1. Podemos escribir toda ecuación diferencial de orden
n
{\displaystyle n}
1
{\displaystyle 1}
n
{\displaystyle n}
x
i
=
x
(
i
−
1
)
{\displaystyle x_{i}=x^{(i-1)}}
i
=
1
,
…
,
n
{\displaystyle i=1,\dots ,n}
x
i
=
x
i
−
1
′
{\displaystyle x_{i}=x'_{i-1}}
Soluciones de la EDO x'-x+t=0 Además,
x
n
=
x
(
n
−
1
)
⟹
x
n
′
=
x
(
n
)
=
r
(
t
)
−
∑
i
=
1
n
a
i
−
1
(
t
)
x
i
(
t
)
.
{\displaystyle x_{n}=x^{(n-1)}\Longrightarrow x'_{n}=x^{(n)}=r(t)-\sum _{i=1}^{n}a_{i-1}(t)x_{i}(t).}
Escrito en forma matricial:
(
x
1
′
x
2
′
⋮
x
n
−
1
′
x
n
′
)
=
A
(
t
)
(
x
1
x
2
⋮
x
n
−
1
x
n
)
+
b
{\displaystyle {\begin{pmatrix}x'_{1}\\x'_{2}\\\vdots \\x'_{n-1}\\x'_{n}\\\end{pmatrix}}=A(t){\begin{pmatrix}x_{1}\\x_{2}\\\vdots \\x_{n-1}\\x_{n}\\\end{pmatrix}}+b}
A
(
t
)
=
(
0
1
0
0
…
0
0
0
1
0
…
0
⋮
⋮
⋮
⋮
⋱
⋮
0
0
0
…
0
1
−
a
0
(
t
)
−
a
1
(
t
)
−
a
2
(
t
)
…
−
a
n
−
2
(
t
)
−
a
n
−
1
(
t
)
)
{\displaystyle A(t)={\begin{pmatrix}0&1&0&0&\dots &0\\0&0&1&0&\dots &0\\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\dots &0&1\\-a_{0}(t)&-a_{1}(t)&-a_{2}(t)&\dots &-a_{n-2}(t)&-a_{n-1}(t)\\\end{pmatrix}}}
n
×
n
{\displaystyle n\times n}
b
=
(
0
⋮
0
r
(
t
)
)
{\displaystyle b={\begin{pmatrix}0\\\vdots \\0\\r(t)\\\end{pmatrix}}}
r
(
t
)
{\displaystyle r(t)}
Si definimos
f
(
t
,
x
)
=
A
(
t
)
x
→
+
b
{\displaystyle f(t,x)=A(t){\vec {x}}+b}
x
→
′
=
f
(
t
,
x
)
{\displaystyle {\vec {x}}'=f(t,x)}
f
{\displaystyle f}
2. Sea
x
,
x
^
∈
C
(
[
a
,
b
]
;
R
n
)
{\displaystyle x,{\hat {x}}\in C([a,b];\mathbb {R} ^{n})}
f
{\displaystyle f}
Teorema de Picard-Lindelöf solo falta comprobar que es Lipschitz respecto a
x
{\displaystyle x}
t
{\displaystyle t}
norma euclídea de un vector, denotada como
‖
v
→
‖
=
∑
j
=
1
n
v
j
2
{\displaystyle \|{\vec {v}}\|={\sqrt {\sum _{j=1}^{n}v_{j}^{2}}}}
v
j
{\displaystyle v_{j}}
j
=
1
,
.
.
.
,
n
{\displaystyle j=1,...,n}
v
{\displaystyle v}
‖
f
(
t
,
x
^
)
−
f
(
t
,
x
)
‖
=
‖
A
(
t
)
x
^
→
+
b
−
A
(
t
)
x
→
−
b
‖
=
‖
(
x
^
2
⋮
x
^
n
∑
j
=
1
n
−
a
j
−
1
x
^
j
)
−
(
x
2
⋮
x
n
∑
j
=
1
n
−
a
j
−
1
x
j
)
‖
=
‖
(
x
^
2
−
x
2
⋮
x
^
n
−
x
n
∑
j
=
1
n
−
a
j
−
1
(
x
^
j
−
x
j
)
)
‖
{\displaystyle \|f(t,{\hat {x}})-f(t,x)\|=\|A(t){\vec {\hat {x}}}+b-A(t){\vec {x}}-b\|={\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}\\\vdots \\{\hat {x}}_{n}\\\sum _{j=1}^{n}-a_{j-1}{\hat {x}}_{j}\\\end{pmatrix}}-{\begin{pmatrix}x_{2}\\\vdots \\x_{n}\\\sum _{j=1}^{n}-a_{j-1}x_{j}\\\end{pmatrix}}\end{Vmatrix}}={\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}-x_{2}\\\vdots \\{\hat {x}}_{n}-x_{n}\\\sum _{j=1}^{n}-a_{j-1}({\hat {x}}_{j}-x_{j})\\\end{pmatrix}}\end{Vmatrix}}}
Sea
M
=
m
a
x
i
∈
[
0
,
n
−
1
]
,
t
∈
[
a
,
b
]
|
a
i
(
t
)
|
{\displaystyle M=max_{i\in [0,{n-1}],t\in [a,b]}\left\vert a_{i}(t)\right\vert }
‖
(
x
^
2
−
x
2
⋮
x
^
n
−
x
n
∑
j
=
1
n
−
a
j
−
1
(
x
^
j
−
x
j
)
)
‖
≤
‖
(
x
^
2
−
x
2
⋮
x
^
n
−
x
n
M
∑
j
=
1
n
(
x
^
j
−
x
j
)
)
‖
{\displaystyle {\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}-x_{2}\\\vdots \\{\hat {x}}_{n}-x_{n}\\\sum _{j=1}^{n}-a_{j-1}({\hat {x}}_{j}-x_{j})\\\end{pmatrix}}\end{Vmatrix}}\leq {\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}-x_{2}\\\vdots \\{\hat {x}}_{n}-x_{n}\\M\sum _{j=1}^{n}({\hat {x}}_{j}-x_{j})\\\end{pmatrix}}\end{Vmatrix}}}
Por la desigualdad triangular de la norma euclídea,
‖
(
x
^
2
−
x
2
⋮
x
^
n
−
x
n
M
∑
j
=
1
n
(
x
^
j
−
x
j
)
)
‖
≤
‖
(
x
^
2
−
x
2
⋮
x
^
n
−
x
n
0
)
‖
+
‖
(
0
⋮
0
M
∑
j
=
1
n
x
^
j
−
x
j
)
‖
=
‖
(
x
^
2
−
x
2
⋮
x
^
n
−
x
n
0
)
‖
+
M
‖
(
0
⋮
0
∑
j
=
1
n
x
^
j
−
x
j
)
‖
{\displaystyle {\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}-x_{2}\\\vdots \\{\hat {x}}_{n}-x_{n}\\M\sum _{j=1}^{n}({\hat {x}}_{j}-x_{j})\\\end{pmatrix}}\end{Vmatrix}}\leq {\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}-x_{2}\\\vdots \\{\hat {x}}_{n}-x_{n}\\0\\\end{pmatrix}}\end{Vmatrix}}+{\begin{Vmatrix}{\begin{pmatrix}0\\\vdots \\0\\M\sum _{j=1}^{n}{\hat {x}}_{j}-x_{j}\\\end{pmatrix}}\end{Vmatrix}}={\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}-x_{2}\\\vdots \\{\hat {x}}_{n}-x_{n}\\0\\\end{pmatrix}}\end{Vmatrix}}+M{\begin{Vmatrix}{\begin{pmatrix}0\\\vdots \\0\\\sum _{j=1}^{n}{\hat {x}}_{j}-x_{j}\\\end{pmatrix}}\end{Vmatrix}}}
‖
(
x
^
2
−
x
2
⋮
x
^
n
−
x
n
0
)
‖
+
M
‖
(
0
⋮
0
∑
j
=
1
n
x
^
j
−
x
j
)
‖
≤
‖
(
x
^
2
−
x
2
⋮
x
^
n
−
x
n
0
)
‖
+
M
∑
j
=
1
n
‖
(
0
⋮
0
x
^
j
−
x
j
)
‖
≤
(
1
+
n
M
)
‖
x
^
−
x
‖
{\displaystyle {\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}-x_{2}\\\vdots \\{\hat {x}}_{n}-x_{n}\\0\\\end{pmatrix}}\end{Vmatrix}}+M{\begin{Vmatrix}{\begin{pmatrix}0\\\vdots \\0\\\sum _{j=1}^{n}{\hat {x}}_{j}-x_{j}\\\end{pmatrix}}\end{Vmatrix}}\leq {\begin{Vmatrix}{\begin{pmatrix}{\hat {x}}_{2}-x_{2}\\\vdots \\{\hat {x}}_{n}-x_{n}\\0\\\end{pmatrix}}\end{Vmatrix}}+M\sum _{j=1}^{n}{\begin{Vmatrix}{\begin{pmatrix}0\\\vdots \\0\\{\hat {x}}_{j}-x_{j}\\\end{pmatrix}}\end{Vmatrix}}\leq (1+nM)\|{\hat {x}}-x\|}
Obtenemos
‖
f
(
t
,
x
^
)
−
f
(
t
,
x
)
‖
≤
(
1
+
n
M
)
‖
x
^
−
x
‖
{\displaystyle \|f(t,{\hat {x}})-f(t,x)\|\leq (1+nM)\|{\hat {x}}-x\|}
Por lo que
f
{\displaystyle f}
constante de Lipschitz
K
=
(
1
+
n
M
)
{\displaystyle K=(1+nM)}
∀
t
∈
[
a
,
b
]
⊂
R
{\displaystyle \forall t\in [a,b]\subset \mathbb {R} }
3. Finalmente, tras haber comprobado que la primera componente,
x
{\displaystyle x}
n
{\displaystyle n}
Teorema de Picard-Lindelöf , por lo que queda demostrado que el problema tiene existencia y unicidad de soluciones dadas unas condiciones iniciales.
Sea el problema de valores iniciales:
{
x
(
4
)
+
3
x
″
−
sin
(
t
)
x
′
+
8
x
=
t
2
,
x
(
0
)
=
0
,
x
′
(
0
)
=
2
,
x
″
(
0
)
=
3
,
x
‴
(
0
)
=
4.
{\displaystyle {\begin{cases}x^{(4)}+3x''-\sin(t)x'+8x=t^{2},\\x(0)=0,x'(0)=2,x''(0)=3,x'''(0)=4.\end{cases}}}
Vamos a transformarlo en un sistema de ecuaciones:
{
x
1
=
x
⟹
x
1
′
=
x
′
=
x
2
,
x
2
=
x
′
⟹
x
2
′
=
x
″
=
x
3
,
x
3
=
x
″
⟹
x
3
′
=
x
‴
=
x
4
,
x
4
=
x
‴
⟹
x
4
′
=
x
(
4
)
=
−
8
x
+
sin
(
t
)
x
′
−
3
x
″
+
t
2
=
−
8
x
1
+
sin
(
t
)
x
2
−
3
x
3
+
t
2
,
{\displaystyle {\begin{cases}x_{1}=x\,\Longrightarrow \ x'_{1}=x'=x_{2},\\x_{2}=x'\Longrightarrow x'_{2}=x''=x_{3},\\x_{3}=x''\Longrightarrow x'_{3}=x'''=x_{4},\\x_{4}=x'''\Longrightarrow x'_{4}=x^{(4)}=-8x+\sin(t)x'-3x''+t^{2}=-8x_{1}+\sin(t)x_{2}-3x_{3}+t^{2},\end{cases}}}
con condiciones iniciales:
x
1
(
0
)
=
1
,
x
2
(
0
)
=
2
,
x
3
(
0
)
=
3
x
4
(
0
)
=
4.
{\displaystyle x_{1}(0)=1,\ x_{2}(0)=2,\ x_{3}(0)=3\,x_{4}(0)=4.}
Sea
x
→
=
(
x
1
x
2
x
3
x
4
)
,
{\displaystyle {\vec {x}}={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}},}
x
→
′
=
A
x
→
+
b
{\displaystyle {\vec {x}}'=A{\vec {x}}+b}
A
=
(
0
1
0
0
0
0
1
0
0
0
0
1
−
8
sin
(
t
)
−
3
0
)
{\displaystyle A={\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\-8&\sin(t)&-3&0\end{pmatrix}}}
b
=
(
0
0
0
t
2
)
{\displaystyle b={\begin{pmatrix}0\\0\\0\\t^{2}\end{pmatrix}}}
x
→
(
0
)
=
(
1
2
3
4
)
{\displaystyle {\vec {x}}(0)={\begin{pmatrix}1\\2\\3\\4\end{pmatrix}}}
Ahora, aplicando el método anteriormente descrito:
Sea
M
=
m
a
x
i
∈
[
0
,
n
−
1
]
,
t
∈
[
a
,
b
]
|
a
i
(
t
)
|
=
8
{\displaystyle M=max_{i\in [0,{n-1}],t\in [a,b]}\left\vert a_{i}(t)\right\vert =8}
En este caso
‖
f
(
t
,
x
^
)
−
f
(
t
,
x
)
‖
=
‖
A
(
t
)
x
^
→
−
A
(
t
)
x
→
‖
{\displaystyle \|f(t,{\hat {x}})-f(t,x)\|=\|A(t){\vec {\hat {x}}}-A(t){\vec {x}}\|}
≤
33
‖
x
^
−
x
‖
{\displaystyle \;\leq 33\|{\hat {x}}-x\|}
Borrador versión 2, por Daniel Ríos Palencia, Asier Fernández Arrazola, Wen Jie Gou Zhou y Lucía Ulecia Rivas.
![{\displaystyle x,{\hat {x}}\in C([a,b];\mathbb {R} ^{n})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cdf836861e3ccdc008eccd3a827fe24e6ab54da)
![{\displaystyle M=max_{i\in [0,{n-1}],t\in [a,b]}\left\vert a_{i}(t)\right\vert }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b872d73c9ae537062ded759279b7ec2023708e7)
![{\displaystyle \forall t\in [a,b]\subset \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f5d4d3130c111caafb357290b8306514a64781f)
![{\displaystyle M=max_{i\in [0,{n-1}],t\in [a,b]}\left\vert a_{i}(t)\right\vert =8}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32c72bd6d45a93a61f6ad69c55cb93d1bfc1ae38)
