Consider a function \(f\) which satisfies the differential equation
$$f^{(n)}=\sum_{j=0}^{n-1} a_{j} f^{(j)}.$$
for certain numbers \(a_{0}, \ldots, a_{n-1} .\) Several special cases have
already received detailed treatment, either in the text or in other problems;
in particular, we have found all functions satisfying \(f^{\prime}=f,\) or
\(f^{\prime \prime}+f=0,\) or \(f^{\prime \prime}-f=0 .\) The trick in Problem
18-42 enables us to find many solutions for such equations, but doesn't say
whether these are the only solutions. This requires a uniqueness result, which
will be supplied by this problem. At the end you will find some (necessarily
sketchy) remarks about the general solution.
(a) Derive the following formula for \(f^{(n+1)}\) (let us agree that "
\(^{\prime \prime} a_{-1} "\) will be 0 ):
$$f^{(n+1)}=\sum_{j=0}^{n-1}\left(a_{j-1}+a_{n-1} a_{j}\right) f^{(j)}.$$
(b) Deduce a formula for \(f^{(n+2)}\)
The formula in part (b) is not going to be used; it was inserted only to
convince you that a general formula for \(f^{(n+k)}\) is out of the question. On
the other hand, as part (c) shows, it is not very hard to obtain estimates on
the size of \(f^{(n+k)}(x)\).
(c) Let \(N=\max \left(1,\left|a_{0}\right|,
\ldots,\left|a_{n-1}\right|\right)\). Then \(\left|a_{j-1}+a_{n-1} a_{j}\right|
\leq 2 N^{2} ;\) this means that
$$f^{(n+1)}=\sum_{j=0}^{n-1} b_{j}^{1} f^{(j)}, \quad \text { where
}\left|b_{j}^{1}\right| \leq 2 N^{2}.$$
Show that
$$f^{(n+2)}=\sum_{j=0}^{n-1} b_{j}^{2} f^{(j)}, \quad \text { where
}\left|b_{j}^{2}\right| \leq 4 N^{3},$$
and, more generally,
$$f^{(n+k)}=\sum_{j=0}^{n-1} b_{j}^{k} f^{(j)}, \quad \text { where
}\left|b_{j}^{k}\right| \leq 2^{k} N^{k+1}.$$
(d) Conclude from part (c) that, for any particular number \(x\), there is a
number \(M\) such that
$$\left|f^{(n+k)}(x)\right| \leq M \cdot 2^{k} N^{k+1} \quad \text { for all }
k.$$
(e) Now suppose that \(f(0)=f^{\prime}(0)=\cdots=f^{(n-1)}(0)=0 .\) Show that
$$|f(x)| \leq \frac{M \cdot 2^{k+1} N^{k+2}|x|^{n+k+1}}{(n+k+1) !} \leq
\frac{M \cdot|2 N x|^{n+k+1}}{(n+k+1) !},$$
and conclude that \(f=0\).
(f) Show that if \(f_{1}\) and \(f_{2}\) are both solutions of the differential
equation
$$f^{(n)}=\sum_{j=0}^{n-1} a_{j} f^{(j)},$$
and \(f_{1}^{(j)}(0)=f_{2}^{(j)}(0)\) for \(0 \leq j \leq n-1,\) then
\(f_{1}=f_{2}\).
In other words, the solutions of this differential equation are determined by
the "initial conditions" (the values \(f^{(j)}(0)\) for \(0 \leq j \leq n-1\) ).
This means that we can find all solutions once we can find enough solutions to
obtain any given set of initial conditions. If the equation
$$x^{n}-a_{n-1} x^{n-1}-\cdots-a_{0}=0$$
has \(n\) distinct roots \(\alpha_{1}, \ldots, \alpha_{n},\) then any function of
the form
$$f(x)=c_{1} e^{\alpha_{1} x}+\cdots+c_{n} e^{\alpha_{n} x}$$
is a solution, and
$$\begin{aligned}
f(0) &=c_{1}+\cdots+c_{n} \\
f^{\prime}(0) &=\alpha_{1} c_{1}+\cdots+\alpha_{n} c_{n} \\
\vdots & \\
f^{(n-1)}(0) &=\alpha_{1}^{n-1} c_{1}+\cdots+\alpha_{n}^{n-1} c_{n}.
\end{aligned}$$
As a matter of fact, every solution is of this form, because we can obtain any
set of numbers on the left side by choosing the \(c\) 's properly, but we will
not try to prove this last assertion. (It is a purely algebraic fact, which
you can easily check for \(n=2\) or \(3 .\) ) These remarks are also true if some
of the roots are multiple roots, and even in the more general situation
considered in Chapter 27.
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