- #1

- 3

- 0

(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?

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- Thread starter bangbangbang
- Start date

- #1

- 3

- 0

(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?

- #2

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- 6,711

- #3

- 3

- 0

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?

- #4

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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).

- #5

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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.

- #6

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