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It is well known that for a system of linear ODE $$x'(t) = A(t) \cdot x(t) + b(t)$$ with initial condition $x(t_0) = x_0$, that for a solution at any other time point $t_1$, $x(t_1) = (z_1, \ldots, z_n)$, all entries of $x(t_1)$ can be written down as a linear combination of $x(t_0) = (y_1, \ldots, y_n)$ plus a constant: $$ z_i = c_i +\sum_{j=1}^{n} b_{ij} y_j,$$ for all $i$ where the $b_i, c_{ij}$ are constants.

My question is now: What are other common systems of ODEs, where we can describe the entries of the solution vector at a different time point (or of a steady state to which the initial condition onverged to) by such a linear combination from above from the initial condition $x(t_0)$?

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  • $\begingroup$ I'm afraid that your claim is wrong even for a single homogeneous equation like $x'=x$. How is $x(t)=e^t$ a constant linear combination of $x(0)=1$? $\endgroup$ Nov 24, 2014 at 18:00
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    $\begingroup$ Well I do not mean it for all time points, but just for a fixed one and each time I am building a new linear combination system for a new time point. But particuclarly the steady state situation is interesting to me, so if there are other system ODEs that hold the abover requirement just for the steady states it would already be very helpful to me. $\endgroup$
    – tobias
    Nov 24, 2014 at 18:05
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    $\begingroup$ @A.D. I think here the time $t_1$ is fixed, and $c_j$ are "constant" w.r. to the $y_j$. $\endgroup$ Nov 24, 2014 at 18:06

2 Answers 2

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What you're saying, I think, is that (for a particular $t_1$) the map $\Psi$ from $x(0)$ to $x(t_1)$ is affine: $\Psi(x) = \Phi x + c$ where $\Phi$ is linear. If this was true for all $t_1$, the differential equation would have to be linear, because $x'(t) = \lim_{h \to 0} \dfrac{x(t+h) - x(t)}{h}$ and a limit of affine maps is affine. But for a single $t_1$, there are many other possibilities. For any invertible $\Phi$ and any $c$, let $\psi(x,t)$ be any smooth function on $\mathbb R^{n+1}$ such that $\psi(x,0) = x$, $\psi(x,t_1) = \Phi x + c$, and $\psi(\cdot,t)$ is a diffeomorphism of $\mathbb R^n$ for each $t \in [0,t_1]$. Let $\phi(\cdot,t)$ be the inverse function of $\psi(\cdot,t)$. Then $X(t) = \psi(x,t)$ satisfies the differential equation $$ \dfrac{dX}{dt} = \psi_t(x,t) = \psi_t(\phi(X,t),t)$$ (the subscript indicating a partial derivative with respect to $t$).

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For a system $\dot x = f(t, x(t))$ with $f\in C^0(I\times \mathbb{R}^n,\mathbb{R}^n)$, the answer is: $f$ is necessarily of the form $f(t,x)=A(t)x+b(t)$, if we assume that the affine dependence of $x(t_1)$ from $x(t_0)$ holds for all pairs $t_0, t_1$ in $I$ (or, for all $t_1$ and some fixed $t_0$, hence for all). Of course, together with the implicit assumption that any IVP for this ODE has unique solution in the whole interval $I$.

Premise. Let $I$ be an open interval and let $W:I\times I\rightarrow M_n (\mathbb{R})$ a $2$-parameter family of matrices verifying the chain law: $$W(r,s)=W(r,t)W(t,s),$$ $$W(t,t)=\mathbb{I},$$ for all $r,s,t$ in $I$. Assume that $W(t,s)$ is continuous in the pair $(t,s)\in I\times I$ and partially differentiable in the first variable.

Then it follows that $\partial_1W(t,s)W(s,t)$ is a matrix $A(t)$ independent from $s$, and that $W$ is continuously differentiable in $I\times I$, with $\partial_1W(t,s)=A(t)W(t,s)$, and $\partial_2W(t,s)=- W(t,s)A(s)$, for all $t,s$ in $I$. In other words, $W(t,s)$ is the transition operator associated with the linear system $\dot u(t)=A(t)u(t).$ (Note that this corresponds to your situation in the special case where $c_j$ are identically zero for all $t_0$ and $t_1$).

Indeed, by the chain law we have that all $W(t,s)$ are invertible, and $W^{-1}(t,s)=W(s,t)$ for all $t,s$, so $W(t,s)$ is differentiable in its second variable too. If we derive w.r.to $t$ the identity $W(r,s)=W(r,t)W(t,s)$ we get $0=\partial_2W(r,t)W(t,s)+W(r,t)\partial_1W(t,s) $, whence multiplying conveniently on the left and on the right we see by separation of variables that both sides of $ \partial_1W(t,s) W(s,t)=- W(t,r) \partial_2W(r,t)$ only depend on $t$. If we denote the common value $A(t):=\partial_1W(t,s) W(s,t)=- W(t,r) \partial_2W(r,t)$, a continuous path of matrices, we have that $W$ solves $\partial_1W(t,s)=A(t)W(t,s)$ (and also verifies $\partial_2W(t,s)=-W(t,s)A(s)$ ), with initial condition $W(t,t)=\mathbb{1}$, that is, it is the transition operator associated with the linear system $\dot u(t)=A(t)u(t).$ (Incidentally, $W$ is actually $C^1$ by the Total Differential Theorem). $$*$$ Now, as in your assumptions, let be given a system of ODE's, all of whose IVP have unique solutions $x:I\to\mathbb{R}^n$, and assume that for all $s,t$ in $I$ the value of the solution $x $ at time $t$ depends affinely from the initial data at time $s$. In other words, assume that for any solution $x:I\to\mathbb{R}^n$ there holds $$x(t)=W(t,s)x(s)+c(t,s),$$ for some $2$-parameter family of matrices $W(s,t)$ and some $2$-parameter family of vectors $c(t,s)$. Since by assumption for any $s\in I$ the values $x(s)$ can be chosen in the whole space, the above form implies that $W(t,t)=\mathbb{I}$ and $c(t,t)=0$, for any $t\in I$. Since any solution $x(t)$ is differentiable, we have that both $W(t,s)$ and $c(t,s)$ are partially differentiable in their first variable; moreover $W$ and $c$ are continuous in the pair $(t,s)$, as the continuous dependence from the initial data is in fact a general consequence of the uniqueness property of solutions of the IVP.

The above form also implies, for any $r,s,t$ in $I$ and for any solution $x$: $$x(r )=W(r,s)x(s)+c(r,s)=W(r,t)x(t)+c(r,t)=W(r,t)\left[ W(t,s)x(s)+c(t,s) \right]+c(r,t) $$ $$ =W(r,t)W(t,s)x(s)+ W(r,t) c(t,s) +c(r,t) . $$ Again this implies two separate identities, because for any fixed $s$, $x(s)$ varies in the whole space $\mathbb{R}^n$: $$W(r,s)=W(r,t)W(t,s)$$ and
$$c(r,s)= W(r,t) c(t,s) +c(r,t),$$ together with $W(t,t)=\mathbb{I}$ and $c(t,t)=0$, for all $r,s,t$.

By the initial remark, we have that $W$ is the transition operator of a linear system $\dot u(t)=A(t)u(t)$ corresponding to a continuous path $A(t)$ of matrices. Finally, differentiating $c(r,s)= W(r,t) c(t,s) +c(r,t)$ w.r.to $r$ we have

$$\partial_1c(r,s)= A(r)W(r,t) c(t,s) +\partial_1c(r,t)$$ $$=A(r )\left[ W(r,t) c(t,s) + c(r,t) \right] -A(r )c(r,t) + \partial_1c(r,t)$$ $$=A(r )c(r,s) -A(r )c(r,t) + \partial_1c(r,t).$$ Separating variables, this tells that $\partial_1c(r,s)-A(r )c(r,s)= \partial_1c(r,t)-A(r )c(r,t) $ is a continuous path $b(t)$ only depending on $t$. We therefore have for all $s,t$ in $I$ and for all solution $x$ $$\dot x(t)=A(t)W(t,s)+\partial_1c(t,s)=A(t)\left[W(t,s)+c(t,s)\right]+\partial_1c(t,s)-A(t)c(t,s)=A(t)x(t)+b(t),$$ proving that the ODE is actually affine.

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