I have some questions about the program you shared

[chenzhaoxingxin]

Dear author，
By running the program you uploaded and learning the documents you shared, it has greatly helped me in my study. When I was studying the document, I saw the sentence:

I have several questions about this place.

1. After studying your documents and other articles, I think the role of Jacobian matrix is to convert the grid from actual coordinates (x, y, z) to reference coordinates（ξ,η,φ)∈[-1,1] ³，and it satisfies the following formula:
   
   When I use the examples you uploaded, square.msh and config.conf. the Jacobian matrix of the first element and the node coordinates of the element are printed out, and the calculation is performed:
   
   Wherein, (-4.278517,1.721583,0) is the coordinate of the first node of the first element (the node tag is 165). (-0.6971,1.45557,0) is the parameter coordinate of the node. I found that the parameter coordinates are outside the range of [-1,1].
   I found that the Jacobian matrix is calculated by using the function of gmsh. I also used the relationship that 2 times the area of the triangle element is equal to detJ to verify, and found that it is consistent. In this way, the determinant of Jacobian matrix seems to have no problem. I wonder why the transformed parameter coordinates are not within the range of [- 1,1], so that the Gaussian integration of formula (5) cannot be performed.
   2 I have some different understandings of Formula 5. I think that the integral of function with respect to K is converted to [- 1,1] ³ Times detJ. For an element, convert to [- 1,1] ³ When, its Jacobian matrix is fixed, and detJ is a fixed constant. When using the Gaussian integral formula, it seems that it is OK to multiply detJ at the outermost. The formula I understand is as follows. I would like to ask the author if I have misunderstood it.
   

3 Formula (9) is obtained after discretization. In this formula, uk is the node solution of each element. I think your program saves the node to the msh file for output. If I want to solve for μ at any point in the element, I need to use the following formula:

If I want to solve the μ。 My idea is: first, locate the element according to the coordinates of m, find the coordinates of the element nodes, and then calculate the area of the triangle according to the coordinates.

According to the definition of Lagrangian basis function, the area coordinates A1, A2, A3 at point m are needed (A1 is the area of small triangle m23, A1 is the area of small triangle m23, and A1 is the area of small triangle m23). S is the area of the element triangle. The basis function can be expressed as N1=A1/S; N2=A2/S; N3=A3/S。At point m, the solution formula of μ is:
μm=N1u1+N2u2+N3*u3
I want to use this method to solve μ at point m in Descartes coordinates. I want to ask you if there is any problem with my thinking.
I thought that the coordinates at point m could also be converted to parameter coordinates for solution, but there might be a problem with the Jacobian matrix in question 1, which made me uneasy to convert to parameter coordinates for solution.

My question may be a little childish. I hope you will give me some advice. Thank you for checking my question in your busy schedule, and look forward to your reply.
Thank you again for sharing the program selflessly. I wish you good health and good luck!