Putnam 1963 B3

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Putnam 1963 B3

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Source: Putnam 1963

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sqrtX

#1 h May 1, 2022, 2:42 PM • 1 Y

Y by ImSh95

Find every twice-differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfies the functional equation
$f(x)^2 -f(y)^2 =f(x+y)f(x-y)$ for all $x,y \in \mathbb{R}.$

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#2 h May 1, 2022, 4:30 PM • 1 Y

Y by ImSh95

If we differentiate twice (once with respect to $x$ , once with respect to $y$ ) the left hand side vanishes, while the right hand side becomes $-f''(x-y)f(x+y)+f'(x-y)f'(x+y)-f'(x-y)f'(x+y)+f(x-y)f''(x+y).$ If we redefine $(x-y,x+y)\rightarrow (x,y)$ , this becomes $f(x)f''(y)=f''(x)f(y) \text{ for all }x,y\in\mathbb{R}.$ If $f(y)=0$ for some real $y$ , then the last equation implies $f\equiv 0$ (which is indeed a solution to the original equation). Otherwise, dividing by $f(x)f(y)$ gives that the function $\frac{f''}{f}$ is constant, leaving us with solving the second-order differential equation $y''=cy.$

In a contest scenario, I guess you could write the solution directly, however I will show the solving steps as well.
If $c=0$ , integrating will give $y=ax+b$ . If $c\neq 0$ , adding $y'\sqrt{c}$ gives $(y'+y\sqrt{c})'=\sqrt{c}(y'+y\sqrt{c}).$ Setting $z=y'+y\sqrt{c}$ implies that $z'=z\sqrt{c}\iff \frac{z'}{z}=\sqrt{c}.$ Integrating yields $\ln(z)=x\sqrt{c}+a\iff z=Ae^{x\sqrt{c}}.$ Substituting back in $z=y'+y\sqrt{c}$ and multiplying by $e^{x\sqrt{c}}$ gives $(ye^{x\sqrt{c}})'=Ae^{2x\sqrt{c}}\iff ye^{x\sqrt{c}}=\frac{A}{2\sqrt{c}}e^{2x\sqrt{c}}+B.$ Finally, this implies that the general solutions of $y''=cy$ are $y=Ax+B \text{ and }Ae^{x\sqrt{c}}+Be^{-x\sqrt{c}}.$
All we now need to do is check which $A,B,c$ satisfy the equation $f^2(x)-f^2(y)=f(x-y)f(x+y)$ for $f=Ax+B$ and $f=Ae^{x\sqrt{c}}+Be^{-x\sqrt{c}}.$ Observe that for $x=y=0$ , we have that $f(0)=0$ thus in the first case we would have $f(x)=Ax$ , which is indeed a valid solution, while in the second case this gives $A+B=0$ . It's a matter of simple algebra to see that all functions $f$ of the form $f(x)=Ae^{x\sqrt{c}}-Ae^{-x\sqrt{c}}$ all verify the hypothesis.

As a side note, it didn't matter that $c$ can be negative. if $c>0$ , the las equation describes the function $2A\sinh(x\sqrt{c})$ , while if $c<0$ , the function represents $2A\sin(x\sqrt{c}).$

This post has been edited 1 time. Last edited by AndreiVila, May 1, 2022, 4:32 PM

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ZETA_in_olympiad

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ZETA_in_olympiad

#4 h May 19, 2022, 8:01 AM

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[This is more of a sketch rather than an actual solution]
Note that $f(0)=0.$ Differentiating in respect to $x,$ then in respect to $y$ and making the substitution $a=x+y$ and $b=x-y,$ we have $f''(a)f(b)=f(a)f''(b).$ For a fixed $b,$ the $f''(b)/f(b)$ is constant. So it is either $k^2, -k^2$ or $0.$ Solving for $f(a)$ we get $f(a)\in \{L\sin(ka), L\sinh(ka), La\}.$

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#5 h Tuesday at 10:44 PM

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This is the weirdest functional equation I've ever done.

Hold $y = c$ constant, and take a derivative with respect to $x$ on both sides to yield $2f(x)f'(x) = f'(x+c)f(x-c) + f(x+c)f'(x-c).$ This equation holds for all $c \in \mathbb R$ , so fixing $x$ , we note that the derivative of the right side with respect to $c$ is equal to zero, i.e. $f''(x+c)f(x-c) = f''(x-c)f(x+c).$ This means that $\frac{f''(x)}{f(x)} = C$ is constant.

The solutions to this differential equation are well-known as functions $f \equiv ax+b$ , $f \equiv \sinh(cx) + \cosh(cx)$ , and $f \equiv A \sin(cx) + B \cos(cx)$ . It is not hard to check that only $f \equiv Ax$ , $f \equiv A\sin(cx)$ , and $f \equiv A\sinh(cx)$ satisfy the original equation, as needed.

This post has been edited 1 time. Last edited by HamstPan38825, Tuesday at 10:44 PM

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