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

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

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