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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:

$$f^{(n)}(t)=F\left(f(t),f'(t),...,f^{(n-1)}(t),t\right)$$

is the same as: $$\frac{d}{dt}\,\begin{pmatrix}g_0(t)\\g_1(t)\\\vdots\\g_{n-1(t)}\end{pmatrix}=\begin{pmatrix}g_1(t)\\\vdots\\g_{n-1}(t)\\F\left(g_0(t),g_1(t),...,g_{n-1}(t),t\right)\end{pmatrix}.$$

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: $$\frac{d}{dt} g(t) = A\cdot g(t)$$ 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 $f=\{C\exp(\frac t2)\mid C\in\mathbb R$ (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 $A\in \mathcal L(X)$.

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    $\begingroup$ I'm still hoping someone to answer this question :) $\endgroup$
    – Our
    Feb 13 '18 at 10:17
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No, in general that is not possible. For instance you can write a delay-differential equation as such an "infinite order" equation using the Taylor expansion, $$ y'(t)=f(t,y(t),y(t-r)) =f\left(t,y(t),\sum_{k=0}^\infty \frac{y^{(k)}(t)}{k!}(-r)^k\right) $$ and it is known that DDE have characteristics that ODE have not.

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  • $\begingroup$ The function $f$ must be linear in $y(t)$ and $y(t-r)$ for this to apply (and additionally not every solution of the delay equation will be a solution of the equation with the Taylor expansion), so there some big restrictions on what DDE you can consider here. Do these restrictions still allow for phenomena incompatible with ODEs? $\endgroup$
    – s.harp
    Sep 14 '19 at 10:39

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