0
$\begingroup$

I'm trying to understand what is the general method for calculating a backwards difference for a mixed partial of $n$ variables. Let's start with one variable:

The forward and backward finite differences are

$$f'(x) =\frac{f(x+h)-f(x)}{h}$$

and

$$f'(x) = \frac{f(x) - f(x-h)}{h}$$

However, when we go to mixed partials for $n=2$ variables discretized on a 2D grid with $h = x_{i+1}-x_{i}$, we can use (a central difference).

$$\left(\frac{\partial^2 u}{\partial x\partial y} \right)_{i,j} = \frac{u_{i+1,j+1} - u_{i+1,j-1} - u_{i-1,j+1} + u_{i-1,j-1}}{4 \Delta x \Delta y} + \mathscr{O}\left((\Delta x)^2, (\Delta y)^2 \right) \qquad (1)$$

How would one go about finding $\mathscr{O}\left((\Delta x)^2, (\Delta y)^2 \right)$? Furthermore, what about backwards difference? Notice the above central difference needs information of the $i+1,j+1$ point which is problematic near or approaching a singularity.

EDIT: Is this approached with a matrix method of the Taylor expansion of the mixed partials?

$\endgroup$
1
$\begingroup$

You can write the Taylor series for the function at the points used in (1), e.g.:

$$ f(x+\Delta x, y+ \Delta y) = \sum_{i=0}^{\infty} \dfrac{(\Delta x)^i}{i!} \dfrac{\partial^{i}}{\partial x^i} f(x, y+ \Delta y) = \sum_{i=0}^{\infty} \dfrac{(\Delta x)^i}{i!} \sum_{j=0}^{\infty} \dfrac{(\Delta y)^j}{j!} \dfrac{\partial^{i+j}}{\partial y^j \partial x^i} f(x, y) $$

, and use them to find the formal accuracy of the scheme.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.