(Why) can we treat a function of a variable as another independent variable?

Question

I'm currently reading my numerical analysis textbook and something's bugging me. To get into it, let's take a look at the following differential equation;

$$u'(x) = f(x, u(x))$$

In order to determine the stability of the equation, one may calculate the Jacobian,

$$J(x, u(x)) = \frac{\partial f}{\partial u}|_{(x, u(x))}$$

Here is a specific differential equation:

$$u'(x) = -\alpha(u(x) - sin(x)) + cos(x)$$

For which the Jacobian is

$$J(x, u(x)) = -\alpha$$

Basically, we treated both $sin(x)$ and $cos(x)$ as constants with respect to $u$, but I don't really understand why. Most of the time, when we take a derivative the variables are independant, which is not the case here as they both depend on the same variable $x$.

This means that the "rate of change of $sin(x)$ with respect to $u(x)$" is zero, but the value of $u(x)$ only changes if the value of x itself changes, so shouldn't the value of $sin(x)$ change aswell?

Thank you!

Answer 1

There is a difference between the partial derivative $\frac{\partial}{\partial x}$ and the total derivative $\frac{d}{dx}$. For example, if we have variables $(u,x)$ and the equation $f=f(x,u(x))=x^2+u^3$ and we take the partial derivative we get $\frac{\partial f}{\partial x}=2x$ but if we take the total derivative we get $\frac{d f}{dx}=2x+3u^2\frac{\partial u}{\partial x}$, applying the chain rule. This distintion is a key point in classical mechanics for example and captures essentially what you are asking.

See: What exactly is the difference between a derivative and a total derivative?

Answer 2

To make things simpler, imagine a very simple autonomous dynamical system

$$ \frac{{\rm d}u}{{\rm d}x} = \alpha (u - u_0) \tag{1} $$

for some constants $\alpha$ and $u_0$. The solutions to this system are of the form

$$ u(x) - u_0 = ce^{\alpha x} \tag{2} $$

The interesting thing to note here is that if $\alpha > 0$ then the distance between $u(x)$ and $u_0$ grows exponentially with increasing $x$, so $u = u_0$ is said to be unstable. Whereas if $\alpha < 0$ the opposite occurs, and the distance between $u$ and $u_0$ shrinks with increasing $x$, in this case $u = u_0$ is stable.

Now let's make things a bit more general. Imagine a system of the form

$$ \frac{{\rm d}u}{{\rm d}x} = f(u) \tag{3} $$

and suppose there exists a point $u_0$ such that $f(u_0) = 0$ (like in Eq. (1)), if you want to understand the local stability of (3) you could Taylor expand $f$ around $u_0$

$$ f(u) = f(u_0) + \left.\frac{{\rm d}f}{{\rm d}u}\right|_{u = u_0}(u - u_0) + \cdots \tag{4} $$

Remember that $f(u_0) = 0$, so at first order $f(u)\approx f'(u_0)(u - u_0)$, and Eq. (3)

$$ \frac{{\rm d}u}{{\rm d}x} \approx \underbrace{f'(u_0)}_{\alpha}(u- u_0) \tag{5} $$

Now compare this with Eq. (1) and you realize that in order to understand the stability of the system around the point $u_0$ you need to know the value of $f'(u_0) $ a.k.a the Jacobian. There's a lot of caveats here you should be aware of, probably your text talks about them (e.g. what happens if $f'(u_0) = 0$, ...)

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EDIT

Now imagine a system in two dimensions, something like

$$ \frac{{\rm d}u}{{\rm d}x} = f(u, v) ~~~~ \frac{{\rm d}v}{{\rm d}x} = g(u, v)\tag{6} $$

You can define vectors ${\bf z} = {u \choose v}$ and ${\bf F} = {f \choose g}$ so that the system above can be written as

$$ \frac{{\rm d}{\bf z}}{{\rm d}x} = {\bf F}({\bf z}) \tag{7} $$

In this case nothing changes much, you can repeat the same analysis as in the first part and realize that the stability of the system around a point ${\bf z} = {\bf z}_0$ is given by the eigenvalues of the Jacobian evaluated at that location. And as before we require that $\color{blue}{{\bf F}({\bf z}_0) = 0}$. I highlight this condition because it will become important later.

Now to the final part. Instead of an autonomous system, consider a system of the form

$$ \frac{{\rm d}u}{{\rm d}x} = f(x, u) \tag{8} $$

You could rename $v = x$ (that is, create a new state), and note that

$$ \frac{{\rm d}u}{{\rm d}x} = f(u, v) ~~~~ \frac{{\rm d}v}{{\rm d}x} = 1 \tag{9} $$

So in theory you could repeat the same analysis all over again, but, you can see from this that the resulting field ${\bf F}$ never vanishes (that is, the blue expression above can never be satisfied)