# Higher-order differential equations, the Laplace transform and exponential order

• Feb 22nd 2009, 03:00 AM
bkarpuz
Higher-order differential equations, the Laplace transform and exponential order
Dear Friends,

I see so many books that show the Laplace transform and its application to linear differential equations for second-order and/or higher-order differential equations.
But as we know the Laplace transform can be applied when the solutions are known to be of exponential order, and for first-order equations we indeed know it by the Gronwall's inequality, but what about higher-order equations?
For instance,
$\displaystyle y^{(n+1)}(t)+py(t)=0\quad\text{for}\ t\geq0,$
where $\displaystyle n\in\mathbb{N}$ and $\displaystyle p\in\mathbb{R}$, do we know that the solutions are of exponential order?

Thanks!
• Feb 22nd 2009, 06:43 AM
Laurent
Quote:

Originally Posted by bkarpuz
Dear Friends,

I see so many books that show the Laplace transform and its application to linear differential equations for second-order and/or higher-order differential equations.
But as we know the Laplace transform can be applied when the solutions are known to be of exponential order, and for first-order equations we indeed know it by the Gronwall's inequality, but what about higher-order equations?
For instance,
$\displaystyle y^{(n+1)}(t)+py(t)=0\quad\text{for}\ t\geq0,$
where $\displaystyle n\in\mathbb{N}$ and $\displaystyle p\in\mathbb{R}$, do we know that the solutions are of exponential order?

Thanks!

In order to prove that solutions of linear equations with constant coefficients grow at most exponentially, you can use a multidimensional version of Gronwall's lemma.

Remember indeed that the solution $\displaystyle y(t)$ of a $\displaystyle n$-th order linear differential equation is a component of the solution $\displaystyle Y(t)$ of a $\displaystyle n$-dimensional linear equation. Then an exponential bound on the euclidean norm of $\displaystyle Y(t)$ gives an exponential bound on $\displaystyle y(t)$. Here's how to obtain such a bound.

If $\displaystyle Y'(t)=AY(t)+B$ where $\displaystyle A,B$ are $\displaystyle n\times n$-matrices, we have $\displaystyle \|Y'(t)\|\leq\|A\|\|Y(t)\|+\|B\|$ for every $\displaystyle t\geq 0$, so that $\displaystyle \|Y(t)-Y(0)\|\leq \|A\| \int_0^t \|Y(s)\| ds + \|B\| t$, hence $\displaystyle \|Y(t)\|\leq \|Y(0)\|+\|B\|t+\|A\|\int_0^t \|Y(s)\| ds$. And now you can apply the one-dimensional Gronwall lemma "in integral form" (quoting the wikipedia) to the function $\displaystyle t\mapsto\|Y(t)\|$. The only online reference I could find for the final result is this french exercise sheet (Exercice 1.4), but you don't need it to find the proof, just apply Gronwall's lemma like I wrote and do an integration by parts to compute the integral. You should get:

$\displaystyle \|Y(t)\|\leq \|Y(0)\|e^{\|A\|t}+\frac{\|B\|}{\|A\|}(e^{\|A\|t}-1)$.

This is what you need.
• Feb 23rd 2009, 05:54 AM
bkarpuz
Quote:

Originally Posted by Laurent
In order to prove that solutions of linear equations with constant coefficients grow at most exponentially, you can use a multidimensional version of Gronwall's lemma.

Remember indeed that the solution $\displaystyle y(t)$ of a $\displaystyle n$-th order linear differential equation is a component of the solution $\displaystyle Y(t)$ of a $\displaystyle n$-dimensional linear equation. Then an exponential bound on the euclidean norm of $\displaystyle Y(t)$ gives an exponential bound on $\displaystyle y(t)$. Here's how to obtain such a bound.

If $\displaystyle Y'(t)=AY(t)+B$ where $\displaystyle A,B$ are $\displaystyle n\times n$-matrices, we have $\displaystyle \|Y'(t)\|\leq\|A\|\|Y(t)\|+\|B\|$ for every $\displaystyle t\geq 0$, so that $\displaystyle \|Y(t)-Y(0)\|\leq \|A\| \int_0^t \|Y(s)\| ds + \|B\| t$, hence $\displaystyle \|Y(t)\|\leq \|Y(0)\|+\|B\|t+\|A\|\int_0^t \|Y(s)\| ds$. And now you can apply the one-dimensional Gronwall lemma "in integral form" (quoting the wikipedia) to the function $\displaystyle t\mapsto\|Y(t)\|$. The only online reference I could find for the final result is this french exercise sheet (Exercice 1.4), but you don't need it to find the proof, just apply Gronwall's lemma like I wrote and do an integration by parts to compute the integral. You should get:

$\displaystyle \|Y(t)\|\leq \|Y(0)\|e^{\|A\|t}+\frac{\|B\|}{\|A\|}(e^{\|A\|t}-1)$.

This is what you need.

Laurent thanks again.
I can do what I want to do by building the matrix form of the equation as you have suggested.
But I need to prove it in other way.
I restrict my problem to the following type equations:
$\displaystyle (1)\rule{5cm}{0cm}x^{\prime\prime}(t)+\sum_{i=1}^{ m}p_{i}x(t-\tau_{i})=0\quad\text{for}\ t\geq0,$
where $\displaystyle p_{i}\in\mathbb{R}$ and $\displaystyle \tau_{i}>0$ for $\displaystyle i=1,2,\ldots, m$.
Existence and uniqueness of such equations are known when the initial function $\displaystyle \varphi\in C([-\tau_{\max},0],\mathbb{R})$, where $\displaystyle \tau_{\max}:=\max\nolimits_{i=1,2,\ldots, m}\{\tau_{i}\}$, is prescribed, and it follows from the method of steps.
The problem looks similar to that Example 1.2(there is a missprint in (ii)) in the interesting exercise sheet you shown to me.
I need to prove that the solutions are of exponential order (or it sufices to prove that its derivative is of exponential order).

$\displaystyle \rule{15cm}{0.01cm}$
The proof for first-order delay equations are very simple.
Consider
$\displaystyle (2)\rule{5cm}{0cm}x^{\prime}(t)+\sum_{i=1}^{m}p_{i }x(t-\tau_{i})=0\quad\text{for}\ t\geq0,$
where the coefficients and the delays are as mentioned previously.
Integrating (2) over $\displaystyle [0,t)\subset[0,\infty)$, we get
$\displaystyle (3)\rule{5cm}{0cm}x(t)=x(0)-\int_{0}^{t}\sum_{i=1}^{m}p_{i}x(s-\tau_{i})\mathrm{d}s\quad\text{for}\ t\geq0.$
Set $\displaystyle \widetilde{x}(t):=\sup\{|x(s)|:s\in[-\tau,t)\}$ for $\displaystyle t\geq0$.
From (3), for all $\displaystyle t\geq0$, we have
$\displaystyle (\ )\rule{5cm}{0cm}|x(t)|\leq|x(0)|+\int_{0}^{t}\sum_ {i=1}^{m}|p_{i}||x(s-\tau_{i})|\mathrm{d}s$
$\displaystyle (\ )\rule{5.95cm}{0cm}\leq|x(0)|+\sum_{i=1}^{m}|p_{i} |\int_{0}^{t}\widetilde{x}(s)\mathrm{d}s,$
which yields
$\displaystyle (4)\rule{5cm}{0cm}\widetilde{x}(t)\leq|x(0)|+\sum_ {i=1}^{m}p_{i}\int_{0}^{t}\widetilde{x}(s)\mathrm{ d}s.$
Finally, performing the Grönwall's inequality to (4), we obtain
$\displaystyle (\ )\rule{5cm}{0cm}\widetilde{x}(t)\leq|x(0)|\exp\Big \{\sum_{i=1}^{m}|p_{i}|t\Big\}\quad\text{for all}\ t\geq0.$
Thus, every solution of (2) is exponentialy bounded since $\displaystyle |x(t)|\leq \widetilde{x}(t)$ for all $\displaystyle t\geq-\tau$.

$\displaystyle \rule{15cm}{0.01cm}$
For (1), instead of proving the exponential order property of all solutions, it suffices for me to prove that positive solutions are of exponential order.
Particular solutions are also welcome, i.e., when all the coefficients have the same sign.

Thanks for the help.

Note. Okay, I built the first-order delay differential system and applied multidimensional Gronwall's inequality.
But I really still wonder if there is an answer without using systems. :S