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I saw in a code for discretization of something like $\frac{d^2T(x)}{d^2x}$ , ( $x = sin(\theta)$ ) tries

   !============== here are some defined variables
   deltaTheta = pi/150
   deltax(i) = cos(theta(i)) * deltaTheta  
   dx = 0.5*(deltax(i) + deltax(i+1))
   dx1 = 0.5*(deltax(i-1)+ deltax(i))
   !============== here is the discretization that I am talking about
   ... = ((T(i+1) - T(i))/dx - (T(i) - T(i-1))/dx1)/(0.5 *(dx + dx1))

this discretization was for non-boundary cites; what does this discretization called?

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  • $\begingroup$ Do you mean beyond being called the three point stencil? Or are you talking about the choice of grid points? $\endgroup$
    – origimbo
    Nov 10, 2017 at 16:16
  • $\begingroup$ @origimbo What I don't understand are defining and usage of "dx" and "dx1" $\endgroup$
    – Mr. Who
    Nov 10, 2017 at 18:32
  • $\begingroup$ The three points might have different separations. $\endgroup$
    – nicoguaro
    Nov 10, 2017 at 19:21

1 Answer 1

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It is just the computation of the second derivative from three non-equidistant points. It is computed by first computing (an approximation of) the derivatives on each of the two intervals adjacent to point $i$ using $$ \frac{T(i+1)-T(i)}{\Delta(i+1)} $$ and $$ \frac{T(i)-T(i-1)}{\Delta(i)} $$ and then computing (an approximation of) the second derivative by taking a finite difference stencil with regard to the midpoints of the two intervals: $$ \frac{ \frac{T(i+1)-T(i)}{\Delta(i+1)} - \frac{T(i)-T(i-1)}{\Delta(i)} }{\frac{\Delta(i+1)+\Delta(i)}{2}} $$ Here, $\Delta(i+1)=x(i+1)-x(i)$ and $\Delta(i)=x(i)-x(i-1)$ are the dx1 and dx terms in the formula of the code you're looking at.

I have no idea how this relates to the dx and dx1 variables -- these look wrong to me since they are not increments but averages of the deltax(i) which themselves are not deltas (differences) but point locations as far as I can see. I can't seem to see that this choice of variable names or expressions makes sense.

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    $\begingroup$ thanks for the reply, since x = sin(\theta) then dx = cos(\theta) d\theta so detlax is really a difference; all the points that you mentioned make sense to me. $\endgroup$
    – Mr. Who
    Nov 11, 2017 at 18:12
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    $\begingroup$ BTW, thanks for your informative website, very glad to know you. $\endgroup$
    – Mr. Who
    Nov 11, 2017 at 18:42

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