Solve a partial differential equations using the explicit method

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Hana Bachi
Hana Bachi 2022-2-11
评论： Hana Bachi 2022-2-11
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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采纳的回答

Abraham Boayue
Abraham Boayue 2022-2-11
%% 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 个评论
显示 3更早的评论
Hana Bachi
Hana Bachi 2022-2-11
Yes, I got unstable resoluts for 1/878 and I was wondring why.
What about dU/dX(1,t)=0. I think this condition is messing in the code
and also the equation of the exact solution for the second code should be Phi(x,t) not the one from the first problem. I got a results after modifying it but they don't look accurate escpecailly with that high error

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更多回答（1 个）

Abraham Boayue
Abraham Boayue 2022-2-11
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
  1 个评论
Hana Bachi
Hana Bachi 2022-2-11
Thank you for making it clearer.
did you check the figure where I changed the exact equation of the second problem? the error is large

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