Iteration of NDSolve

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...

• Hi ! You once asked a question here and I am sure you noticed the proper code formatting. Please, edit your post and format it properly. Just head to the help centre to read more about code formatting. Dec 3, 2014 at 23:31
• The differential equations described above could be solved analytically as well. Dec 4, 2014 at 2:59

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} .

• @Dmitri you are saying "thank you" 2nd time already for 2nd question you asked. You should start accepting the answers May 15, 2015 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.

• 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. Dec 4, 2014 at 5:38