coupled differential equation with constant coefficients

2020 4 22
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更新时间：2020 4 23
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here 'kappa' and 'sigma' are constants. The boundary conditions are;
R(-1)=1
S(1)=0
kappa=1.
I tried it using dsolve but the graph obtained werenot correct. If anyone can solve it using bvp4c in the region (0,2) , it would be of great help.

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Ameer Hamza
Ameer Hamza 2020-4-22
编辑：Ameer Hamza 2020-4-22

0 个投票

This code solves the given BV problem. See the documentation for details.
x = linspace(-1, 1, 100);
init = bvpinit(x, [0; 0]);
sol = bvp4c(@odeFun, @bvFun, init);
subplot(2,1,1);
plot(sol.x, real(sol.y));
title('real(sol)');
legend({'R', 'S'})
subplot(2,1,2);
plot(sol.x, imag(sol.y));
title('imaginary(sol)');
legend({'R', 'S'})
function dRSdz = odeFun(z, RS)
sigma = 1;
kappa = 2;
dRdz = 1i*sigma*RS(1) + 1i*kappa*RS(2);
dSdz = -1i*sigma*RS(2) - 1i*kappa*RS(1);
dRSdz = [dRdz; dSdz];
end
function res = bvFun(RSa, RSb)
res = [RSa(1)-1;
RSb(2)-0];
end

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vipul kumar
vipul kumar 2020-4-23
@Ameer hamza,
Please have a look at the following picture.
vipul kumar
vipul kumar 2020-4-23
编辑：vipul kumar 2020-4-23
Here kappa and L can be chosen but i have to generate a plot between normalised wavelength and the reflectivity.
The norm wavelegth is also a function of sigma given by,
norm wave=1/(1+L*sigma/pi*10000).
What i did earlier was create an array for sigma.
Calculated values of norm wave at these discrete sigma values.
By the solution of above diff eqn, calculated discrete values of reflectivity.
And plotted them. For kappa*L=2,8.
Ameer Hamza
Ameer Hamza 2020-4-23
I am not sure what is the question here?

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