I have a problem with iteration of the result of NDSolve. Namely, the following code works fine

a[x_] := Exp[-x^2]

uo[x_] := Exp[-2 x^2]

sol := Module[{x, t}, NDSolve[{D[u[x, t], t] == 1 – u[x, t] a[x], u[x, 0] == uo[x]},

u, {x, -10, 10}, {t, 0, 10}]]

d[x_, t_] := Evaluate[u[x, t] /. sol]

d[1, 1]

Out[58]= {0.942309}

However, when I try to use the function d(x,t), I get an error, namely:

newsol := Module[{y, s}, NDSolve[{D[v[y, s], s] == 1 – v[y, s] d[y, s], v[y, 0] == uo[y]},

v, {y, -10, 10}, {s, 0, 10}]]

Evaluate[v[1, 1] /. newsol]

Thread::tdlen: Objects of unequal length in {0} {1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.,1.} cannot be combined. >>

and so on…

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2

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– Sektor

Dec 3 ’14 at 23:31

The differential equations described above could be solved analytically as well.

– bbgodfrey

Dec 4 ’14 at 2:59

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2 Answers

2

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The problem is, that NDSolve always returns a list of solutions, in this case a list of length 1.

You can see it here:

d[1, 1]

Out[58]= {0.942309}

If you know there is only one solution, you can use

d[x_, t_] := Evaluate[u[x, t] /. First@sol]

to make d[1,1] evaluate to 0.942309 instead of {0.942309}

.

Thank you very much!

– Dmitri

Dec 4 ’14 at 5:35

1

@Dmitri you are saying “thank you” 2nd time already for 2nd question you asked. You should start accepting the answers

– Vitaliy Kaurov

May 15 ’15 at 9:50

Although @DaveStrider has answered the question above fully, I think it worth noting that these equations can be solved analytically. For instance,

a = Exp[-x^2]; uo = Exp[-2 x^2];

d = u[t] /. DSolve[{u'[t] == 1 – a u[t], u[0] == uo}, u, t][[1]]

with solution

E^(-(t/E^x^2) – 2*x^2)*(1 – E^(3*x^2) + E^(t/E^x^2 + 3*x^2))

Interestingly, N[d /. {x -> 1, t -> 1}] evaluates to 0.930365, which differs slightly from the value obtained with NDSolve.

To continue, the second equation is solved by

f = v[t] /. DSolve[{v'[t] == 1 – v[t] d, v[0] == uo}, v, t][[1]]

which has a more complicated solution

-(E^(-E^(-x^2) + E^(-(t/E^x^2) – x^2) – E^(-(t/E^x^2) + 2*x^2) – E^x^2*t –

2*x^2)*(-E^E^(2*x^2) + E^(E^(-x^2) + 2*x^2)*

Integrate[E^(-E^(-x^2 – K[1]/E^x^2) + E^(2*x^2 – K[1]/E^x^2) + E^x^2*K[1]),

{K[1], 1, 0}] – E^(E^(-x^2) + 2*x^2)*

Integrate[E^(-E^(-x^2 – K[1]/E^x^2) + E^(2*x^2 – K[1]/E^x^2) + E^x^2*K[1]),

{K[1], 1, t}]))

Finally, N[f /. {x -> 1, t -> 1}] is 0.795889. Analytical solutions, when available, often provide more insight than do numerical solutions.

1

I understood this, surely. The idea is that I need to solve much more complicate nonlinear equation with convolutions which is considered by Mathematica as an equation with delay and hence non-solvable. I know that solution exists by a fixed point theorem and iterate. I got similar mistakes in that difficult situation and to ask a question I ‘cut’ the equation to a simple (now linear) form.

– Dmitri

Dec 4 ’14 at 5:38