3
$\begingroup$

I am trying to solve the following problem of the free fall dynamics under gravity of a inextensible horizontal string attached at its end, in a 2D vertical plane. If I'm right, that is the governing equations, for $x=[x1,x2]\in\mathbb{R}^2$ and $s$ the curvilinear coordinate along the string:

\begin{align} \|x'(s,t)\| &=1 &\text{(inextensibility)}\\ \rho \ddot x(s,t) &= \begin{bmatrix} 0 \\ -\rho g\end{bmatrix} + \lambda(s,t) x'(s,t)^\top x''(s,t) x'(s,t) &\text{(dynamics)}\\ x(0,t)&=[0,0], x(1,t)=[0.5,0] & \text{(boundary cond.)}\\ x(s,0)&=[0,0],\quad \dot x(s,0) = [0,0] & \text{(initial cond.)} \end{align}

Implementation:

x[s_, t_] = {x1[s, t], x2[s, t]};
rho = g = 1;
const = D[x[s, t], s] . D[x[s, t], s] - 1 == 0;
eq1 = rho*D[x1[s, t], {t, 2}] == lambda[s,t]*D[x[s, t], s] . D[x[s, t], {s, 2}] D[x1[s, t], s];
eq2 = rho*D[x2[s, t], {t, 2}] == -rho*g + lambda[s,t]*(D[x[s, t], s] . D[x[s, t], {s, 2}]) D[x2[s, t], s];
bc = {x1[0, t] == 0, x1[1, t] == 0.5, x2[0, t] == 0, x2[1, t] == 0};
ic = {x1[s, 0] == 0, (D[x1[s, t], t] /. t -> 0 ) == 0};

Then,

NDSolve[Flatten@{eq1, eq2, const, bc, ic}, {x1[s, t], x2[s, t], 
  lambda[s,t]}, {s, 0, 1}, {t, 0, 10}]

returns

The maximum derivative order of the nonlinear PDE coefficients for the Finite Element Method is larger than 1. It may help to rewrite the PDE in inactive form.

So I changed the code to (note that I added some NeumannValue):

eq1 = rho*D[x1[s, t], {t, 2}] == lambda[s, t]*D[x[s, t], s] . Inactive[D][D[x[s, t], {s, 1}], {s, 1}] D[x1[s, t], s] + NeumannValue[0, t == 0];
eq2 = rho*D[x2[s, t], {t, 2}] == -rho*g + lambda[s,t]*(D[x[s, t], s] . Inactive[D][D[x[s, t], {s, 1}], {s, 1}]) D[x2[s, t], s] + NeumannValue[0, t == 0];
bc = {x1[0, t] == 0, x1[1, t] == 0.5, x2[0, t] == 0, x2[1, t] == 0}
ic = {x1[s, 0] == 0, x2[s, 0] == 0, lambda[s, 0] == 0};

and now I have:

enter image description here

Any idea on how to overcome this?

Edit As per xzczd's link, I rewrote the equations using Inactive[Grad] instead of Inactive[D] which is presently not implemented in the FEM package.

eqs = Thread[rho*D[x[s, t], {t, 2}] == {0, -rho*g} + lambda[s,t]
      *(D[x[s, t], s] . Inactive[Grad][D[x[s, t], {s, 1}], {s}])
      *D[x[s, t], s] + NeumannValue[0, t == 0]];
NDSolveValue[Flatten@{eqs, const, ic}, {x1[s, t], x2[s, t], lambda[s, t]},
  {s, 0, 1}, {t, 0, 10}]

I now get the NDSolve:dgsvars error: "The differentiation variables {s} given for Inactive[Grad] should be the spatial independent variables {s,t}". How not to make it believe t is a space variable?

$\endgroup$
6
  • 2
    $\begingroup$ 1. Inactive@D[...] isn't the correct syntax, should be Inactive[D][...]. 2. Even if you write it right, FiniteElement won't be able to handle it at least for now: mathematica.stackexchange.com/q/217169/1871 $\endgroup$
    – xzczd
    Jul 2, 2022 at 10:03
  • $\begingroup$ 3. t is not spatial variable, and you've forgotten 2 i.c.s. $\endgroup$
    – xzczd
    Jul 2, 2022 at 11:31
  • $\begingroup$ @xzczd 3. Still no luck with eqs = Thread[rho*D[x[s, t], {t, 2}] == {0, -rho*g} + lambda[s, t]*(D[x[s, t], s] . Inactive[Grad][D[x[s, t], {s, 1}], {s}])* D[x[s, t], s]];, dirichlet = {DirichletCondition[x[s, t] == {0, 0}, s == 0], DirichletCondition[x[s, t] == {0.5, 0}, s == 1]}; and ic = {x1[s, 0] == 0, x2[s, 0] == 0, (D[x1[s, t], t] /. t -> 0) == 0, (D[x2[s, t], t] /. t -> 0) == 0, lambda[s, 0] == 0}: NDSolveValue returns The dependent variable in {x1,x2}=={0,0} in the b.c. DirichletCondition[{x1,x2}=={0,0},s==0] needs to be linear $\endgroup$
    – anderstood
    Jul 2, 2022 at 11:49
  • $\begingroup$ 4. It's clear that equations in vector form isn't supported by DirichletCondition, use the traditional one instead. 5. I don't think you can have terms like Inactive[Grad][D[x[s, t], {s, 1}], {s}] in code, see this post for more info: mathematica.stackexchange.com/q/225711/1871 $\endgroup$
    – xzczd
    Jul 2, 2022 at 12:13
  • 1
    $\begingroup$ Also, a quick test via FDM suggests the convergency of the system is rather bad, according to my (limited) experience, a well-posed IBVP won't be like this. Are you sure the system is correct? $\endgroup$
    – xzczd
    Jul 2, 2022 at 13:44

1 Answer 1

2
$\begingroup$

To get the integration started you need to specify sufficient initial conditions:

x[s_, t_] = {x1[s, t], x2[s, t]};
rho = g = 1;
const = D[x[s, t], s] . D[x[s, t], s] - 1 == 0;
eq1 = rho*D[x1[s, t], {t, 2}] == 
   lambda[s, t]*D[x[s, t], s] . D[x[s, t], {s, 2}] D[x1[s, t], s];
eq2 = rho*D[x2[s, t], {t, 2}] == -rho*g + 
    lambda[s, t]*(D[x[s, t], s] . D[x[s, t], {s, 2}]) D[x2[s, t], s];
bc = {x1[0, t] == 0, x1[1, t] == 0.5, x2[0, t] == 0, x2[1, t] == 0};
ic = {x1[s, 0] == 0, (D[x1[s, t], t] /. t -> 0) == 0, 
   x2[s, 0] == 0, (D[x2[s, t], t] /. t -> 0) == 0,
   lambda[s, 0] == 0
   };

Monitor[NDSolve[
  Flatten@{eq1, eq2, const, bc, ic}, {x1[s, t], x2[s, t], 
   lambda[s, t]}, {s, 0, 1}, {t, 0, 10}, 
  EvaluationMonitor :> (currentTime = Row[{"t = ", CForm[t]}])]
 , currentTime]

This still gives messages and I did not run this to the end; but this should be a starting point. Forget FEM for this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.