Problem with pdetoode for two coupled PDEs

Question

I tried to adapt a code for a single equation to solve the following system using 'pdetoode'

Updated answer

domain = {0, 1};
difforder = 4;
points = 25; 
grid = Array[# &, points, domain]
With[{p = p[t, x], q = q[t, x]},
sy = {D[q, t] - D[q, x, x] + p == 0, D[p, t] + D[p, x, x] + q == 0};
ic = {q == Sin[Pi x] /. t -> 0, q == 0 /. t -> 0.4};
bc = {{p == 0, q==0} /. x -> 0, 
{p==0, q==0} /. x -> 1};]
ptoofunc = pdetoode[{p, q}[t, x], t, grid, difforder];
removeredundant1 = #[[2 ;; -2]] &;
removeredundant2 = #[[3 ;; -3]] &;
ode@1 = removeredundant1@ptoofunc@sy[[1]];
ode@2 = removeredundant2@ptoofunc@sy[[2]];
odeic = ptoofunc@ic;
odebc = ptoofunc@With[{sf = 1}, diffbc[t, sf]@bc];
var = Outer[#[#2] &, {p, q}, grid];
sollst = NDSolveValue[{ode /@ {1, 2}, odeic, odebc}, var, {t, 0, 0.4}];

I got the error

There are more dependent variables..., than equations, so the system is underdetermined

Any hint is appreciated.

Edit as mentioned by @xzczd, this post is close to my case. I adapted it accordingly but still got the 2nd error "There are more dependent variables .. than equations, so the system is underdetermined."

It worked quite easily without 'pdetoode' as:

System = {D[q[x, t], t] - D[q[x, t], x, x] + p[x, t] == 0, 
D[p[x, t], t] + D[p[x, t], x, x] + q[x, t] == 0, 
p[x, t] == 0 /. x -> 0, p[x, t] == 0 /. x -> 1, 
q[x, t] == 0 /. x -> 0, q[x, t] == 0 /. x -> 1, 
q[x, 0] == Sin[Pi x], q[x, 0.4] == 0};
{q, p} = NDSolveValue[System, {q, p}, {x, 0, 1}, {t, 0, 0.4}]

{InterpolatingFunction[{{0., 1.}, 0., 0.4}}, InterpolatingFunction[{{0., 1.}, 0., 0.4}} }..

Plot3D[q[x, t], {t, 0, 0.4}, {x, 0, 1}]

…You're using pdetoode blindly. Please re-read my answer to your question 4 years ago, then read the document of Part ([[]]), Span (;;) and Map (/@) and think about how to make room for your i.c.s/b.c.s. Also, you're setting q == 0 /. t -> 0.4 as a constraint, this looks like an ill-posed problem, are you sure it is well-posed? If not, then please notice pdetoode is just auxiliary tool for implementation of FDM, and naive FDM won't be able to handle ill-posed problem. — xzczd, Oct 31 '23 at 01:50
@xzczd thanks for your reply. Okay, I'll re-try. The problem is well-posed, it worked without 'pdetoode' but I'd like to compare.. — S. Maths, Oct 31 '23 at 06:03
@S.Maths I am curious, could you share how you have done this without pdetoode? — user21, Oct 31 '23 at 08:08
@S.Maths Looking at your solution without pdetoode plot of q[x,t] shows numerical problems near x==0. Your statement "It worked quite easily" does not hold I think — Ulrich Neumann, Nov 01 '23 at 12:17
@UlrichNeumann which problems do you mean? I added the plot. — S. Maths, Nov 01 '23 at 12:52
@S.Maths Sorry I meant Plot3D[p[x, t], {t, 0, 0.4}, {x, 0, 1}] — Ulrich Neumann, Nov 01 '23 at 13:12
@S.Maths Ok, but one point should be mentioned. If you eliminate p[x,t] you get one pde for q[x,t] $-q^{(0,1)}(x,t)+q^{(4,0)}(x,t)+q(x,t)=0$ , but this new pde isn't fullfilled by your solution q[x,t] ! — Ulrich Neumann, Nov 01 '23 at 16:44
Still, you're coding blindly. You need to know more about list manipulation of Mathematica, and finite difference method. See the figure in e.g. https://mathematica.stackexchange.com/a/28011/1871 and think carefully about which function you should use (pdetoode or pdetoae?) and where you should remove the equations. — xzczd, Nov 02 '23 at 05:35
@S.Maths Your solution "without pdetoode" isn't trustable. q looks fine , p looks critical. How can you conclude that one solution part is correct? — Ulrich Neumann, Nov 02 '23 at 07:23

Problem with pdetoode for two coupled PDEs

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