# Linear Algebra Response

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 $i$.

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Maybe $P$ and $Q$ are 4x4 and 4x1 matrices of functions of $y$ instead of 3x3 and 3x1. –  Carl Brannen Mar 24 '11 at 1:34
@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. –  picakhu Mar 24 '11 at 1:36
@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. –  Carl Brannen Mar 24 '11 at 1:43
Another thing that comes to mind, does $\vec{r}$ depend on time? –  Carl Brannen Mar 24 '11 at 1:44
@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 $y$. –  picakhu Mar 24 '11 at 1:46

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: $$0 = \vec{r}\cdot\frac{dP(y)}{dy}\cdot\int_0^t\vec{x}(\tau)\;d\tau + \vec{r}\cdot P(y)\cdot\int_0^t\frac{\partial \vec{x}(\tau)}{\partial y}\;d\tau + \vec{r}\cdot\frac{d\;\vec{q}(y)}{dy}\;t$$

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I added that $y$ is a constant. Does that help with anything? by constant, I mean $y(t)=y(0)$ –  picakhu Mar 24 '11 at 2:17
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). –  Carl Brannen Mar 24 '11 at 2:19
@Carl, Hey, I realized that I had misread the question. It wanted only the steady state response to be independent of $y$. –  picakhu Mar 26 '11 at 0:38
@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. –  Carl Brannen Mar 26 '11 at 0:47
@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) –  picakhu Mar 26 '11 at 1:27