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Solving Interpolating Function

Posted 10 years ago
Hi, 

I've solved a system of three differential equations to produce three interpolating functions. I need to find the value of t when the first interpolating function is equal to 1. Any idea's on how to do this? Thanks.

Tha
POSTED BY: Amy Gordon
5 Replies
Oh, now I see the original question was about crossing 1, not 0.
In[10]:= FindRoot[sol[t] == 1, {t, 0, 20}, WorkingPrecision -> 200] // N

Out[10]= {t -> 0.00176324}

In[11]:= sol[t /. %]

Out[11]= 1.
Alternatively, could use WhenEvent:
In[12]:= Reap[NDSolveValue[{eq1, eq2, eq3, h[0] == 1000000, z[0] == 1,
      r[0] == 0, WhenEvent[h[t] == 1, Sow[t]]}, h, {t, 0, 20},
     WorkingPrecision -> 200, Method -> "StiffnessSwitching"]][[2, 1]] // N

Out[12]= {0.00176324}
POSTED BY: Ilian Gachevski
The multiple zero crossings are just an artifact of using machine precision. It looks like h is monotonically decreasing and 0 is a horizontal asymptote that it can get arbitrarily close to, so one just needs to decide at what tolerance h will be considered to vanish.

 In[1]:= A = 1/100;
 B = 1/100000;
 eq1 = h'[t] == -A*h[t]*z[t]*r[t];
 eq2 = z'[t] == A*h[t]*z[t] - B*h[t]*z[t];
 eq3 = r'[t] == B*h[t]*z[t];
 sol = NDSolveValue[{eq1, eq2, eq3, h[0] == 1000000, z[0] == 1,
     r[0] == 0}, h, {t, 0, 20}, WorkingPrecision -> 200,
    Method -> "StiffnessSwitching"];
 
In[7]:= sol /@ {3/1000, 4/1000, 5/1000} // N

Out[7]= {1.85407*10^-11, 3.89607*10^-20, 8.19073*10^-29}
POSTED BY: Ilian Gachevski
 A = 0.01;
 B = 0.00001;
 eq1 = h'[t] == -A*h[t]*z[t]*r[t];
 eq2 = z'[t] == A*h[t]*z[t] - B*h[t]*z[t];
 eq3 = r'[t] == B*h[t]*z[t];
 sol = First[
    h /. NDSolve[{eq1, eq2, eq3, h[0] == 1000000, z[0] == 1,
       r[0] == 0}, {h, r, z}, {t, 0, 20}]];
 

FindRoot[sol[t] == 0, {t, 10^-3}]
{t -> 0.00288353}
There seem to be a number of points where the solution is zero, looking at a graph of the log of the solution.
POSTED BY: Frank Kampas
Posted 10 years ago
 A = 0.01
 B = 0.00001
 eq1 = h'[t] == -A *h[t]* z[t]*r[t]
 eq2 = z'[t] == A *h[t]*z[t] - B *h[t]*z[t]
 eq3 = r'[t] == B *h[t] *z[t]
 % sol = First[
   h /. NDSolve[{eq1, eq2, eq3, h[0] == 1000000, z[0] == 1,
      r[0] == 0}, {h, r, z}, {t, 0, 20}]]
 sol = First[
  h /. NDSolve[{eq1, eq2, eq3, h[0] == 99.9, z[0] == 0.1,
     r[0] == 0}, {h, r, z}, {t, 0, 20}]]
sol2 = First[
  z /. NDSolve[{eq1, eq2, eq3, h[0] == 99.9, z[0] == 0.1,
     r[0] == 0}, {h, r, z}, {t, 0, 20}]]
I put the text above in and got three interpolating functions out. I need to solve for t when the first interpolating function is equal to 0.
POSTED BY: Amy Gordon
If you copy your code into a "spikey" windows instead of posting a screen shot, other people can copy it.
POSTED BY: Frank Kampas
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