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I have a differential equation that looks like this:

 DSolve[(D[R[r, t], t])^2/(2*c) - (G*M1[r, z])/(
   c^2*R[r, t]) - (λ*(R[r, t])^2)/3 - E1[r] == 0, ... "boundary conditions etc go here"]

I need to get R[r, t] using my values of E1[r] and M1[r]. However, my M1[r] is defined as:

M1[r_] :=  NIntegrate[(1 + f*Exp[-(r/y)^2])*(r)^2, r]

Herein lies the problem. I realise that DSolve uses Integrate internally, but the function I want it to solve has an NIntegrate in it and this causes problems. I'm not sure how else to evaluate an error function.

Any ideas?

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  • $\begingroup$ This gives a solution in terms of an unresolved integral. Eventually. DSolve[(D[R[r, t], t])^2/(2*c) - (G*M1[r, z])/(c^2* R[r, t]) - (\[Lambda]*(R[r, t])^2)/3 - E1[r] == 0, R[r, t], t] $\endgroup$
    – Daniel Lichtblau
    Feb 9, 2013 at 22:33
  • $\begingroup$ What is the definition of f in your function M1[r_]? a unknown function? $\endgroup$
    – Enter
    Feb 13, 2015 at 7:51

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Not knowing what the actual differential equation is, I can only point out one obvious problem: you should define 

Clear[r,f,y];
M1[r_] = Integrate[(1 + f*Exp[-(r/y)^2])*(r)^2, r]

instead of using := NIntegrate because you seem to be looking for an indefinite integral there. For numerical integration you have to specify integration limits. The Clear is just added for safety before I define the function with = so that the integral will be done at that time, once and for all.

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  • $\begingroup$ Wrapping the integral in Block[{r,f,y}, ...] would work, also, and limit the impact on the rest of the notebook. $\endgroup$
    – rcollyer
    Feb 8, 2013 at 21:41
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    $\begingroup$ For all I know, f in the question could in fact have a value - so it may be a moot point... $\endgroup$
    – Jens
    Feb 8, 2013 at 21:52
  • $\begingroup$ True, very true. $\endgroup$
    – rcollyer
    Feb 8, 2013 at 22:06

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