# What does it mean for a functional equation to have a unique solution?

It is my thinking that unique conventionally means special or one of its kind. But in the context of solving functional equations*, I am confused what it means to have a unique solution...

*e.g. Find solution(s) to $f(x+y)+f(x-y)=2x^2+2y^2$

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A unique solution means it is the only solution. In your example: if $(x_0, y_0)$ solves your equation then it is called unique iff the following statement is true: if $(x_1, y_1)$ solves the equation, then this implies $x_0= x_1, y_0 = y_1$. –  user20266 Dec 31 '11 at 17:35
@Thomas: As it is a functional equation, one would be looking for functions $f$ that satisfy that property for all $x$ and $y$, such as $f(t)=t^2$. –  Zev Chonoles Dec 31 '11 at 17:40
How do we prove that a solution is "unique" then? Also, am I right in understanding that x=-1 is a unique solution to (x+1)^2=0 then? –  MathMathCookie Dec 31 '11 at 17:42

It means just what you'd think; if $f$ and $g$ are two solutions to the functional equation, then $f=g$, i.e. $f(t)=g(t)$ for all $t$ in the domain.
In the problem you've given, we can prove that there is a unique solution as follows: suppose $f$ is some solution to the functional equation$$f(x+y)+f(x-y)=2x^2+2y^2,$$ i.e. $f(x+y)+f(x-y)=2x^2+2y^2$ for all $x$ and $y$. Then, in particular, we have that $$f(x+0)+f(x-0)=2x^2+2\cdot 0^2$$ $$2f(x)=2x^2$$ and therefore $f(x)=x^2$. Thus the function $f(t)=t^2$ is the only solution to the functional equation $f(x+y)+f(x-y)=2x^2+2y^2$.
Here's a functional equation that does not have a unique solution: $f(x)^2=x^2$. For example, all of the functions $$f(t)=t,\qquad g(t)=-t,\qquad h(t)=|t|$$ satisfy the function equation, i.e. $f(x)^2=x^2$ for all $x$, $g(x)^2=(-x)^2=x^2$ for all $x$, and $h(x)^2=|x|^2=x^2$ for all $x$. However, $f$, $g$, and $h$ are different functions: for example, $$f(1)=1\neq-1=g(1)$$ so $f\neq g$, and similarly $f\neq h$ and $g\neq h$.