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I'm currently learning about linear inhomogenous difference equations and I completely understand why knowing a particular solution and all solutions of the appropriate homogenous equation is enough. However, I don't quite understand how we know that we have captured all the solutions of the homogenous equation. For example, consider the general form of second order linear homogenous difference equations:

$$u_{n+1} = au_n + bu_{n-1}$$

So, the auxiliary equation is $$\lambda^2 - a\lambda - b = 0$$ with roots $\lambda_1$, $\lambda_2$. Now I see that every function of the form $A\lambda_1^n + B\lambda_2^n$ is a solution to the given homogenous difference equation, assuming that $\lambda_1 \neq \lambda_2$. However, why is every solution of that form?

I know that the solutions form a vector space over the real numbers, but why is the dimension of the vector space equal to 2?

frgt
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    You know that if you fix the initial conditions, there exists a unique solution (since you are $C^\infty$). You have two ICs, thus at max 2 degrees of freedom. You already found 2 independent solutions to the general problem, so there cannot be more since these would imply the existence of more degrees of freedom/ICs to choose. – NeitherNor Nov 30 '20 at 21:15
  • @NeitherNor Now, I see that the vector space is just a subspace of $\mathbb{R}^2$, and so its dimension is obviously $\leq2$ and in fact equal to 2 since we've already found two independent solutions. However, is it possible to prove the same result without any linear algebra? And what do you exactly mean by IC? – frgt Nov 30 '20 at 21:49
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    IC=initial conditions. Basically, you have to fix 2 values to uniquely define your solution, e.g. $u_0$ and $u_1$. Assume you have 3 hom. solutions. Now, how can the 3 factors in front of them be uniquely determined if you only have 2 equations/IC? – NeitherNor Nov 30 '20 at 22:07
  • @NeitherNor Yes, I see your point, but it is not a formal proof and it seems that it couldn't be formalized without a tad of linear algebra. So, is there a way to prove that these are the only solutions with no linear algebra at all? – frgt Nov 30 '20 at 22:21
  • What exactly is not formal with this? If you know that you can parametrize all solutions by at most 2 parameters, and if you already found two independent solutions, you know that you are done. All I am saying is that if $x\leq 2$ and $x\geq 2$, then $x=2$. – NeitherNor Nov 30 '20 at 22:54
  • @NeitherNor How do you define independence? – frgt Dec 01 '20 at 09:31
  • A(x) and B(x) are independent if $a A(x)+ b B(x)=0$ implies a=b=0. – NeitherNor Dec 02 '20 at 14:41

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If we can solve these equations directly we will have a better understanding of the roles of the underlying characteristic equation.

By using the forward shift operator $\mathbb{E}$ such that $\mathbb{E} a_n=a_{n+1}, \mathbb{E}^2 a_n=a_{n+2}$, etc., a linear difference equation can then be written as $f(\mathbb{E})a_n=0$, where $f$ is a polynomial.

First order LDE: $(\mathbb{E} - r)a_n=0$. The solution is simply $a_n = r^n a_0.$

Second order LDE: $f(\mathbb{E})a_n=(\mathbb{E}-r)(\mathbb{E} - s)a_n=0$. From the first order case we have $$(\mathbb{E}-s)a_n = r^n (\mathbb{E}-s)a_0 \tag1$$ or equivalently

$$a_{n+1} - s a_n = r^n (a_1-s a_0)\tag 2$$ You can then solve this recursively (denote $C_1=a_1-sa_0$) $$a_{n}=C_1 r^{n-1}+sa_{n-1} = C_1 r^{n-1} + s(C_1 r^{n-2}+sa_{n-2}) = C_1 (r^{n-1}+sr^{n-2}) +s^2 a_{n-2} = \cdots \tag{A}$$ And that's why we see the $r^n$ and $s^n$ terms in our final solution. This can be tedious but fortunately we have a shortcut method to solve it much more quickly:

By symmetry $$(\mathbb{E}-r)a_n = s^n (\mathbb{E}-r)a_0 \tag3$$ Or equivalently $$a_{n+1} - r a_n = s^n (a_1-ra_0) \tag4$$

Subtract (4) from (2) you get $$a_n=\frac{a_1-s a_0}{r-s} r^n - \frac{a_1-r a_0}{r-s} s^n$$

You now see clearly that from first order to second order, a new geometric sequence component via the second root (assuming distinct roots) is "added" to the geometric sequence solution to the first order, regardless of the initial values. They always have the same form, the only differences are the coefficients which can be determined from the initial values.

We can apply this factoring technique repeatedly to a higher order LDE. There won't be a shortcut method like in the second order case but you can solve recursively as in $(A)$.

Another clean way to solve is via linear algebra. It's easy to show that big transition matrix has the same rank as the order of the LDE.

Neat Math
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