# exponential differential integral equation

I have a following equation, which has a singular kernel,

0.282095 Integrate[
Derivative[A11][k]/Sqrt[-k + t] , {k, 0,
t}] + ((-1 + E^A11[t]) Derivative[A11][t])/(Sqrt[
2] Sqrt[-1 + E^A11[t] - A11[t]]) ==
0.5 (0.2 - 0.0433792711308264 Sqrt[-1 + E^A11[t] - A11[t]] +
A11[t] -
0.015336865356479351 Integrate[(2 Sqrt[
1/(-k + t)] Derivative[A11][k])/Sqrt[π] , {k, 0, t}])


and the Initial condition is A11==0

I used following series expansion for the nonlinear parts

Series[Sqrt[-1 + E^A11[t] - A11[t]], {A11[t], 0, 3}]
Series[(-1 + E^A11[t]), {A11[t], 0, 3}]


The above expressions are replaced in the equation, Then I used Laplace transform to solve the equation, where the answer is

A11[s_] = 0.1/(
s (-0.48466313872191963 + 0.5153368612780803 Sqrt[s] + s))


Finally, the Laplace inverse package was used to obtain the solution The solution is not so bad, but I am wondering is it possible to solve these kind of questions in original form without any assumptions. I do not know how to write this equation in the standard Volterra integral equation.

The nonlinear integral-differential equation in the question can be solved iteratively as follows. Begin with the expression in the question, but with 2/Sqrt[π] factored out of the second integral (because Simplify does not automatically remove constants from integrands).

eq1 = Simplify[0.282095 Integrate[Derivative[A11][k]/Sqrt[-k + t], {k, 0, t}] +
((-1 + E^A11[t]) Derivative[A11][t])/(Sqrt Sqrt[-1 + E^A11[t] - A11[t]]) ==
0.5 (0.2 - 0.0433792711308264 Sqrt[-1 + E^A11[t] - A11[t]] + A11[t] -
0.015336865356479351 Integrate[Sqrt[1/(-k + t)] Derivative[A11][k], {k, 0, t}]
2/Sqrt[π]), 0 < k < t];


and solve it for A11'[t].

ap = Simplify[A11'[t] /. First@Solve[eq1, A11'[t]]
(* (0.03067377677953718 - 0.03067377677953718*E^A11[t] +
0.1414213562373095*Sqrt[-1. + E^A11[t] - 1.*A11[t]] +
(0.03067377677953718 + 0.7071067811865475*Sqrt[-1. + E^A11[t] -
1.*A11[t]])*A11[t] - 0.4111796229566854*Sqrt[-1. + E^A11[t] -
1.*A11[t]]*Integrate[A11'[k]/Sqrt[-k + t], {k, 0, t}])/(-1. + 1.*E^A11[t]) *)


The question further specifies that A11 == 0. Inserting this into ap yields

(Series[ap, {A11[t], 0, 0}, Assumptions -> A11[t] >= 0] /. t -> 0) // Normal
(* 0.1 *)


In other words, the corresponding value of A11' is 0.1. Now, NDSolve could solve A11'[t] == ap without difficulty, were it not for the Abel integral, Integrate[A11'[k]/Sqrt[-k + t], {k, 0, t}]. So, let us approximate A11'[k] in the integral by A11', use NDSolve to obtain a better approximation for A11'[k], and continue iterating until A11[k] converges. Define

int[n_, t_?NumericQ] := NIntegrate[sp[n - 1][k]/Sqrt[t - k], {k, 0, t},
Method -> {Automatic, "SymbolicProcessing" -> False}]
apn[n_] = ap /. Integrate[A11'[k]/Sqrt[t - k], {k, 0, t}] -> int[n, t];


and iterate

tmin = 1/1000; sp[-1][t_] = .1;
Do[{s[n], sp[n]} = Quiet@NDSolveValue[{A11'[t] ==
Piecewise[{{apn[n], t > tmin}}, 1/10], A11 == 0}, {A11, A11'}, {t, 0, 1}], {n, 0, 5}]


Fortunately, this process actually converges, and rapidly.

Join[{.1}, Table[s[n], {n, 0, 5}]]
(* {0.1, 0.0799404, 0.0885934, 0.0854804, 0.0864669, 0.0861837, 0.0862587} *)

Plot[{0.1 t, s[t], s[t]}, {t, 0, 1}]
` Shown are the initial guess (blue), the result of the first iteration (tan), and the result of the sixth iteration (green). The entire computation takes only seconds.

• Thank you, it is very interesting, but A11 == 0, not A11 == 1. The solution is correct. I think it is just a typo. – hesamaero May 5 '17 at 21:20
• @hesamaero Yes, a typo, now corrected. Thanks for pointing it out, as well as for accepting the answer. By the way, your equation can be recast as a Volterra equation of the second kind coupled to a first order ODE. However, it is not clear to me that doing so would be helpful. – bbgodfrey May 5 '17 at 21:26