D[x[t] - x[t - 1]/(2 E), {t, 3}] + Integrate[E^(-δ)*x[t - δ]/5^t, {δ, 2, 2.5}] == 0
I found solve this problem is hard with Mathematica.
I also find a article here about how to deal with Integral equation with Mathematica code.
But my function is implicit function in the equation,so how to solve this problem?
Large t Approximation
Because this is a neutral delay integral-differential equation, solving it may seem very difficult at first glance. However, the term 1/5^t makes the integral negligible for t greater than about 2. So, with the integral ignored for large t, the equation can be reduced to
x[t] - x[t - 1]/(2 E) = c[2] t^2 + c[1] t + c[0]
where the constants c are determined by boundary conditions. The solution to this difference equation is
x[t] = Sum[f[t - n] (2 E)^-n, {n, 1, N}] + x[t - N] (2 E)^-N
Because this series converges very rapidly for N>3, N can be set to infinity for large t, yielding
(t^2*c[2])/(-1 + 2*E) + (t*((-1 + 2*E)*c[1] - 4*E*c[2]))/(1 - 2*E)^2
+ ((1 - 2*E)^2*c[0] + 2*E*(c[1] - 2*E*c[1] + c[2] + 2*E*c[2]))/(-1 + 2*E)^3
Approximate Solution Including Integral
According to the Wolfram discussion of delay differential equations, NDSolve cannot solve the complete original equation. However, if x[t - δ] in the integrand is approximated by linear interpolation, then the integral can be performed. For instance,
n = 5;
Piecewise[Table[{x[t - 2 - 0.5 i/n] + (x[t - 2 - 0.5 (i + 1)/n] -
x[t - 2 - 0.5 i/n]) 2 n δ, 0.5 i/n <= δ <= 0.5 (i + 1)/n}, {i, 0, n - 1}]];
int = Simplify[5^-t Integrate[Exp[-(δ + 2)] %, {δ, 0, .5}]]
to yield
(0.03877636669329322*x[-2.5 + t] + 0.003170210316893817*x[-2.4 + t] + 0.0025002000402758024*x[-2.3 + t]
+ 0.001654193122766361*x[-2.2 + t] + 0.0006025810391796899*x[-2.1 + t] + 0.006546733400304792*x[-2. + t])/5^t
and NDSolve can solve the resulting equation:
NDSolve[{x'''[t] - x'''[t - 1]/(2 E) + int == 0, x[t /; t <= 0] == ϕ[t]}, x, {t, -3, 100}]
where ϕ[t] is the initial condition for the equation. Choosing
ϕ[t] = -HeavisidePi[t] + 2 HeavisidePi[t + 1] - HeavisidePi[t + 2]
which is a challenging initial condition for many delay differential equations, gives
Note that this curve can be fitted to a quadratic polynomial at an accuracy of better than 0.000001 for t > 2, validating the large t approximation presented in the first half of this Answer.
1 Dec 14 update: Integral treated with improved accuracy. c[3] deleted.
x[t] must be specified numerically for 0>t>-2.5 in order to evaluate x[t] for t>0. Please provide your best estimates. Also, please consider my solution above, if you have not done so already. It should be quite general and accurate for large t. If you know the asymptotic behavior of the solution, it may be possible to constrain f[t].
– bbgodfrey
Nov 28 '14 at 16:50
N -> Infinity. The rest of the code is provided, except for the process of fitting the curve shown by a cubic. The code for this is fit = FindFit[Table[{i, x[i] /. First[sol]}, {i, 92, 100, .1}], a + b y + c y^2 + d y^3, {a, b, c, d}, y]. As you can see, all this is quite simple, once one realizes how to proceed.
– bbgodfrey
Nov 29 '14 at 13:33
RSolve[x[n] - x[n - 1]/(2 E) - f[n] == 0, x[n], n]. At first glance, the result may not look like the second equation, but it is. Hope this helps.
– bbgodfrey
Dec 03 '14 at 02:57
x[t]byy[t]=x[t] - x[t - 1]/(2 E)as the dependent variable. – bbgodfrey Nov 28 '14 at 14:04