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?

  • $\begingroup$ You might have a look at the several responses to prior questions on integral equations in MSE. $\endgroup$ Commented Nov 28, 2014 at 13:38
  • $\begingroup$ Three boundary condition are needed to solve this equation. What are they? Also, you may find it beneficial to replace x[t] by y[t]=x[t] - x[t - 1]/(2 E) as the dependent variable. $\endgroup$
    – bbgodfrey
    Commented Nov 28, 2014 at 14:04
  • $\begingroup$ Then how to solve this? in fact the solution will go to 0 when t-> infinity ,but don't always be a constant.And I don't know how to transform the equation to x(t)=... like the equations in that article. $\endgroup$
    – partida
    Commented Nov 28, 2014 at 14:18

1 Answer 1


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

enter image description here

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.

  • $\begingroup$ I am confused,how to obtain the equation "x[t] = Sum[f[t - n - 1/2] (2 E)^-n, {n, 1, N}] + x[t - N] (2 E)^-N"? I don't uderstand.And t solution is not sufficient,you think it may have a numerical solution? Does the article above makes some help? $\endgroup$
    – partida
    Commented Nov 28, 2014 at 15:31
  • $\begingroup$ p.s:the solution is a Non-oscillatory solution,so x(infinity)->0 maybe most important.Can you explian use a boundary conition use what's you think? $\endgroup$
    – partida
    Commented Nov 28, 2014 at 15:38
  • $\begingroup$ @user15961, the article you mentioned is about solving integral equations. Your equation is much more general, so I do not believe that the article will help much. However, I do believe that the equation can be solved numerically. To help you make progress, I need to know the range of t in which you are interested and the boundary conditions. $\endgroup$
    – bbgodfrey
    Commented Nov 28, 2014 at 15:38
  • $\begingroup$ the equation is experimental.so boundary condition is also a experiment.the t range from 0 to 100.i guess it is difficult. $\endgroup$
    – partida
    Commented Nov 28, 2014 at 16:01
  • $\begingroup$ @user15961, NDSolve will require numerical boundary conditions to solve this equation. Additionally, 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]. $\endgroup$
    – bbgodfrey
    Commented Nov 28, 2014 at 16:50

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