Podemos escribir toda ecuación diferencial de orden
n
{\displaystyle n}
1
{\displaystyle 1}
n
{\displaystyle n}
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{\displaystyle x_{i}=x^{(i-1)}}
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n
{\displaystyle i=1,\dots ,n}
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{\displaystyle x_{i}=x'_{i-1}}
Soluciones de la EDO x'-x+t=0 Además,
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{\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:
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{\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
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{\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
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r
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{\displaystyle b={\begin{pmatrix}0\\\vdots \\0\\r(t)\\\end{pmatrix}}}
r
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t
)
{\displaystyle r(t)}
Si expresamos
f
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x
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x
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b
{\displaystyle f(t,{\textbf {x}})=A(t){\textbf {x}}+b}
x
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1
⋮
x
n
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{\displaystyle {\textbf {x}}={\begin{pmatrix}x_{1}\\\vdots \\x_{n}\\\end{pmatrix}}}
x
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f
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t
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{\displaystyle x'=f(t,{\textbf {x}})}
f
{\displaystyle f}
Puesto que
f
{\displaystyle f}
Teorema de Picard-Lindelöf , faltaría comprobar que es Lipschitz respecto a
x
{\displaystyle x}
t
{\displaystyle t}
norma euclídea de un vector, denotada como
‖
v
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∑
j
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1
n
v
j
2
{\displaystyle \|{\textbf {v}}\|={\sqrt {\sum _{j=1}^{n}v_{j}^{2}}}}
v
j
{\displaystyle v_{j}}
j
=
1
,
.
.
.
,
n
{\displaystyle j=1,...,n}
v
{\displaystyle {\textbf {v}}}
x
,
x
^
∈
R
n
{\displaystyle {\textbf {x}},{\hat {\textbf {x}}}\in \mathbb {R} ^{n}}
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{\displaystyle \|f(t,{\hat {\textbf {x}}})-f(t,{\textbf {x}})\|=\|A(t){\hat {\textbf {x}}}-A(t){\textbf {x}}\|={\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
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max
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0
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−
1
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[
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{\displaystyle M=\max _{i\in [0,{n-1}],t\in [a,b]}\left\vert a_{i}(t)\right\vert }
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{\displaystyle {\begin{aligned}{\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}\\0\\\end{pmatrix}}\end{Vmatrix}}+{\begin{Vmatrix}{\begin{pmatrix}0\\\vdots \\0\\\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}\\0\\\end{pmatrix}}\end{Vmatrix}}+\sum _{j=1}^{n}{\begin{Vmatrix}{\begin{pmatrix}0\\\vdots \\0\\-a_{j-1}({\hat {x}}_{j}-x_{j})\\\end{pmatrix}}\end{Vmatrix}}\\&\leq \|{\hat {\textbf {x}}}-{\textbf {x}}\|+\sum _{j=1}^{n}|a_{j-1}||{\hat {x}}_{j}-x_{j}|\leq \|{\hat {\textbf {x}}}-{\textbf {x}}\|+\sum _{j=1}^{n}M|{\hat {x}}_{j}-x_{j}|\leq (1+nM)\|{\hat {\textbf {x}}}-{\textbf {x}}\|.\end{aligned}}}
Obtenemos
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≤
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{\displaystyle \|f(t,{\hat {\textbf {x}}})-f(t,{\textbf {x}})\|\leq (1+nM)\|{\hat {\textbf {x}}}-{\textbf {x}}\|}
por lo que
f
{\displaystyle f}
constante de Lipschitz
K
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{\displaystyle K=(1+nM)}
∀
t
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⊂
R
{\displaystyle \forall t\in [a,b]\subset \mathbb {R} }
Puesto que se cumplen las hipótesis del Teorema de Picard-Lindelöf , queda demostrado que hay una única solución
x
{\displaystyle {\textbf {x}}}
x
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{\displaystyle x_{i}(t_{0})=\xi _{i}}
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n
{\displaystyle i=1,\dots ,n}
t
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{\displaystyle t_{0}\in [a,b]}
Si
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{\displaystyle {\textbf {x}}\in C^{1}([a,b];\mathbb {R} ^{n})}
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{\displaystyle {\begin{aligned}&x_{n-1}'=x_{n},x_{n}\in C^{1}([a,b];\mathbb {R} )&&\implies x_{n-1}\in C^{2}([a,b];\mathbb {R} ),\\&x_{n-2}'=x_{n-1},x_{n-1}\in C^{2}([a,b];\mathbb {R} )&&\implies x_{n-2}\in C^{3}([a,b];\mathbb {R} ),\\&\vdots &&\\&x_{1}'=x_{2},x_{2}\in C^{n-1}([a,b];\mathbb {R} )&&\implies x_{1}\in C^{n}([a,b];\mathbb {R} ),\end{aligned}}}
implicando que
x
1
{\displaystyle x_{1}}
n
{\displaystyle n}
x
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{\displaystyle x=x_{1}}
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{\displaystyle \;x_{1}^{(n)}=x_{n}'\in C([a,b];\mathbb {R} )\implies x=\in C^{n}([a,b];\mathbb {R} )\implies x^{(n)}=r(t)-\sum _{i=1}^{n}a_{i-1}(t)x^{(i-1)}(t)}
Sea el problema de valores iniciales:
{
x
(
4
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+
3
x
″
−
sin
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+
8
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‴
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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
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4
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−
8
x
+
sin
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′
−
3
x
″
+
t
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=
−
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x
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+
sin
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,
{\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:
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{\displaystyle x_{1}(0)=1,\ x_{2}(0)=2,\ x_{3}(0)=3\,x_{4}(0)=4.}
Sea
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=
(
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,
{\displaystyle {\textbf {x}}={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}},}
x
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{\displaystyle {\textbf {x}}'=A{\textbf {x}}+b}
A
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sin
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{\displaystyle A={\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\-8&\sin(t)&-3&0\end{pmatrix}}}
b
=
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{\displaystyle b={\begin{pmatrix}0\\0\\0\\t^{2}\end{pmatrix}}}
x
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{\displaystyle {\textbf {x}}(0)={\begin{pmatrix}1\\2\\3\\4\end{pmatrix}}}
Ahora, aplicando el método anteriormente descrito:
Sea
M
=
max
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∈
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t
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=
8
{\displaystyle M=\max _{i\in [0,{n-1}]t\in [a,b]}\left\vert a_{i}(t)\right\vert =8}
En este caso
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{\displaystyle \|f(t,{\hat {\textbf {x}}})-f(t,{\textbf {x}})\|=\|A(t){\hat {\textbf {x}}}-A(t){\textbf {x}}\|}
≤
33
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^
−
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{\displaystyle \;\leq 33\|{\hat {\textbf {x}}}-{\textbf {x}}\|}
![{\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/9d4741728a82c34c9c7163ee9affa8528bcef8af)
![{\displaystyle \forall t\in [a,b]\subset \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f5d4d3130c111caafb357290b8306514a64781f)
![{\displaystyle t_{0}\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25c9abe498a595aa49b749c92b744e127800c140)
![{\displaystyle {\textbf {x}}\in C^{1}([a,b];\mathbb {R} ^{n})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/648f5b3ac9947b59161d98c5504ebb7a7cae3a6f)
![{\displaystyle {\begin{aligned}&x_{n-1}'=x_{n},x_{n}\in C^{1}([a,b];\mathbb {R} )&&\implies x_{n-1}\in C^{2}([a,b];\mathbb {R} ),\\&x_{n-2}'=x_{n-1},x_{n-1}\in C^{2}([a,b];\mathbb {R} )&&\implies x_{n-2}\in C^{3}([a,b];\mathbb {R} ),\\&\vdots &&\\&x_{1}'=x_{2},x_{2}\in C^{n-1}([a,b];\mathbb {R} )&&\implies x_{1}\in C^{n}([a,b];\mathbb {R} ),\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/872df04df2f944804f411d8d98446f72a5238355)
![{\displaystyle \;x_{1}^{(n)}=x_{n}'\in C([a,b];\mathbb {R} )\implies x=\in C^{n}([a,b];\mathbb {R} )\implies x^{(n)}=r(t)-\sum _{i=1}^{n}a_{i-1}(t)x^{(i-1)}(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2046460fd6cb465e5f32fb725e171d65d2f8d741)
![{\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/4ab90fd3e8a19d58ef5e5ffee09f4079f13f5a8e)
