Solve a partial differential equations using the explicit method
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Hello,
I have two codes. Can anyone help me to check them and tell me what's wrong with these codes.
the first code, I'm not getting the error results neither its plot
the second code, the exacct solution and the error are not appearing, and the results of the explict equation are doubtful.
I attached the problems and the codes
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Accepted Answer
Abraham Boayue
on 11 Feb 2022
%% Cp2ex2nd
% 1. Space steps
beta = sqrt(0.03); % sqrt(0.003); sqrt(1/878);
xa = 0;
xb = 1;
dx = 1/40;
N = (xb-xa)/dx ;
x = xa:dx:xb;
%2.Time steps
ta = 0;
tb = 0.5;
dt = 1/3300;
M = (tb-ta)/dt ;
t = ta:dt:tb;
%3. Controling Parameters
R1 = (beta/dx)^2*dt;
R2 = dt/(2*dx);
%4. Define equations A, B , C and phi(x,t)
A = @(x,t) (50/3)*(x-0.5+4.95*t);
B = @(x,t) (250/3)*(x-0.5+0.75*t);
C = @(x,t) (500/3)*(x-0.375);
phi = @(x,t)(0.1*exp(-A(x,t)) +0.5*exp(-B(x,t))+exp(-C(x,t)))./...
(exp(-A(x,t)) +exp(-B(x,t))+exp(-C(x,t)));
% 5 Initial and boundary conditions
f = @(x) phi(x,t(1)); % Initial condition
g1 = @(t)phi(x(1),t); % Left boundary condition
g2 = @(t)phi(x(N+1),t); % Right boundary condition
% 6 Implementation of the explicit method
u = zeros(N+1,M+1);
u(2:N,1) = f(x(2:N)); % Put in the initial condition
u(1,:) = g1(t); % The boundary conditions, g1 and g2 at
u(N+1,:) = g2(t); % x = 0 and x = 1
for i = 2:N
u(i,2) = R1*(u(i+1,1)-2*u(i,1)+u(i-1,1))-R2*(u(i+1,1)-u(i-1,1))*u(i,1)+u(i,1);
end
for j=2:M % Time Loop
for i= 2:N % Space Loop
u(i,j+1) = R1*(u(i+1,j)-2*u(i,j)+u(i-1,j))-R2*(u(i+1,j)-u(i-1,j))*u(i,j)+u(i,j);
end
end
% Make some plots
T= 0:.1:M;
V = [];
for i= 1: length(T)
P = find(t==T(i));
V = [V P];
end
figure
subplot(131)
for i = 1:length(V)
hold on
plot(x,u(:,V(i)),'*-','linewidth',2.5,'DisplayName',sprintf('t = %1.4f',t(V(i))))
end
legend('-DynamicLegend','location','bestoutside');
a = ylabel('Heat');
set(a,'Fontsize',14);
a = xlabel('x');
set(a,'Fontsize',14);
a=title(['Using The Explicit Method - Lamda =' num2str(R1)]);
set(a,'Fontsize',16);
grid;
% disp(u(:,V)'); % Each row corresponds to a particular value of t and
% Each column corresponds to a particular value of x
% Implement the exact solution and compare it to the exact solution
Exact = @(x,t) 0.5*erfc((0.5*(x-t))./(beta*sqrt(t)))+0.5*exp(x/beta^2).*...
erfc((0.5*(x+t))./(beta*sqrt(t)));
subplot(132)
for j = 1:length(V)
hold on
plot(x,Exact(x,t(V(j))),'*-','linewidth',2.5,'DisplayName',sprintf('t = %1.4f',t(V(j))))
end
legend('-DynamicLegend','location','bestoutside');
a = ylabel('Heat');
set(a,'Fontsize',14);
a = xlabel('x');
set(a,'Fontsize',14);
a=title(' Exact solution');
set(a,'Fontsize',16);
grid;
% 6 Create the error function
[X,T] = meshgrid(x,t);
Exact2 = 0.5*erfc((0.5*(X-T))./(beta*sqrt(T)))+0.5*exp(X/beta^2).*...
erfc((0.5*(X+T))./(beta*sqrt(T)));
Error =abs(Exact2'-u);
subplot(133)
for j = 1:length(V)
hold on
plot(x,Error(:,(V(j))),'*-','linewidth',2.5,'DisplayName',sprintf('t = %1.4f',t(V(j))))
end
legend('-DynamicLegend','location','bestoutside');
a = ylabel('Heat');
set(a,'Fontsize',14);
a = xlabel('x');
set(a,'Fontsize',14);
a=title(' Error ');
set(a,'Fontsize',16);
grid;
%% Cp2Exo12nd
beta = sqrt(0.03); % sqrt(1/878);
xa = 0;
xb = 1;
dx = 1/40;
N = (xb-xa)/dx ;
x = xa:dx:xb;
%2.Time steps
ta = 0;
tb = 0.5;
dt = 1/3300;
M = (tb-ta)/dt ;
t = ta:dt:tb;
%3. Controling Parameters
R1 = (beta/dx)^2*dt;
R2 = dt/(2*dx);
% 5 Initial and boundary conditions
f = @(x) 0; % Initial condition
g1 = @(t)1; % Left boundary condition
g2 = @(t)0;
% 6 Implementation of the explicit method
u = zeros(N+1,M+1);
u(2:N,1) = f(x(2:N)); % Put in the initial condition
u(1,:) = g1(t); % Insert the left boundary condition at beginning
% Perform the inner computations
for j=2:M % Time Loop
for i= 2:N % Space Loop
u(i,j+1) = R1*(u(i+1,j)-2*u(i,j)+u(i-1,j))-R2*(u(i+1,j)-u(i-1,j))+u(i,j);
end
end
% Insert the right boundary condition at end
for j = 1:M
U = u(1,j)-2*dx*g2(t(j));
u(1,j+1) = R1*(u(2,j)-2*u(1,j)+U)-R2*(u(2,j)-U)+u(1,j);
end
% Make some plots
T= 0:.1:M;
V = [];
for j= 1: length(T)
P = find(t==T(j));
V = [V P];
end
figure
subplot(131)
for j = 1:length(V)
hold on
plot(x,u(:,V(j)),'*-','linewidth',2.5,'DisplayName',sprintf('t = %1.4f',t(V(j))))
end
legend('-DynamicLegend','location','bestoutside');
a = ylabel('Heat');
set(a,'Fontsize',14);
a = xlabel('x');
set(a,'Fontsize',14);
a=title(['Using The Explicit Method - Lamda =' num2str(R1)]);
set(a,'Fontsize',16);
grid;
% disp(u(:,V)'); % Each row corresponds to a particular value of t and
% Each column corresponds to a particular value of x
% Implement the exact solution and compare it to the exact solution
Exact = @(x,t) 0.5*erfc((0.5*(x-t))./(beta*sqrt(t)))+0.5*exp(x/beta^2).*...
erfc((0.5*(x+t))./(beta*sqrt(t)));
subplot(132)
for j = 1:length(V)
hold on
plot(x,Exact(x,t(V(j))),'*-','linewidth',2.5,'DisplayName',sprintf('t = %1.4f',t(V(j))))
end
legend('-DynamicLegend','location','bestoutside');
a = ylabel('Heat');
set(a,'Fontsize',14);
a = xlabel('x');
set(a,'Fontsize',14);
a=title(' Exact solution');
set(a,'Fontsize',16);
grid;
% 6 Create the error function
[X,T] = meshgrid(x,t);
Exact2 = 0.5*erfc((0.5*(X-T))./(beta*sqrt(T)))+0.5*exp(X/beta^2).*...
erfc((0.5*(X+T))./(beta*sqrt(T)));
Error =abs(Exact2'-u);
subplot(133)
for j = 1:length(V)
hold on
plot(x,Error(:,(V(j))),'*-','linewidth',2.5,'DisplayName',sprintf('t = %1.4f',t(V(j))))
end
legend('-DynamicLegend','location','bestoutside');
a = ylabel('Heat');
set(a,'Fontsize',14);
a = xlabel('x');
set(a,'Fontsize',14);
a=title(' Error ');
set(a,'Fontsize',16);
grid;
5 Comments
More Answers (1)
Abraham Boayue
on 11 Feb 2022
What about dU/dX(1,t)=0. I think this condition is messing in the code.
No, it is not. Remember the last attachment I emailed you where I derived the discretized FDM equations of the PDE? The boundary condition was incorporated in the equations that I placed in the rectangular box. It is in this section of the code:
for j = 1:M
U = u(1,j)-2*dx*g2(t(j));
u(1,j+1) = R1*(u(2,j)-2*u(1,j)+U)-R2*(u(2,j)-U)+u(1,j);
end
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