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Finite Difference method solver

조회 수: 12 (최근 30일)
Jose Aroca
Jose Aroca 2020년 11월 6일
댓글: Jose Aroca 2020년 11월 9일
I have the following code in Mathematica using the Finite difference method to solve for c1(t), where . However, I am having trouble writing the sum series in Matlab. The attatched image shows how the plot of real(c(t) should look like.
\[CapitalOmega] = 0.3;
\[Alpha][\[Tau]] := Exp[I \[CapitalOmega] \[Tau]] ;
dt = 0.1;
Ns = 1000;
ds = dt/Ns;
Ttab = Table[T, {T, 0, 10, dt}];
Stab = Table[s, {s, 0, dt - ds, ds}];
c[0] = 1;
Do[corrSum[n] = Sum[c[nn - 1]*Sum[\[Alpha][n dt - m ds]*ds, {m, Ns (nn - 1), Ns nn , 1}], {nn, 1, n}];
c[n] = c[n - 1] - dt*corrSum[n](*c[n-1]*\[Alpha][n dt]*), {n, 1, 100}]
cTtab = Table[{n*dt, c[n]}, {n, 0, 100}]
FDiff = ListPlot[Re[cTtab], PlotStyle -> Orange, PlotLegends -> {"Finite Difference"}]

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Alan Stevens
Alan Stevens 2020년 11월 7일
By differentiating c' again you can solve the resulting second order ode as follows
c0 = 1;
v0 = 0;
IC = [c0 v0];
tspan = [0 10];
[t, C] = ode45(@odefn, tspan, IC);
c = C(:,1);
plot(t,real(c),'ro'),grid
xlabel('t'), ylabel('real(c)')
function dCdt = odefn(~,C)
Omega = 0.3;
c = C(1);
v = C(2);
dCdt = [v;
-1i*Omega*v - c];
end
This results in
  댓글 수: 7
Alan Stevens
Alan Stevens 2020년 11월 8일
Well, here is an explicit finite difference version that uses an outer and an inner loop, with the same values of dt and ds as in the Mathematica version. However, it is probably not what you want still, as it isn't a line by line translation of the Mathematica code. I'll leave further development to you!
Omega = 0.3;
dt = 0.1;
Ns = 1000;
ds = dt/Ns;
t = 0:dt:10;
N = numel(t);
c = zeros(1,N);
c(1) = 1;
v = 0;
for n = 1:N-1 % Outer loop; c is updated each iteration of this loop
% Inner loop: c is taken to be constant through this loop, the same
% approximation as is made in section 5.2.1 of the pdf.
for nn = 1:Ns
v = v - (1i*Omega*v + c(n))*ds;
end
c(n+1) = c(n) + v*dt;
end
plot(t,real(c),'.'),grid
xlabel('t'),ylabel('real(c)')
Jose Aroca
Jose Aroca 2020년 11월 9일
Hi, I think that I should be able to work with this. Thank you very much for your help!

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