if h < 0. In case h > 0, the integrand is less than
t^{k+1}
|ϕ(t)|
e^{−st}≤ t^{k+1}Ce^{(λ−s)
t}, a function in L^{1} since s > λ. In the other case, for
|h|
small enough, the
integrand is dominated by t^{k+1}Ce^{(λ−(s+|h|))
t}. Letting
|h|
< ε where s−λ < ε, the integrand is dominated
by t^{k+1}Ce^{(λ−(s+ε))
t}< Ct^{k+1}e^{−εt}, also a function in L^{1}. By the dominated convergence theorem, one can
pass to the limit and obtain
∫ ∞
f(k+1)(s) = (− t)k+1e−stϕ(t)dt
0
Now consider the final claim about Res > λ. The difference quotient looks the same. However, s will
now be complex as will h. From the properties of the complex exponential, Section 1.9,
Now it follows from that section that the integrand converges to −te^{−st}ϕ
(t)
. Therefore, it only is required
to obtain an estimate and use the dominated convergence theorem again. Say h = re^{iθ}. Then the integrand
is
−(s+h)t1−-eht
e h ϕ(t)
Thus its magnitude satisfies
| | | | | |
|| −(s+h)t1−-eht || λt||1-−-eht|| −(s+h) −(s+h−λ)t||1-−-eht||
|e h ϕ(t)| ≤ Ce | h |e ≤ Cte | th |
Now e^{−}
(s+h−λ)
t = e^{−(Re(s+h)− λ)
t}≤ e^{−εt} whenever
|h|
is small enough because of the assumption that
Res > λ. e^{ht}− 1 = ∫_{0}^{t}he^{hu}du and so the last expression is
|∫ | |∫ |
||-t0 hehudu-|| 1 || t hu ||
|| th || ≤ t |0 e du|
There is ω,
|ω|
= 1,ω = e^{iα} and ω∫_{0}^{t}e^{hu}du =
||∫t hu ||
| 0 e du|
and so the above is no larger than
∫ ∫ ∫
1 t hu 1 t iαhu 1 t ( iα hu)
tω 0 e du = t 0 e e du = t 0 Re e e du
1∫ t| | 1 ∫ t
≤ t |ehu|du = t euRehdu
∫0t 0
≤ 1 eu|h|du ≤ et|h|
t 0
and so for
|h|
small enough, say smaller than ε∕2, the integrand is no larger than Cte^{−εt}e^{ε
2
t} = Cte^{−ε
2
t} which is
in L^{1} and so the dominated convergence theorem applies and it follows that f^{′}
(s)
= ∫_{0}^{∞}
(− t)
e^{−st}ϕ
(t)
dt
whenever Re
(s)
> λ. In fact all the derivatives will exist, by the same argument, but this will
follow more easily as a special case of more general results when we get around to using this.
■
The whole approach for Laplace transforms in differential equations is based on the assertion that if
ℒ
(f)
= ℒ
(g)
, then f = g. However, this is not even true because if you change the function on a
set of measure zero, you don’t change the transform. However, if f,g are continuous, then it
will be true. Actually, it is shown here that if ℒ
(f)
= 0, and f is continuous, then f = 0.
The approach here is based on the Weierstrass approximation theorem or rather a case of
it.
Lemma 10.2.3Suppose q is a continuous function defined on
[0,1]
. Also suppose that for alln = 0,1,2,
⋅⋅⋅
,
∫ 1
q(x)xndx = 0
0
Then it follows that q = 0.
Proof: By assumption, for p
(x)
any polynomial, ∫_{0}^{1}q
(x)
p
(x)
dx = 0. Now let
{pn(x)}
be a sequence
of polynomials which converge uniformly to q
du = 0. By fundamental theorem of calculus, at every t not a
jump, h
(t)
= 0. That is, f
(t)
= g
(t)
other than for t a point of discontinuity of either f or g.
■
With the Lebesgue fundamental theorem of calculus which has not been presented, the above argument
will show that it is only necessary to assume the functions are in L^{1} and conclude they are equal a.e. if
their Laplace transforms are equal.
∥ ∥
lim ∥pn ∘ l−1 − f ∥∞ = 0
n→ ∞
and it is clear that p_{n}∘ l^{−1} is a polynomial in x ∈
[a,b]
because l^{−1} is just a linear function like l.
■