@Preben Alsholm I will address each of the issues that you have brought up.

1. By de2, v cannot be zero on any interval no matter what value is assigned to signum(0).

You are right.  That's why I said it remains a mystery to me how Maple gets that solution.

But as I also explained, Maple's solution agrees with the hand-calculated exact soluition  of the problem (not the solution of the DE) which I found by splitting the motion into a sequence of stick and slip phases.  The exact solution for the absolute displacement of the house, that is x(t)+X(t), is a piecewise function consisting of sinosoidal and parabolic patches.

2. If you look at an interval where v appears to be zero ... we see rapid oscillation.

Those oscillations are numerical artifacts.  Their amplitude is roughly 0.004, which is in effect zero considering the choice of stepsize=1e-3.  If we set stepsize=1e-4, the amplitude is cut down by a factor of 10.  And if we set stepsize=1e-5, the amplitude is cut down by a factor of 100.

We see that Maple is attempting to produce an identically zero function v(t) on the interval 3.25..3.9.  As you have observed, an identically zero function is inconsistent with the differential equation de2.  But mysteriously, it is consistent with the exact solution which one obtains by hand by splitting the motion into a sequence of stick and slip phases.

3. Making changes in order to match the original problem I tried using g = 9.81, mu=0.8, and cos instead of sin...The results where disturbing.

The results are just right.  With your choice of parameters there will be no slippage, and therefore the solution v(t) should be identically zero.   Maple shows v(t) oscillating with an amplitude of roughly 0.0002, which is in effect zero plus numerical noise.

To elaborate, let f be the horizontal force exerted by the ground on the house.  Newton's law says f = m*(a + A), where as before, A is the acceleration of the ground, and a is the house's acceleration relative to the ground.  In particular, if the house does not slide, then f = m*A.  The house will begin to slide only when |m*A| exceeds the force of friction, which is  mu*m*g.  With your choice of parameters, |m*A(t)| is always less than mu*m*g, therefore the house never slides.

4. Thus I don't trust the results. And I should add: The mathematical model consisting of just de2 (and de1).

I agree.  de2 is definitely inadequate becasue as you have observed, it is inconsistent with v(t) being zero on an interval.  I do not understand how Maple "solves" that incorrect model and magically obtains the  correct solution  solution to the correct model.

@Oloyemike I understand your model a little better now, but things are still unclear.  At places you have Q(r, x, tau), at other places you have Q(r, tau), and yet at some other places you have just Q.  Shouldn't there always be three arguments to each Q?

Similarly, at places you have U(x,tau),  and at some other places you have just U.  Shouldn't there always be two arguments to each U?

Mathematics is a precision instrument.  Sloppiness is not acceptable.

Furthermore, based on what you have shown, I can tell you the solution of your system of PDEs, and I don't need Maple to do it.  The solution is U(x,tau) = 0 and Q(r,x,tau) = 0.  Just substitute these in the equations to verify. If this is not what you had expected, then your model needs further adjustments.  Oops, sorry, I take that back -- U(x,tau) = 0 and Q(r,x,tau) = 0 is not a solution.

@Carl Love Let's say we have a ball of radius R, and let r be the distance from the center of the ball.  As we move from the center to the ball's boundary, r goes from 0 to R.  Let's introduce rho = r/R.  Then as we move from the center to the boundary, rho goes from 0 to 1.  In this context, rho is called the dimensionless radius.  Similarly one defines dimensionless time, dimensionless force, etc.  The terminology is quite common in engineering and physics.

@tomleslie The equation shown is slight modification of a standard difference scheme for a parabolic PDE.  There is no "step j" and "step j+1".  The OP's equation expresses a transformation from the array a[] to the array A[], where a[] is the known solution at a certain time step, and A[] is the solution at the next time step.  One needs boundary conditions (not an initial condition) to complete the definition of that transformation  The initial condition only enters at a later stage when that transformation is applied recursively to generate the solution of the PDE.

I can't make sense of the phrase "compute A[j] for a[j]".  It will help if you reword.

Furthermore, what are the boundary conditions?  What is the range of the j index?

@Carl Love I haven't used Maple's PDE solver in any substantive way and I know very little about it.  I made a naive attempt, just for learning purposes, to solve a much simpler version of the OP's equations.  Specifically, let's look at Burgers equation

@Oloyemike I stopped responding because I just don't trust those equations, and therefore I don't want to waste my time.

If you clarify by explaining the physics behind the problem, the geomertry of the domain, and the meanings of the various variables, then I may consider taking another look.

@MDD This begins to smell like a homework assignment, so here is a hint.  Write a proc which will do the equivalent of ArrayTools:-IsZero(). Such a proc is quite short and straightforward.

The elements of a basis set should be linearly independent, by definition.  So if A, B, C, D form a basis, then you cannot express C as a linear combination of A and B.  Reformulate your question.

@Markiyan Hirnyk I have not thought at all about the existence of solutions of such an equation, or the convergence of Euler's scheme.

@Preben Alsholm I suspect that Maple's algorithm is not equipped to deal with this kind of equation, which is actually not a delay differential equation in the conventional sense, since the delay itself is unknown.  It appears that it applies an algorithm to it blindly and just by luck picks up some parts of the correct solution.  As you have noted, its "solution" extends to x = 1.0058, which is probably wrong, since all evidence suggests that the solution should break down before reaching x = 0.85.

As to the incorrect plot in the x<0 region, that's not something particular to that specific equation.   This happens even in the simplest case:

eq:=diff(y(x),x)= -y(x-1);
res:=dsolve({eq,y(0)=1},numeric);
plots:-odeplot(res,[x,y(x)],-1..10);

The graph in the x>0 region is correct, and actually corresponds to y(x)≡1 in x<0, as intended.  The graph in the x<0 region is incorrect, however, in the same way as the graph for the more complex equation that you have shown.  I have no idea how Maple comes up with that graph.

@NickB Alejandro's and my results are equivalent.  You will get his result by applying the asympt() function to mine.  Choose whichever form best suits your needs.

@Preben Alsholm The equation's variable delay is tau = x-y(x)+2, as you have noted. If we plot the delay, as in:

plots:-odeplot(res,[x,x-y(x)+2],0..1);

we see that it becomes negative after around x=0.85.  Is the solution meaningful after that point?