1
$\begingroup$

If response is defined as $Res=\sum_{i=1}^{4} r_i x_i$ and $x=(x_1;x_2;x_3;x_4)$. $y$ is constant.

And

$$\frac{d}{dt} x = P(y) \cdot x + Q(y)$$

Then, how can we pick $r_i$'s such that the steady state Response is independent of $y$?

I have no idea how to even begin on this question. Note: $P$ and $Q$ are 4x4 and 4x1 matrices which have some elements as a function of $y$.

Additional note: $\sum x_i=1$ but we are told not to pick an obvious choice for r's. So I am pretty sure that the obvious choice is $r_i=1$ for all $i$.

  • 0
    Maybe $P$ and $Q$ are 4x4 and 4x1 matrices of functions of $y$ instead of 3x3 and 3x1.2011-03-24
  • 0
    @Carl, fixed. The reason I had 3x3 before was because I used $x_4=1-x_1-x_2-x_3$ to reduce one term. But I am not sure if that is needed.2011-03-24
  • 0
    @picakhu, It seems like you need to solve the differential equation. I'm thinking that it's easily solved at least formally, by dividing both sides by $P(y)\cdot x + Q(y)$ and integrating. You might need that $P$ has an inverse. (Physicists do this sort of thing all the time.) And a funny thing is that the "additional note" greatly increased my interest in the problem as I've been messing around with something similar in quantum mechanics.2011-03-24
  • 0
    Another thing that comes to mind, does $\vec{r}$ depend on time?2011-03-24
  • 0
    @Carl, I am highly skeptical about solving the equation. If you can solve for $x$ directly, then the exercise seems quite useless. The idea is to simplify the situation by getting rid of $y$.2011-03-24
  • 0
    @Carl, It is stated that the response is a linear combination. I think that implies that r's are constants.2011-03-24
  • 0
    @picakhu; Maybe you don't need to actually solve it, but can instead put it into a different form. For example: $\vec{x} = P(y)\cdot\int_0^t\vec{x}(t)\;dt + \vec{Q}(y)\;t$, then take the dot product with $\vec{r}^t$, and then differentiate with respect to $y$. And maybe you can ignore the nasty integral. (Just think'n'.)2011-03-24
  • 0
    @picakhu; the only thing to do is to try it, and see if it gets downvoted.2011-03-24
  • 0
    I managed to figure it out! The thing I did was to split $\sum r_i x_i$ into the numerator and denominator, and then compare all the terms. So if say that I figured out that the sum was $\frac{r_1+r_1 r_2 y+r_3 y^2 +r_4 y^3}{1+y+y^2+y^3}$ then I can say that $r_1=k/1,r_2 r_1 = k/1, r_3=k/1, r_4=k/1$ where $k$ is any constant.2011-03-26

1 Answers 1

0

Not a solution! I think this leads nowhere:
We want to eliminate $y$ from $R = \vec{r}\cdot\vec{x}$ given that $x$ follows the following differential equation:
$$\dot{\vec{x}} = P(y)\cdot\vec{x} + \vec{q}(y)$$
To do this, let's first eliminate the derivative with respect to time by integration:
$$\vec{x}(t) = P(y)\cdot\int_0^t\vec{x}(\tau)\;d\tau + \vec{q}(y)\;t$$ So we need to choose $\vec{r}$ such that the following does not depend on $y$:
$$R = \vec{r}\cdot\vec{x} = \vec{r}P(y)\cdot\int_0^t\vec{x}(\tau)\;d\tau + \vec{r}\cdot\vec{q}(y)\;t$$ To ensure that the above does not depend on $y$, we differentiate it with respect to $y$ and set it equal to zero. We get: $$0 = \vec{r}\cdot\frac{dP(y)}{dy}\cdot\int_0^t\vec{x}(\tau)\;d\tau + \vec{r}\cdot P(y)\cdot\int_0^t\frac{\partial \vec{x}(\tau)}{\partial y}\;d\tau + \vec{r}\cdot\frac{d\;\vec{q}(y)}{dy}\;t$$

  • 0
    I added that $y$ is a constant. Does that help with anything? by constant, I mean $y(t)=y(0)$2011-03-24
  • 0
    I'm still hosed. But I'll leave this on my computer and if I get indigestion and wake up in the middle of the night with time on my hands (which is how I solve physics problems), I'll take another look at it. By the way, are you sure that an answer exists? It's late at night and we should likely see a real answer tomorrow (whence I will delete this answer).2011-03-24
  • 0
    @Carl, Hey, I realized that I had misread the question. It wanted only the steady state response to be independent of $y$.2011-03-26
  • 0
    @picakhu; So you can take the derivative with respect to $y$ of $\vec{r}\cdot\vec{x} = \vec{r}P^{-1}(y)\vec{Q}(y)$ and assign it to be zero.2011-03-26
  • 0
    @Carl, I tried your method, it did not work.. Unless I did something wrong. (I am working with a lot of variables. So its not practical to post my mathematica file here)2011-03-26
  • 0
    @picakhu; Hmmm. You're looking for $\vec{r}$ that is orthogonal to the vector $d_y\;(P^{-1}(y)\cdot\vec{Q}(y)\;)$. (I suspect this simplifies.) But this isn't enough to give a single value for $\vec{r}$. For that matter, why not choose $\vec{r}=0$?2011-03-26
  • 0
    @Carl, because we are explicitly told not to pick an obvious $\vec{r}$. However, MAYBE, $r_i$=constant is not an obvious $\vec{r}$2011-03-26
  • 0
    But that doesnt make sense either, because then the response is a constant. (we want the response at eqbm to be constant)2011-03-26
  • 0
    What I'm saying is that I believe the question, as stated, doesn't have a single answer. There are a lot of solutions.2011-03-26
  • 0
    There may be a family of solutions, but I need to find just one. ($r_i$ being constant is not an acceptable one)2011-03-26