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I have had no experience with differential equations before I was presented with this problem on a homework. The equation is:

$ -u'' + \beta u' = 0 $

$ u(0) = 0 $

$ u(1) = 1 $

I have found the general solution to be $ u = \frac{-1}{e^\beta - 1} + \frac{e^{\beta x}}{e^{\beta} - 1}$ and am now asked to derive the central, forward, and backward finite difference schemes (I am only worried about understanding central for now though).

Bear with me now, like I said I have not had any experience with differential equations. From my understanding, the central finite difference yields:

$ u''(x) \approx \frac{2u(x) - u(x+h) - u(x-h)}{h^2} $

and therefore $ -u'' + \beta u' = 0 $ becomes:

$ \frac{2u(x) - u(x+h) - u(x-h)}{h^2} + \beta u'(x) = 0 $

$ 2u(x) + u(x + h) - u(x - h) = -\beta u'(x)h^2 $

Am I on the right track so far? I am not sure what a complete "finite difference scheme" even is, so when should I stop? I have seen examples where the end result is a large matrix equation - is that required to answer this question?

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up vote 2 down vote accepted

It looks like your overall approach is correct. Your second derivative central difference seems to be off by a sign though, the numerator should be $u(x+h)-2u(x)+u(x-h)$, see .

You still need to expand the first derivative of $u$ as a finite difference. If the problem is asking for first order central differences as well, you would write $u'(x)\approx \frac{u(x+h/2)-u(x-h/2)}{h}$, or most likely for this problem is wanted $u'(x)\approx \frac{u(x+h)-u(x-h)}{2h}$.

Notice that once you write all derivatives out, you get an equation that involves only $u(x)$, $u(x+h)$ and $u(x-h)$. This is effectively your "scheme" or rule for updating your $u$. If say your $u$ is a function of space, then you can write out a matrix equation. Here's how: if you are on the interval $[0,1]$, then you can write down the vector $[u(0),u(h),u(2h),\ldots,u(1-h),u(1)]$. Since your $u$ satisfies a linear equation involving $u(x)$, $u(x+h)$ and $u(x-h)$, you can think of this equation as an appropriate matrix multiplying the aforementioned vector.

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Okay thanks. It does not specifically ask for a first order central difference, it just asks for "the central difference scheme for computing the numerical solution of the equation". Do you know what this means, exactly? – Logan Serman Apr 29 '12 at 16:51
There are multiple central difference formulas, but the simplest and most popular one is what Sam said: $u'(x) = (u(x+h)-u(x-h)) / (2h)$. That's the one I would use unless the question specifies otherwise. – Jitse Niesen Apr 30 '12 at 7:29

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