Plotting Solution for Different Times: Hyperbolic PDE

Question

Matlab12345 2020-4-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/518998-plotting-solution-for-different-times-hyperbolic-pde

评论： darova 2020-4-19

I am trying to solve the Advection-Diffusion PDE delu/delt - 2*t*delu/delx = 0 using the Maccormack method. The domain is 0 <= x <= 1, t > 0, and I am given the initial conditions u(x,0) = sin(2*pi*x) for 0 <=x<=1 and the periodic boundary condition u(0, t) = u(1,t). The length step is deltax = 0.001, and the time step is deltat = 0.0005. So far, I have the following:

clear;
clc;
%% problem definition and discretization 
xstep = 0.001;
tstep = 0.0005; 
xdomain= [0 1];
tdomain= [0 1];
nx = round((xdomain(2)-xdomain(1))/xstep);
nt = round((tdomain(2)-tdomain(1))/tstep);
alpha=zeros((nx+1),1);
xt0 = zeros((nx+1),1); % initial condition 
xold = zeros((nx+1),1); % solution at timestep k
xnew = zeros((nx+1),1);% solution at timestep k+1
xnew1 = xnew; % intermediate solution 
for h = 1:nx+1
alpha(h) =-2*h;
end
%x0 = 0.0; % left boundary condition 
%xl = 0.0; % right boundary condition
damping = -0.00; %-0.001;
% initial condition
for i=1:nx+1 
    xi = (i-1)*xstep; 
    xt0(i) = 0.0;
    
    if(xi>0 && xi<=1)
        xt0(i)=sin(2*3.1416*xi);
    end
end 
CFL = alpha*tstep/xstep;
xold = xt0;
%xold(1) = x0;
%xold(nx+1) = xl; 
for k=1:nt
    i = 0;
    time = k*tstep;
    for i=1:nx+1
        % Use periodic boundary condition, u(nx+1)=u(1)
       if(i==nx+1) 
            dudx = xold(1)-xold(nx+1);
            du2dx2 = xold(i-1)-2*xold(i)+xold(1);
        elseif(i==1) 
            dudx = xold(i+1)-xold(i);
            du2dx2 = xold(nx+1)-2*xold(i)+xold(i+1);
        else  
            dudx = xold(i+1)-xold(i); % C/2*(U(i+1,k)-U(i-1,k)
            du2dx2 = xold(i+1)-2*xold(i)+xold(i-1);
        end 
        % Predictor step 
        xnew1(i) = xold(i) - CFL(i)*dudx + tstep*damping*xold(i); % U(i,k+1)= U(i,k)-C/2*(U(i+1,k)-U(i-1,k)
        % correction step 
        if(i==1) 
            dudx1 = xnew1(1) - xnew1(nx+1);
        else        
            dudx1 = xnew1(i) - xnew1(i-1); 
        end
        xnew(i) = 0.5*(xold(i) + xnew1(i) - CFL(i)*dudx1) + 0.5*tstep*damping*xnew1(i);
    end 
    
%    xold= xnew;
end
x=linspace(xdomain(1),xdomain(2),nx+1);
t=linspace(tdomain(1),tdomain(2),(nx+1));
close all
hold on;
h1=plot(x,xt0,'r','linewidth',1);
hold on;
h2=plot(x,xnew,'g','linewidth',1);
hold on;
legend('Exact','Initital','t=0', 't=0.5','t=1')
xlabel('x [m]');
ylabel('Displacement, u(x,t) [m]');
ax=gca;
ax.FontSize= 12;
ax.XLim= [0 1];
ax.YLim= [-20 20];

I am trying to obtain solutions at time = 0, 0.5, and 1. However, I am not sure how to change the value of the time so that the value of the analytical solution changes, and so that the solutions can be stored and plotted. Any help would be appreciated.