Hii Friends,
Could you give me your suggestion regarding this problem.
I have the following equation where in the initial, boundary conditions and other parameters have been specified.
I was able to solve in matlab using the following code
%---pdepe function-----% function [c,f,s] = gvdpde(t,z,u,DuDt) i = sqrt(-1); beta2 = -2.907e-25; T0 = 35e-12; %Ld = T0^2/abs(beta2); Ld = 4500; c = (2*i*Ld)./T0^2; f = DuDt; s = 0;
%---Initial condition----% function u0 = gvdic(t) T0 = 35e-12; u0 = exp(-0.5.*(t/T0).^2);
%-----Boundary condition-----% function [pl,ql,pr,qr] = gvdbc(tl,ul,tr,ur,z) pl = 0; ql = 1; pr = ur - 1; qr = 1;
%-----Main function-----% function gvd
clc; clear all; close all;
m = 0; T0 = 35e-12; t = linspace(-30*T0,30*T0,150); z = linspace(0,1000,100); z1 = linspace(0,4500,100); z2 = linspace(0,45000,100);
sol = pdepe(m,@gvdpde,@gvdic,@gvdbc,t,z); sol1 = pdepe(m,@gvdpde,@gvdic,@gvdbc,t,z1); sol2 = pdepe(m,@gvdpde,@gvdic,@gvdbc,t,z2); % Extract the first solution component as u. u = sol(:,:,1); u1 = sol1(:,:,1); u2 = sol2(:,:,1);
% A surface plot is often a good way to study a solution. surf(t,z,abs(u2)); title('Numerical solution computed with 20 mesh points.'); xlabel('Time t'); ylabel('Distance z'); % A solution profile can also be illuminating. figure; plot(t,abs(u(end,:)),t,abs(u1(end,:)),t,abs(u2(end,:)),'LineWidth',2); legend('z<<Ld','z=Ld','z>Ld'); xlabel('Time t'); ylabel('Intensity'); grid on % --------------------------------------------------------------
Now in the next step i need to add 3'rd order flux term as shown in the snap
Is it possible to solve it using the same pdepe function. Can anyone give me an hint to solve this.
Awaiting your kind replies.
pa1
