Proving Nonhomogeneous ODE is Bounded

Solution 1:

Monday 9 June 2014 11:38 PM PST

THIS ANSWER has been sitting around, deleted, in an open tab in my browser for at least nine or ten weeks. Long enough. Time to kick it the hell out of there. And to where do we kick it? Why, to Math Stack Exchange, of course!

This is a pretty long answer in which I have attempted to address not only many of the mathematical points essential to the solution, but also to clarify certain erroneous assumptions which are often made in trying to solve ODEs of the form $\dot x(t) = A(t) x(t) + h(t)$. I know they are often made, for I have made them myself many times . . .

These things being said:

Before wading into the deeper waters implicit in this question, which by and large concerns equations of the form

$\dot x = A(t)x + h(t), \tag{0}$

I would like to make a few (hopefully) clarifying remarks. First of, let us examine the one-dimensional case of the given equation, which is

$\dot x = a(t) x + h(t), \tag{1}$

where I have written $a(t)$ in lieu of $A(t)$ to emphasize that the coefficient of $x(t)$ is a "mere" function and not a matrix. Let us further restrict our attention to the homogeneous case $h(t) = 0$; then (1) becomes

$\dot x = a(t) x. \tag{2}$

It is evident that the solution to (2) taking the value $x(\tau)$ at time $t = \tau$ is given by

$x(t) = e^{\int_\tau^t a(s) ds} x(\tau); \tag{3}$

indeed differentiating (3) yields

$\dot x(t) = \dfrac{d}{dt}(\int_\tau^t a(s) ds) e^{\int_\tau^t a(s) ds} x(\tau) = a(t) x(t), \tag{4}$

and furthermore, setting $t = \tau$ on the right of (3) returns

$e^{\int_\tau^\tau a(s) ds} x(\tau) = e^0 x(\tau) = x(\tau), \tag{5}$

showing $x(t)$ as specified in (3) satisfies the requisite initial condition. Taking things a step further and allowing $h(t) \ne 0$, we have

$x(t) = e^{\int_\tau^t a(s) ds}(x(\tau) + \int_\tau^t e^{-\int_\tau^s a(u) du} h(s) ds), \tag{6}$

as may also be validated by direct differentiation of (6):

$\dot x(t) = \dfrac{d}{dt}(e^{\int_\tau^t a(s) ds})(x(\tau) + \int_\tau^t e^{-\int_\tau^s a(u) du} h(s) ds)$ $+ e^{\int_\tau^t a(s)ds} \dfrac{d}{dt} (x(\tau) + \int_\tau^t e^{-\int_\tau^s a(u) du} h(s) ds), \tag{7}$

or

$\dot x(t) = a(t)e^{\int_\tau^t a(s) ds}(x(\tau) + \int_\tau^t e^{-\int_\tau^s a(u) du} h(s) ds)$ $+ e^{\int_\tau^t a(s)ds} (e^{-\int_\tau^t a(u) du} h(t)), \tag{8}$

or

$\dot x(t) = a(t)x(t) + h(t); \tag{9}$

it is also easy to see from (6) that $x(t)$ respects the assigned initial condition $x(\tau)$. (3) and (6) are of course the well-known variation of parameters or variation of constants formulas for equations (2) and (1), respectively. They take the particularly simple explicit forms presented above by virtue of the fact that we have restricted the dimension of the system to be one. When we remove this restriction, however, it is no longer possible, in general, to write the solution in terms of a simple matrix exponential of $\int_\tau^t A(s) ds$; when $A$ is a constant matrix, however, this may always be done. For constant $A$, we have

$x(t) = e^{A(t - \tau)}(x(\tau) + \int_\tau^t e^{-A(s - \tau)} h(s) ds), \tag{10}$

which may also be validated by direct differentiation, using the fact that, for constant $A$,

$\dfrac{d}{dt}(e^{A(t - \tau)}) = A(e^{A(t - \tau)}); \tag{11}$

we also note that for such $A$

$A(t - \tau) = \int_\tau^t A ds, \tag{12}$

so that if we use (12) then (10) takes the form

$x(t) = e^{\int_\tau^t A ds}(x(\tau) + \int_\tau^t e^{-\int_\tau^s A du} h(s) ds). \tag{13}$

Comparing (6) and (13), it is tempting to assume that, for time-dependent $A(t)$ with $\text{size}(A) > 1$, the formula analogous to (13) solves (0), that is, the solution of (0) may be expressed in the form

$x(t) = e^{\int_\tau^t A(s) ds}(x(\tau) + \int_\tau^t e^{-\int_\tau^s A(u) du} h(s) ds); \tag{14}$

however, succuambing to such temptation leads on straight down the wide and perilous road of error! (14) does not in general solve (0) when $A$ is non-constant. Further light may be shed on such a situation via examination of the power series defining $e^{B(t)}$ for a differentiable, time-dependent matrix $B(t)$. We have

$e^{B(t)} = \sum_0^\infty \dfrac{(B(t))^j}{j!}, \tag{15}$

and if we differentiate (15) term-by-term we obtain

$\dfrac{d}{dt}e^{B(t)} = \sum_0^\infty \dfrac{1}{j!} ((B(t))^j)'; \tag{16}$

in (16) and other expressions throughout this answer, the "${}'$" and "$\dot {}$" are used interchangably to denote the $t$-derivative as notational convenience may demand. We have:

$((B(t))^j)' = (B(t))'(B(t))^{j -1} + B(t)(B(t))'(B(t))^{j - 2}$ $+ (B(t))^2(B(t))'(Bt))^{j - 3} + \ldots + (B(t))^{j -1}(B(t))', \tag{17}$

which may also be expressed in the more compact form

$((B(t))^j)' = \sum_0^{j -1} (B(t))^k(B(t))'(B(t))^{j -k -1}; \tag{18}$

(17) and (18) both follow from the Leibniz product rule for derivatives applied to $(B(t))^j$. We see from these equations that we cannot in general move the factor $(B(t))'$ to the front of the expansions, allowing us to write

$((B(t))^j)' = j(B(t))'(B(t))^{j - 1} \tag{19}$

and hence

$(e^{B(t)})' = (B(t))'e^{B(t)} \tag{20}$

unless $(B(t))' = \dot B(t)$ and $B(t)$ commute, that is $\dot B(t) B(t) = B(t) \dot B(t)$ or $[B(t), \dot B(t)] = 0$. But in fact (20) plays a central role in validating the formulas (6) and (10); indeed, we see that, taking $B(t) = A(t - \tau)$ we have $\dot B(t) = A$ which manifestly commutes with $A(t - \tau)$; thus (20) may be applied in the case $A$ is constant; but since (20) in the main will not apply to (14), we cannot affirm that (14) solves the equation

$\dot x = A(t) x + h(t). \tag{21}$

Taking $B(t) = \int_\tau^t A(s)ds$, we see that the preceding criteria translate to $A(t) = \dot B(t)$ and $\int_\tau^t A(s)ds$ commuting for all $t$, that is, $[A(t), \int_\tau^t A(s) ds] = 0$. And though there are important classes of matrices for which this holds, it is certainly not true for general $A(t)$.

We nevertheless observe that, under the hypotheses given on $A(t)$ and $h(t)$, it is in fact a relatively simple matter to show that $x(t)$ as given in (14) is bounded, whether or not it solves (21), for if $\tau \ge 1$ then

$\vert e^{\int_\tau^t A(s)ds} \vert \le e^{\vert \int_\tau^t A(s) ds \vert} \le e^{\int_\tau^t \vert A(s) \vert ds} \le e^{\int_1^\infty \vert A(s) \vert ds}, \tag{22}$

where the right-hand exponential is finite by virtue of the hypothesis placed upon $A(t)$. In addition, the integral $e^{-\int_\tau^t A(s)ds}$ submits to an identical bound, for

$\vert e^{-\int_\tau^t A(s)ds} \vert \le e^{\vert -\int_\tau^t A(s) ds \vert} = e^{\vert \int_\tau^t A(s) ds \vert} \le e^{\int_1^\infty \vert A(s) \vert ds}, \tag{23}$

using (22). The bounds on these exponentials, when used in concert with the hypothesis on $h(t)$, viz. $\int_1^\infty \vert h(t) \vert dt < \infty$, easily show $x(t)$ given by (14) is bounded as $t \to \infty$:

$\vert x(t) \vert = \vert e^{\int_\tau^t A(s) ds}(x(\tau) + \int_\tau^t e^{-\int_\tau^s A(u) du} h(s) ds) \vert$ $\le \vert e^{\int_\tau^t A(s) ds} \vert (\vert x(\tau) \vert + \vert \int_\tau^t e^{-\int_\tau^s A(u) du} h(s) ds \vert)$ $\le \vert e^{\int_\tau^t A(s) ds} \vert (\vert x(\tau) \vert + \int_\tau^t \vert e^{-\int_\tau^s A(u) du} \vert \vert h(s) \vert ds)$ $\le e^{\int_\tau^\infty \vert A(s) \vert ds} (\vert x(\tau) \vert + e^{\int_\tau^ \infty \vert A(s) \vert ds}\int_\tau^\infty \vert h(s) \vert ds); \tag{24}$

we have used (22), (23) and the integral hypotheses on $A(t)$ and $h(t)$ in arriving at (24); we see that (24) bounds $\vert x(t) \vert$ globally by a constant, hence $x(t)$ is bounded as $t \to \infty$. Finally, the assumption that $\tau \ge 1$ is not much of a restriction, since for $0 < \tau < 1$ we can simply propagate solutions forward to $t = 1$ and then apply the present results.

Hopefully the above comments do in fact add some clarity to the situation.

Well! I have certainly gone on at length about a hypothetical and erroneous attempt at estimating solutions to (21), a most likely reason being that I myself have fallen into this trap so many times. So perhaps I am taking the opportunity to exploit this near-rant as a form of much-needed mathematical psychotherapy, and I can only express my thanks to my MSE audience for hearing, or better I should write, reading me out on this. The relative ease of preparing estimates of $x(t)$ from (14) serves to compound the false hope that it in fact solves (21), but such misplaced faith is indeed the first step on the road to mathematical perdition, wherein many wander until the grace of illumination finds them. I shall try to be a worthy bearer of such light in what follows. And if I sound like Jonathan Edwards, preacher of renown in the pre-revolution days of the North Atlantic States, it is only because I fear becoming a sinner in the hands of an angry goddess, Our Queen of Sciences, Her Majesty Ms. Mathematics.

To continue, back to the main line . . .

It appears that our OP user75514 was attempting to deploy (14) or a similar formula in the remarks made near the end of the text of the question. However, the formula

$x(t,\tau,\xi)=exp(\int_\tau^t A(s)ds)\xi+\int_\tau^t exp(\int_s^t A(w)dw)h(s)ds \tag{25}$

can't be correct even in the event that $[\int_\tau^t A(s) ds, A(t)] = 0$, for if this holds and we differentiate (25) we obtain

$\dot x(t, \tau, \xi) = A(t)exp(\int_\tau^t A(s)ds)\xi + exp(\int_s^s A(w)dw)h(s)ds$ $= A(t)exp(\int_\tau^t A(s)ds)\xi + h(t) \ne A(t)x(t, \tau, \xi) + h(t). \tag{26}$

when $x(t, \tau, \xi)$ is as in (25). I suspect our OP user75514 was groping for something akin to (14) when (25) was written in the original post; but as we have seen, (14), though easy enough to manipulate for the purposes of estimation, is still itself erroneous as a solution to (21); however, this mistake is an easy one to make, as I have tried to explain; indeed, I have made it many times myself. But it is not the correct approach to (21).

There is in fact a formula analogous to (21) which correctly expresses solutions to (14). Let $X(t, \tau)$ be a fundamental matrix solution to

$\dot x = A(t)x, \tag{27}$

the homogeneous counterpart to (14); that is, $X(t, \tau)$ is an $n \times n$ differentiable matrix function of $t$ such that

$\dot X(t, \tau) = A(t)X(t,\tau), \; X(\tau, \tau) = I. \tag{28}$

It is well-known that such an $X(t, \tau)$ uniquely exists and that it is non-singular for all $t$ (see for example Jack K. Hale's Ordinary Differential Equations, 2009 Dover Publications, Inc., chapter III and especially section III.1); hence $X^{-1}(t, \tau)$ also exists for all $t$. We can in fact use $X(t, \tau)$ and $X^{-1}(t, \tau)$ to present a formula which correctly solves (21) with the initial condition $x(\tau) = \xi$; it is

$x(t) = X(t, \tau)(\xi + \int_\tau^t X^{-1}(s, \tau) h(s) ds). \tag{29}$

It is easy to see, by direct differentiation, that $x(t)$ given by (29) solves (21); indeed,

$\dot x(t) = \dot X(t, \tau)(\xi + \int_\tau^t X^{-1}(s, \tau) h(s) ds) + X(t, \tau) X^{-1}(t, \tau)h(t)$ $= A(t)X(t, \tau)(\xi + \int_\tau^t X^{-1}(s, \tau) h(s) ds) + h(t) = A(t)x(t) + h(t) \tag{30};$

furthermore it is evident that

$x(\tau) = X(\tau, \tau)(\xi + \int_\tau^\tau X^{-1}(s, \tau) h(s) ds) = I(\xi + 0) = \xi. \tag{31}$

It appears from the above, (28)-(31) in particular, that $X(t, \tau)$ is the appropriate replacement/generalization for $e^{\int_\tau^t A(s)ds}$, since it satisfies (28) and since (30) actually solves (21). We note that in the event of $[A(t), \int_\tau^t A(s) ds] = 0$ we can in fact write $X(t, \tau) = e^{\int_\tau^t A(s) ds}$ which solves (28) in this case; but under the more typical scenario $[A(t), \int_\tau^t A(s) ds] \ne 0$ we must simply accept $X(t, \tau)$ obeying (28) as correct.

We use (28) to prepare an estimate of $\Vert X(t, \tau) \Vert$; integrating (28) we find

$X(t, \tau) - I = X(t, \tau) - X(\tau, \tau) = \int_\tau^t A(s)X(s, \tau) ds, \tag{32}$

or

$X(t, \tau) = I + \int_\tau^t A(s)X(s, \tau) ds, \tag{33}$

and upon taking norms we see that

$\Vert X(t, \tau) \Vert = \Vert I + \int_\tau^t A(s)X(s, \tau) ds \Vert \le \Vert I \Vert + \Vert \int_\tau^t A(s)X(s, \tau) ds \Vert$ $\le 1 + \int_\tau^t \Vert A(s) \Vert \Vert X(t, \tau) \Vert ds. \tag{34}$

We apply the integral form of Gronwall's inequality to (34). For the present purposes, it may be taken to assert that, for continuous functions $u(t)$ and $\beta(t) \ge 0$ on an interval $I \subset \Bbb R$ and constant $\alpha \ge 0$ with

$u(t) \le \alpha + \int_{t_0}^t \beta(s) u(s) ds \tag{35}$

we have

$u(t) \le \alpha e^{\int_{t_0}^t \beta(s) ds}. \tag{36}$

Applying (35)-(36) to (34) with $u(t) = \Vert X(t, \tau) \Vert$ and $\beta(t) = \Vert A(t) \Vert$ ande $t_0 = \tau$ we find

$\Vert X(t, \tau) \Vert \le e^{\int_\tau^t \Vert A(s) \Vert ds}, \tag{37}$

and since

$\int_\tau^t \Vert A(s) \Vert \le \int_\tau^\infty \Vert A(s) \Vert ds < \infty \tag{38}$

by our hypothesis on $A(t)$, it follows that $X(t, \tau)$ is bounded for all $t$:

$\Vert X(t, \tau) \Vert \le e^{\int_\tau^\infty \Vert A(s) \Vert ds}. \tag{39}$

We also need to establish a bound for $X^{-1}(t, \tau)$, which occurs under the integral sign in (29); as opposed to the bound for $\vert e^{-\int_\tau^t A(s)ds} \vert$, which follows easily from the fact that it is a matrix exponential, the lack of such a representation for $X^{-1}(t, \tau)$ creates the need for an separate argument. To this end we note that since

$X^{-1}(t, \tau) X(t, \tau) = I \tag{40}$

differentiation with respect to $t$ yields

$\dot X^{-1}(t, \tau) X(t, \tau) + X^{-1}(t, \tau) \dot X(t, \tau) = 0, \tag{41}$

and thus

$\dot X^{-1}(t, \tau) = -X^{-1}(t, \tau) \dot X(t, \tau) X^{-1}(t, \tau), \tag{42}$

and using (28) shows that

$\dot X^{-1}(t, \tau) = -X^{-1}(t, \tau) A(t) X(t, \tau) X^{-1}(t, \tau) = -X^{-1}(t, \tau) A(t), \tag{43}$

which is in fact well-known. From (43) we derive the inequality

$\Vert X^{-1}(t, \tau) \Vert \le 1 + \int_\tau^t \Vert A(s) \Vert \Vert X^{-1}(t, \tau) \Vert ds \tag{44}$

in a manner completely analogous to the derivation of (34) from (28). Once again, an application of the Gronwall inequality shows that

$\Vert X^{-1}(t, \tau) \Vert \le e^{\int_\tau^\infty \Vert A(s) \Vert ds}, \tag{45}$

just as in the case of $\Vert X(t, \tau) \Vert$. We note that the application of Gronwall's integral inequality to (44) yields the same bound, $e^{\int_\tau^\infty \Vert A(s) \Vert ds}$, as we obtained for $\Vert X(t, \tau)$ from (34). We are at last in position to form an estimate of $\vert x(t) \vert$. From (29),

$\vert x(t) \vert = \vert X(t, \tau)(\xi + \int_\tau^t X^{-1}(s, \tau) h(s) ds)) \vert$ $\le \Vert X(t, \tau) \Vert \vert \xi + \int_\tau^t X^{-1}(s, \tau) h(s) ds \vert \le \Vert X(t, \tau) \Vert (\vert \xi \vert + \vert \int_\tau^t X^{-1}(s, \tau) h(s) ds \vert)$ $\le \Vert X(t, \tau) \Vert (\vert \xi \vert + \int_\tau^t \Vert X^{-1}(s, \tau) \Vert \vert h(s) \vert ds), \tag{46}$

and, since $\Vert X(t, \tau) \Vert, \Vert X^{-1}(t, \tau) \Vert \le e^{\int_\tau^\infty \Vert A(s) \Vert ds} < \infty$, (46) yields

$\vert x(t) \vert \le e^{\int_\tau^\infty \Vert A(s) \Vert ds}(\vert \xi \vert + e^{\int_\tau^\infty \Vert A(s) \Vert ds} \int_\tau^t \vert h(s) \vert ds)$ $\le e^{\int_\tau^\infty \Vert A(s) \Vert ds}(\vert \xi \vert + e^{\int_\tau^\infty \Vert A(s) \Vert ds} \int_\tau^\infty \vert h(s) \vert ds), \tag{47}$

where by the hypothesis on $A(t), h(t)$ the right-hand side is finite. Thus we see $\vert x(t) \vert < \infty$ for all $t$, and the requisite bound is established. QED.

Whew!!!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Solution 2:

Let me try to provide a shorter answer:

First observation. It suffices to show that the solution $\varPhi(t;\tau)$ of the initial value problem $$ \left\{\begin{array}{lc}X'=A(t)X, \\ X(\tau)=I,\end{array}\right. \qquad\qquad (\star) $$ where $I$ is the identity matrix in $\mathbb R^{n\times n}$, and $X\in\mathbb R^{n\times n}$ is uniformly bounded for $1\le \tau <\infty$.

Indeed, if $\|\varPhi(t;\tau)\|\le M$, and as the solution of $$ x'=A(t)x, \quad x(1)=\xi, $$ is expressed as $$ x(t)=\varPhi(t;1)\xi+\int_1^t \varPhi(t;s) \,h(s)\,ds, $$ then $$ \|x(t)\|\le M\|\xi\|+M\int_1^\infty \|h(s)\|\,ds. $$

Proof of the uniform boundedness of $\varPhi(t;\tau)$. Let $a(t)=\|A(t)\|$, the $L^2-$norm (operator norm) and for fixed $\tau\ge 1$, $y(t)=\|\varPhi(t;\tau)\|$. Then $\varPhi$ satisfies the the equivalent to the $(\star)$ integral equation $$ \varPhi(t;\tau)=I+\int_\tau^t A(s)\,\varPhi(s;\tau)\,ds, $$ implies that $$ y(t)\le 1+\int_\tau^t a(s)\,y(s)\,ds, \qquad\qquad (\star\star) $$ and as in dealing with Gronwald's type inequalities, multiplying both sides of the above by $a(t)\exp\big(-\int_\tau^t a(s)\,ds\big)$, we obtain $$ a(t)y(t)\exp\big(-\int_\tau^t a(s)\,ds\big)-a(t)\exp\big(-\int_\tau^t a(s)\,ds\big)\int_\tau^t a(s)\,y(s)\,ds\\ \le a(t)\exp\big(-\int_\tau^t a(s)\,ds\big) $$ or $$ \left(\exp\big(-\int_\tau^t a(s)\,ds\big)\int_\tau^t a(s)\,y(s)\,ds+ \exp\big(-\int_\tau^t a(s)\,ds\big)\right)' \le 0, $$ and integrating in $[\tau,t]$ we get $$ \exp\big(-\int_\tau^t a(s)\,ds\big)\int_\tau^t a(s)\,y(s)\,ds+ \exp\big(-\int_\tau^t a(s)\,ds\big)\le 1, $$ or $$ \int_\tau^t a(s)\,y(s)\,ds+ 1\le \exp\big(\int_\tau^t a(s)\,ds\big) \qquad\qquad (\star\star\star). $$ Combining $(\star\star)$ with $(\star\star\star)$ we obtain that $$ \|\varPhi(t;\tau)\|=y(t) \le \exp\big(\int_\tau^t a(s)\,ds\big)\le \exp\big(\int_\tau^\infty \|A(s)\|\,ds\big), $$ and hence $\varPhi(t;\tau)$ is uniformly bounded.