# Solve integro-differential equations

I am trying to numerically solve the following integro-differential equation and get some plots for $$n(s)$$ vs $$s$$:

$$\dot{n}(s)-\frac{\dot{w}(s)}{2w(s)}\int_{-\infty}^s ds'\frac{\dot{w}(s')}{w(s')}(1+2n(s'))\,\cos[2\int_{s'}^s ds'' w(s'')]=0$$

with $$\dot{w}(x)=\frac{dw(x)}{dx}$$. The initial condition for $$n(s=-\infty)=0$$ can be considered and $$w(s)$$ is a general function of $$s$$ which one can consider as an example a function like $$w(s)=a^2+\sqrt{c-d(1+Tanh[s])}$$ with $$a,c,d$$ constants. I am new in Mathematica that is why I do not have a clear idea where to start. There are some different methods to solve these type of problems in the literature like Adomian, Galerkin methods, etc... but what I have here is a bit more involved than what people used to consider based on those methods...so I thought a numerical solution would be good idea to start with.

MMA code:

w[s_] := a^2 + Sqrt[c - d (1 + Tanh[s])]
Inactivate[n'[s] - w'[s]/(2 w[s])*Integrate[Cos[2 Integrate[w[x], {x, t, s}]]*(w'[t]/
w[t])*(1 + 2 n[t]), {t, -Infinity, s}] == 0, Integrate] // ExpandAll// TeXForm


$$\frac{d \text{sech}^2(s) \int _{-\infty }^s\left(-\frac{d \cos \left(2 \int _t^s\left(a^2+\sqrt{c-d-d \tanh (x)}\right)dx\right) \text{sech}^2(t)}{2 \sqrt{c-d-d \tanh (t)} a^2+2 c-2 d-2 d \tanh (t)}-\frac{2 d \cos \left(2 \int _t^s\left(a^2+\sqrt{c-d-d \tanh (x)}\right)dx\right) n(t) \text{sech}^2(t)}{2 \sqrt{c-d-d \tanh (t)} a^2+2 c-2 d-2 d \tanh (t)}\right)dt}{4 a^2 \sqrt{c-d \tanh (s)-d}+4 c-4 d \tanh (s)-4 d}+n'(s)=0$$

• What have value of constant a,c,d  ? Oct 8 '18 at 12:21
• Let's put them all equal to one. Oct 8 '18 at 14:24

UPDATE 11.10.2018 for speed-up the code.

w[s_] := a^2 + Sqrt[c - d (1 + Tanh[s])]
D[w[s],s]/ w[s]

(* -((d Sech[t]^2)/(2 Sqrt[c - d (1 + Tanh[t])] (a^2 + Sqrt[c - d (1 + Tanh[t])]))) *)


Simplifying and putting to integro-differential equation, and next compute integral:

Integrate[w[x], {x, t, s}, Assumptions -> {a > 0, c > d, d > 0, c > 0, t ∈ Reals, s ∈ Reals, s > t}]

(* ? *)


Mathematica tires to solve and does not give a solution within 1 hour computation.

With Maple 2018.2 I have the answer:

$Version (* 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018) *) ClearAll["Global*"]; Remove["Global*"]; inf = -17; B = 2; (* inf is Infinity. With -18 value - > error occurs!! *) a = 1; c = 3; d = 1; (* At some c,d values, the integral is divergent *) steps = 1/100; tolerance = 10^-8; SOL = FixedPointList[Function[n, Module[{INT, int1, n1}, INT = ParametricNDSolveValue[{int1'[t] == Cos[2*(a^2 s - a^2 t + Sqrt[-c + 2 d] ArcTan[Sqrt[c + c E^(2 s) - 2 d E^(2 s)]/( Sqrt[-c + 2 d] Sqrt[1 + E^(2 s)])] - Sqrt[-c + 2 d] ArcTan[Sqrt[c + c E^(2 t) - 2 d E^(2 t)]/( Sqrt[-c + 2 d] Sqrt[1 + E^(2 t)])] - Sqrt[c] ArcTanh[Sqrt[c + c E^(2 s) - 2 d E^(2 s)]/( Sqrt[c] Sqrt[1 + E^(2 s)])] + Sqrt[c] ArcTanh[Sqrt[c + c E^(2 t) - 2 d E^(2 t)]/( Sqrt[c] Sqrt[1 + E^(2 t)])])]*(-((d Sech[t]^2)/( 2 Sqrt[c - d (1 + Tanh[t])] (a^2 + Sqrt[ c - d (1 + Tanh[t])]))))*(1 + 2*n[t]), int1[inf] == 0}, int1, {t, inf, B}, {s}, PrecisionGoal -> 20]; NDSolve[{n1'[ s] - (-((d Sech[s]^2)/( 4 Sqrt[c - d (1 + Tanh[s])] (a^2 + Sqrt[c - d (1 + Tanh[s])]))))* Evaluate[INT[s][s]] == 0, n1[inf] == 0}, n1, {s, inf, B}, PrecisionGoal -> 20][[1, 1, 2]]]], Function[s, s], SameTest -> (Max[Abs[Table[#1[s] - #2[s], {s, inf, B, steps}]]] < tolerance &)]; Plot[Evaluate[SOL[[-1]][s]], {s, inf, B}, PlotRange -> All, PlotLegends -> {"n[s]"}, AxesLabel -> {"s", "n[s]"}]  Numeric errors check: M[s_] := (SOL[[-1]][s]); CHECK[s_?NumericQ] := M'[s] - (-((d Sech[s]^2)/( 4 Sqrt[c - d (1 + Tanh[s])] (a^2 + Sqrt[c - d (1 + Tanh[s])]))))* NIntegrate[ Cos[2*(a^2 s - a^2 t + Sqrt[-c + 2 d] ArcTan[Sqrt[c + c E^(2 s) - 2 d E^(2 s)]/( Sqrt[-c + 2 d] Sqrt[1 + E^(2 s)])] - Sqrt[-c + 2 d] ArcTan[Sqrt[c + c E^(2 t) - 2 d E^(2 t)]/( Sqrt[-c + 2 d] Sqrt[1 + E^(2 t)])] - Sqrt[c] ArcTanh[Sqrt[c + c E^(2 s) - 2 d E^(2 s)]/( Sqrt[c] Sqrt[1 + E^(2 s)])] + Sqrt[c] ArcTanh[Sqrt[c + c E^(2 t) - 2 d E^(2 t)]/( Sqrt[c] Sqrt[1 + E^(2 t)])])]*(-((d Sech[t]^2)/( 2 Sqrt[c - d (1 + Tanh[t])] (a^2 + Sqrt[c - d (1 + Tanh[t])]))))*(1 + 2*M[t]), {t, inf, s}, Method -> {Automatic, "SymbolicProcessing" -> 0}, PrecisionGoal -> 20] Plot[CHECK[s], {s, inf, B}, PlotRange -> All, PlotLegends -> {"n[s]"}, AxesLabel -> {"s", "n[s]"}] // Quiet  • Looks good Mariusz, thanks. Just what I expected was positive values for n(s), and can you please tell me what are A and B in your expressions? Oct 10 '18 at 13:30 • @Naser .Function of s from A to B, from s=-10 to s=0. Oct 10 '18 at 15:29 • @Naser. Maybe you know how a graph should look at given parameters? Oct 10 '18 at 20:45 • This new one looks much better and indeed very close to what it has to be, as you mentioned Mathematica is very slow and it gives a very different solution. Oct 12 '18 at 0:52 • @ Mariusz. I tried to use your new codes for the plot but I can't get it, apparently there is a non-numerical value which Mathematica complains about it but after setting a value for$A\$ I still do not get the plot. Any idea what is going wrong? Oct 15 '18 at 2:22