{"id":1679,"date":"2026-10-20T10:51:29","date_gmt":"2026-10-20T02:51:29","guid":{"rendered":"https:\/\/people.utm.my\/arazin\/?page_id=1679"},"modified":"2026-11-07T20:57:53","modified_gmt":"2026-11-07T12:57:53","slug":"2d-ndm-explorer","status":"publish","type":"page","link":"https:\/\/people.utm.my\/arazin\/2d-ndm-explorer\/","title":{"rendered":"2D NDM Explorer"},"content":{"rendered":"
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Interactive tool · Meshfree methods · NURBS<\/p>\n

2D NDM Explorer<\/h1>\n
\nAbout this tool<\/summary>\n

The NURBS-Divergence-Meshless method: the divergence theorem turns the heat equation over each cell into four line integrals, and NURBS make every cell a rectangle in the parameter space, so each normal is simply (0, ±1) or (±1, 0). Inspect the equation of any cell, switch between MLS and RBF, let the cells overlap, and compare the accuracy with finite elements.<\/p>\n<\/details>\n<\/div>\n

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Alatan interaktif · Meshfree methods · NURBS<\/p>\n

2D NDM Explorer<\/h1>\n
\nTentang alatan ini<\/summary>\n

Kaedah NURBS-Divergence-Meshless: teorem divergence menukar persamaan haba bagi setiap cell kepada empat line integral, dan NURBS menjadikan setiap cell segi empat tepat dalam ruang parameter, jadi setiap normal hanyalah (0, ±1) atau (±1, 0). Teliti persamaan mana-mana cell, tukar antara MLS dan RBF, biarkan cell bertindih, dan bandingkan ketepatannya dengan finite element.<\/p>\n<\/details>\n<\/div>\n

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Loading the interactive tool…<\/span>Memuatkan alatan interaktif…<\/span><\/p>\n<\/div>\n

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\nHow the NDM method works<\/summary>\n
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The NURBS-Divergence-Meshless method (Rasin et al., 2025) starts from steady heat conduction, k∇²T + Q = 0, and does not use a weak form. Each node gets a cell, the equation is integrated over the cell, and the divergence theorem turns the area integral into the flux through the cell boundary:<\/p>\n

∮Γc<\/sub><\/sub> ∇T · n ds = −∫Ωc<\/sub><\/sub> Q \/ k dΩ = F̂<\/div>\n

The cells are drawn in the NURBS parameter space, where they are rectangles. On each side one component of the outward normal is ±1 and the other is 0, so the boundary integral splits into four line integrals along the sides (the paper’s Eq. 8):<\/p>\n

−∫b<\/sub> (T,ξ<\/sub> G12<\/sub> + T,η<\/sub> G22<\/sub>) dξ + ∫r<\/sub> (T,ξ<\/sub> G11<\/sub> + T,η<\/sub> G12<\/sub>) dη + ∫t<\/sub> (T,ξ<\/sub> G12<\/sub> + T,η<\/sub> G22<\/sub>) dξ − ∫l<\/sub> (T,ξ<\/sub> G11<\/sub> + T,η<\/sub> G12<\/sub>) dη = F̂
G11<\/sub> = (xη<\/sub>² + yη<\/sub>²) \/ |J|    G12<\/sub> = −(xξ<\/sub>xη<\/sub> + yξ<\/sub>yη<\/sub>) \/ |J|    G22<\/sub> = (xξ<\/sub>² + yξ<\/sub>²) \/ |J|<\/div>\n

The G terms carry the whole geometry, taken from the derivatives of the NURBS map. Each side integral is one-dimensional and uses 1, 3 or 5 Gauss–Legendre points. The cell inspector beside the parameter space shows the four integrals for the selected node, evaluated with the solution, and their sum.<\/p>\n