Solving a functional equation and general theory of functional equations

Question

$f(x-1)$ + $f(x+1)$ = $kf(x)$ given that $k>1$ then I need to find $f(x)$.

I am clueless about solving functional equations please help me this . Also give some general advice about how to solve functional equations.

Answer 1

The answer might be a bit more complicated, as it depends a little on what you actually want to solve. In particular, is this a functional equation on the real variable $x$ or is $x$ restricted to be an integer? In any case the solution is not unique without specifying more details.

If the latter is the case, we are simply dealing with a recurrence relation: $$ f(n) = k f(n-1) - f(n-2) $$ with the characteristic polynomial $$ x^2 - k x + 1 =0 $$ which has the roots $$ \lambda_\pm = \frac{k \pm \sqrt{k^2-4}}{2} $$ and the general solution $$ f(n) = c_+ \lambda_+^n + c_- \lambda_-^n $$ for some constants $c_+$ and $c_-$. Their values would follow from two given functional values, which need to be stated in order to have a unique solution. Note that for $k^2 < 4$ the roots are complex and that there is a special case for $k=\pm 2$ where things are slightly modified.

If the problem is actually a functional equation that is to be solved for $x \in {\mathbb R}$, it gets more interesting and there are many different types of solutions that can be continuous and/or discontinous.

For instance we can find the solution $$ f(x) = g(x) ~\left(\lambda_+\right)^x $$ which is valid for any periodic function $g(x)=g(x+1)$. This is easier to see if we introduce constants $\alpha_\pm$ such that $\lambda_\pm = e^{\alpha_\pm}$, because inserting this in the functional equations gives: $$ f(x-1) + f(x+1) = g(x-1)e^{\alpha_+ (x-1)} + g(x+1)e^{\alpha_+ (x+1)} \\= g(x) e^{\alpha_+ (x-1)} \left( 1 + e^{2 \alpha_+} \right) \\= g(x) e^{\alpha_+ (x-1)} \left( k e^{\alpha_+} \right) \\= k f(x) $$ A whole family of real, continuous solutions is for instance given by $$ f(x) = \cos(2 \pi k x + \delta) \left( e^{\alpha_+ x} + e^{\alpha_- x} \right) $$ for any constant $\delta$ and integer $k$. This also means that there is no unique answer unless additional restrictions to the solution are given.

As to solving these type of problems that depends a bit on whether it is a sequence of numbers or a problem on functions. For functional equations is often looking for fixed points $f(c)=c$ or symmetries such as $f(x)=-x$. Sometimes it is possible to take suitable derivatives or special combinations in multi-variable equations. Another approach is to make an expansion in the vicinity of a fixed point that might reveal a clue to the solution. In general I am inclined to say that these problems are rather tough and that there is no generic strategy that will work, but that it requires creativity to solve them and of course practice helps a lot.

Answer 2

If we write $\displaystyle F(x) = \sum_{n=0}^\infty f(n)x^n$, then we have

$$f(n-1) + f(n+1) = kf(n)$$ $$\sum_{n=0}^\infty f(n-1)x^n + \sum_{n=0}^\infty f(n+1)x^n = kF(n)$$ $$\sum_{n=-1}^\infty f(n)x^{n+1} + \sum_{n=1}^\infty f(n)x^{n-1} = kF(x)$$ $$f(-1)+\sum_{n=0}^\infty f(n)x^{n+1} -f(0)x^{-1}+\sum_{n=0}^\infty f(n)x^{n-1} = kF(x)$$

$$f(-1)+xF(x) -f(0)x^{-1}+ x^{-1}F(x) = kF(x)$$ $$f(-1) - f(0)x^{-1} = F(x)(k - x-x^{-1})$$ $$F(x) = \frac{f(0) - xf(-1)}{x^2 - kx +1}$$

If we use $U_n$ for Chebyshev polynomials of the second kind, we find that $\displaystyle\sum_{n=0}^\infty U_n(y/2)x^n = \frac{1}{x^2 -yx +1}$, and

$$F(x)= \Big(f(0) - xf(-1))\Big) \sum_{n=0}^\infty U_n(k/2)x^n$$

$$\sum_{n=0}^\infty f(n)x^n = \sum_{n=0}^\infty \Big [f(0)U_n(k/2) - f(-1) U_{n-1}(k/2)\Big ]x^n$$

$$f(n) = f(0)U_n(k/2) - f(-1) U_{n-1}(k/2)$$

Now with

$$U_n(x) = \frac{(x + \sqrt{x^2-1})^{n+1}- (x - \sqrt{x^2-1})^{n+1}}{2\sqrt{x^2-1}}$$

You have a (messy) closed form.