Matlab---can pdepe solve 1st order pde set like this?

Question

Equation：

variable: z, t

want: S = S(z,t), A = A(z,t)

parameter: a,b,c

i is sqrt(-1)

d is the partial derivative operator

dS/dt = -i*a*z*S - i*b*A

(d/dz +c*d/dt)A = -i*S

Boundary condition:

sigma is the pulse width

tm is center of the pulse

at z = 0,

A(1,t) = (0.398942/sigma)*exp(-((t-tm).^2)/(2*sigma^2)) (Gaussian pulse)

at z = 1, boundary condition is not clear. Expected to be near 0. If I use Mathematica, then it is not necessary to specify this boundary condition. Actually this boundary is the aim of solving this pde.

Initial condition:

at t = 0,

S = 0

A = 0(near 0. It should be the rim of the above Gaussian pulse)

I was trying to solve this pde set with pdepe. The result shows:

??? Error using ==> daeic12 at 77 This DAE appears to be of index greater than 1.

My question is: can pdepe solve peds like these?

×××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××

Details: change the equation into the form of:

[1; c].* d/dt [S; A] = d/dz [0; -A] + [-i*a*z*S-i*b*A; -i*S]

Then my code for equation: (parameter name is changed, not long called a, b and c)

function [ c,f,s ] = myPdeNoDecay( z,t,u,dudz)
LoverzetacT = 6.712414322355693e-05;
betaLT = 66.713366022882500;
omegayetaksbarLT = 4.183966095613130e+02;

c = [1;LoverzetacT];
f = [0;-u(2)];
s = -1i*[betaLT.*z.*u(1)+omegayetaksbarLT.*u(2);u(1)];
end

Boundary condition:

function [ pl,ql,pr,qr ] = myPdeNoDecayBC( xl,ul,xr,ur,t )
tm = 0.5; %ns, Normalized central time of the pulse
sigma = 0.1; %ns, Normalized width of the pulse, 10^-7s

pl = [ul(2)-iniShape(t,sigma,tm);0];
ql = [0;0];
pr = [0;0];
qr = [0;0];
end

function [G] = iniShape(t,sigma,tm)
G = (0.398942/sigma)*exp(-((t-tm).^2)/(2*sigma^2));
end

Initial condition:

function [ u0 ] = myPdeNoDecayIC( x )
u0 = [0;0];
end

main function:

clear
close all
tlist = 0:0.05:1;
zlist = 0:0.05:1;

LoverzetacT = 6.712414322355693e-05;
betaLT = 66.713366022882500;
omegayetaksbarLT = 4.183966095613130e+02;

tm = 0.5; %ns, Normalized central time of the pulse
sigma = 0.1; %ns, Normalized width of the pulse, 10^-7s

m = 0;
sol = pdepe(m,@myPdeNoDecay,@myPdeNoDecayIC,@myPdeNoDecayBC,zlist,tlist);
figure('numbertitle','off','name','my PDE with no decay or two-ph detuning')
subplot(211);
surf(zlist,tlist,sol(:,:,1));
title('The solution of S');
xlabel('z');
ylabel('t');
zlabel('S(z,t)');
title('The solution of a');
xlabel('z');
ylabel('t');
zlabel('a(z,t)');

error is:

??? Error using ==> daeic12 at 77

This DAE appears to be of index greater than 1.

Answer 1

No, pdepe can't solve equations where the DAE index is greater than one.