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I have a set of data, $x, y$ and $ z$, each with length n:

$x \rightarrow \{x_{1}...x_{n}\}$

$y \rightarrow \{y_{1}...y_{n}\}$

$z \rightarrow \{z_{1}...z_{n}\}$

$y$ and $z$ are parameterised by $x$:

$y = y(x)$

$z = z(x)$

Problem 1

I wish to find derivatives $\Large \frac{dy}{dx}$ and $\Large \frac{dz}{dx}$.

The 'second-order error' numerical scheme: $\Large \frac{dy}{dx} = \frac{y(x+dx)-y(x-dx)}{2dx} + \mathcal{O}({dx}^{2})$

woud usually work, but in my case, my 'x' data is not uniformly spaced (it is nonlinear) i.e. '$dx$' is not constant (hence my problem turns into a first order error one). The problem applies for the derivative of z.

Can anyone suggest a good method to find these derivatives? (for anyone interested in meteorology, $x$ are pressure coordinates in atmosphere, $y$ is the potential temperature, $z$ is the geopotential height).

One person suggested that I interpolate $x$, $y$ and $z$ onto a new $x$ whereby this new list of $x$ values are uniformly spaced. However, I also want to see how the error introduced here propagates to the end result.

Problem 2

I wish to find the derivative: $\Large \frac{dy}{dz}$. However, as my $z$ are not necessarily spaced evenly between each other, my numerical scheme is not second order. Hence, what scheme should I use and what is my resulting error, preferably as a function of $x$. I guess it's a similar problem to above.

SomePhysicsStudent
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  • You can use Taylor expansion to determine appropriate weights to get the best possible estimate for $\frac{dy}{dx}$ given $y(x+h_1),y(x-h_2)$. – Ian Nov 09 '16 at 15:15
  • (Or using even more data, if you want, but this is probably the most straightforward because it lets you estimate the derivative at a point even when none of your data points are there). – Ian Nov 09 '16 at 15:29

1 Answers1

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By Taylor expansion, for $h_1 \neq h_2$ we have

$$a y(x+h_1) + by(x+h_2) =a(y(x)+y'(x)h_1 +y''(x)h_1^2/2+o(h_1^2)) +b(y(x)+y'(x)h_2 +y''(x)h_2^2/2+o(h_2^2)).$$

To make this approximate $y'(x)$, you want to eliminate the zeroth order terms and make the net coefficient on $y'(x)$ be $1$. This means

$$a+b=0 \\ ah_1 + b h_2=1.$$

The solution to this is $a=-\frac{1}{h_2-h_1},b=\frac{1}{h_2-h_1}$.

Unfortunately, the net second order term is now

$$\frac{1}{h_2-h_1}(y''(x)h_2^2-y''(x)h_1^2)$$

which does not vanish unless $h_1=-h_2$ (in which case you recover the familiar centered difference scheme).

You can do similar techniques to come up with other schemes for approximating derivatives on irregular grids.

Ian
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