Use second-order and fourth-order Runge-Kutta methods
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The one-dimensional linear convection equation for a scalar T is,
๐๐/๐๐ก + ๐*๐๐/๐๐ฅ = 0 0 โค ๐ฅ โค 1
With the initial condition and boundary conditions of, (๐ฅ, 0) = ๐ ^(โ200(๐ฅโ0.25)^ 2) , ๐(0,๐ก) = 0, ๐(1,๐ก): outflow condition Take ๐ = 1, โ๐ฅ = 0.01
Plot the results for t = 0.25, .5, 0.75 for both of them.
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darova
on 9 Mar 2021
I'd try simple euler scheme first. The scheme i used
๐๐/๐๐ก + ๐*๐๐/๐๐ฅ = 0
clc,clear
a = 1;
x = 0:0.01:1;
t = 0:0.01:0.75;
U = zeros(length(t),length(x));
U(1,:) = exp(-200*(x-0.25).^2);
U(:,1) = 0;
dx = x(2)-x(1);
dt = t(2)-t(1);
for i = 1:size(U,1)-1
for j = 1:size(U,2)-1
U(i+1,j) = U(i,j) + a*dt/dx*(U(i,j+1)-U(i,j));
end
end
surf(U)
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