## finite element method – Solving differential equation system with FEM when the equations depend on the mesh cell size

I am trying to implement Large eddy simulation with FEM for solving air flow simulation using Smagorinsky sub-grid stress model. In order to implement this there is a parameter (filter size) in the differential equation system that has to be calculated based on the grid size: $$Delta=(Delta_xDelta_yDelta_z)^{1/3}$$ where $$Delta_x,Delta_y,Delta_z$$ is the size of the mesh cell in the x, y and z direction, respectively. Is there any possibility to implement this with the NDsolve function? I did not find any solution in the documentation of the FEM.

## differential equations – How to solve NDSolve dependent on two NIntegrates?

`````` Kab(n_, z_) := NIntegrate((1 + (t*n)/(f0(z))^2), {t, 0, 1});
S4(n_, z_) := NIntegrate((t/(n*(f0(z))^2)*Exp(-2*Kab(n, z)*z)), {t, 0, 1});
Com(n_, z_) := NDSolve({(f0'(z)) == n*Exp(-2*Kab(n, z)*z)*S4(n, z), f0(0) == 1, f0'(0) == 0}, f0, {z, 0, 3});
ca(n_, z_) = Com(2, z)
Plot(Evaluate(f0(z) /. ca(n, z)), {z, 0, 3})
``````

My code is given above. I am unable to get the plot and getting some errors.

## differential equations – how we can extract the value of the solution of a PDE in a point x? (NDSolve)

Please I need your help, I calculate the solution of heat equation using methode of line
This is my code:

``````n = 10
grid = 1/n  Range[0, n];
d1 = NDSolve`FiniteDifferenceDerivative[Derivative[1], grid];
d2 = NDSolve`FiniteDifferenceDerivative[Derivative[2], grid];
M1 = d1["DifferentiationMatrix"];
M2 = d2["DifferentiationMatrix"];
y00[t_] := Sqrt[2] Sin[Pi t];
T= 0.02;
tab = Table[u[i]
tab1 = Table[u[i], {i, Length[grid]} ];
ux = M1.tab;
uxx = M2.tab;
solu1   = D[u[1]
solu2 = D[u[n + 1]
solu3 = Table[D[u[i]
solu4 = Table[u[i][0] == y00[grid[[i]]], {i, Length[grid]}];
sol1 = NDSolve[{solu1, solu2, solu3, solu4}, tab1, {t, 0, T},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MaxPoints" -> 25}}];
``````

Now that I have calculated the solution “sol1”. I need to approximate the value of this solution in $$t = T$$ in the space $$grid$$ something like

``````h = Table [sol1 [grid[[i]],T],{i,Length[grid]}]
``````

but this is not working for me, can anyone help me with that please??

## real analysis – How to get the reverse Hölder inequality for the weak solution of elliptic equations?

real analysis – How to get the reverse Hölder inequality for the weak solution of elliptic equations? – MathOverflow

## differential equations – Solving an ODE with paramters and taking the limit of the solution

I am very new to Mathematica and already spent a lot of time trying to do this but failed.

I am trying to solve an ODE:

``````solution = DSolve({-((m (1 + m) + 4/(9 (-2/3 + t) t)) y(t)) +
2 (-1/3 + t) y'(t) + (-2/3 + t) t y''(
t) == (-4 (1 + C/2))/(9 (-2/3 + t) t), y(1) == 1, y'(1) == C},
y, t)
``````

where $$m$$ is a nonnegative integer and $$C$$ is a real number. (I am not sure how to incorporate that $$m$$ is a nonnegative integer in the code. )

I want to show that there exists a $$C$$ such that the solution is $$0$$ at infinity.
When I try that code:

``````Limit(y(t) /. solution((1)), t -> Infinity, m (Element) Integers)
``````

it just spits out the same thing.

What should I do?
(Note that I don’t need to find that value of $$C$$; I just need to show that for every $$m$$, there is a number $$C$$ in which the solution vanishes at infinity. )

## Any reference on solving functional and differenial equations involving power towers of a function?

I am curious if this is covered in any textbooks or papers. I am interested in finding out how to solve functional and differential (even pde) equations involving power towers in them. For example if tow(f(x),n)= g(x) where tow meant a diagonal power tower of f with n determining it’s height. Also, can this be solved if some of the f’s in the tower (not necessarily all of them) were derivatives , or even partial derivatives in a multivariable case? I’m also interested in what happens if n goes to infinity. Thanks.

## differential equations – Numerically Solving Fourth Order PDE with Boundary Condition at Infinity

I am trying to numerically simulate (using NDSolve) the following differential equation with the following boundary conditions:

$$partial_x^2psi +cpartial_x^4psi – b(apsi + psi^4)= lambdapartial_tpsi$$
$$partial_tpsi(x,t=infty)= 0$$
$$psi(x,t=0)=f(x)$$

where $$a,b,c,lambda$$ are all constants. I have been encountering a couple of problems. For example, it seems like Mathematica cannot simulate fourth order partial differential equations(?) Also when working with simpler versions of the problem, I was running into trouble with the boundary condition at $$tto infty$$. The code I’ve been working with is something along the lines of this:

``````a = 1;
b = 1;
c = 1;
l = 1;

NDSolveValue({D((Psi)(x, t), {x, 2}) + D((Psi)(x, t), {x, 4}) -
b*(a*(Psi)(x, t) + (Psi)(x, t)^3) ==
l*D((Psi)(x, t), t), (Psi)(x, 0) ==
Tanh(x), (D((Psi)(x, t), t) /. t -> 1000) == 0}, (Psi), {x, -20,
20}, {t, 0, 1000}
)
``````

I tried making the time boundary condition some large number, rather than putting in Infinity. But this entire scheme seems to be too simple.

## How to solve this kind of ordinary differential equations \$t y y^{prime prime}-2 tleft(y^{prime}right)^{2}+3 y y^{prime}=0\$

Often, to solve an ODE, it if enough to classify it correctly and apply correspondent well-known method. But in this case I struggle

$$t y y^{prime prime}-2 tleft(y^{prime}right)^{2}+3 y y^{prime}=0$$

It is non-linear homogeneous second-order ODE with variable coefficients. In textbooks, people talk about similar equations where one variable is missing (x or y or y’). In that case, we use substitution like y = z(z) or y’ = p(x). But in my ODE, I do not see what to do.

## differential equations – Modified Heat Transfer in Fluid Flow

I am trying to simulate Modified Heat Transfer in Fluid Flow in MMA 12.02 (based on Buoyancy-Driven Flow in a Square Cavity ).

The modified heat transfer takes the form:

with the solid volume fraction:

The whole model is defined as:

``````ClearAll("Global`*")

Needs("NDSolve`FEM`")
sizes = {length -> 1, hight -> 1};
(CapitalOmega) = Rectangle({0, 0}, {length, hight}) /. sizes;
em = ToElementMesh((CapitalOmega) , MaxCellMeasure -> .0001,
"MeshOrder" -> 2);

Pr = 50;
Ra = 2.27*10^5;
timetol = 0.00001;
Th = 1.0;
Tc = -0.01;

ClearAll((Nu))
op = {

!(*SuperscriptBox((u),
TagBox(
RowBox({"(",
RowBox({"1", ",", "0", ",", "0"}), ")"}),
Derivative),
MultilineFunction->None))(t, x, y) +
Inactive(Div)((- Inactive(Grad)(u(t, x, y), {x, y})), {x,
y}) + {u(t, x, y), v(t, x, y)} .
Inactive(Grad)(u(t, x, y), {x, y}) +
!(*SuperscriptBox((p),
TagBox(
RowBox({"(",
RowBox({"0", ",", "1", ",", "0"}), ")"}),
Derivative),
MultilineFunction->None))(t, x, y),

!(*SuperscriptBox((v),
TagBox(
RowBox({"(",
RowBox({"1", ",", "0", ",", "0"}), ")"}),
Derivative),
MultilineFunction->None))(t, x, y) +
Inactive(Div)((-Inactive(Grad)(v(t, x, y), {x, y})), {x,
y}) + {u(t, x, y), v(t, x, y)} .
Inactive(Grad)(v(t, x, y), {x, y}) +
!(*SuperscriptBox((p),
TagBox(
RowBox({"(",
RowBox({"0", ",", "0", ",", "1"}), ")"}),
Derivative),
MultilineFunction->None))(t, x, y) - Ra/Pr*T(t, x, y),

!(*SuperscriptBox((u),
TagBox(
RowBox({"(",
RowBox({"0", ",", "1", ",", "0"}), ")"}),
Derivative),
MultilineFunction->None))(t, x, y) +
!(*SuperscriptBox((v),
TagBox(
RowBox({"(",
RowBox({"0", ",", "0", ",", "1"}), ")"}),
Derivative),
MultilineFunction->None))(t, x, y),

!(*SuperscriptBox((T),
TagBox(
RowBox({"(",
RowBox({"1", ",", "0", ",", "0"}), ")"}),
Derivative),
MultilineFunction->None))(t, x,
y)*(1.0 + 1/0.045*20* Sech(40* (0.01 - T(t, x, y)))^2 ) +
Inactive(Div)((-Inactive(Grad)(T(t, x, y), {x, y})), {x,
y}) + {u(t, x, y), v(t, x, y)} .

wall = DirichletCondition({u(t, x, y) == 0, v(t, x, y) == 0}, True);
reference = DirichletCondition(p(t, x, y) == 0, x == 0 && y == 0);
temperatures = {DirichletCondition(T(t, x, y) == Th, x == 0),
DirichletCondition(T(t, x, y) == 0, x == length)};
bcs = {wall, reference, temperatures} /. sizes;
ic = {u(0, x, y) == 0, v(0, x, y) == 0, p(0, x, y) == 0,
T(0, x, y) == 0};

Monitor(AbsoluteTiming({xVel, yVel, pressure, temperature} =
NDSolveValue({op == {0, 0, 0, 0}, bcs, ic}, {u, v, p,
T}, {x, y} (Element) em, {t, 0, timetol},
Method -> {"TimeIntegration" -> {"IDA",
"MaxDifferenceOrder" -> 2},
"PDEDiscretization" -> {"MethodOfLines",
"SpatialDiscretization" -> {"FiniteElement",
"InterpolationOrder" -> {u -> 2, v -> 2, p -> 1,
T -> 2}}}},
EvaluationMonitor :> (currentTime =
Row({"t = ", CForm(t)})));), currentTime)
``````

However, this modified simulation model produces some numerical errors:

How to set the solver parameters that we can fix the bug “Matrix SparseArray(<1395840>, {101404, 101404}) is singular …”

## differential equations – Different answer between HeatTransferPDEComponent and Matlab

I’ve been working on a 2-layer HeatTransferPDE Model.
The diagram of the model is

The two materials have an initial temperature of 37℃, and the outside is 75℃. The model is for heat insulation so it can simplify to a one-dimensional HeatTransferPDE.

Mathematica codes are:

``````outsideTemp = 273.15 + 75;
boundary1 = 0.6/1000;
boundary2 = (0.6 + 6)/1000;
Ω1 = Line({{0}, {boundary1}});
Ω2 = Line({{boundary1}, {boundary2}});
vars = {T(t, x), t, {x}};

pars1 = <|"ThermalConductivity" -> 0.082, "AmbientTemperature" -> 273.15 + 37, "Density" -> 300, "SpecificHeatCapacity" -> 1377|>;

B1 = HeatTransferValue(x == 0, vars, <|"HeatTransferCoefficient" -> 114, "AmbientTemperature" -> outsideTemp|>);
pde1 = {HeatTransferPDEComponent(vars, pars1) == B1, T(0, x) == 273.15 + 37};
Tfun1 = NDSolveValue(pde1, T, {t, 0, 100}, {x} ∈ Ω1);

pars2 = <|"ThermalConductivity" -> 0.37, "Density" -> 862, "SpecificHeatCapacity" -> 2100, "SurfaceTemperature" -> Tfun1(t, boundary1)|>;
B2 = HeatTemperatureCondition(x == boundary1, vars, pars2);
pde2 = {HeatTransferPDEComponent(vars, pars2) == 0, B2, T(0, x) == 273.15 + 37};
Tfun2 = NDSolveValue(pde2, T, {t, 0, 100}, {x}∈ Ω2);

Plot3D({Tfun1(t, x), Tfun2(t, x)}, {t, 0, 100}, {x, 0, boundary2}, PlotRange -> All)

``````

The output is

However, I use finite difference with same parameters and the output is

The matlab code is too long to put in here(because it’s for a 4-layer model).

I checked the code for a day and still have no clue why there is a difference between them, the material properties is in the table below.

Thanks!