258
IX – Multivariate Differential and Integral Calculus
Δ n+2 (t) ≤ rA|t| + A
I(t)
du
rA|u| + A
I(u)
Δ n (v)dv
= r
A|t| + A
2
|t|
2 /2!
+ A
2
I(t)
du 1
I(u1)
Δ n (u 2 )du 2 ,
and so on. As Δ n (u) ≤ Δ n (t) for u ∈ I(t), it follows that
I(t)
du 1
I(u1)
du 2 . . .
I(up−1)
Δ n (u p )du p ≤
≤ Δ n (t)
I(t)
du 1
I(u1)
du 2 . . .
I(up−1)
du p = Δ n (t) (|t|)
p /p! .
Hence iterating (18) gives
Δ n+p (t) ≤ r (A|t| + . . . + A
p
|t|
p /p!) + Δ n (t)A
p
|t|
p /p!
≤ r [exp (Aa
) − 1] + Δ n (t) (Aa
)
p /p!
for n ≥ N (r), p ≥ 1 and |t| ≤ a
. (The reader will recognize these to be
arguments given by Liouville already outlined in Chap. VII, n
◦ 18). On the
right hand side, the first term is arbitrarily small if r is conveniently chosen,
and the second one tends to 0 as p increases. Hence lim Δ n (t) = 0 for all t
and in particular for t = ±a
. So D 3 y n (t, z) is uniformly convergent. Hence,
the function y(t, z) is indeed C
1 .
It remains to show that it is C
k if L is C
k . For k = 2, the right hand side
of the differential equation
Dy(t, z) = L [t, y(t, z), z] , y(0, z) = 0
is C
1 like L and y, so that D
2 y(t, z) and D 3 D 2 y(t, z) exist and are continuous.
On the other hand, D 3 y(t, z) = Y (t, z) satisfies (16), i.e.
DY (t, z) = U (t, z)Y (t, z) + V (t, z) , Y (0, z) = 0 .
(15.19)
As U (t, z) and V (t, z) are C
1 by (15) if L is C
2 and if y is C
1 , solution
Y (t, z) of (19) is C
1 . The function y(t, z) is, therefore, C
2 . And so on, which
finishes the proof of statements (a), (b) and (c). I give up turning them into
a theorem whose statement would take half a page.
(v) Matrix exponential. As seen above, the derivative Y (t, z) = D 3 y(t, z)
satisfies a differential equation (19) whose right hand side is an affine linear
function of Y . Equations of the form
x
(t) = A(t)x(t) + b(t) , x(0) = ξ ,
where x(t), b(t) ∈ R
n and A(t) ∈ M n (R), are dealt with by using the general
method, but there is an additional result in this case: the integrals are defined
on every interval on which A(t) and b(t) are defined and continuous.
IX – Multivariate Differential and Integral Calculus
Δ n+2 (t) ≤ rA|t| + A
I(t)
du
rA|u| + A
I(u)
Δ n (v)dv
= r
A|t| + A
2
|t|
2 /2!
+ A
2
I(t)
du 1
I(u1)
Δ n (u 2 )du 2 ,
and so on. As Δ n (u) ≤ Δ n (t) for u ∈ I(t), it follows that
I(t)
du 1
I(u1)
du 2 . . .
I(up−1)
Δ n (u p )du p ≤
≤ Δ n (t)
I(t)
du 1
I(u1)
du 2 . . .
I(up−1)
du p = Δ n (t) (|t|)
p /p! .
Hence iterating (18) gives
Δ n+p (t) ≤ r (A|t| + . . . + A
p
|t|
p /p!) + Δ n (t)A
p
|t|
p /p!
≤ r [exp (Aa
) − 1] + Δ n (t) (Aa
)
p /p!
for n ≥ N (r), p ≥ 1 and |t| ≤ a
. (The reader will recognize these to be
arguments given by Liouville already outlined in Chap. VII, n
◦ 18). On the
right hand side, the first term is arbitrarily small if r is conveniently chosen,
and the second one tends to 0 as p increases. Hence lim Δ n (t) = 0 for all t
and in particular for t = ±a
. So D 3 y n (t, z) is uniformly convergent. Hence,
the function y(t, z) is indeed C
1 .
It remains to show that it is C
k if L is C
k . For k = 2, the right hand side
of the differential equation
Dy(t, z) = L [t, y(t, z), z] , y(0, z) = 0
is C
1 like L and y, so that D
2 y(t, z) and D 3 D 2 y(t, z) exist and are continuous.
On the other hand, D 3 y(t, z) = Y (t, z) satisfies (16), i.e.
DY (t, z) = U (t, z)Y (t, z) + V (t, z) , Y (0, z) = 0 .
(15.19)
As U (t, z) and V (t, z) are C
1 by (15) if L is C
2 and if y is C
1 , solution
Y (t, z) of (19) is C
1 . The function y(t, z) is, therefore, C
2 . And so on, which
finishes the proof of statements (a), (b) and (c). I give up turning them into
a theorem whose statement would take half a page.
(v) Matrix exponential. As seen above, the derivative Y (t, z) = D 3 y(t, z)
satisfies a differential equation (19) whose right hand side is an affine linear
function of Y . Equations of the form
x
(t) = A(t)x(t) + b(t) , x(0) = ξ ,
where x(t), b(t) ∈ R
n and A(t) ∈ M n (R), are dealt with by using the general
method, but there is an additional result in this case: the integrals are defined
on every interval on which A(t) and b(t) are defined and continuous.