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| In a current version of Mathematica 12.1 there is no special solver for integrodifferencial equations. [Here][1] we show solver with using Haar wavelets for dynamic system presented in a paper M.A. Khan, A. Atangana, Modeling the dynamics of... |
| I used the standard code to triangulate the mesh coord = CountryData["USA", "Polygon"][[1, 1, 1]]; bmr = BoundaryMeshRegion[coord, Line[Append[Range[Length[coord]], 1]]]; usatri = TriangulateMesh[bmr, MaxCellMeasure -> .1]; ... |
| There is a cylinder, and you use plane geometry. Equation `QE` is therefore erroneous. You do not use FEM at all. To use FEM, call `Needs["NDSolve`FEM`"]`, then read the tutorial on a page FEMDocumentation/guide/FiniteElementMethodGuide. |
| You changed the code and changed the question. My answer now looks out of place. Which question should I answer? |
| This can be used as a regular analytical solution, for example s = DSolve[{i[t] == Cr vc'[t], vc[t] == vco[t] (1 + m) - Lr i'[t], -(1 + m) i[t] - vco[t]/RL == Co vco'[t], vc[0] == Vcrmax, i[0] == 0, vco[0] == Vo}, {i, vc, ... |
| To answer this question, we can compare two methods for solving the heat equation: 1) the `"MethodOfLines"` in which boundary condition `Derivative[1, 0][u][5, t] == -h (u[5, t] - u1)` is used; and 2) FEM in which boundary condition `NeumannValue[-h... |
| There are many ways, I will show one of them. Let a random process be described by a function matRealizations = Table[Sin[RandomReal[-1, 1] t + RandomReal[]], {81}]; You can display them all together or selectively Show[Table[ ... |
| In this problem, it is necessary to coordinate the initial and boundary conditions, as well as use the solution method and increase the time interval to 200 for clarity. eq = { 2423750 Derivative[0, 1][T][x, t] - 50 Derivative[2,... |
| Thanks Vitaliy! I decided to check out what they think about Wolfram Mathematica. The result was unexpected, but reasonable. ![fig1][1] [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=WM.jpg&userId=1218692 |
| gamma[beta_] := 1/Sqrt[1 - beta^2] c = 299792458 ; (*speed of light, m/s*) Mile = 1600 ;(*Meter*) Hour = 3600 ;(*Second*) speedF35 = 1200.*Mile/Hour; betaF35 = speedF35/c; gammaF35 = gamma[betaF35]; ... |