Can (linear) differential equations of infinite order be recast into equations of first order?












10














In most analysis courses one sees that differential equations of order $n$ are basically a subset of higher dimensional differential equations of order $1$, for example the equation:



$$f^{(n)}(t)=Fleft(f(t),f'(t),...,f^{(n-1)}(t),tright)$$



Is the same as:
$$frac{d}{dt},begin{pmatrix}g_0(t)\g_1(t)\vdots\g_{n-1(t)}end{pmatrix}=begin{pmatrix}g_1(t)\vdots\g_{n-1}(t)\Fleft(g_0(t),g_1(t),...,g_{n-1}(t),tright)end{pmatrix}$$



This is especially useful, as it allows one to write down explicitly the solutions to linear differential equations of finite order, if we have:
$$f^{(n)}(t)=sum_{k=0}^{n-1}a_k, f^{(k)}(t)$$
Then the corresponding $n$-dimensional equation is of the form:
$$frac{d}{dt} g(t) = Acdot g(t)$$
for some matrix $A$ and the solution is $g(t)=exp(A,t)g(0)$. It is possible to generalise to time dependent coefficients $a_k$.



Is there a way to implement this trick for differential equations that are essentially of infinite order? For example the equation
$$f=sum_{k=1}^infty f^{(k)}(t),$$
of which the solution space is $f={Cexp(frac t2)mid Cinmathbb R$ (or $mathbb C$)$}$. More generally I would like to put something of the form
$$sum_{k=0}^infty a_k, f^{(k)}(t)=0$$
(where $a_k$ are as regular as needed (but with infinite non-zero terms)) into the form
$$frac{d}{dt} u = A(u)$$
Where $u$ is a map $C^infty(mathbb R,X)$ with $X$ a Banach space and $Ain mathcal L(X)$.










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  • I'm still hoping someone to answer this question :)
    – onurcanbektas
    Feb 13 at 10:17


















10














In most analysis courses one sees that differential equations of order $n$ are basically a subset of higher dimensional differential equations of order $1$, for example the equation:



$$f^{(n)}(t)=Fleft(f(t),f'(t),...,f^{(n-1)}(t),tright)$$



Is the same as:
$$frac{d}{dt},begin{pmatrix}g_0(t)\g_1(t)\vdots\g_{n-1(t)}end{pmatrix}=begin{pmatrix}g_1(t)\vdots\g_{n-1}(t)\Fleft(g_0(t),g_1(t),...,g_{n-1}(t),tright)end{pmatrix}$$



This is especially useful, as it allows one to write down explicitly the solutions to linear differential equations of finite order, if we have:
$$f^{(n)}(t)=sum_{k=0}^{n-1}a_k, f^{(k)}(t)$$
Then the corresponding $n$-dimensional equation is of the form:
$$frac{d}{dt} g(t) = Acdot g(t)$$
for some matrix $A$ and the solution is $g(t)=exp(A,t)g(0)$. It is possible to generalise to time dependent coefficients $a_k$.



Is there a way to implement this trick for differential equations that are essentially of infinite order? For example the equation
$$f=sum_{k=1}^infty f^{(k)}(t),$$
of which the solution space is $f={Cexp(frac t2)mid Cinmathbb R$ (or $mathbb C$)$}$. More generally I would like to put something of the form
$$sum_{k=0}^infty a_k, f^{(k)}(t)=0$$
(where $a_k$ are as regular as needed (but with infinite non-zero terms)) into the form
$$frac{d}{dt} u = A(u)$$
Where $u$ is a map $C^infty(mathbb R,X)$ with $X$ a Banach space and $Ain mathcal L(X)$.










share|cite|improve this question
























  • I'm still hoping someone to answer this question :)
    – onurcanbektas
    Feb 13 at 10:17
















10












10








10


4





In most analysis courses one sees that differential equations of order $n$ are basically a subset of higher dimensional differential equations of order $1$, for example the equation:



$$f^{(n)}(t)=Fleft(f(t),f'(t),...,f^{(n-1)}(t),tright)$$



Is the same as:
$$frac{d}{dt},begin{pmatrix}g_0(t)\g_1(t)\vdots\g_{n-1(t)}end{pmatrix}=begin{pmatrix}g_1(t)\vdots\g_{n-1}(t)\Fleft(g_0(t),g_1(t),...,g_{n-1}(t),tright)end{pmatrix}$$



This is especially useful, as it allows one to write down explicitly the solutions to linear differential equations of finite order, if we have:
$$f^{(n)}(t)=sum_{k=0}^{n-1}a_k, f^{(k)}(t)$$
Then the corresponding $n$-dimensional equation is of the form:
$$frac{d}{dt} g(t) = Acdot g(t)$$
for some matrix $A$ and the solution is $g(t)=exp(A,t)g(0)$. It is possible to generalise to time dependent coefficients $a_k$.



Is there a way to implement this trick for differential equations that are essentially of infinite order? For example the equation
$$f=sum_{k=1}^infty f^{(k)}(t),$$
of which the solution space is $f={Cexp(frac t2)mid Cinmathbb R$ (or $mathbb C$)$}$. More generally I would like to put something of the form
$$sum_{k=0}^infty a_k, f^{(k)}(t)=0$$
(where $a_k$ are as regular as needed (but with infinite non-zero terms)) into the form
$$frac{d}{dt} u = A(u)$$
Where $u$ is a map $C^infty(mathbb R,X)$ with $X$ a Banach space and $Ain mathcal L(X)$.










share|cite|improve this question















In most analysis courses one sees that differential equations of order $n$ are basically a subset of higher dimensional differential equations of order $1$, for example the equation:



$$f^{(n)}(t)=Fleft(f(t),f'(t),...,f^{(n-1)}(t),tright)$$



Is the same as:
$$frac{d}{dt},begin{pmatrix}g_0(t)\g_1(t)\vdots\g_{n-1(t)}end{pmatrix}=begin{pmatrix}g_1(t)\vdots\g_{n-1}(t)\Fleft(g_0(t),g_1(t),...,g_{n-1}(t),tright)end{pmatrix}$$



This is especially useful, as it allows one to write down explicitly the solutions to linear differential equations of finite order, if we have:
$$f^{(n)}(t)=sum_{k=0}^{n-1}a_k, f^{(k)}(t)$$
Then the corresponding $n$-dimensional equation is of the form:
$$frac{d}{dt} g(t) = Acdot g(t)$$
for some matrix $A$ and the solution is $g(t)=exp(A,t)g(0)$. It is possible to generalise to time dependent coefficients $a_k$.



Is there a way to implement this trick for differential equations that are essentially of infinite order? For example the equation
$$f=sum_{k=1}^infty f^{(k)}(t),$$
of which the solution space is $f={Cexp(frac t2)mid Cinmathbb R$ (or $mathbb C$)$}$. More generally I would like to put something of the form
$$sum_{k=0}^infty a_k, f^{(k)}(t)=0$$
(where $a_k$ are as regular as needed (but with infinite non-zero terms)) into the form
$$frac{d}{dt} u = A(u)$$
Where $u$ is a map $C^infty(mathbb R,X)$ with $X$ a Banach space and $Ain mathcal L(X)$.







real-analysis sequences-and-series functional-analysis differential-equations analysis






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edited Nov 24 at 14:53

























asked Aug 6 '16 at 14:06









s.harp

8,39312049




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  • I'm still hoping someone to answer this question :)
    – onurcanbektas
    Feb 13 at 10:17




















  • I'm still hoping someone to answer this question :)
    – onurcanbektas
    Feb 13 at 10:17


















I'm still hoping someone to answer this question :)
– onurcanbektas
Feb 13 at 10:17






I'm still hoping someone to answer this question :)
– onurcanbektas
Feb 13 at 10:17

















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