# Practical system with the following ODE form

I was wondering if anyone is familiar with an example of a practical / real system with the following ODE form:

$\dot{x}_1= a_{11} x_1$

$\dot{x}_2=a_{21} x_1 + a_{22} x_2 + b u$,

where $u$ is a control input to the system, $a_{11}, a_{21}, a_{22}, b$ are known constants.

I really appreciate examples such a system so that I can read up more about it. A link or paper would help too. Thank you.

UPDATE: I have seen papers in control systems, where examples of above ODE type are used. Refer: http://www.ijcas.org/admin/paper/files/4-3-2.pdf

• Could you say more about u, e.g. is $u=u(x_1,x_2)$ or $u=u(x_1,x_2,t)$ or something else?
– PM.
Mar 24, 2017 at 16:26
• @scitamehtam $u$ could be either state feedback or time varying as well. That is something upto the designer to achieve the desired behavior.
– ems
Mar 24, 2017 at 20:51

This video about matrix methods for inhomegeneous systems sets up a real example for $u=u(t)$ concerning 2 connected vessels containing a chemical solution where the vessels have both inflowing and outflowing solution. The example begins at 27 minutes in.

The constant coefficients in the matrix arise from the physical situation modelled and that in the video is slightly more general in that respect than your system. However, if for example, considering the upper connecting pipe, there was a filter fitted so there was no flow of chemical from the right hand to the left hand tank, but still flow of solvent to maintain a filled left hand tank, then the corresponding coefficient in the matrix $a_{12}$ would be zero.

• Thanks for the video. But the system described at 27 minutes in, the mixing problem, doesn't yield the system I have mentioned since $x_1$ and $x_2$ are present in both $\dot{x}_1$ and $\dot{x}_2$ dynamics. In other words, $a_{12} \ne 0$ in mixing problem. I am looking at real system where $a_{12}=0$ in the form I have mentioned! Please let me know.
– ems
Mar 24, 2017 at 23:52
• @ems I updated my answer so i hope that helps with the example you were looking for for. See other answers here for further algebraic discussions.
– PM.
Mar 25, 2017 at 9:46
• If there was no flow from right to left, for example, then both $a_{12}$ and $a_{21}$ are $0$. In other words, if there is no outflow from right to left which is same as no inflow to left from right. So, both above coefficients would disappear which is not what I want in the ODE system.
– ems
Mar 25, 2017 at 17:04
• @ems There was some imprecision in my edit which I fixed but that was not related to your last comment. I am not sure how $a_{21} =0$ as you say, aren't the vessels still connected from left to right by the lower pipe meaning the concentration in the right hand tank increases because some chemical flows in to it from the left hand tank, and decreases on account of the exterior exit pipe implying also $a_{22} \neq 0$?
– PM.
Mar 25, 2017 at 21:21
• I think your edit now clarifies it better in fact. The key was adding "a filter" to the whole mixing problem, so that the salt or chemical can only travel in one direction and not the other, while the solvent flows unhindered. This results in the corresponding off-diagonal coefficient to be $0$. Thank you!
– ems
Mar 25, 2017 at 22:32

## Non-homogeneous Linear System

I'm gonna consider $$u = u(t)$$ (and $$b \in \mathbb{R}$$) as a function of time. We let $$\dot{x}(t) := \mathrm dx/\mathrm dt$$ then

$$\left\{\begin{matrix}\dot{x}_1 = a_{11}x_1\\\dot{x}_2 = a_{21}x_1+a_{22}x_2 + bu(t)\end{matrix}\right.$$

Which can be transformed in a linear vector differential equation

$$\dot{X} = AX + U(t) \tag{1}$$

Where we have that

$$X := \left(\begin{matrix}x_1\\x_2\end{matrix}\right) \hspace{3cm}A := \left(\begin{matrix}a_{11}&0\\a_{21} & a_{22}\end{matrix}\right)$$ $$U(t):=\left(\begin{matrix}0\\bu(t)\end{matrix}\right)$$

We consider $$A$$ constant and $$U = U(t)$$ time-dependent. So we have to solve the equation $$(1)$$.

## Existence and Uniqueness

Consider that a initial condition $$X_0 := X(0)$$ is given, then the equation $$(1)$$ has one and only one solution $$X(t)$$ that satisfies the initial condition. This is known as the Cauchy problem. The proof for that is so beautiful, in my opinion, that, even a little long, deserves to be shown. Let us first consider the homogeneous differential equation with the below Cauchy problem:

$$\dot{X} = AX \hspace{3cm}X(0):=0$$

Where $$A$$ is a $$n\times n$$ matrix and $$X$$ is a $$n$$-dimensional vector. If we integrate both sides of this equation we have that

$$X(t) = \int_0^tAX(t_1)\mathrm dt_1$$

Using this fact interactively for $$m$$ times we have we have

$$X(t) = \int_0^t\int_0^{t_1}\dots\int_0^{t_{m-1}}A^mX(t_m)\mathrm dt_m\dots\mathrm dt_1$$

If we look for the $$i$$-coordinate of the vector $$X$$ this equation is saying

$$x_i(t) = \int_0^t\int_0^{t_1}\dots\int_0^{t_{m-1}}\sum_{j=1}^n\left(\underbrace{A\cdot A\cdot A\dots A}_{\text{m times}}\right)_{ij}x_j(t_m)\mathrm dt_m \dots \mathrm dt_1$$

Where I used that

$$(AX)_i = \sum_{j=1}^nA_{ij}X_j = \sum_{j=1}^nA_{ij}x_j$$

For product of matrix we have that

$$(AB)_{ij} = \sum_kA_{ik}B_{kj}$$

Then a product of $$m$$ matrices we have $$m-1$$ sums so

$$\left(A^m\right )_{ij} = \sum_{k_1 = 1}^n\sum_{k_2 = 1}^n \dots \sum_{k_{m-1}=1}^nA_{ik_1}A_{k_1k_2}\dots A_{k_{m-1}j}$$

This implies that

$$x_i(t) = \sum_{j=1}^n\sum_{k_1 = 1}^n\sum_{k_2 = 1}^n \dots \sum_{k_{m-1}=1}^n\int_0^t\int_0^{t_1}\dots\int_0^{t_{m-1}}A_{ik_1}A_{k_1k_2}\dots A_{k_{m-1}j}x_j(t)\mathrm dt_m \dots \mathrm dt_1$$

This implies that the absolute value of $$x_i(t)$$ respects the inequality

$$\vert x_i(t)\vert \leq \sum_{j=1}^n\sum_{k_1 = 1}^n\sum_{k_2 = 1}^n \dots \sum_{k_{m-1}=1}^n\int_0^t\int_0^{t_1}\dots\int_0^{t_{m-1}}\vert A_{ik_1}\vert \vert A_{k_1k_2}\vert \dots\vert A_{k_{m-1}j}\vert \vert x_j(t)\vert \mathrm dt_m \dots \mathrm dt_1$$

Now suppose (in this moment) that $$t \in [0,T]$$ for some finite $$T \in \mathbb{R}$$ then we define $$\alpha := \max_{i,j\in\{1,\dots,n\}}\vert A_{ij} \vert$$ and $$M := \max_{t \in [0,T]}\max_{i \in \{1,\dots,n\}}\vert x_i(t)\vert$$ then the inequality becomes

$$\vert x_i(t) \vert \leq \alpha^mM\frac{t^m}{m!}n^m$$

Then we have that this inequality must hold for $$t \in [0,T]$$ for every $$i$$-component of $$X$$. Note that the right-side of the equation doesn't depends on $$m$$, this is just the number of interaction made. If we set $$m \rightarrow \infty$$ we have that $$x_i(t)=0$$ for $$t \in [0,T]$$ for all $$i$$ and for all $$t$$. We set $$T$$ arbitrarily so this is true for all $$t$$ and $$i$$. We conclude that the only solution is the vector $$X = 0$$ for the homogeneous equation with initial condition $$X_0 = 0$$. Now take equation $$(1)$$ and let $$X_1$$ and $$X_2$$ be two different solutions of equation $$(1)$$ with initial condition $$X_1(0) = X_2(0) = X_0$$. So let define $$Z := X_1 - X_2$$ so we have that $$Z(0)=0$$ and

$$\dot{Z} = AX_1 + \dot{U} - AX_2 - \dot{U} = A(X_1 - X_2) = AZ$$

But then we have that $$Z = Z(t)$$ is a solution to the homogeneous equation with initial condition $$Z(0)=0$$ so $$Z = 0$$ and then $$X_1 = X_2$$ which proves uniqueness. Now, with existence, we have just to find a solution than it will prove that solutions exists and are unique.

## Solution

Let define the Dyson Series, given an matrix $$A = A(t)$$, $$I$$ is $$n\times n$$ identity matrix,

$$D(t,s):= I + \sum_{n=1}^{\infty}\int_s^{t}\int_s^{t_1}\dots\int_s^{t_{n-1}}A(t_1)\dots A(t_n)\mathrm dt_n \dots \mathrm dt_1$$

Then I affirm (now without a proof) that the solution to the equation $$\dot{X} = A(t)X(t) + F(t)$$ with initial condition $$X(0) = X_0$$ is

$$X(t) = D(t)X_0 + \int_0^tD(t,s)F(s)\mathrm ds$$

If $$A$$ is a constant matrix as in your problem we have that (try to prove that ) $$D(t,s) = e^{A(t-s)}$$. So the solution to your problem will be, given an initial condition $$X_0 = X(0)$$

$$X(t) = e^{At}X_0+\int_0^te^{A(t-s)}U(s)\mathrm ds$$

So, for the complete solution you have to find $$e^A$$. This is not a very symple task. Just because of the element $$a_{21}$$. If you make an $$n$$ product of the matrix $$A$$ you'll have

$$A^n = \left(\begin{matrix}a_{11}^n & 0 \\ f^n(a_{11},a_{21},a_{22}) & a_{22}^n\end{matrix}\right)$$

Where

$$f^1(a_{11}, a_{21}, a_{22}) = a_{21}$$

$$f^2(a_{11}, a_{21}, a_{22}) = a_{21}(a_{11}+a_{22})$$

$$f^3(a_{11}, a_{21}, a_{22}) = a_{21}(a_{11}^2+a_{22}(a_{11}+a_{22}))$$

• Hope this can help you. It is not an simple problem that you asked, but the answer given is a solution always. Note it is not a very simple solution because that you have to find $e^A$ the matrix exponential.
– R.W
Mar 25, 2017 at 0:07
• I appreciate your answer Rafael. To clarify, I am basically not looking at the solution to the ODE, but a practical or real system that satisfies the above ODE. Do you have an example of such as system?
– ems
Mar 25, 2017 at 0:12
• Sorry I thought that you asked about solutions. I'll delete the answer then and no, I don't know about any example of such a system. Usually physical sistems of interest have an $A$ of the form $(\begin{matrix}0&x\\y&z\end{matrix})$
– R.W
Mar 25, 2017 at 1:07
• @ Rafael. I have seen papers in control systems, where examples of above ODE type are used. Refer: ijcas.org/admin/paper/files/4-3-2.pdf
– ems
Mar 25, 2017 at 1:54
• Here is a very vast application list -theorem 23 is the solution given in my answer and look pg 544 - also in HERE and in HERE seems to exist a very common use of this equation to caracterize certain aspecs of population dynamics in biology. In the papper of your comment they seem to use a very similar solution with the solution I give here also. (+1)
– R.W
Mar 25, 2017 at 18:11

I found tumor cells (virus)-immune system dynamics, which are a variant of predator-prey models, can satisfy the above ODE under some conditions.

Refer: "Differential Equation Techniques for Modeling a Cycle-Specific Oncolytic Virotherapeutic" by Joanna R. Wares, Joseph J. Crivelli, and Peter S. Kim. This paper discusses about tumor growth models in quiescent and active phases and transition between them, which satisfy the above type of ODE under certain conditions.