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?


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 http://en.wikipedia.org/wiki/Finite_difference .

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.

  • $\begingroup$ 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? $\endgroup$ – Logan Serman Apr 29 '12 at 16:51
  • $\begingroup$ 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. $\endgroup$ – Jitse Niesen Apr 30 '12 at 7:29

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.