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5
1.2 The Solution and Its Properties
solution of (1.5) and (1.6) is given by
−t
u(t)= u 0 e .
This is easily checked by inspection;
u(0) = u 0 e = u 0 ,
0
and
−t
= −u(t).

u (t)= −u 0 e
Faced with a problem posed by a differential equation and some initial
or boundary conditions, we can generally check a solution candidate by
determining whether both the differential equation and the extra conditions
are satisfied. The tricky part is, of course, finding the candidate. 4
The motivation for studying differential equations is—to a very large
extent—their prominent use as models of various phenomena. Now, if (1.5)
is a model of some process, say the density of some population, then u 0
is a measure of the initial density. Since u 0 is a measured quantity, it is
only determined to a certain accuracy, and it is therefore important to
see if slightly different initial conditions give almost the same solutions. If
small perturbations of the initial condition imply small perturbations of
the solution, we have a stable problem. Otherwise, the problem is referred
to as unstable.
Let us consider the problem (1.5)–(1.6) with slightly perturbed initial
conditions,

v (t)= −v(t), (1.7)
v(0) = u 0 + , (1.8)
for some small . Then
−t
v(t)=(u 0 + )e ,
and
−t
|u(t) − v(t)| = | |e . (1.9)
We see that for this problem, a small change in the initial condition leads to
small changes in the solution. In fact, the difference between the solutions
is reduced at an exponential rate as t increases. This property is illustrated
in Fig. 1.1.

4
We will see later that it may also be difficult to check that a certain candidate is in
fact a solution. This is the case if, for example, the candidate is defined by an infinite
series. Then problems of convergence, existence of derivatives etc. must be considered
before a candidate can be accepted as a solution.