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:


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 $A\in \mathcal L(X)$.

  • $\begingroup$ I'm still hoping someone to answer this question :) $\endgroup$ – onurcanbektas Feb 13 '18 at 10:17

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