Linearization and differentials

[Mårten]

If you have a water tank with an inflow u, and an outflow v, you have that

$$\frac{dV}{dt} = A \frac{dh}{dt} = u - v.$$

You can now linearize this expression so that you get

$$A \frac{d}{dt}(\Delta h) = \Delta u - \Delta v = (u-u_0) - (v-v_0),$$

where $$\Delta h = h-h_0.$$

I think I know the Taylor mechanism behind this pretty well, but what I can't figure out is how d/dt(h-h0) becomes (u-u0) - (v-v0) (for simplicity, I assume A=1 here). I mean, h0 is a constant, so the derivative of (h-h0) should therefore be the same as the derivative of h, i.e., equal to u-v. What am I doing wrong here - could someone explain?

 

[Jitse Niesen]

I don't understand what you want to do. As you say, the equation you get after "linearization" is very different from the original equation, so there must be a mistake in this linearization process that you're doing. I would say that the original equation is already linear, so the linearized equation is just the same:

$$A \frac{d}{dt} (\Delta h) = u-v.$$

But perhaps you're thinking about a different kind of linearization.

 

[Mårten]

I'll try to give some more details. The reason you want to do a linearization here, is because the equation is non-linear from the beginning, with, according to Bernoulli, the outflow $v=a \sqrt{2gh}=4.43a \sqrt{h}$, with a = the area of the outflow hole. So the equation then really is

$$\frac{dV}{dt} = A \frac{dh}{dt} = u - 4.43a \sqrt{h}.$$

The question is how you come from here to the following equation below

$$A \frac{d}{dt}(\Delta h) = \Delta u - \Delta v = (u-u_0) - (v-v_0),$$

when at the same time you know that $A \frac{d}{dt}(h-h0) = A \frac {dh}{dt}$.

 

[Jitse Niesen]

I still don't quite understand what you're doing. So instead of answering directly, let me try to explain how I would linearize that equation and perhaps you find the answer to your question in my explanation.

In general, linearization means that you assume that something is so small that you can neglect its square. In this case, you assume that h is close to h0, so that $\Delta h = h-h_0$ is small.

Now, let's look at your differential equation
$$A \frac{dh}{dt} = u - 4.43a \sqrt{h}.$$
On the left-hand side, you get
$$A \frac{dh}{dt} = A \frac{d}{dt} (h_0 + \Delta h) = A \frac{d}{dt} (\Delta h)$$
because $h_0$ is a constant. That's also what you got. Now for the right-hand side:
$$u - 4.43a \sqrt{h} = u - 4.43a \sqrt{h_0 + \Delta h}.$$
To get any further, you have to develop this in a Taylor series around $h=h_0$:
$$u - 4.43a \sqrt{h} = u - 4.43a \sqrt{h_0} - \frac{4.43a}{2\sqrt{h_0}} (h-h_0) + \frac{4.43a}{8h_0\sqrt{h_0}} (h-h_0)^2 - \frac{4.43a}{16h_0^2\sqrt{h_0}} (h-h_0)^3 + \cdots$$
Then you neglect the terms with $(h-h_0)^2$ and $(h-h_0)^3$ and higher powers. Of course, in practise you wouldn't even compute these terms. So altogether you get
$$A \frac{d}{dt} (\Delta h) = u - 4.43a \sqrt{h_0} - \frac{4.43a}{2\sqrt{h_0}} \Delta h.$$
That is what I would call the linearization of the differential equation.

Now back to your equation. The term
$$\frac{4.43a}{2\sqrt{h_0}} \Delta h$$
on the right-hand side is how much the outflow changes (up to first order) when the height changes from $h_0$ to $h_0+\Delta h$, so you can denote it by $\Delta v$. So you can write the equation
$$A \frac{d}{dt} (\Delta h) = u - 4.43a \sqrt{h_0} - \frac{4.43a}{2\sqrt{h_0}} \Delta h.$$
as
$$A \frac{d}{dt} (\Delta h) = u_0 - v_0 - \Delta v$$
where $u_0$ and $v_0$ are the inflow and outflow when the height is $h=h_0$. There is no term $\Delta u$ because the inflow u does not depend on the height h, but in general you would have
$$A \frac{d}{dt} (\Delta h) = u_0 + \Delta u - v_0 - \Delta v.$$
That's another form in which you can write the linearized equation.

Finally, why do you want to linearize the equation in the first place? One reason is to assess the stability of some equilibrium point of the equation. The height $h_0$ is an equilibrium point if the inflow and the outflow are equal. In that case, $u_0 = v_0$ and thus the linearized equation becomes
$$A \frac{d}{dt} (\Delta h) = \Delta u - \Delta v.$$
That is the only way I can see that you can get to that equation, and in that case, there is no problem.

 

[Mårten]

Thank you very much for that thorough answer!

One of the points that I missed from the beginning, was just the fact that a precondition is that you have equilibrium in the system, i.e., inflow equals outflow, so the water level doesn't change, $\frac{dh}{dt}=0$, so u_0=v_0. So all in all you get

$$A \frac{d}{dt}(\Delta h) = \Delta u - \Delta v = (u-u_0) - (v-v_0) =$$

$$A \frac{dh}{dt} = u - v.$$

That these two equations above are equal was the first thing I couldn't see from the beginning.

Now, these equations are non-linear, because $v=a \sqrt{2gh}=4.43 \sqrt{h}$ (I presume a=A=1 here). So I tried to do a twodimensional Taylorexpansion, since water level change depends on both inflow, u, and outflow, v:

$$\frac{dh}{dt} = \frac{d}{dt}(\Delta h) = f(u,h) = u -4.43\sqrt{h} \approx$$

$$f(u_0,h_0) + f'_u(u_0,h_0)(u-u_0) + f'_h(u_0,h_0)(h-h_0) =$$

$$\left(u_0 - 4.43\sqrt{h_0}\right) + \Bigl( 1\cdot (u-u_0)\Bigr) + \left(-4.43\frac{1}{2\sqrt{h_0}}(h-h_0)\right).$$

As we said, just at (u_0,h_0) we had equilibrium, so $v_0=4.43\sqrt{h_0}$ equals u_0, and so the leftmost parenthesis in the lowest equation above disappears (this was also something I failed to see at first). So finally we get the approximated linearization as

$$\frac{d}{dt}(\Delta h) \approx \Delta u -4.43\frac{1}{2\sqrt{h_0}}\Delta h .$$

This equation could be solved, for instance by Laplace transformation methods, and you receive a function which approximates $\Delta h(t)$ pretty well around (u_0,h_0).

I hope I got it all right now - hopefully someone else could have use of this, which I myself had a hard time with.