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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 = &lt; ;+ 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 initial-boundary value problem for (4.4.20a) with m = 5 when the exact solution is a \running wave&quot; 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 initial-boundary 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 &gt; 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 constant-coe 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. Variable-Coe 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 time-centered space scheme (4.4.2a) is consistent. 2. Consider the initial-boundary value problem for Burgers' equation ut + uux = uxx u(x 0) = (x) 0&lt;x&lt;1 u(0 t) = gL(t) t&gt;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 Crank-Nicolson 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. Variable-Coe cient and Nonlinear Problems 47 where, as usual, 2 denotes the second-centered di erence. Develop a procedure for solving the Crank-Nicolson system using the tridiagonal algorithm and Newton's iteration. Give formulas for evaluating any Jacobians. 2.2. Write a program to solve the Crank-Nicolson 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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