# Uniqueness result in linear differential equation of degree $n$.

Suppose that $f$ is such that

$$f^{(n)}=\sum_{j=0}^{n-1}a_jf^{(j)}$$

Some little work is needed to get to ($a_j=0$ if $j<0$)

$${f^{(n + 1)}} = \sum\limits_{j = 0}^{n - 1} {\left( {{a_{j - 1}} + {a_{n - 1}}{a_j}} \right)} {f^{(j)}}$$

and $${f^{(n + 2)}} = \sum\limits_{j = 0}^{n - 1} {\left( {{a_{j - 2}} + {a_{n - 1}}{a_{j - 1}} + {a_{n - 2}}a{ _j} + {a_j}a_{n - 1}^2} \right)} {f^{(j)}}$$

This evidences the increased difficulty in finding a closed form for $f^{n+k}$. However, we can prove by induction that -setting $N=\max(1,|a_0|,\dots,|a_{n-1}|)$ - we have

$${f^{(n + k)}} = \sum\limits_{j = 0}^{n - 1} {{b_{jk}}} {f^{(j)}}$$

with each $b_{jk}\leq 2^kN^{k+1}$. Just as an example:

$$\left| {{a_{j - 1}} + {a_{n - 1}}{a_j}} \right| \leqslant \left| {{a_{j - 1}}} \right| \cdot 1 + \left| {{a_{n - 1}}} \right|\left| {{a_j}} \right| \leqslant N \cdot N + N \cdot N = 2{N^2}$$

\eqalign{ & \left| {{a_{j - 2}} + {a_{n - 1}}{a_{j - 1}} + {a_{n - 2}}{a_j} + {a_j}a_{n - 1}^2} \right| \leqslant \cr & \left| {{a_{j - 2}}} \right| \cdot 1 \cdot 1 + \left| {{a_{n - 1}}} \right|\left| {{a_{j - 1}}} \right| \cdot 1 + \left| {{a_{n - 2}}} \right|\left| {{a_j}} \right| \cdot 1 + \left| {{a_j}} \right|\left| {a_{n - 1}^2} \right| \leqslant \cr & {N^3} + {N^3} + {N^3} + {N^3} = 4{N^3} \cr}

Now, I need to show that for each particular $x$ there exists some $M$ such that, $$\left| {{f^{(n + k)}}\left( x \right)} \right| \leqslant {2^k}{N^{k + 1}}M$$

for each $k$.

This with some linear algebra establishes a uniqueness theorem and provides the general solution to this equations.

Could you hint me so I can finish this?

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If I understood it correct, for it it seems that the bound $M$ is what only matters in your proof. Am I going right? – Sangchul Lee Oct 24 '12 at 1:49
Yes. I need to find $M$. – Pedro Tamaroff Oct 24 '12 at 2:36
You may restrict your attention to $[-R, R]$ for $R > 0$. Since $M(x)$ is continuous, $M = \sup_{|x| \leq R} M(x)$ works, yielding $f \equiv 0$ on $[-R, R]$. Since $R$ is arbitrary, the conclusion follows. – Sangchul Lee Oct 24 '12 at 2:48
@sos440 My bad, after a lot of time I re-read the exercise and see it says "prove for each $x$ there exists an $M$ such that...". So we have $\forall x\exists M_x$ not $\exists M\forall x$... The textbook says, "Deduce (...) that for each particular $x$ there exists a number $M$ such that $$\left| {{f^{(n + k)}}\left( x \right)} \right| \leqslant {2^k}{N^{k + 1}}M$$ for all $k$." – Pedro Tamaroff Apr 14 '13 at 1:02
I'm voting to close because the OP noted in a comment that the problem was just copied incorrectly. :) (It happens to the best of us, doesn't it?) – apnorton Nov 8 '13 at 19:30