How to solve the system of time dependent coupled PDE's?

The system this paper (DOI: 10.1017/S0022112003003835) Thank you.
Fig10a(0.2,2050)
function Fig10a(delta, Reynolds)
% Main function to solve for F and G and plot G(1/4, t)

% Initialize variables
R = Reynolds;
dt = 1/10;
nmax = 82001;
dy = 1/100;
yTarget = 1/4;
gValues = [];

Delta = delta;
H = @(t) 1 + Delta * cos(2 * t);
dH = @(t) -2 * Delta * sin(2 * t);

% Create the grid and differentiation matrices
ygrid = 0:dy:1;
ny = length(ygrid);

% Finite difference differentiation matrices (for dy1 and dy2)
dy2 = fdcoeffFDM2(ny, dy); % Second-order finite difference matrix
dy1 = fdcoeffFDM1(ny, dy); % First-order finite difference matrix

% Initialize variables for F and G
fvar0 = zeros(ny, nmax);
gvar0 = zeros(ny, nmax);

% Time-stepping loop
for i = 1:nmax-1
% Update variables for F and G at the current time step
fvar = fvar0(:, i);
gvar = gvar0(:, i);

% Define the equations
eqf = dy2 * fvar + (dy1 * fvar)./ygrid’ – fvar./(ygrid’.^2) + …
gvar * H((i + 1) * dt)^2;
eqg = (gvar – gvar0(:, i)) / dt – …
dH((i + 1) * dt) / H((i + 1) * dt) * (dy1 * gvar)./ygrid’ – …
(dy1 * gvar) .* fvar / H((i + 1) * dt) + …
1/H((i + 1) * dt) * (dy1 * fvar) .* gvar + …
2./(ygrid’ * H((i + 1) * dt)) .* fvar .* gvar – …
(1/(R * H((i + 1) * dt)^2)) * (dy2 * gvar + dy1 * gvar./ygrid’ – gvar./(ygrid’.^2));

% Apply boundary conditions
eqf(1) = fvar(1); % F(0, t) = 0 at y = 0
eqf(end) = fvar(end) + dH((i + 1) * dt); % F(1, t) = -H'(t) at y = 1
eqg(1) = gvar(1); % G(0, t) = 0 at t = 0
eqg(end) = dy1(end, 🙂 * fvar – dH((i + 1) * dt); % F'(1, t) = H'(t) at y = 1

% Combine eqf and eqg into a single system
eqns = [eqf; eqg];

% Solve the system using backslash operator
sol = eqns; % Since we are already evaluating eqns as the result

% Check that the solution vector matches the expected size
if length(sol) ~= 2 * ny
error(‘The number of equations does not match the number of unknowns’);
end

% Update variables for the next step
fvar0(:, i+1) = sol(1:ny);
gvar0(:, i+1) = sol(ny+1:end);

% Interpolate G(y, t) and store G(1/4, t)
if i >= 81000 && i <= 82001
gInterp = interp1(ygrid, gvar0(:, i+1), yTarget);
gValues = [gValues; i, gInterp];
end

% Display debugging information every 1000 iterations
if mod(i, 1000) == 0 & i>=81000
disp([‘Iteration: ‘, num2str(i), ‘ | G(1/4, t): ‘, num2str(gInterp)]);
end
end

% Plot the values of G(1/4, t)
figure;
if isempty(gValues)
disp(‘No values of G(1/4, t) were recorded.’);
else
plot(gValues(:,1), gValues(:,2), ‘r-‘, ‘LineWidth’, 2);
grid on;
xlabel(‘t’);
ylabel(‘G(1/4, t)’);
title(‘G(1/4, t) from t = 8100 to t = 8200’);
end
end

% Auxiliary function for second-order finite difference matrix
function D2 = fdcoeffFDM2(ny, dy)
e = ones(ny, 1);
D2 = spdiags([e -2*e e], -1:1, ny, ny) / (dy^2);
D2(1, 🙂 = 0; D2(end, 🙂 = 0; % Apply boundary conditions
end

% Auxiliary function for first-order finite difference matrix
function D1 = fdcoeffFDM1(ny, dy)
e = ones(ny, 1);
D1 = spdiags([-e e], [-1 1], ny, ny) / (2*dy);
D1(1, 🙂 = 0; D1(end, 🙂 = 0; % Apply boundary conditions
endThe system this paper (DOI: 10.1017/S0022112003003835) Thank you.
Fig10a(0.2,2050)
function Fig10a(delta, Reynolds)
% Main function to solve for F and G and plot G(1/4, t)