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NDSolve and equation with a NeumannValue?

Posted 2 years ago
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I would like some tips to implement, in the NDSolve function, the following expression below with a NeumannValue:

 Derivative[1, 0][u][5, t] == -h*(u[5, t] - u1)

Was it something like this?

NeumannValue[Derivative[1, 0][u][x, t] + h*(u[x, t] - u1), x == 5]
2 Replies
Posted 2 years ago

Thanks for the attention and tips. I am also consulting the Tutorial: Finite Element Method. Regards. Sinval

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 (u[x, t] - u1), x == 5] is used.

eq = D[u[x, t], t] - D[u[x, t], x, x];
ic = u[x, 0] == 0;
bc = {u[0, t] == 0, 
   Derivative[1, 0][u][5, t] == -h (u[5, t] - u1) (1 - Exp[-10 t])};
bcn = DirichletCondition[u[x, t] == 0, x == 0];
h = 2; u1 = 2.5;
sol = NDSolveValue[Flatten[{eq == 0, ic, bc}], 
  u, {t, 0, 5}, {x, 0, 5}, 
  Method -> {"MethodOfLines", 
    "SpatialDiscretization" -> {"TensorProductGrid", 
      "MinPoints" -> 100, "MaxPoints" -> 100, "DifferenceOrder" -> 4}}]
   Needs["NDSolve`FEM`"];
mesh = ToElementMesh[ImplicitRegion[0 <= x <= 5, {x}], 
   "MaxCellMeasure" -> 0.001];
solN = NDSolveValue[
  Flatten[{eq == NeumannValue[-h (u[x, t] - u1), x == 5], ic, bcn}], 
  u, {t, 0, 5}, x \[Element] mesh, 
  Method -> {"FiniteElement", 
    "MeshOptions" -> {"MaxCellMeasure" -> 0.001}}]

Comparing the two solutions in Fig. 1, we see their identity

{Plot3D[sol[x, t], {t, 0, 5}, {x, 0, 5}, Mesh -> None, 
  ColorFunction -> Hue, AxesLabel -> Automatic, 
  PlotLabel -> "Method Of Lines"], 
 Plot3D[solN[x, t], {t, 0, 5}, {x, 0, 5}, Mesh -> None, 
  ColorFunction -> Hue, PlotRange -> All, AxesLabel -> Automatic, 
  PlotLabel -> "Finite Element"]}

Figure 1

Note that in the Method Of Lines a damping factor is used to agree on the initial and boundary conditions (1 - Exp[-10 t]).

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