First Order Differential Equation with Reflected Argument

[bangbangbang]

I am trying to solve:

(x + 1 + f(-x) )(1 - f ' (x) ) = x+1
f(0) = x_0
x in (-1,1)

I approximated it numerically but any analytic method I try fails. Any ideas?

 

[haruspex]

Hint: What do you get if you differentiate (x + 1 + f(-x) )? It should make you think of the chain rule.

 

[bangbangbang]

Ok, so I can make a substitution:

y(x) = x + 1 + f(-x). Then

y ' (x) = 1 - f ' (-x).

I don't see where to go from there, since y ' (x) does not appear in the original equation. However

y ' (-x) = 1 - f ' (x) does appear in the original equation. If i make that substitution I get

y(x) * y ' (-x) = x+1

But I still do now know how to solve that. Is my reasoning correct so far?

 

[haruspex]

Hmm.. yes I was too quick - I thought the minus sign would not be a problem.
However, you can at least try making an assumption about y(-x) and see if it produces a solution consistent with the assumption. I.e., try y(-x) = - y(x), then try y(-x) = y(x).

 

[bangbangbang]

Judging from the numeric approximation, neither of those seem to be the case. I will try it and see what happens though.

I have tried many things. One thing I was considering was to use the cross correlation of f(x) with a conveniently chosen function, make a substitution, and use the Laplace transform to force the negative sign inside, then hope the inverse Laplace transform works out. That sounds really horribly complicated though.

I have tried a series solution but it did not seem to result in anything worthwhile.

Perhaps a solution does not exist.

 

[haruspex]

I finaly realized that f(x) might as well be separate functions for x> 0 and x < 0. so write g( x) = f(-x). obtain (1-x+f(x))(-g'(x)) = 1-x by swapping sign of x. Write orig equation as g(x) = expression in x and f'(x), then differentiate that to obtain expression for g'(x) and substitute in above. Now have 2nd ord (nasty) equation in only f and x. Not sure it helps much though.