I am studying differential-algebraic equations, that is, loosely speaking, equations of the form
$$E\mathbf x'(t)=Ax(t)+f(t)$$
where $E,A\in\mathbb C^{n\times n}$, $t\in[0,\infty)$ and $f:[0,\infty)\to\mathbb C^n$, subject to some initial condition $\mathbf x(t_0)=\mathbf c$. Here, the system can be expressed as a system involving an ODE and an algebraic equation, as in the following example.
Let $\mathbf x(t)=(x_1(t),x_2(t))$ for $t\in[0,\infty)$, and let $\mathbf c=(c_1,c_2)\in\mathbb C^2$. Consider the following differential-algebraic system (or equation),
\begin{align*}
x_1'(t)+x_2'(t)+x_1(t) & = 0\quad \text{when $t\in(0,\infty)$} \\
x_2(t)&= 0\quad \text{when $t\in(0,\infty)$}\\
\mathbf x(0)&= \mathbf c,
\end{align*}
where the last equation corresponds to the initial condition. The unique solution here is given by $\mathbf x(t)=(c_1e^{-t},0)$ for $t\in[0,\infty)$.
In differential-algebraic equations, the notion of a consistent initial value is central, and is used to describe when the corresponding initial value problem has at least one solution. I have read that for differential-algebraic equations, a key difference to ODEs is that, for a given initial condition, the system might not be solvable.
I'm trying to see how this is showcased in the above example – is it simply that for any initial value $\mathbf c=(c_1,c_2)\in\mathbb C^2$ where $c_2\neq0$, the corresponding initial value problem is not solvable?
Furthermore, what is the connection between this and the fact that for differential-algebraic equations, the matrix $E$ is allowed to have nontrivial kernel? How is it that $E$ having a nontrivial kernel enables the above?
Best Answer
For the example, note that since $x_2 \equiv 0, x_2'\equiv 0$ also, so the first equation becomes $x_1' + x_1 = 0$, which has the unique continuous solution $x_1 = c_1e^{-t}$ for any $c_1$. $x_2$ also must be a specific function: $$x_2(t) = \begin{cases}c_2,&t = 0\\0,&t > 0\end{cases}$$ The issue here is that we demand the solution be continuous at the boundaries - in this case, at $t = 0$. So this problem is only solvable when $c_2 = 0$, exactly because that is the only time $x_2$ can be continuous at $0$.
For the general problem, if $E$ has a trivial kernal, then it is invertible, so the equation can be rewritten as $$\mathbf x' = B\mathbf x + g$$ Where $B = E^{-1}A, g = E^{-1}f$. Thus when $E$ has trivial kernal, the problem can be recast as an ordinary multidimensional differential equation, which under sufficient regularity conditions will have local solutions. This case offers nothing new that requires a broader theory.
But when $E$ has a non-trivial kernal, $E$ will not have an inverse, so the previous results no longer guarantee a solution, no matter how regular $f$ is forced to be. This is demonstrated by the example, where $f \equiv 0$, which is as regular as any map can be.
The theory of differential-algebraic equation has been developed exactly to investigate the case when $E$ has non-trivial kernal, because the other case had already been covered.