NDSolve Encountered non-numerical value

Question

Below a sample of some first-order differential equations I try to solve. Although there are no non-numerical values involved in the equations, Mathematica systematically spits out "Encountered non-numerical value at $\tau$ == 0".

And I also get the error: "Cannot solve to find an explicit formula for the derivatives. NDSolve
will try solving the system as differential-algebraic equations"... What does that mean exactly and how can I remedy this?

Might the problem lie somewhere else?

Thanks in advance for your help !

eqs = {0 == (1.38 D[r[5/6, 0][\[Tau]], \[Tau]] == -10.18 + 
   9.61 r[5/6, 0][\[Tau]] - 3.1 r[5/6, \[Pi]/4][\[Tau]]), 
  0 == (2.95 (D[r[5/6, \[Pi]/4][\[Tau]], \[Tau]]) == -21.7 + 
   13.8 r[5/6, \[Pi]/4][\[Tau]] + 
   1.66 (r[5/6, 0][\[Tau]] - 1. r[5/6, \[Pi]/2][\[Tau]])^2 - 
   1.67 (4.64 .3 r[5/6, \[Pi]/4][\[Tau]]) (r[5/6, 0][\[Tau]] - 
       1. r[5/6, \[Pi]/2][\[Tau]])^2 - 
   3.33 (r[5/6, 0][\[Tau]] - 2 r[5/6, \[Pi]/4][\[Tau]] + 
      r[5/6, \[Pi]/2][\[Tau]])), 
  0 == (4.53 (D[r[5/6, \[Pi]/2][\[Tau]], \[Tau]]) == -33.2 - 
   5.10 (2 r[5/6, \[Pi]/4][\[Tau]] - 2 r[5/6, \[Pi]/2][\[Tau]]) + 
   21.14 r[5/6, \[Pi]/2][\[Tau]])};

Unknowns = {r[5/6, 0], r[5/6, \[Pi]/4], r[5/6, \[Pi]/2]};
Unknowns\[Tau] = Unknowns /. r[a_, b_] :> r[a, b][\[Tau]];

Init = Join[Map[(0 == #) &, Unknowns\[Tau] /. \[Tau] -> 0], 
Map[(0 == #) &, D[Unknowns\[Tau], \[Tau]] /. \[Tau] -> 0]];

sols = NDSolve[Flatten[{eqs, Init}], Unknowns, {\[Tau], 1, 3}][[1]];

---------- Edit--------

So the context is that I am trying to solve some differential equation for a function $r(x,y,t)$ that I have discretised on a grid along the x-y direction leaving the time dependence continuous. So I end up with a set of differential equations for t with unknowns the point in my grid. The code above is for a sample of points $r(a,b)$ in this grid (and the corresponding equations.)

$\left\{0=\left(1.38 r\left(\frac{5}{6},0\right)'\tau =9.61 r\left(\frac{5}{6},0\right)\tau -3.1 r\left(\frac{5}{6},\frac{\pi }{4}\right)\tau -10.18\right),0=\left(2.95r\left(\frac{5}{6},\frac{\pi }{4}\right)'\tau =-2.32464r\left(\frac{5}{6},\frac{\pi }{4}\right)\tau \left(r\left(\frac{5}{6},0\right)\tau -1.r\left(\frac{5}{6},\frac{\pi }{2}\right)\tau \right)^2+1.66\left(r\left(\frac{5}{6},0\right)\tau -1.r\left(\frac{5}{6},\frac{\pi }{2}\right)\tau \right)^2+13.8r\left(\frac{5}{6},\frac{\pi }{4}\right)\tau -3.33\left(r\left(\frac{5}{6},0\right)\tau -2r\left(\frac{5}{6},\frac{\pi }{4}\right)\tau +r\left(\frac{5}{6},\frac{\pi }{2}\right)\tau \right)-21.7\right),0=\left(4.53r\left(\frac{5}{6},\frac{\pi }{2}\right)'\tau =-5.1\left(2r\left(\frac{5}{6},\frac{\pi }{4}\right)\tau -2r\left(\frac{5}{6},\frac{\pi }{2}\right)\tau \right)+21.14r\left(\frac{5}{6},\frac{\pi }{2}\right)\tau -33.2\right)\right\}$

Answer 1

Your code is a bit hard to parse and the math seems wrong. You have a system of 3 first order diff-eqs but 6 initial conditions (all variables and their derivatives are 0 at $t=0$). That's overconstrained, and if the variables and their derivatives are all 0, then the solution is trivial. They all stay 0 forever.

What's up with the 0==(a==b) formatting of the equations? If that's really what you mean, then you again just get the trivial solution $r(5/6,0)=r(5/6,\pi/4)=r(5/6,\pi/2)=0$ for all $\tau$.

Let's remove the initial condition on the derivatives and the 0== terms in the equations. Let's also rename $\tau$ to $t$ since it's ugly to have \[Tau]'s floating around here. And for conciseness let's also call $r(5/6,0)=r1, \ r(5/6,\pi/4)=r2, \ r(5/6,\pi/2)=r3$. Then we get

eqs = {1.38 r1'[t] == -10.18 + 9.61 r1[t] - 3.1 r2[t], 
  2.95 r2'[t] == -21.7 + 13.8 r2[t] + 1.66 (r1[t] - 1. r3[t])^2 - 2.32464 r2[t] (r1[t] - 1. r3[t])^2 - 3.33 (r1[t] - 2 r2[t] + r3[t]), 
  4.53 r3'[t] == -33.2 - 5.1 (2 r2[t] - 2 r3[t]) + 21.14 r3[t]} 
vars[t_] = {r1[t], r2[t], r3[t]};
init = Thread[vars[0] == 0];
sols = NDSolve[Flatten[{eqs, init}], vars[t], {t, 1, 3}][[1]]

This only goes to about $t=2.35$ where the solution appears to run into divergence issues. You can see this from a plot of the solution

Plot[Evaluate[{r1[t], r2[t], r3[t]} /. sols], {t, 1, 2.35}, PlotRange -> All]

Hope that helps!

Answer 2

Try this:

eqs = {(1.38 D[r[5/6, 0][\[Tau]], \[Tau]] == -10.18 + 
  9.61 r[5/6, 0][\[Tau]] - 
  3.1 r[5/6, \[Pi]/4][\[Tau]]), (2.95 (D[
    r[5/6, \[Pi]/4][\[Tau]], \[Tau]]) == -21.7 + 
  13.8 r[5/6, \[Pi]/4][\[Tau]] + 
  1.66 (r[5/6, 0][\[Tau]] - 1. r[5/6, \[Pi]/2][\[Tau]])^2 - 
  1.67 (4.64 .3 r[5/6, \[Pi]/4][\[Tau]]) (r[5/6, 0][\[Tau]] - 
      1. r[5/6, \[Pi]/2][\[Tau]])^2 - 
  3.33 (r[5/6, 0][\[Tau]] - 2 r[5/6, \[Pi]/4][\[Tau]] + 
     r[5/6, \[Pi]/2][\[Tau]])), (4.53 (D[
    r[5/6, \[Pi]/2][\[Tau]], \[Tau]]) == -33.2 - 
  5.10 (2 r[5/6, \[Pi]/4][\[Tau]] - 2 r[5/6, \[Pi]/2][\[Tau]]) + 
  21.14 r[5/6, \[Pi]/2][\[Tau]])};

Unknowns = {r[5/6, 0], r[5/6, \[Pi]/4], r[5/6, \[Pi]/2]};
Unknowns\[Tau] = Unknowns /. r[a_, b_] :> r[a, b][\[Tau]];
Init = Map[(0 == #) &, Unknowns\[Tau] /. \[Tau] -> 0]

sols = NDSolve[Flatten[{eqs, Init}], Unknowns, {\[Tau], 1, 3}][[1]];

Now the integration proceeds until $\tau\approx 2.28$