W.D. Henshaw Math 6840: Solutions for Problem Set 4

1. (10 points) von Neumann analysis in two-dimensions.

Consider the two-dimensional parabolic problem for u = u(x, y, t),

∂tu = κ1∂2xu+ κ2∂

2yu− αu, x ∈ [0, 2π]2,

u(x, y, 0) = u0(x, y),

with periodic boundary conditions. Here κ1 > 0, κ2 > 0 and α > 0.

(a) Write down the method-of-lines (MOL) scheme for this problem using second-order accuratecentral differences (CD2).

(b) Determine the time-stepping eigenvalue λk for the MOL scheme. Expand λk for small kx∆xand ky∆y to determine the order of accuracy of the scheme in space.

(c) Determine the amplification factor, A, when using the theta-scheme in time to solve the test-equation found in part (b).

(d) Determine the time-step restriction based on absolute stability when using the theta-scheme intime. Assume 0 ≤ θ ≤ 1.

Solution:

(a) Here is the MOL approximation (for periodic BC’s we add a ghost point at −1).

dVµνdt

=(κ1D+xD−x + κ2D+yD−y

)Vµ,ν − αVµ,ν , µ = 0, 1, . . . , Nx − 1, ν = 0, 1, . . . , Ny − 1,

VNx,ν(t) = V0,ν(t), V−1,ν(t) = VNx−1,ν(t), ν = −1, 0, 1, 2, . . . , Ny,

Vµ,Ny(t) = Vµ,0(t), Vµ,−1(t) = Vµ,Ny−1(t), µ = −1, 0, 1, 2, . . . , Nx,

Vµ,ν(0) = u0(xµ,ν , yµ,ν), µ = −1, 0, 1, 2, . . . , Nx − 1, ν = −1, 1, 2, . . . , Ny − 1,

(b) Let

Vµν = V̂ (t)eik·xµν

Using

D+xD−xeik·xµν =

−4 sin2(ξx/2)

∆x2,

D+yD−yeik·xµν =

−4 sin2(ξy/2)

∆y2,

where ξx = kx∆x and ξy = ky∆y, gives

dV̂

dt= λkV̂ ,

λk = −(κ1

4 sin2(ξx/2)

∆x2+ κ2

4 sin2(ξy/2)

∆y2

)− α.

1

For small kx∆x and small ky∆y we have

4 sin2(kx∆x/2)

∆x2= k2

x +O(∆x2),

4 sin2(ky∆y/2)

∆y2= k2

y +O(∆y2),

and thus

λk = −(κ1k

2x + κ2k

22

)+O(∆x2 + ∆y2)

and the scheme is second order accurate since the true symbol is λek = −(κ1k2x + κ2k

22)− α.

(c) Using the θ scheme gives (from the previous problem set)

A =1 + (1− θ)λk∆t

1− θλk∆t

(d) Following the previous problem set, the time-step restriction is as follows.For θ ≥ 1

2 there is no time-step restriction.For θ < 1

2 , we require

∆t(κ1

4

∆x2+ κ2

4

∆y2+ α

)≤ 2

1− 2θ

2

2. AFS

(a) The AFS scheme solves[I − ∆t

2D+xD−x

] [I − ∆t

2D+yD−y

]Un+1ij

=

[I +

∆t

2D+xD−x

] [I +

∆t

2D+yD−y

]Unij ,+

∆t

2

[f(xij , t

n+1) + f(xij , tn)],

def= Gij , i = 1, 2, . . . , Nx − 1, j = 1, 2, . . . , Ny − 1

with boundary and initial conditions

Un+10,j = gax(yj , t

n+1), Un+1Nx,j

= gbx(yj , tn+1), j = 0, 1, 2, . . . , Ny,

Un+1i,0 = gay(xi, t

n+1), Un+1i,Ny

= gby(xi, tn+1), i = 0, 1, 2, . . . , Nx,

U0i,j = u0(xij), i = 0, 1, 2, . . . , Nx, j = 0, 1, 2, . . . , Ny.

In matrix form this looks like[I − ∆t

2A

] [I − ∆t

2B

]Wn+1 =

[I +

∆t

2A

] [I +

∆t

2B

]Wn +

∆t

2

[Fn+1 + Fn

]These equations can be efficiently solved using a set of 1D tridiagonal solves. Define

U∗ij =

[I − ∆t

2D+yD−y

]Un+1ij .

Stage I: Solve the 1D tridiagonal systems in i (for each fixed j = 1, 2, . . . , Ny − 1)[I − ∆t

2D+xD−x

]U∗ij = Gij , i = 1, 2, . . . , Nx − 1,

with boundary conditions on the left and right,

U∗0,j =

[I − ∆t

2D+yD−y

]gax(yj , t),

U∗Nx,j =

[I − ∆t

2D+yD−y

]gbx(yj , t).

Stage II: Solve the 1D triagonal systems in j (for each fixed i = 1, 2, . . . , Nx − 1)[I − ∆t

2D+yD−y

]Un+1ij = U∗

ij , j = 1, 2, . . . , Ny − 1 (1)

with boundary conditions on the bottom and top,

Un+1i,0 = gay(xi, t

n+1), Un+1i,Ny

= gby(xi, tn+1).

Also set boundary conditions on the left and right

Un+10,j = gax(yj , t

n+1), Un+1Nx,j

= gbx(yj , tn+1), j = 0, 1, 2, . . . , Ny

(b)

Polynomial manufactured solution results:

3

>> heat2dAFSMMS

poly: t=0.50 CPU=2.00e-02 Nx= 20 Ny= 20 Nt= 10 dt=5.00e-02 max-Err=3.55e-15

poly: t=0.50 CPU=3.00e-02 Nx= 40 Ny= 40 Nt= 20 dt=2.50e-02 max-Err=1.42e-14 ratio=0.25

poly: t=0.50 CPU=1.00e-01 Nx= 80 Ny= 80 Nt= 40 dt=1.25e-02 max-Err=4.75e-14 ratio=0.30

poly: t=0.50 CPU=4.80e-01 Nx=160 Ny=160 Nt= 80 dt=6.25e-03 max-Err=1.07e-13 ratio=0.45

poly: t=0.50 CPU=2.95e+00 Nx=320 Ny=320 Nt=160 dt=3.13e-03 max-Err=1.35e-13 ratio=0.79

True solution results:

>> heat2dAFSMMS

true: t=0.50 CPU=2.00e-02 Nx= 20 Ny= 20 Nt= 10 dt=5.00e-02 max-Err=2.69e-03

true: t=0.50 CPU=3.00e-02 Nx= 40 Ny= 40 Nt= 20 dt=2.50e-02 max-Err=6.56e-04 ratio=4.10

true: t=0.50 CPU=1.00e-01 Nx= 80 Ny= 80 Nt= 40 dt=1.25e-02 max-Err=1.63e-04 ratio=4.02

true: t=0.50 CPU=4.70e-01 Nx=160 Ny=160 Nt= 80 dt=6.25e-03 max-Err=4.07e-05 ratio=4.01

true: t=0.50 CPU=2.52e+00 Nx=320 Ny=320 Nt=160 dt=3.13e-03 max-Err=1.02e-05 ratio=4.00

-0.05

2

1.5 1

0u

Heat Eqn true (AFS) t=5.000e-01 Nx=40

y

1

x

0.05

0.50.5

0 0

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

-1

2

-0.5

1.5 1

0

10-3

err

or

Error - Heat Eqn true (AFS) t=5.000e-01 Nx=40

0.5

y

1

x

1

0.50.5

0 0

-6

-4

-2

0

2

4

6

10-4

Figure 1: Solution and error for Nx = 40.

Listing 1: heat2dAFSMMS.m

1 %

2 % Solve the 2D heat equation using an Approximate Factored Scheme (AFS)

3 % and manufactured solutions

4 %

5 % u_t = kappa * (u_xx + u_yy ), ax < x < bx, ay < y < by, 0 < t <= tFinal

6 % u(ax,y,t) = gax(y,t), u(bx,y,t)=gbx(y,t), ay < y < by

7 % u(x,ay,t) = gay(x,t), u(x,by,t)=gby(x,t), ax < x < ay

8 % u(x,y,0) = u_0(x,y)

9 %

10 clear; clf; fontSize=14;

11 rainbow; % load rainbow colour table

1213 debug=0; % set to 1 for debugging

14 plotOption=1; % set to 1 for plotting

15 kappa=.05; % coefficient of diffusion

16 kx=3.*pi; % x-wave number in the IC and exact solution

17 ky=2.*pi; % y-wave number in the IC and exact solution

18 ax=0.; bx=1.; ay=0.; by=2.; % space interval interval

19 tFinal=.5; % final time

20

4

21 % exact solution function:

22 % ue = @(x,y,t) cos(kx*x).*cos(ky*y)*exp(-kappa*(kx^2+ky^2)*t);

2324 % Choose the manufactured solution:

25 ms = ’poly’;

26 % ms = ’trig’; % un-comment for trig MS

27 % ms = ’true’;

2829 if( strcmp(ms,’poly’) )

30 % Polynomial manufactured solution:

31 % b0=.5; b1=.7; b2=.0; % time coefficients

32 b0=.5; b1=.7; b2=.9; % time coefficients

33 c00=1.; c10=.8; c20=.6; c01=.3; c11=.25; c02=.2; % space coefficients

34 % b1=1.; b2=1.; c10=1; c20=1; c01=1; c11=.0; c02=.0;

3536 ue = @(x,y,t) (b0+b1*t+b2*t^2)*(c00 +c10*x +c20*x.^2 + c01*y + c11*x.*y + c02*y.^2);

37 uet = @(x,y,t) (b1+2.*b2*t )*(c00 +c10*x +c20*x.^2 + c01*y + c11*x.*y + c02*y.^2);

38 uexx = @(x,y,t) (b0+b1*t+b2*t^2)*(2.*c20);

39 ueyy = @(x,y,t) (b0+b1*t+b2*t^2)*(2.*c02);

40 f = @(x,y,t) uet(x,y,t) - kappa*(uexx(x,y,t)+ueyy(x,y,t)); % forcing

4142 elseif( strcmp(ms,’trig’) )

43 % Trigonometric manufactured solution:

44 kt = pi; kx=5*pi; ky=4.*pi;

45 ue = @(x,y,t) cos(kt*t)*cos(kx*x).*cos(ky*y);

46 uet = @(x,y,t) -kt*sin(kt*t)*cos(kx*x).*cos(ky*y);

47 uexx = @(x,y,t) (-kx^2)*ue(x,y,t);

48 ueyy = @(x,y,t) (-ky^2)*ue(x,y,t);

49 f = @(x,y,t) uet(x,y,t) - kappa*(uexx(x,y,t)+ueyy(x,y,t)); % forcing

5051 elseif( strcmp(ms,’true’) )

52 % exact solution function:

53 kx=3.*pi; ky=2.*pi;

54 ue = @(x,y,t) sin(kx*x).*cos(ky*y)*exp(-kappa*(kx^2+ky^2)*t);

55 f = @(x,y,t) 0.; % forcing

56 end

575859 % Choose BC’s and IC to match the true solution

60 gax = @(y,t) ue(ax,y,t); % BC RHS at x=ax

61 gbx = @(y,t) ue(bx,y,t); % BC RHS at x=bx

62 gay = @(x,t) ue(x,ay,t); % BC RHS at y=ay

63 gby = @(x,t) ue(x,by,t); % BC RHS at y=by

64 u0 = @(x,y) ue(x,y,0); % IC function

6566 numResolutions=5;

67 for m=1:numResolutions % Loop over grid resolutions

6869 % --- Setup the grid ---

70 Nx=10*2^m; Ny=Nx; % number of space intervals

71 dx=(bx-ax)/Nx; dy=(by-ay)/Ny; % grid spacing

7273 iax=1; iay=1; % index of boundary point at x=xa, y=ya

74 ibx=iax+Nx; iby=1+Ny; % index of boundary point at x=xb, y=yb

75 Ngx=ibx; Ngy=iby; % number of grid points in x and y

7677 i1x=iax+1; i1y=iay+1; % first interior pt in x and y

78 i2x=ibx-1; i2y=iby-1; % last interior pt in x and y

79

5

80 % -- form the 2D grid points ---

81 x = zeros(Ngx,Ngy); y = zeros(Ngx,Ngy); % grid

82 for( iy=1:Ngy )

83 for( ix=1:Ngx )

84 x(ix,iy)=ax + (ix-iax)*dx;

85 y(ix,iy)=ay + (iy-iay)*dy;

86 end;

87 end;

8889 % allocate space for the solution

90 un = zeros(Ngx,Ngy); % holds U_i^n

9192 dt = dx; % time step (adjusted below)

93 Nt = round(tFinal/dt); % number of time-steps

94 dt = tFinal/Nt; % adjust dt to reach tFinal exactly

9596 t=0.;

97 un = u0(x,y); % initial conditions

9899 % ----- Form the implicit matrices ------

100101 % Allocate the sparse matrix, vector-solution and RHS

102 Ax = sparse(Ngx,Ngx); % implicit matrix

103 Wx = zeros(Ngx,1); % solution as a vector

104 rhsx = zeros(Ngx,1); % right-hand-side

105106 Ay = sparse(Ngy,Ngy); % implicit matrix

107 Wy = zeros(Ngy,1); % solution as a vector

108 rhsy = zeros(Ngy,1); % right-hand-side

109110 % Fill in the interior equations: I - (.5*kappa*dt)*( D+x D-x )

111 for( ix=i1x:i2x )

112 ie = ix; % eqn number

113 Ax(ie,ix-1) = - (.5*kappa*dt)*( 1/dx^2 );

114 Ax(ie,ix ) = 1. - (.5*kappa*dt)*( -2*( 1/dx^2 ) );

115 Ax(ie,ix+1) = - (.5*kappa*dt)*( 1/dx^2 );

116 end;

117 % Boundary conditions for the matrix

118 ix=iax; ie=ix; Ax(ie,ie)=1.; % BC at x=xa

119 ix=ibx; ie=ix; Ax(ie,ie)=1.; % BC at x=xb

120121 % Fill in the interior equations: I - (.5*kappa*dt)*( D+y D-y )

122 for( iy=i1y:i2y )

123 ie = iy; % eqn number for pt (ix,iy)

124 Ay(ie,iy-1) = - (.5*kappa*dt)*( 1/dy^2 );

125 Ay(ie,iy ) = 1. - (.5*kappa*dt)*( -2*( 1/dy^2 ) );

126 Ay(ie,iy+1) = - (.5*kappa*dt)*( 1/dy^2 );

127 end;

128 % Boundary conditions for the matrix

129 iy=iay; ie=iy; Ay(ie,ie)=1.; % BC at y=ya

130 iy=iby; ie=iy; Ay(ie,ie)=1.; % BC at y=yb

131132 DpxDmx = @(u,I,J) (u(I+1,J)-2*u(I,J)+u(I-1,J))/(dx^2);

133 DpyDmy = @(u,I,J) (u(I,J+1)-2*u(I,J)+u(I,J-1))/(dy^2);

134135 % --- Start time-stepping loop ---

136 cpu0=cputime;

137138 dAx =decomposition(Ax,’lu’); % factor the matrix

6

139 dAy =decomposition(Ay,’lu’); % factor the matrix

140141 I1=i1x:i2x; I2=i1y:i2y; % interior points

142 J1=iax:ibx; J2=iay:iby; % includes boundary points

143144 for( n=1:Nt )

145146 t = n*dt; % new time

147148 % ======== STAGE I =====

149 % -- evaluate the right-hand-side at interior pts: (over-write un)

150 un(J1,I2) = un(J1,I2) + (.5*kappa*dt)*DpyDmy(un,J1,I2); % NOTE J1, include boundary pts

151 un(I1,I2) = un(I1,I2) + (.5*kappa*dt)*DpxDmx(un,I1,I2) ...

152 + (.5*dt)*( f(x(I1,I2),y(I1,I2),t) + f(x(I1,I2),y(I1,I2),t-dt) );

153154 % Boundary conditions for U^* = [I - .5*kappa*dt*D+yD-y] U^{n+1}

155 ix=iax; un(ix,I2)=gax(y(ix,I2),t) ...

156 - (.5*kappa*dt/dy^2)*( gax(y(ix,I2+1),t) -2.*gax(y(ix,I2),t) + gax(y(ix,I2-1),t) );

157 ix=ibx; un(ix,I2)=gbx(y(ix,I2),t) ...

158 - (.5*kappa*dt/dy^2)*( gbx(y(ix,I2+1),t) -2.*gbx(y(ix,I2),t) + gbx(y(ix,I2-1),t) );

159160 % --- solves in x-direction (for U^*) ---

161 un(J1,I2) = dAx\un(J1,I2);

162163 % ========== STAGE II =============

164 iy=iay; un(J1,iy)=gay(x(J1,iy),t); % BC at y=ya

165 iy=iby; un(J1,iy)=gby(x(J1,iy),t); % BC at y=yb

166167 % --- solves in y-direction ---

168 un(I1,J2) = (dAy\(un(I1,J2)’))’;

169170 ix=iax; un(ix,J2)=gax(y(ix,J2),t); % BC at x=xa

171 ix=ibx; un(ix,J2)=gbx(y(ix,J2),t); % BC at x=xb

172173 if debug==1

174 surf( x,y,un );

175 title(sprintf(’Solution t=%9.3e’,t));

176 uexact = ue(x,y,t); % eval exact solution:

177 errMax = max(max(abs(un-uexact))); % max-norm error

178 fprintf(’%s: t=%10.4e: Nx=%d Ny=%d Nt=%d dt=%9.3e max-Error=%8.2e\n’,ms,t,Nx,Ny,Nt,dt,errMax);

179 pause

180 % plot errors

181 surf( x,y,un-uexact );

182 title(sprintf(’Error - Heat Equation (AFS) t=%9.3e (Nx=%d) cfl=%4.2f’,t,Nx,cfl));

183 xlabel(’x’); ylabel(’y’); zlabel(’error’);

184 pause

185 end

186187 end;

188 % --- End time-stepping loop ---

189 cpu = cputime-cpu0;

190191 uexact = ue(x,y,t); % eval exact solution:

192 errMax = max(max(abs(un-uexact))); % max-norm error

193 err(m)=errMax;

194 fprintf(’%s: t=%4.2f CPU=%8.2e Nx=%3d Ny=%3d Nt=%3d dt=%8.2e max-Err=%8.2e’,ms,t,cpu,Nx,Ny,Nt,dt,

errMax);

195 if( m>1 ) fprintf(’ ratio=%4.2f\n’,err(m-1)/err(m)); else fprintf(’\n’); end;

196

7

197 % plot results

198 if( plotOption==1 && m==2 )

199 figure(1)

200 surf( x,y,un ); colormap(rainbowMap); colorbar;

201 title(sprintf(’Heat Eqn %s (AFS) t=%9.3e Nx=%d’,ms,t,Nx));

202 xlabel(’x’); ylabel(’y’); zlabel(’u’); set(gca,’FontSize’,fontSize);

203 % grid on;

204 plotName=sprintf(’heat2d_AFS_MS%s_N%i.eps’,ms,Nx);

205 print(’-depsc2’,plotName); % save as an eps file

206 % fprintf(’Wrote plot file %s\n’,plotName);

207208 figure(2)

209 % plot errors

210 surf( x,y,un-uexact ); colormap(rainbowMap); colorbar;

211 title(sprintf(’Error - Heat Eqn %s (AFS) t=%9.3e Nx=%d’,ms,t,Nx));

212 xlabel(’x’); ylabel(’y’); zlabel(’error’); set(gca,’FontSize’,fontSize);

213 plotName=sprintf(’heat2d_AFS_MS%s_N%i_err.eps’,ms,Nx);

214 print(’-depsc2’,plotName); % save as an eps file

215 % fprintf(’Wrote plot file %s\n’,plotName);

216217 pause(1);

218 end;

219220 end; % end loop over grid resolutions

8

3. (20 points) Advection equation. Consider the initial value problem for the advection equation

ut + cux = 0, 0 ≤ x < 2π, 0 ≤ t ≤ tfinal,

u(x, 0) = u0(x),

u(x+ 2π, t) = u(x, t),

with periodic boundary conditions. Discretize on a uniform grid xj = j∆x, j = −1, 0, . . . , Nx with∆x = 2π/Nx. Use one ghost point on the left to implement the periodic boundary conditions.

Solution:

(a) The exact solution is

u(x, t) = u0(x− ct)

(b) The Lax-Wendroff scheme is

Un+1i = Uni − c∆tD0U

ni + c2 ∆t2

2D+xD−xU

ni

Substituting Unj = Aneikxj leads to

A = 1− c∆t

∆xi sin(ξ)− 2

c2∆t2

∆x2sin2(ξ/2),

= 1− λi sin(ξ)− 2λ2 sin2(ξ/2),

= 1− 2λi sin(ξ/2) cos(ξ/2)− 2λ2 sin2(ξ/2),

where ξ = k∆x. For absolute stability we require |A| ≤ 1 or |A|2 ≤ 1. Now |A|2 is

|A|2 = (1− 2λ2 sin2(ξ/2))2 + 4λ2 sin2(ξ/2) cos2(ξ/2),

= 1− 4λ2 sin2(ξ/2) + 4λ4 sin4(ξ/2) + 4λ2 sin2(ξ/2)− 4λ2 sin4(ξ/2),

= 1 + 4λ2 sin4(ξ/2)(λ2 − 1)

Thus |A|2 ≤ 1 implies

1 + 4λ2 sin4(ξ/2)(λ2 − 1) ≤ 1,

=⇒ sin4(ξ/2)(λ2 − 1) ≤ 0,

=⇒ λ2 ≤ 1

The time-step restriction from absolute stability is thus

|c|∆t∆x

≤ 1

(c) The code is given below. Here are results for the sine initial condition

9

>> advSchemes

Solve the advection equation with scheme=LF

LF:sine: t=20.000: Nx= 20 Nt= 141 dt=1.42e-01 errMax=1.46e+00 errL1=5.94e+00

LF:sine: t=20.000: Nx= 40 Nt= 283 dt=7.07e-02 errMax=1.80e+00 errL1=7.33e+00 ratio=[0.81,0.81] (max,L1)

LF:sine: t=20.000: Nx= 80 Nt= 566 dt=3.53e-02 errMax=5.06e-01 errL1=2.05e+00 ratio=[3.56,3.57] (max,L1)

LF:sine: t=20.000: Nx=160 Nt=1132 dt=1.77e-02 errMax=1.26e-01 errL1=5.04e-01 ratio=[4.01,4.07] (max,L1)

LF:sine: t=20.000: Nx=320 Nt=2264 dt=8.83e-03 errMax=3.14e-02 errL1=1.25e-01 ratio=[4.02,4.02] (max,L1)

>> advSchemes

Solve the advection equation with scheme=LW

LW:sine: t=20.000: Nx= 20 Nt= 141 dt=1.42e-01 errMax=9.96e-01 errL1=4.05e+00

LW:sine: t=20.000: Nx= 40 Nt= 283 dt=7.07e-02 errMax=1.17e+00 errL1=4.75e+00 ratio=[0.85,0.85] (max,L1)

LW:sine: t=20.000: Nx= 80 Nt= 566 dt=3.53e-02 errMax=4.68e-01 errL1=1.87e+00 ratio=[2.50,2.54] (max,L1)

LW:sine: t=20.000: Nx=160 Nt=1132 dt=1.77e-02 errMax=1.24e-01 errL1=4.97e-01 ratio=[3.78,3.76] (max,L1)

LW:sine: t=20.000: Nx=320 Nt=2264 dt=8.83e-03 errMax=3.13e-02 errL1=1.25e-01 ratio=[3.96,3.97] (max,L1)

0 1 2 3 4 5 6

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LF t=2.000e+01 (Nx=20) dt=1.42e-01

computed

true

0 1 2 3 4 5 6

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LF t=2.000e+01 (Nx=40) dt=7.07e-02

computed

true

0 1 2 3 4 5 6

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LF t=2.000e+01 (Nx=80) dt=3.53e-02

computed

true

0 1 2 3 4 5 6

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LF t=2.000e+01 (Nx=160) dt=1.77e-02

computed

true

Figure 2: Leap-frog.

10

0 1 2 3 4 5 6

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LW t=2.000e+01 (Nx=20) dt=1.42e-01

computed

true

0 1 2 3 4 5 6

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LW t=2.000e+01 (Nx=40) dt=7.07e-02

computed

true

0 1 2 3 4 5 6

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LW t=2.000e+01 (Nx=80) dt=3.53e-02

computed

true

0 1 2 3 4 5 6

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LW t=2.000e+01 (Nx=160) dt=1.77e-02

computed

true

Figure 3: Lax-Wendroff.

11

(c) Periodic top-hat

>> advSchemes

Solve the advection equation with scheme=LF

LF:topHat: t=20.000: Nx= 20 Nt= 141 dt=1.42e-01 errMax=7.38e-01 errL1=1.55e+00

LF:topHat: t=20.000: Nx= 40 Nt= 283 dt=7.07e-02 errMax=5.55e-01 errL1=1.35e+00 ratio=[1.33,1.15] (max,L1)

LF:topHat: t=20.000: Nx= 80 Nt= 566 dt=3.53e-02 errMax=8.43e-01 errL1=1.02e+00 ratio=[0.66,1.32] (max,L1)

LF:topHat: t=20.000: Nx=160 Nt=1132 dt=1.77e-02 errMax=6.74e-01 errL1=7.42e-01 ratio=[1.25,1.38] (max,L1)

LF:topHat: t=20.000: Nx=320 Nt=2264 dt=8.83e-03 errMax=7.18e-01 errL1=6.53e-01 ratio=[0.94,1.14] (max,L1)

>> advSchemes

Solve the advection equation with scheme=LW

LW:topHat: t=20.000: Nx= 20 Nt= 141 dt=1.42e-01 errMax=5.48e-01 errL1=1.01e+00

LW:topHat: t=20.000: Nx= 40 Nt= 283 dt=7.07e-02 errMax=5.79e-01 errL1=8.46e-01 ratio=[0.95,1.20] (max,L1)

LW:topHat: t=20.000: Nx= 80 Nt= 566 dt=3.53e-02 errMax=6.57e-01 errL1=6.01e-01 ratio=[0.88,1.41] (max,L1)

LW:topHat: t=20.000: Nx=160 Nt=1132 dt=1.77e-02 errMax=6.41e-01 errL1=3.98e-01 ratio=[1.03,1.51] (max,L1)

LW:topHat: t=20.000: Nx=320 Nt=2264 dt=8.83e-03 errMax=6.73e-01 errL1=2.70e-01 ratio=[0.95,1.48] (max,L1)

0 1 2 3 4 5 6

x

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

u

Advection Eqn. LF t=2.000e+01 (Nx=20) dt=1.42e-01

computed

true

0 1 2 3 4 5 6

x

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

u

Advection Eqn. LF t=2.000e+01 (Nx=40) dt=7.07e-02

computed

true

0 1 2 3 4 5 6

x

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

u

Advection Eqn. LF t=2.000e+01 (Nx=80) dt=3.53e-02

computed

true

0 1 2 3 4 5 6

x

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

u

Advection Eqn. LF t=2.000e+01 (Nx=160) dt=1.77e-02

computed

true

Figure 4: Leap-frog.

12

0 1 2 3 4 5 6

x

-0.2

0

0.2

0.4

0.6

0.8

1

u

Advection Eqn. LW t=2.000e+01 (Nx=20) dt=1.42e-01

computed

true

0 1 2 3 4 5 6

x

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

u

Advection Eqn. LW t=2.000e+01 (Nx=40) dt=7.07e-02

computed

true

0 1 2 3 4 5 6

x

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

u

Advection Eqn. LW t=2.000e+01 (Nx=80) dt=3.53e-02

computed

true

0 1 2 3 4 5 6

x

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

u

Advection Eqn. LW t=2.000e+01 (Nx=160) dt=1.77e-02

computed

true

Figure 5: Lax-Wendroff.

13

Listing 2: advSchemes.m

1 %

2 % Solve the advection equation on a periodic domain with different schemes

3 %

4 % u_t = -c*u_x , 0 < x < 2*pi , 0 < t <= tFinal

5 % u(x+2*pi,t) = u(x,t),

6 % u(x,0) = u_0(x)

7 %

8 clear; clf;

91011 scheme=’LF’;

12 scheme=’LW’;

1314 solution=’sine’;

15 % solution=’topHat’;

1617 plotOption=0; % 1=plot movie

1819 c=2.; % wave speed

20 kx=4.; % wave number in the IC and exact solution

21 a=0.; b=2*pi; % space interval interval

22 tFinal=20.; % final time

23 cfl=.9; % cfl scale factor

2425 % define the exact solution function

26 if( strcmp(solution,’sine’) )

27 uexact = @(x,t) sin(kx*(x-c*t));

28 else

29 % top hat

30 uexact = @(x,t) periodicTopHat( x-c*t, 1.,3. );

31 end

3233 u0 = @(x) uexact(x,0); % initial condition function

3435 fprintf(’Solve the advection equation with scheme=%s\n’,scheme);

36 % Loop over grid resolutions:

37 for m=1:5

38 Nx=10*2^m; % number of space intervals

39 dx=(b-a)/Nx; % grid spacing

4041 nga=1; % number of ghost on left

42 ia=1+nga; % index of boundary point at x=a

43 ib=ia+Nx; % index of boundary point at x=b

44 i1=ia; % first index to apply equation at.

45 i2=ib-1; % last index to apply equation

46 Ng=ib; % number of grid points

4748 x = linspace(a-nga*dx,b,Ng)’; % spatial grid (include ghost points)

4950 % allocate space for the solution at two levels

51 un = zeros(Ng,1); % holds U_i^n

52 unp1 = zeros(Ng,1); % holds U_i^{n+1}

5354 dt = cfl*dx/c; % time step (adjusted below)

55 Nt = round(tFinal/dt); % number of time-steps

56 % fprintf(’dt=%9.3e, adjusted=%9.3e, diff=%8.2e\n’,dt,tFinal/Nt,abs(dt-tFinal/Nt));

57 dt = tFinal/Nt; % adjust dt to reach tFinal exactly

14

58 I = i1:i2;

5960 Dz = @(u,I) (u(I+1) -u(I-1))/(2.*dx);

61 DpDm = @(u,I) (u(I+1)-2*u(I)+u(I-1))/(dx^2);

6263 t=0.;

64 un = u0(x); % initial conditions

65 if( strcmp(scheme,’LF’) )

66 um = uexact(x,t-dt); % Leap Frog needs u(-dt)

67 end

6869 if( plotOption > 0 ) % plot initial conditions

70 plot(x,un,’b-+’); title(sprintf(’%s: t=%9.3e’,scheme,t)); xlabel(’t’); xlim([0,2*pi]); set(gca

,’FontSize’,14);

71 drawnow;

72 pause;

73 end

7475 % --- Start time-stepping loop ---

76 for( n=1:Nt )

77 t = n*dt; % new time

7879 if( strcmp(scheme,’LF’) )

8081 % Leap-frog

82 unp1(I) = um(I) + (-2*c*dt )*Dz(un,I);

83 um = un;

8485 else

8687 % Lax-Wendroff:

88 unp1(I) = un(I) + (-c*dt)*Dz(un,I) + (.5*c^2*dt^2)*DpDm(un,I);

8990 end

9192 unp1(ia-1)=unp1(ia-1+Nx); % periodic BC on left

93 unp1(ib )=unp1(ib -Nx); % periodic BC at x=b

9495 un=unp1; % Set un <- unp1 for next step

9697 if( plotOption>0 ) % make a movie

98 plot(x,un,’b-+’); title(sprintf(’%s: t=%9.3e’,scheme,t)); xlabel(’t’); xlim([0,2*pi]); set(gca

,’FontSize’,14);

99 drawnow; pause(.1);

100 end

101 end;

102 % --- End time-stepping loop ---

103104 ue = uexact(x,t); % eval exact solution:

105 err = un(ia:ib-1)-ue(ia:ib-1); % error (do not include periodic ghost);

106 errMax(m) = max(abs(err)); % max-norm error

107 errL1(m) = norm(err,1)*dx; % discrete L1-norm

108 fprintf(’%s:%s: t=%6.3f: Nx=%3d Nt=%4d dt=%8.2e errMax=%8.2e errL1=%8.2e’,scheme,solution,t,Nx,Nt

,dt,errMax(m),errL1(m));

109 if( m==1 ) fprintf(’\n’); else fprintf(’ ratio=[%3.2f,%3.2f] (max,L1)\n’,errMax(m-1)/errMax(m),

errL1(m-1)/errL1(m)); end;

110111 % plot results

112 plot( x(ia:ib),un(ia:ib),’b-o’, x(ia:ib),ue(ia:ib),’k-’,’Linewidth’,2);

15

113 legend(’computed’,’true’); set(gca,’FontSize’,14);

114 title(sprintf(’Advection Eqn. %s t=%9.3e (Nx=%d) dt=%8.2e’,scheme,t,Nx,dt));

115 xlabel(’x’); ylabel(’u’); xlim([0,2*pi]);

116 grid on;

117 plotFile=sprintf(’adv_%s_%s_Nx_%d.eps’,scheme,solution,Nx);

118 print(’-depsc2’,plotFile); % fprintf(’Saved file %s\n’,plotFile);

119120 if( plotOption>0 ) pause; end;

121122 end; % end for m

16