Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Please can someone please tell me the intuition behind solving a functional equation?
For example, $$f(x)+f(y)=2f\left(\frac{x+y}2\right)f'\left(\frac{x-y}2\right)$$
Now at the first look, it seems like $f(x)=\sin x$, but how to solve it rigorously?
P.S: As written in the title, I am a beginner and I have no experience at solving a functional equation. I know little bit about series expansions and I know calculus upto 12th grade. (Derivatives, Integrals (Impropers-a bit), Limits)
$$f(x)+f(y)=2f\left(\frac{x+y}2\right)f'\left(\frac{x-y}2\right)$$
Let's define $t=\frac{x+y}{2}$, $u=\frac{x-y}{2}$
$$f(t+u)+f(t-u)=2f\left(t\right)f'\left(u\right)$$
for $u=0$
$$f(t)+f(t)=2f\left(t\right)f'\left(0\right)$$ $$2f(t)=2f\left(t\right)f'\left(0\right)$$ $$f'\left(0\right)=1$$
To derivate both side for $u$ $$f'(t+u)-f'(t-u)=2f\left(t\right)f''\left(u\right)$$
Then for $u=0$ again
$$f'(t)-f'(t)=2f\left(t\right)f''\left(0\right)$$
$$0=2f\left(t\right)f''\left(0\right)$$ $$f''\left(0\right)=0$$
To derivate both side for $u$ again $$f''(t+u)+f''(t-u)=2f\left(t\right)f'''\left(u\right)$$
Then for $u=0$ again
$$2f''(t)=2f\left(t\right)f'''\left(0\right)$$
$$f'''\left(0\right)=c^2$$
if $c=0$ then $$f''(t)=0$$ and $$f(t)=a+bt$$ and we know that $$f'\left(0\right)=1$$
Thus $b=1$ $$f(t)=a+t$$
to put into the function equation to find $a$
$$a+x+a+y=2(a+\frac{x+y}2)$$ $$2a+x+y=2a+x+y$$
thus
$$f(t)=a+t$$ is general solution for $c=0$ where $a∈C$ .
if $c \neq0$ then
$$f''(t)=c^2f\left(t\right)$$
The solution of the equation
$$f(t)=k_1e^{ct}+k_2e^{-ct}$$
we know $$f'\left(0\right)=1$$ $$f''\left(0\right)=0$$ and $$f'''\left(0\right)=c^2$$
To solve all $$f'(t)=k_1ce^{ct}-k_2ce^{-ct}$$ $$f'(0)=c(k_1-k_2)$$ $$c(k_1-k_2)=1$$
$$f''(t)=k_1c^2e^{ct}+k_2c^2e^{-ct}$$ $$f''(0)=c^2(k_1+k_2)$$ $$c^2(k_1+k_2)=0$$ $$k_1=-k_2$$
Thus
$$c(k_1-k_2)=1$$ $$2ck_1=1$$ $$k_1=\frac{1}{2c}$$ $c \neq 0$
Our general solution is for $c \neq 0$
$$f(t)=\frac{1}{2c}(e^{ct}-e^{-ct})$$
To confirm the solution
$$\frac{1}{2c}(e^{cx}-e^{-cx})+\frac{1}{2c}(e^{cy}-e^{-cy})=2\frac{1}{2c}(e^{c\frac{x+y}2}-e^{-c\frac{x+y}2})\frac{1}{2}(e^{c\frac{x-y}2}+e^{-c\frac{x-y}2})$$
$$\frac{1}{2c}(e^{cx}-e^{-cx}+e^{cy}-e^{-cy})=\frac{1}{2c}(e^{c\frac{x+y}2}-e^{-c\frac{x+y}2})(e^{c\frac{x-y}2}+e^{-c\frac{x-y}2})$$
$$\frac{1}{2c}(e^{cx}-e^{-cx}+e^{cy}-e^{-cy})=\frac{1}{2c}(e^{cx}+e^{cy}-e^{-cy}-e^{-cx})$$
and you are right that $c=i$ ,$f(t)=\sin(t)$ is a solution of the function equation but not general solution. it is just particular solution
$$f_i(t)=\frac{1}{2i}(e^{it}-e^{-it})=\frac{1}{2i}(\cos{}t +i\sin{t} -\cos{t}+i \sin{t})=\sin{t}$$
The general formula is :
for $c \neq 0$; $$f(t)=\frac{1}{2c}(e^{ct}-e^{-ct})$$
for $c = 0$; $$f(t)=a+t$$ ;where $a∈C$
and also $f(t)=0$ is a solution of the function equation.
Taking $x=y$ gives $$ 2f(x) = 2f(x)f'(0), \tag{1} $$ so either $f(x) \equiv 0$ or $f'(0)=1$. Suppose it is the latter (else we have found $f$). Setting $x=-y$ gives $$ f(x)+f(-x) = 2f(0)f'(x). \tag{2} $$ Differentiating the original equation with respect to $x$ gives $$ f'(x) = f'\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) + f\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right), \tag{3} $$ and setting $y=-x$ gives $$ f'(x) = f'(0)f'(x)+f(0)f''(x) = f'(x)+f(0)f''(x). $$ This says that either $f''(x) \equiv 0$ or $f(0)=0$. It is easy to check that the only functions with $f''(x) \equiv 0$ which satisfy the equation are $f(x) =0$ and $f(x)=x$, so now assume $f(0)=0$. (2) then gives that $$ f(-x) = -f(x), $$ so we also have $f''(0)=0$. Differentiating (3) with respect to $y$ gives $$ 0 = f''\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) - f'\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right) + f\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right) - f\left( \frac{x+y}{2} \right)f'''\left( \frac{x-y}{2} \right) \\ = f''\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) - f\left( \frac{x+y}{2} \right)f'''\left( \frac{x-y}{2} \right). $$ Now, if we take $y=x$, this reduces to $$ f'''(0) f(x) = f''(x) f'(0) = f''(x). $$ At this point, we can write down the general solution to this equation: setting $f'''(0)=A^2$ gives $$ f(x) = B\sinh{Ax}+C\cosh{Ax}, $$ and imposing the boundary conditions $f(0)=0$, $f'(0)=1$ gives general solution $$ f(x) = \frac{1}{A}\sinh{Ax}. $$ It is simple to check that this always satisfies the functional equation, and further, you can recover the $f(x)=x$ solution by taking the limit as $A \to 0$.
(Remark: this does include the sine solution you expected, since $$ \sin{ax} = \frac{\sinh{iax}}{i}, $$ so taking $A=ia$ gives the sine solution.
Therefore we have shown
If $f(x)$ is thrice-differentiable and satisfies the functional equation $$ f(x)+f(y) = 2f\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right), $$ then $f(x)$ is one of
- $0$
- $x$
- $A^{-1}\sinh{Ax}$, some $A \in \mathbb{C}$.
EDIT: In fact, we only need twice-differentiability, because setting $x=0$,$y=2z$ in (3) gives $$ 1 = (f'(z))^2-f(z)f''(z), $$ which is a thoroughly rotten-looking DE, but we can solve it by noticing that $$ \left(\frac{f'}{f}\right)' = \frac{ff''-f'^2}{f^2}, $$ so an equivalent is $$ -\frac{1}{f^2} = \left(\frac{f'}{f}\right)' $$ Multiply by $2f'/f$: $$ -2\frac{f'}{f^3} = 2\left(\frac{f'}{f}\right)\left(\frac{f'}{f}\right)' $$ Integrating this once, we have $$ a^2+\frac{1}{f^2} = \frac{f'^2}{f^2}, $$ and rearranging this gives $$ 1 = \frac{f'^2}{a^2 f^2+1} $$ Taking the positive square root (since $f'(0)=1$) and integrating, $$ z-0 = \int_0^{f(z)} \frac{dt}{\sqrt{1+a^2 t^2}} $$ Now, this is easy to integrate by setting $u=a^{-1}\sinh{t}$: then the integral just simplifies to $a^{-1}\int_0^{\arg\sinh{(af(z))}} du =\arg\sinh{(af(z))}, $ and hence $f(z)=\frac{1}{a}\sinh{az}$ as before. (And $a=0$ gives the $f(x)=x$ solution.)
