How to find solutions to equation $f'(x)+f(x)-Kf(-x)=0$

Question

I want to find solutions to the ordinary differential equation

$$f'(x)+f(x)-Kf(-x)=0$$

when $K$ is a nonzero real constant. When $K=1,$ I can find the solution $f(x)=C(x-\tfrac{1}{2}),$ but for more general $K$ it is more challenging. I am able to find power series solutions simply by taking

$$f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$$

but I would really like to have something more insightful here, such as a closed-form solution in terms of whatever functions they may be, or even just some properties that we can say about solutions to this differential equation.

Alternatively, if anyone can give me suggestions on how to proceed with finding out these things for myself, I would welcome any help. The main issue comes from that $f(-x)$ term anyway. Thanks.

Answer 1

Let $g(x)=\frac12(f(x)+f(-x))$ and $h(x)=\frac12(f(x)-f(-x))$ be the even and the odd part of $f$. Then $f(x)=g(x)+h(x)$ and $f(-x)=g(x)-h(x)$, so we have $$ g'(x)+h'(x)+\frac{1-K}2g(x)+\frac{1+K}2h(x)=0.$$ Separating odd and even parts again, $$g'(x)+\frac{1+K}2h(x)=0, \qquad h'(x)+\frac{1-K}2g(x)=0.$$ Taking derivatives once more $$g''(x)+\frac{1+K}2h'(x)=0, \qquad h''(x)+\frac{1-K}2g'(x)=0.$$ From this by elimination, $$ g''(x)-\frac{1-K^2}4g(x)=0, \qquad h''(x)-\frac{1-K^2}4h(x)=0$$ and by addition $$ f''(x)-\frac{1-K^2}4f(x)=0.$$

Answer 2

More simply you can work like this: you derive your equation and get

$$f''(x) + f'(x) +Kf'(-x)=0$$

then you substitute $f'(x)$ and $f'(-x)$ and get:

$$ f''(x)+Kf(-x) -f(x) +K^2f(x) -Kf(-x)=0$$

so it becomes just:

$$ f''(x) -(1-K^2)f(x)=0$$

which is something you know how how to solve. The idea is to derive in order to get some extra information, but once you solve that, you need to put that solution back to the original equation: for example, $$f(x)=0$$ has only a trivial functional solution, but if you derive you obtain $$ f'(x)=0$$ which would imply that $f(x)$ is a constant, but the first equation was more restrictive.

Answer 3

You have 2 locations $x$ and $-x$ in the equation. Thus it is natural to check what the equation says about the derivative at $-x$ and what you can do about separating the locations or writing it as a coupled system. Thus set $g(x)=f(-x)$ to find $$g'(x)=-f'(-x)=f(-x)-Kf(x)=g(x)-Kf(x)$$ and use it together with $$f'(x)=-f(x)+Kf(-x)=-f(x)+Kg(x)$$

This now is an ordinary $2×2$ system, $$ \begin{bmatrix} f'(x)\\g'(x) \end{bmatrix} = \begin{bmatrix} -1&K\\-K&1 \end{bmatrix} · \begin{bmatrix} f(x)\\g(x) \end{bmatrix}, $$ and the answer of Hagen von Eitzen is about decoupling it into order-2 scalar ODE. You can also directly diagonalize this system via eigenvalues and eigenvectors.

Answer 4

Define $$f(x)=\sum _{k=0}^{\infty } a_k x^k;\;f'(x)=\sum _{k=1}^{\infty } k\,a_k x^{k-1};f(-x)=\sum _{k=0}^{\infty } (-1)^ka_k x^k;\;f(0)=1$$ We want to solve $$f'(x)+f(x)-hf(-x)=0;\;f(0)=1$$ Plug the power series into the equation $$\sum _{k=1}^{\infty } k\,a_k x^{k-1}+\sum _{k=0}^{\infty } a_k x^k-h\sum _{k=0}^{\infty } (-1)^ka_k x^k;\;a_0=1$$ We get $$a_1-h+1=0\to a_1=-(1-h);\;a_1 h+a_1+2 a_2=0\to a_2=\frac{1}{2} \left(1-h^2\right)$$ and so on $$a_3=-\frac{(k-1)^2 (k+1)}{3!};\;a_4=\frac{(k-1)^2 (k+1)^2}{4!};\;\ldots$$ The general expression is $$f(x)= \sum _{k=0}^{\infty } \frac{(-1)^k x^k (1-h)^{\left\lceil \frac{k}{2}\right\rceil } (1+h)^{\left\lfloor \frac{k}{2}\right\rfloor }}{k!}$$ Hope this helps

Answer 5

Inspired by the solution of Hagen van Eitzen:

$$f'(x)+f(x)-Kf(-x)=0$$ $$f'(-x)+f(-x)-Kf(x)=0$$

Subtract both equations:

$$[f(x)-f(-x)]'+[f(x)-f(-x)]+K[f(x)-f(-x)]=0.$$

Use the substitution $u(x)=f(x)-f(-x)$ to obtain:

$$u'+(1+k)u=0\implies u(x)=u(0)\exp\left[-(1+k)x \right].$$

Now, split $u(x)$ into its odd and even part and do the same with $f(x)-f(-x)$ and compare both expressions.