Ndsolve with NIntegerate

Question

N1 = 5000;
th = Pi/4;
s1 = 0.01 Sin[th];
s0 = 0.01 Sin[th];
sig = 0.01;
h = 0.5;
nst1 = NDSolve[{
   (x (1 - x) )/2  D[T1[x], x, x] +  s0 N1  x (1 - x) D[T1[x], x] + 
  sig Cos[th] x (1 - x) == 
 NIntegrate[x2 NIntegrate[x1, {x1, 0, x2}], {x2, 0, x}],
T1[0.000000001] == 0, T1[.99999999] == 0
}, {T1}, {x, 0.000000001, .99999999}];

I have attached the code below, the main error is I am not able to use double Nintegerate in the ndsolve where NIntegerate has x dependence. Please don't simplify the double integration by hand because, in the original code, the double integration can not be done analytically.

Daniel Huber · Accepted Answer

You can not do an numerical integral with symbolic bounds. You must ensure that the integral is only called with numeric values. Towards this aim, you can define a function for each integral and constrain the undefined bound to be numeric. Here is one way of doing this:
N1 = 5000;
th = Pi/4;
s1 = 0.01 Sin[th];
s0 = 0.01 Sin[th];
sig = 0.01;
h = 0.5;

int1[x_] /; NumericQ[x] := NIntegrate[x2 int2[x2], {x2, 0, x}]
int2[x2_] /; NumericQ[x2] := NIntegrate[x1, {x1, 0, x2}];

nst1 = NDSolve[{(x (1 - x))/2 D[T1[x], x, x] + 
     s0 N1 x (1 - x) D[T1[x], x] + sig Cos[th] x (1 - x) == int1[x], 
   T1[0.000000001] == 0, T1[.99999999] == 0}, {T1}, {x, 
   0.000000001, .99999999}][[1]]

Ndsolve with NIntegerate

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