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I would like to construct a matrix $G$ composed of block matrices $G^{(i)}$ as defined entry-wise below stacked vertically as $$G = \begin{bmatrix} G^{(1)} \\ G^{(2)} \\ \vdots \\ G^{(i)} \\ \vdots \\ G^{(n)} \end{bmatrix}$$ for some natural number $n$.

$G^{(i)}_{j,k}$ stands for the element of the $j$'th row and the $k$'th column of the $i$'th block matrix. \begin{align} G^{(i)}_{j,k} :=& \frac{t_{i+1}-t}{h_i}\delta_{k,i}+\frac{t-t_i}{h_i}\delta_{k,i+1} -\frac16(t-t_i)(t_{i+1}-t)\Big(1+\frac{t_{i+1}-t}{h_i}\Big)\delta_{k,n+i} \\ &-\frac16(t-t_i)(t_{i+1}-t)\Big(1+\frac{t-t_i}{h_i}\Big)\delta_{k,n+i+1} \ \Bigg|_{t=t_{j,i}} \end{align} where $i\in\{1,2,\cdots,n\}$, $j\in\{1,2,\cdots,m(j)\}$, $m(j)$ is a natural numbered function of $j$, and $k\in\{1,2,\cdots,2n-2\}$, $\delta_{i,j}$ is the Kronecker delta function, and the vertical bar at the right end indicates $t$ is to be evaluated at $t_{j,t}$.

What is the most elegant and convenient way to do this?

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  • $\begingroup$ ` t is to be evaluated at t_j,t` - explain please. You have a mixture of $t,t_i,t_{i+1}$ in your problem, but is $t$ a function? And if so why does it take either 1 (i or i+1) or two arguments (j,i)? $\endgroup$
    – flinty
    Commented Aug 4, 2020 at 17:49
  • $\begingroup$ @flinty: It just means substituting $t$ by $t_{j,i}$. $\endgroup$
    – Hans
    Commented Aug 4, 2020 at 19:23
  • $\begingroup$ Use SparseArray. Or DiagonalMatrix. What are the limits on j and k? Are the G's square matrices? Can n be anything or is it related to the bounds on i and j? More details would be helpful. $\endgroup$
    – march
    Commented Aug 4, 2020 at 21:08
  • $\begingroup$ You have a $\delta_{k,i}$ and a $\delta_{n+k,n+i}$, but $\delta_{n+k,n+i}=\delta_{k,i}$, so these actually are part of the same matrix element. I suspect that you meant something else, however, since you wrote them separately. Can you clarify? I think editing your post with a small example output would be very helpful, with explicit (small) limits on j and j like 3 or something, i = 2 (so, two blocks), and n = something. As it stands, $n$ actually doesn't do anything in this expression, as far as I can tell. $\endgroup$
    – march
    Commented Aug 4, 2020 at 21:19
  • $\begingroup$ @march: You are absolutely right. I have edited my question. Please review. Thank you. $\endgroup$
    – Hans
    Commented Aug 4, 2020 at 23:18

2 Answers 2

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Assuming you have definitions already for n, single argument t[x], and two-argument t[x,y], then this is fairly straightforward:

g[i_, n_] := 
 Table[(t[i + 1] - t[j, i])/h[i] KroneckerDelta[k, 
     i] + (t[j, i] - t[i])/h[i] KroneckerDelta[k, i + 1] - 
   1/6 (t[j, i] - t[i]) (t[i + 1] - 
      t[j, i]) (1 + (t[i + 1] - t[j, i])/h[i]) KroneckerDelta[k, 
     n + i] - 
   1/6 (t[j, i] - t[i]) (t[i + 1] - 
      t[j, i]) (1 + ((t[j, i] - t[i])/h[i])) KroneckerDelta[k, 
     n + i + 1], {j, 1, 3}, {k, 1, 2*n-2}]

result = With[{n = 4}, Join @@ Array[g[#, n] &, n]]
result // Dimensions (*expected {12,3}*)
result // MatrixForm
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  • $\begingroup$ Is there a command to show the final explicit matrix form without the explicit KroneckerDelta function? $\endgroup$
    – Hans
    Commented Aug 4, 2020 at 19:32
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    $\begingroup$ @Hans can you explain what you're after? MatrixForm is just a presentational thing - you shouldn't use it in computations. $\endgroup$
    – flinty
    Commented Aug 4, 2020 at 22:05
  • $\begingroup$ Your solution is good. I understand I can use it for computation. I am wondering if there is a command to see the matrix in the explicit more conventional form as a sanity check. $\endgroup$
    – Hans
    Commented Aug 4, 2020 at 22:55
  • $\begingroup$ Yes that's what the //MatrixForm is for. It just looks weird because the expressions in the matrix are very wide. $\endgroup$
    – flinty
    Commented Aug 4, 2020 at 22:57
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    $\begingroup$ @Hans Array[g[#, n] &, n] generates a list of {g[1,n],g[2,n],g[3,n],..., g[n,n]} and @@ splices this list into a sequence of arguments for Join , so Join[g[1,n],g[2,n],g[3,n]...]. This is stacking them 'vertically' though Mathematica isn't like matlab; it works on lists, so it doesn't have a notion of vertical/horizontal, just lists and list nesting. If you wanted them horizontal you would use Join[g[1,n],g[2,n],g[3,n]....,{2}] with a level spec {2} at the end. $\endgroup$
    – flinty
    Commented Aug 5, 2020 at 11:26
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Is this a formula with the Einstein sum convention. If so then the summation over two same indices has to be carried through.

g[i_, j_, k_, n_] := 
 Sum[(t[i + 1] - t[j, i])/h[ii] KroneckerDelta[k, 
     ii] + (t[j, i] - t[ii])/h[ii] KroneckerDelta[k, ii + 1] - 
   1/6 (t[j, i] - t[ii]) (t[ii + 1] - 
      t[j, i]) (1 + (t[ii + 1] - t[j, i])/h[ii]) KroneckerDelta[n + k,
      n + ii] - 
   1/6 (t[j, i] - t[ii]) (t[ii + 1] - 
      t[j, i]) (1 + ((t[j, i] - t[ii])/h[ii])) KroneckerDelta[n + k, 
     n + ii + 1], {ii, 1, 3}]

g[1, j, k, n]

(KroneckerDelta[1, k] (t[2] - t[j, 1]))/h[1] + (
 KroneckerDelta[2, k] (t[2] - t[j, 1]))/h[2] + (
 KroneckerDelta[3, k] (t[2] - t[j, 1]))/h[3] + (
 KroneckerDelta[2, k] (-t[1] + t[j, 1]))/h[1] - 
 1/6 KroneckerDelta[1 + n, 
   k + n] (1 + (t[2] - t[j, 1])/h[1]) (t[2] - t[j, 1]) (-t[1] + 
    t[j, 1]) + (KroneckerDelta[3, k] (-t[2] + t[j, 1]))/h[2] - 
 1/6 KroneckerDelta[2 + n, 
   k + n] (1 + (t[3] - t[j, 1])/h[2]) (t[3] - t[j, 1]) (-t[2] + 
    t[j, 1]) + (KroneckerDelta[4, k] (-t[3] + t[j, 1]))/h[3] - 
 1/6 KroneckerDelta[3 + n, 
   k + n] (1 + (t[4] - t[j, 1])/h[3]) (t[4] - t[j, 1]) (-t[3] + 
    t[j, 1]) - 
 1/6 KroneckerDelta[2 + n, 
   k + n] (t[2] - t[j, 1]) (-t[1] + t[j, 1]) (1 + (-t[1] + t[j, 1])/
    h[1]) - 1/
  6 KroneckerDelta[3 + n, 
   k + n] (t[3] - t[j, 1]) (-t[2] + t[j, 1]) (1 + (-t[2] + t[j, 1])/
    h[2]) - 1/
  6 KroneckerDelta[4 + n, 
   k + n] (t[4] - t[j, 1]) (-t[3] + t[j, 1]) (1 + (-t[3] + t[j, 1])/
    h[3])

The interpretation distinguishes between the indices i of the Einstein summation and the indices i from the discretization of time i in t_ij.

Example for displaying this as a matrix is

Table[h[i, j, k, n], {i, 1, 3}] // MatrixForm

enter image description here

This is much to lengthy for g.

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  • $\begingroup$ Actually, there is no summation over 'ii' here. 'ii' should be the index of the block matrix. $\endgroup$
    – Hans
    Commented Aug 4, 2020 at 19:29

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