How can I create a function with several summations and Kronecker delta symbol in two dimension?

I have tried to create a several summations function including Kronecker delta symbol.

The function is the follwing.

where

I am working with n=2 which is two dimensional space and theta=-1/2.

Here alpha and beta are two dimensional point. For example, alpha=(2,0), beta=(2,0) and ei's are standard unit normal vector in two dimension.

I have tried to create the Kronecker delta symbol in two dimension, but everything didn't work.

I would really appreciate it, if you can help me.

[Bruno Luong]

What is the subscribe

alpha(2*ek)

does that means taking the scalar (dot) product between alpha and (2*ek) ?

Same question with alpha(ei+ej) but I guess the answer is also the same.

Or does the (a*ek) is not on subscribe but argument of the kronecker delta function, meaning result is 1 if (and only if) alpha == 2*ek?

[matlab beginner]

Thank you for your question. For example, if alpha=(2,0), and e_{k}=e_{2}=(0,1) then alpha is not equal to 2e_{2}=(0,2) which means delta_{alpha (2 ek)} =0, but if alpha=(2,0), and e_{k}=e_{1}=(1,0) then apha is equal to 2e_{1}=(2,0) which yields delta_{alpha (2ek)}=1. So, this delta literally means the Kronecker delta symbol in two dimension.

[Bruno Luong]

Then your notation is not standard: the parenthesis should not be a subsribe but in the same level than the delta symbol.

[matlab beginner]

The argument that you added is exactly correct!

Answer 1

[Bruno Luong]

Here is the code, I do not check careful though

% Dummy data
alpha = [0 1;
    1, 0;
    1, 1;
    2, 0;
    0, 2];
beta = [1, 1;
    2, 0;
    0, 2];
theta = [1; 2; 3; 4];
n = 2;
[i, j] = ndgrid(1:n);
ij = [i(:) j(:)];

% Tensor calculation start here
m = size(alpha,1);
n = size(beta,1);
p = size(theta,1);
Res = zeros(m,n,p);
for k=1:p
    Pa = GetPTesnsor(alpha, ij, theta(k));
    Pb = GetPTesnsor(beta, ij, theta(k));
    T = tensorprod(Pa,Pb,[3 4],[3 4]);
    Res(:,:,k) = reshape(T,[m n]);
end
Res

Res = 
Res(:,:,1) =

     0     0     0
     0     0     0
     2     0     0
     0     5     4
     0     4     5


Res(:,:,2) =

     0     0     0
     0     0     0
     2     0     0
     0    13    12
     0    12    13


Res(:,:,3) =

     0     0     0
     0     0     0
     2     0     0
     0    25    24
     0    24    25


Res(:,:,4) =

     0     0     0
     0     0     0
     2     0     0
     0    41    40
     0    40    41

function P = GetPTesnsor(alpha, ij, theta)
ALPHA = reshape(alpha,[size(alpha),1,1,1,1]);
IJ = reshape(ij,[1,1,size(ij),1,1]);
I = IJ(1,1,:,1,1,1);
J = IJ(1,1,:,2,1,1);
THETA = reshape(theta,[1,1,1,1,size(theta)]);
n = max(ij,[],'all');
delta = @(x1,x2) x1==x2;
k = reshape((1:n),[1 1 n]);
P = deltav(ALPHA,getei(I,n)+getei(J,n)) + ...
    THETA.*delta(I,J).*sum(deltav(ALPHA,2*getei(k,n)),3);
end

function ei = getei(i,n)
ei = accumarray([(1:length(i))' i(:)],1,[length(i) n]);
szei = [size(i) n];
ei = reshape(ei, szei);
end

function x = deltav(a, b)
sza = size(a);
szb = size(b);
szar = [sza(1:end-1) ones(1,length(szb)-length(sza)), sza(end)];
ar = reshape(a, szar);
x = all(ar == b, length(szar));
end

[matlab beginner]

Wow.... it looks so complicated and super nice! I really appreciate your help!

I will try with this code. I would never gonna be able to do like this.

Thank you very much again!!!

[matlab beginner]

Hi Bruno,

I really apprecaite this code perfectly worked!!!

I have tried this to expand a little bit but my code is not working now.

Could you please look at my code if you don't mind?

Here is the code:

% Dummy data
format rat
alpha = [2 0 0 ;
    0 2 0 ;
    0 0 2 ;
    1 1 0 ;
    1 0 1 ;
    0 1 1];
beta = [2 0 0 ;
    0 2 0 ;
    0 0 2 ;
    1 1 0 ;
    1 0 1 ;
    0 1 1];
n = 3;
theta = -1/4;
c=1/(1+2*theta+n*theta.^2);
[i, j, l] = ndgrid(1:n);
ijl = [i(:) j(:) l(:)];
% Tensor calculation start here
m = size(alpha,1);
n = size(beta,1);
p = size(theta,1);
Res = zeros(m,n,p);
for k=1:p
    Pa = GetPTesnsor(alpha, ijl, theta(k));
    Pb = GetPTesnsor(beta, ijl, theta(k));
    T = tensorprod(Pa,Pb,[3 4],[3 4]);
    Res(:,:,k) = c*reshape(T,[m n]);
end
Res
function P = GetPTesnsor(alpha, ijl, theta)
ALPHA = reshape(alpha,[size(alpha),1,1,1,1]);
IJL = reshape(ijl,[1,1,size(ijl),1,1]);
I = IJL(1,1,:,1,1,1);
J = IJL(1,1,:,2,1,1);
L = IJL(1,1,:,3,1,1);
THETA = reshape(theta,[1,1,1,1,size(theta)]);
n = max(ijl,[],'all');
delta = @(x1,x2) x1==x2;
k = reshape((1:n),[1 1 n]);
P = deltav(ALPHA,getei(I,n)+getei(J,n)+getei(L,n)) + ...
    THETA.*delta(I,J).*sum(deltav(ALPHA,2*getei(k,n)+getei(L,n)),3)+ ...
    THETA.*delta(J,L).*sum(deltav(ALPHA,2*getei(k,n)+getei(I,n)),3)+ ...
    THETA.*delta(I,L).*sum(deltav(ALPHA,2*getei(k,n)+getei(J,n)),3);
end
function ei = getei(i,n)
ei = accumarray([(1:length(i))' i(:)],1,[length(i) n]);
szei = [size(i) n];
ei = reshape(ei, szei);
end
function x = deltav(a, b)
sza = size(a);
szb = size(b);
szar = [sza(1:end-1) ones(1,length(szb)-length(sza)), sza(end)];
ar = reshape(a, szar);
x = all(ar == b, length(szar));
end

[Bruno Luong]

Try this change

function P = GetPTesnsor(alpha, ijl, theta)
ALPHA = reshape(alpha,[size(alpha),1,1,1,1]);
IJL = reshape(ijl,[1,1,size(ijl),1,1]);
I = IJL(1,1,:,1,1,1);
J = IJL(1,1,:,2,1,1);
L = IJL(1,1,:,3,1,1);
THETA = reshape(theta,[1,1,1,1,size(theta)]);
n = max(ijl,[],'all');
delta = @(x1,x2) x1==x2;
k = reshape((1:n),[1 1 1 1 n]); % change
P = deltav(ALPHA,getei(I,n)+getei(J,n)+getei(L,n)) + ...
    THETA.*delta(I,J).*sum(deltav(ALPHA,2*getei(k,n)+getei(L,n)),5)+ ...
    THETA.*delta(J,L).*sum(deltav(ALPHA,2*getei(k,n)+getei(I,n)),5)+ ...
    THETA.*delta(I,L).*sum(deltav(ALPHA,2*getei(k,n)+getei(J,n)),5); % change
end

Answer 2

[Sam Chak]

@matlab beginner

Have you tried this kroneckerDelta() function?

https://www.mathworks.com/help/symbolic/sym.kroneckerdelta.html

[matlab beginner]

Thank you very much for your comment.

Yes, I did try it but I have no idea how to apply it in two dimension. For example, delta_{(2,0),(2,1)} like this.

I first tried to use delta_{(2,2)} delta_{(0,1)} but this doesn't work when I work with index like delta_{alpha,(e_{i}+e_{j})}.

Answer 3

[Walter Roberson]

Your expressions involve delta subscript something that is a multiplication, rather than delta subscript something, all multiplied by something. That means that the entire subscript is the arguments to the delta function.

The below defines a multidimensional delta

https://functions.wolfram.com/IntegerFunctions/KroneckerDelta2/introductions/TensorFunctions/ShowAll.html

in short it is 1 if and only if all of the arguments are 0.

But your second arguments in each case are vectors in 3 space (not two space like you said) and it is not obvious that you can compare a vector such as e_j to a scalar 0. The definition does not say anything about comparing magnitude, only about comparing values.

You could potentially rewrite your expressions in terms of magnitude in each of 4 dimensions (scalar, i, j, k) and then it might make sense to take the delta... but I am not convinced that your expressions are meaningful in their current form.

[matlab beginner]

Thank you for your comment. I don't understand what you mean by the multiplication. For example, delta_{alpha(e_{i}+e_{j})} means it is 1 only when alpha=e_{i}+e_{j} where e_{i}'s are standard unit normal vectors in R2 and alpha is also the vector in R2. By these reasons, my second arguments are also vector in two dimensional space. I have checked your link and I don't understand how it works for this problem because I think it only works for one dimensional case.

I really appreciate your comment again!

[Walter Roberson]

delta_{alpha(e_{i}+e_{j})} means it is 1 only when alpha=e_{i}+e_{j}

However, you indicated

alpha=(2,0)

That is an ordered tuple of values, each in a non-directional space, not a vector or list of vectors in 2-space or 3-space.

What definition of δ are you using? With links to the definition, please. The definition I linked to at Wolfram would say that should be the same as which would be defined as 1 if alpha == 0 and == 0. The only delta notation I have seen for "1 only when alpha=e_{i}+e_{j}" would be or

[Bruno Luong]

@Walter Roberson

The right notation is

\delta_\alpha (e_i + e_j). Meaning

e_{i}+e_{j} 

is a vector to be compared with

alpha

[Walter Roberson]

@Bruno Luong

Reference link?

[Bruno Luong]

My math school teacher. This notation allows to interprets

\delta_y(x) := \delta_(x,y) = \delta_x^y

as a series of function of x, parametrized by y.

[Bruno Luong]

@Walter Roberson " which would be defined as 1 if alpha == 0 and ei+ej== 0. "

Not what I read, according to Wolfram link

"KroneckerDelta[n1,n2,…] gives the Kronecker delta , equal to 1 if all the ni are equal, and 0 otherwise. "

it is equal to 1 if alpha = ei+ej.

Anyway OP has stated exactly what he wants.

[Walter Roberson]

@Bruno Luong

You are right, I overlooked part of the description in the link I posted.