I am trying to solve an optimization problem with constraints that leads to the following integro-differential equation

Log[f'[x0]]*Exp[f[x0]] + Integrate[ (f'[x]*Log[f'[x]] - f'[x] + 1)*Exp[f[x]], {x,x0,xmax}] - c == 0 

The goal is to find f[x] such that this equation holds for all x0 between 0 and xmax=10. The parameter c comes from a Lagrange multiplier that imposes the constraint f[xmax] == 10, and we have the initial condition f[0] == 0. In an analytical approach, one would solve the integro-differential equation for f[x] in terms of c, plug the result in the constraint equation to find the value for c, and eventually substitute the resulting value in the expression for f.

I am not sure how I can go about solving this numerically (or analytically, but I doubt that'd be possible). I have seen a few posts dealing with integral equations and integro-differential equations (like this or this). However, it appears to me that these approaches do not directly apply to my case, since (a) differentiating the equation doesn't seem to simplify it much and (b) I have an undetermined parameter c that needs to be determined through the constraint equation.


1 Answer 1


А ларчик просто открывался. Let us differentiate this equation with respect to x0.

D[Log[f'[x0]]*Exp[f[x0]], x0] +  
D[Integrate[(f'[x]*Log[f'[x]] - f'[x] + 1)*Exp[f[x]], {x, x0,   xmax}], x0] == 0

E^f[x0] Log[Derivative[1][f][x0]] Derivative[1][f][x0] - E^f[x0] (1 - Derivative[1][f][x0] + Log[Derivative[1][f][x0]] Derivative[1][f][x0]) + ( E^f[x0] (f^\[Prime]\[Prime])[x0])/Derivative[1][f][x0] == 0

DSolve[%, f[x0], x0]

{{f[x0]->Subscript[\[ConstantC], 2]+Log[E^x0+E^Subscript[\[ConstantC], 1]]}}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.