# Odd Differencing Method for Solving DE

1. Jul 12, 2011

### Kreizhn

Hey all,

I have a friend working in Earth Sciences who appears to be doing something with DEs. If I understand his data correctly, it appears as though he's solving a differential equation of the form
$$\frac{dy}{dx} = f(x)$$
Let [0,L] be the interval over which this is to be solved, $y(0) = y_0$ and take a partition $0 =x_0 < x_1 < \cdots < x_{n-1} < x_n = L$. Then it appears that his differencing method is giving the approximation of the $(i+1)^{st}$ value as
$$y(x_{i+1}) = \frac{ f(x_{i+1}) (x_{i+1} - x_i) - \left(\sum_{j=0}^i y(x_j)\right) (x_i - x_{i-1}) }{x_{i+1}}$$
I don't recognize the formula. The summation term would be theoretically reminiscent of an integral yes? Does anyone recognize this?

Edit: Sorry, that summation might actually only be $y(x_i)$. I'm not quite sure yet since the data is a little fuzzy. If that's the case, this is almost an Euler method right? But it still doesn't quite seem there.

2. Jul 12, 2011

### pmsrw3

Are you sure it's a y(x_j) in the sum and not f(x_j)? Given the DE, y(x_j) makes no sense at all.

EDIT: No, actually, it doesn't make sense either way.

3. Jul 12, 2011

### Kreizhn

Just checked the data. That summation is wrong, but it does represent the solution at the previous partition point. This would be $y(x_i)$.

It almost looks like an Euler method with some sort of bizarre scaling. We can re-arrange it to get

$$y(x_{i+1}) x_{i+1} = y(x_i)(x_i -x_{i-1}) + f(x_{i+1}) (x_{i+1} - x_i)$$

My friend now tells me that if we drop the $x_{i-1}$ term from the y(x_i) he gets better results. This would leave

$$y(x_{i+1}) x_{i+1} = y(x_i)x_i + f(x_{i+1}) (x_{i+1} - x_i)$$

Which looks exactly like an Euler method with some sort of weird scaling on the solution.

4. Jul 12, 2011

### Studiot

Looks more like Adams-Bashforth than Euler to me.

5. Jul 13, 2011

### Kreizhn

Isn't one-step AB just Euler?