| Author |
Comment/Response |
Michael
|
 11/17/09 10:03am
I have a question as to an apparent inconsistency I noticed while using Mathematica to solve differential equations, I have 2 ODE like this, and I solve them symbolically using Mathematica 7.0:
system = {p'[t] == -(rho + d)*p[t],
n'[t] == rho*p[t] + (rho - d)*n[t]};
sol = DSolve[system, {p, n}, t][[1]]
I get the following solution:
{p -> Function[{t}, E^((-d - rho) t) C[1]],
n -> Function[{t},
1/2 (-E^((-d - rho) t) + E^((-d + rho) t)) C[1] +
E^((-d + rho) t) C[2]]}
It looks to be correct.
Then, manually, I know p(0)=C1, n(0)=0, so C1=p0, C2=0.
I know the values for rho and d. So for t from 0 to 120, I can calculate and plot the values for p(t) and n(t).
Up to now, this is all good.
Now I want to solve the system of ODEs numerically (even though I have already solved it, I have good reasons to want to learn to solve it numerically, as I am going to add more complexity to this model), so I did the following:
sol4 = NDSolve[{p'[t] == -(0.0258 + 0.0123)*p[t],
n'[t] == 0.0258*p[t] + (0.0258 - 0.0123)*n[t], p[0] == 30000,
n[0] == 0}, {p, n}, {t, 0, 120}]
Now I get the following solution:
{{p-> InterpolatingFunction[{{0.`,120.`}},"<>"], n->InterpolatingFunction[{{0.`,120.`}},"<>"]}}
It may be correct, but I then plotted it using
Plot[Evaluate[{n[t]} /. First[sol4]], {t, 0, 120},
PlotRange -> All]
Now my question is:
I generated plot #1 of n(t) by solving the system of ODEs first symbolically, then manually inputting in values (C1, C2, rho, d).
I generated plot #2 of n(t) by solving the system of ODEs numerically, then had mathematica plot the solutions of and n(t) for me.
Why does plot #1 and #2 look so different? By that I mean, n(t) in plot 2 is much lower than n(t) in plot 1.
Could you please explain what I did wrong? Thank you. I have attached the 2 plots for your convenience.
Attachment: attached plot.doc, URL: , |
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