :
n
1
(4.4.22a)
x2 i
@Uin+1 : 0
otherwise
In matrix form
FUn+1 (Vk ) =
3
2
2(U1n+1 )m;1 ;(U2n+1 )m;1
7
t 6 ;(U1n+1 )m;1 2(U2n+1 )m;1 ;(U3n+1 )m;1
7
6
:
I + m x2 6
7
...
5
4
+1
+1
;(UJn;2 )m;1 2(UJn;1 )m;1 (Un+1 =Vk )
(4.4.22b) 44 Parabolic PDEs Thus, to solve this problem using Newton's method, we would use (4.4.21a) with F(Vk )
given by (4.4.20b  4.4.20d) and the Jacobian given by (4.4.22b). The Jacobian is tridiagonal thus, one tridiagonal solution is required per iterative step.
Richtmyer and Morton 17], p. 201) solve an initialboundary value problem for
(4.4.20a) with m = 5 when the exact solution is a \running wave" u(x t) = (v(vt ; x + x0 ))
where v and x0 are prescribed constants. The function is implicitly determined as the
solution u of
5 (u ; u )4 + 20 u (u ; u )3 + 15u2(u + u )2 + 20u3(u ; u )+
0
0
0
0
0
0
4
30
5u4 ln(u ; u0) = v(vt ; x + x0)
0
where u0 is another prescribed constant. Computationally, Richtmyer and Morton 17]
solve an initialboundary value problem with data that is consistent with the exact running wave solution. Thus, the initial and boundary conditions for (4.4.20b) are Uj0 = (v(x0 ; xj ))
U0n = (v(vtn + x0)) j = 0 1 ::: J n
UJ = (v(vtn ; 1 + x0 )) n > 0: Richtmyer and Morton 17] solve a problem with = 0:4, v t= x = 0:75, and
5u4 t= x2 = 0:005. They only perform one Newton iteration per time step thus, they
0
take V1 = Un+1 . (This not good practice. The convergence of Newton's method should
be checked before proceeding to the next time step.) Their results for the numerical and
exact solutions as functions of x (j ) are shown for several times t (time steps n) in Figure
4.4.2.
The weighted average scheme (4.4.20b) is expected to have a stability restriction with
= 0:4. Thus, there is no logical reason for this choice relative to, e.g., the CrankNicolson scheme with = 0:5 Richtmyer and Morton 17] choose = 0:4 to show that a
constantcoe cient stability analysis su ces for this example since its solution is smooth.
To do this, they write (4.4.20a) ut = (mum;1 ux)x: 4.4. VariableCoe cient and Nonlinear Problems 45 Figure 4.4.2: Richtmyer and Morton's 17] solution of Example 4.4.6. The solid curves
are the exact solution and the dots show the numerical solution obtained by the weighted
average scheme (4.4.20b) with m = 5 and = 0:4.
In this form, it has the appearance of a heat conduction equation ut = ( ux)x
with the di usion coe cient = mum;1: Based on the constant coe cient analysis of Section 4.1, we would expect this weighted
average scheme to be absolutely stable when (4.1.20) is satis ed, i.e., when
t
1
2 2(1 ; 2 ) :
x
With m = 5, we have = 5u4 thus, the stability condition is
1:
5u4 t
2
x
2(1 ; 2 )
Using the indicated parameter values
1
u4
0:005 u
2(1 ; 2(0:4)) = 2:5
0
or
u 4:73:
u
0 46 Parabolic PDEs The results of Figure 4.4.2 indicate that oscillations and a loss of absolute stability set
in shortly after the solution at x = 0 exceeds the above value. Problems 1. Use (4.4.2b  4.4.2d) with (4.4.7) to show that the forward timecentered space
scheme (4.4.2a) is consistent.
2. Consider the initialboundary value problem for Burgers' equation ut + uux = uxx
u(x 0) = (x) 0<x<1 u(0 t) = gL(t) t>0
u(1 t) = gR(t): where is a positive constant. Burgers proposed this equation to model certain
aspects of turbulent ows. It is also used as a proving ground for numerical methods
because when is small the solution exhibits a shock like structure that is di cult
to calculate.
2.1. Consider the two possibilities for discretizing the nonlinear term:
n
n
n U n Uj +1 ; Uj ;1
(uux)j
j
2x
and (u2)x]n (Ujn+1)2 ; (Ujn;1)2
j
:
2
4x
Both formulas are O( x2) approximations of uux. Rather than decide which
one is best, let's use the linear combination
Q(Ujn)
(uux) 2 x where (uux)n =
j (Ujn+1)2 ; (Ujn;1)2
+ (1 ; p)Ujn (Ujn+1 ; Ujn;1)
2
with p selected on 0 1].
Suppose that we discretize Burgers' equation using the CrankNicolson scheme Q(Ujn ) = p Ujn+1 ; Ujn Q((Ujn+1 + Ujn )=2)
=
t+
2x 2 ((U n+1 + U n )=2)
j
j
2
x 4.4. VariableCoe cient and Nonlinear Problems 47 where, as usual, 2 denotes the secondcentered di erence.
Develop a procedure for solving the CrankNicolson system using the tridiagonal algorithm and Newton's iteration. Give formulas for evaluating any
Jacobians.
2.2. Write a program to solve the CrankNicolson system developed in Part 1.
Choose the initial and Dirichlet boundary data so that the exact solution is u(x t) = 1 ; tanh x2; t :
Solve the problem with = 0:01 for t 1 using p = 0, 2/3, and 1. Select
x = 0:05, 0.02, and 0.01 and compute with t= x = 0:5 and 1. Plot the
solutions as functions of x at t = 0, 0.5, and 1. Tabulate or plot the error
in the discrete L2 norm at t = 1 as a functi...
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 Spring '14
 JosephE.Flaherty
 The Land

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