Question on heat equation 1D Forward in Time Centered in Space
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    Janvier Solaris
 on 5 Jun 2018
  
    
    
    
    
    Answered: Rauhussaba Rauhy
 on 24 Mar 2022
            Hi all I have the following question I am trying to solve the PDE with forward time centered in space with the following parameters:

My code so far
   function E=Expheat(h,k)
tmax = 1; 
t = linspace(0,tmax);
N = 1/h;
M = 1/k; 
x = linspace(-pi/2, pi/2);
g(x) = sin*(pi/2*x) + 1/2*sin*(2*pi*x);
alpha = 1;
mu = alpha^2*k/h^2;
%loop for IC
for j = 1 : M
    U( 1, j) = 0
end
%loop for 
for n = 1: tmax
    U(end, j)= exp(pi^2*t/4)
end
for i =  1 : M+1
    U(g(x), 0) = (1 - 2*U(n,j))-U(n+1)+U(n-1,j+1)
    for j = 2 : N 
           U(g(x), 0) = (1 - 2*U(n,j))-U(n+1)+U(n-1,j+1)
    end
end
I am getting errors and am tying to figure out how to solve this any help would be greatly appreciated.
Thank you
4 Comments
  Abraham Boayue
      
 on 6 Jun 2018
				Here is a code that you may find useful to help solve your problem. I can't really say much about the solution since you did not post the original problem; my code is based on the numerical equations you posted. I would have like to see how the boundary conditions were given instead of the two boundary functions you provided. For example, you may have u(x=0,t) = O1(t), u(x=1,t)= O2(t) or u(0,t) = O1(t), ux(1,t) = O2(t) or something else. If this isn't right, post the original problem as it was given.
clear all
close all
n = 500;
Lx = 1;
dx = Lx/(n-1);
x = 0:dx:Lx;
% 2. Parameters for the t vector
m = 100; 
tf = 5;
dt = tf/(m-1);
t = 0:dt:tf;
% 3. Other parameters needed for the solution
% The value of alpha
Fo = 1/4;   % a mutiplicative constant that should be < = 1/2
            % to insure stability
% Initial and boundary conditions
f  =  @(x)sin(pi/2*x)+(1/2)*sin(2*pi*x);   
g1 = @(t)0;           
g2 = @(t)exp(-pi^2/4*t);
u  =  zeros(n,m);
u(2:n,1) =  f(x(2:n));    % Put in the initial condition starting from
                              % 2 to n-1  since f(0) = 0 and f(N) = 1 for
                              % N = n-1
u(1,1:m)   =  g1(t);       % The boundary conditions, g1 and g2 at
u(n,1:m)   =  g2(t);       % x = 0 and x = 1
% Implementation of the explicit method
for k = 2:m-1   % Time Loop
      for i= 2:n-1  % Space Loop
        u(i,k+1) = Fo*(u(i-1,k)+u(i+1,k))+(1-2*Fo)*u(i,k);
      end
end
plot(x,u,'--','linewidth',2);
a = ylabel('Pressure');
set(a,'Fontsize',14);
a = xlabel('x');
set(a,'Fontsize',14);
a=title(['Using The Explicit Method - Fo =' num2str(Fo)]);
set(a,'Fontsize',16);
grid;
figure
[X, T] = meshgrid(x,t);
s2 = mesh(X',T',u);
title(['3-D plot of the 1D Heat Equation using the Explicit Method - Fo =' num2str(Fo)])
set(s2,'FaceColor',[0 0 1],'edgecolor',[0 0 0],'LineStyle','--');
a = title('Exact solution of the 1D Diffusivity Equation');
set(a,'fontsize',14);
a = xlabel('x');
set(a,'fontsize',20);
a = ylabel('y');
set(a,'fontsize',20);
a = zlabel('z');
set(a,'fontsize',20);
% disp(u');
Accepted Answer
  Abraham Boayue
      
 on 7 Jun 2018
        
      Edited: Abraham Boayue
      
 on 7 Jun 2018
  
      Here is an updated version of the code. I also used matlab pdepe function to validate the results which seem to agree with one another. However, the result obtained from matlab pdepe is more superior than the finite difference method.
clear variables
close all
n = 20;
Lx = 1;
dx = Lx/(n-1);
x = 0:dx:Lx;
% 2. Parameters for the t vector
m = 100; 
tf = 1;
dt = tf/(m-1);
t = 0:dt:tf;
% 3. Other parameters needed for the solution
% The value of alpha
Fo = .5;   % a mutiplicative constant that should be < = 1/2
            % to insure stability
% Initial and boundary conditions
f  = @(x)sin(pi/2*x)+(1/2)*sin(2*pi*x);   
g1 = @(t)0;           
g2 = @(t)exp(-pi^2/4*t);
u  =  zeros(n,m);
u(2:n,1) =  f(x(2:n));    % Put in the initial condition starting from
                          % 2 to n-1  since f(0) = 0 and f(N) = 1 
u(1,:)   =  g1(t);       % The boundary conditions, g1 and g2 at
u(n,:)   =  g2(t);       % x = 0 and x = 1
% Implementation of the explicit method
for k = 2:m-1   % Time Loop
      for i= 2:n-1  % Space Loop
        u(i,k+1) = Fo*(u(i-1,k)+u(i+1,k))+(1-2*Fo)*u(i,k);
      end  
end
figure
hold all 
for i=1:4:numel(t)
    plot(x,u(:,i),'linewidth',1.5,'DisplayName',sprintf('t = %1.2f',t(i)))
end
a = ylabel('Heat');
set(a,'Fontsize',14);
a = xlabel('x');
set(a,'Fontsize',14);
a=title(['Using The Explicit Method - Fo =' num2str(Fo)]);
set(a,'Fontsize',16);
legend('-DynamicLegend','location','bestoutside');
grid;
figure
[X, T] = meshgrid(x,t);
s2 = mesh(X',T',u);
title(['3-D plot of the 1D Heat Equation using the Explicit Method - Fo =' num2str(Fo)])
set(s2,'FaceColor',[0 0 1],'edgecolor',[0 0 0],'LineStyle','--');
a = title('Exact solution of the 1D Diffusivity Equation');
set(a,'fontsize',14);
a = xlabel('x');
set(a,'fontsize',20);
a = ylabel('y');
set(a,'fontsize',20);

a = zlabel('z');
set(a,'fontsize',20);
% disp(u');
Here is the result from the pdepe function.
function PDEPE_diff
m = 0;
n = 20;
Lx = 1;
dx = Lx/(n-1);
x = 0:dx:Lx;
N = 100; 
tf = 1;
dt = tf/(N-1);
t = 0:dt:tf;
sol = pdepe(m,@pdepfun,@icfun,@bcfun,x,t);
u = sol(:,:,1);
figure
hold all % the use of all makes the colors cycle with each plot
for i=1:4:numel(t)
    plot(x,sol(i,:),'linewidth',1.5,'DisplayName',sprintf('t = %1.2f',t(i)))
end
title('Numerical solution computed using PDEPE function.')
xlabel('Distance x')
ylabel('Time t')
legend('-DynamicLegend','location','bestoutside');
grid
figure
s2 = surf(x,t,u);
set(s2,'FaceColor',[0 0 1],'edgecolor',[0 0 0],'LineStyle','--');
title('u1(x,t)')
xlabel('Distance x')
ylabel('Time t')
function [g,f,s] = pdepfun(x,t,u,DuDx)
g  = 1;
f = DuDx;
s = 0;
      function u0 = icfun(x)
   u0 = sin(pi/2*x)+(1/2)*sin(2*pi*x);
   function [pl,ql,pr,qr] = bcfun(xl,ul,xr,ur,t)
  pl = ul;
  ql = 0;
  pr = ur-exp(-pi^2/4*t);
  qr = 0;

2 Comments
More Answers (1)
  Rauhussaba Rauhy
 on 24 Mar 2022
        How to Build an algorithm for piecewise linear Rayleigh ritz method? _d/dx(p(x)d/dx)+q(x)y=f(x), for x E [0,1] with y(0)=y(1)=0 with piece wise linear function
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