Expand by Taylor in to get , with the obvious choices. Invoking we obtain and the result follows by using Gronwall's inequality.
Dear MHF members,
consider the PDE IVP
for
under the initial condition
for , where is a function of exponential order,
i.e., there exists such that for all .
Prove that for each fixed , the solution is of exponential order, i.e.,
there exists such that for all .
Note. Do not use the solution directly to show it.
Rebesques thanks for the reply, but it seems that there is a mistake with your Taylor's formula, it should read as follows.
.
However, as the function has first order derivatives, we can not expand it up to its second-order derivative, i.e., we can only use
..........._ .
But I dont know if we can move somewhere from here...
If has the series expansion
satisfying for all ,
where , then is of exponential order.
One can show by direct substitution that
,
which yields
for all .
Also if is a nonnegative sequence, then we obtain
for all ,
which implies is of exponential order too.
But I dont know how we can show this only with .