Linearize and then substitute vs. Substitute and then linearize

Question

If I have some nonlinear function $f(x,u)$, I can linearize it as $$ f(x,u) \approx f(x_{ss},u_{ss})+\left.\frac{\partial f(x,u)}{\partial x}\right|_{x_{ss},u_{ss}}(x - x_{ss}) + \left.\frac{\partial f(x,u)}{\partial u}\right|_{x_{ss},u_{ss}}(u - u_{ss}) $$

Let's say I want $u$ to be some function of $x$. For instance, $u = -Kx$. I can see two ways to proceed with linearization:

1. Linearize as above, and then substitute $u = -Kx$ into the linearized function
2. Substitute $u=-Kx$ into the original nonlinear equation, and then linearize only about $x$

My question is: are these two approaches equivalent? If so, how can this be shown? If not, is one more valid than the other?

Answer 1

Define

$$\delta u = u - \bar u,\quad \delta x = x - \bar x$$

where $(\bar x, \bar u)$ is the operating point.

Indeed under the standard linearization procedure, wherein

$$\delta u = K\,\delta x,$$

the two approaches are the same under the right linearizations. To see why, define $g(\delta x) = f(a, b) := f(\bar x + \delta x, \bar u + K\,\delta x)$ to avoid confusion of terms. Linearize $g$ about $\delta x = 0,$

$$\begin{aligned} g(\delta x) &= g(0) + \left.\frac{\partial g}{\partial \delta x}\right|_{\delta x = 0}\left( \delta x \right) + \mathcal{O}(\delta x^2)\\ &=g(0) + \left(\left.\frac{\partial f}{\partial a}\right|_{a = \bar x} \left.\frac{\partial a}{\partial \delta x} \right|_{\delta x = 0} + \left.\frac{\partial f}{\partial b}\right|_{b = \bar u} \left.\frac{\partial b}{\partial \delta x} \right|_{\delta x = 0}\right) \left( \delta x \right) + \mathcal{O}(\delta x^2)\\ &=g(0) + \left(\left.\frac{\partial f}{\partial a}\right|_{a = \bar x} + \left.\frac{\partial f}{\partial b}\right|_{b = \bar u}\,K\right) \left( \delta x \right) + \mathcal{O}(\delta x^2)\\ &=g(0) + \left.\frac{\partial f}{\partial a}\right|_{a = \bar x}\,\delta x + \left.\frac{\partial f}{\partial b}\right|_{b = \bar u}\,K\,\delta x + \mathcal{O}(\delta x^2)\\ \end{aligned} $$

Since $\delta u = K\,\delta x$ and $g(0)= f(\bar x, \bar u)$,

$$ g(\delta x) = f(\bar x,\bar u) + \left.\frac{\partial f}{\partial a}\right|_{a = \bar x}\,\delta x + \left.\frac{\partial f}{\partial b}\right|_{b = \bar u}\,\delta u + \mathcal{O}(\delta x^2) $$

The claim follows from substituting $g(\delta x)$ with $f(x, u)$ and replacing all the $\delta x$ and $\delta u.$ Note that if $u=K x$ the claim is not true (or if you linearize $g$ about $x$). It is important that the linearization happens about the deviation once you've applied the feedback since that is the new system (the dynamics of $\delta x(t)$) that are linear and locally mimics the behaviour of $x(t)$ about $\bar x.$