# Some conceptual doubts about steady and transient solvers of the NS + energy equations

Recently, I have been solving some transient and steady, flow and heat transfer problems in Mathematica. The transient problem is essentially the reciprocating (i.e., fully reversing) flow of a fluid over a thick heated block until the system reaches a cyclic steady-state (i.e., the system temperature oscillates around a mean when this point is reached). I received help regarding this in one of my earlier question on this site. Following is the code:

Needs["NDSolveFEM"]

{f = 0.5; L = 0.040, d = 0.003, e = 0.005, kf = 0.026499, ks = 16,
rho = 1.1492, rhos = 7860, mu = 18.923*10^-6, cp = 1.0069*10^3,
cps = 502.4}; u0 = 3; nu = mu/rho; om = 2 Pi f;
tflow = 100;
t0 = .3;
NV = 2 f tflow;
nn = Round[NV \[Pi]/(om t0)]
Ti = 307; q = 1000/Ti;

reg1 = ImplicitRegion[0 <= x <= L && 0 <= y <= d, {x, y}]; reg2 =
ImplicitRegion[0 <= x <= L && -e <= y <= d, {x, y}];
UX[0][x_, y_] := 0;
VY[0][x_, y_] := 0;
P[0][x_, y_] := 0;
Tfs[0][x_, y_] := 307/Ti; appro =
With[{k = 2. 10^6}, ArcTan[k #]/Pi + 1/2 &];
ade[y_] := (ks + (kf - ks) appro[y])
rde[y_] := (cps rhos + (cp rho - cps rhos) appro[y]);

Monitor[Do[{UX[i], VY[i], P[i]} =
NDSolveValue[{{Inactive[
u[x, y], {x, y}]), {x, y}] + D[p[x, y], x] +
UX[i - 1][x, y]*D[u[x, y], x] +
VY[i - 1][x, y]*D[u[x, y], y] + (u[x, y] - UX[i - 1][x, y])/
t0, Inactive[

v[x, y], {x, y}]), {x, y}] + D[p[x, y], y] +
UX[i - 1][x, y]*D[v[x, y], x] +
VY[i - 1][x, y]*D[v[x, y], y] + (v[x, y] - VY[i - 1][x, y])/
t0, D[u[x, y], x] + D[v[x, y], y]} == {0, 0, 0} /. \[Mu] ->
nu, {DirichletCondition[{u[x, y] == u0*Sin[om*i*t0],
v[x, y] == 0},
x == L (1 - Sign[Sin[om*i*t0]])/2 && 0 < y < d],
DirichletCondition[{u[x, y] == 0, v[x, y] == 0},
y == 0 || y == d]},
DirichletCondition[p[x, y] == 0,
x == L (1 + Sign[Sin[om*i*t0]])/2 && 0 < y < d]}, {u, v,
p}, {x, y} \[Element] reg1,
Method -> {"FiniteElement",
"InterpolationOrder" -> {u -> 2, v -> 2, p -> 1},
"MeshOptions" -> {"MaxCellMeasure" -> 0.0000005}}];
ux = If[y <= 0, 0, UX[i][x, y]]; vy = If[y <= 0, 0, VY[i][x, y]];
Tfs[i] =
NDSolveValue[{rde[
y] ((ux*D[T[x, y], x] +
vy*D[T[x, y], y]) + (T[x, y] - Tfs[i - 1][x, y])/t0) -
Inactive[Div][
NeumannValue[q, y == -e],
DirichletCondition[{T[x, y] == 1},
x == L (1 - Sign[Sin[om*i*t0]])/2 && 0 <= y <= d]},
T, {x, y} \[Element] reg2,
Method -> {"FiniteElement", "InterpolationOrder" -> {T -> 2},
"MeshOptions" -> {"MaxCellMeasure" -> 0.0000001}}] //
Quiet;, {i, 1, nn}],ProgressIndicator[i,{1,nn}]]; // AbsoluteTiming

ListLinePlot[
Table[{i t0, Tfs[i][0.5*L, -e/2]*Ti - 273.16}, {i, 0, nn}],
AxesLabel -> {"t(s)", "T(K)"}, PlotRange -> Full]


The above code runs the simulation for 100s flow-time for a flow oscillating with a frequency of 0.5Hz using the velocity profile $$u = 3\sin(2*\pi*0.5*t)$$. As can be seen in the above code the Do loop marches forward in time. Inside the loop the flow and temperature field is being solved at each iteration nn.

I modified the above code for a uni-directional flow, i.e., find a steady-state solution for a flow from left to right over the heated block. I removed the time-dependent terms inside the loop and the nn, instead of a time-stepper becomes an iterator to reach a converged solution. The code is:

Needs["NDSolveFEM"]

{f = 0.5; L = 0.040, d = 0.003, e = 0.005, kf = 0.026499, ks = 16,
rho = 1.1492, rhos = 7860, mu = 18.923*10^-6, cp = 1.0069*10^3,
cps = 502.4}; u0 = 3; nu = mu/rho; om = 2 Pi f;
Ti = 307; q = 1000/Ti;
uavg = (1/(Pi/om)) Integrate[u0*Sin[om*t], {t, 0, Pi/om}] // N
nn = 10;

reg1 = ImplicitRegion[0 <= x <= L && 0 <= y <= d, {x, y}]; reg2 =
ImplicitRegion[0 <= x <= L && -e <= y <= d, {x, y}];
UX[0][x_, y_] := 0;
VY[0][x_, y_] := 0;
P[0][x_, y_] := 0;
Tfs[0][x_, y_] := 307/Ti; appro =
With[{k = 2. 10^6}, ArcTan[k #]/Pi + 1/2 &];
ade[y_] := (ks + (kf - ks) appro[y])
rde[y_] := (cps rhos + (cp rho - cps rhos) appro[y]);

Do[{UX[i], VY[i], P[i]} =
NDSolveValue[{{Inactive[
u[x, y], {x, y}]), {x, y}] + D[p[x, y], x] +
UX[i - 1][x, y]*D[u[x, y], x] +
VY[i - 1][x, y]*D[u[x, y], y],
Inactive[
v[x, y], {x, y}]), {x, y}] + D[p[x, y], y] +
UX[i - 1][x, y]*D[v[x, y], x] +
VY[i - 1][x, y]*D[v[x, y], y],
D[u[x, y], x] + D[v[x, y], y]} == {0, 0, 0} /. \[Mu] ->
nu, {DirichletCondition[{u[x, y] == uavg, v[x, y] == 0},
x == 0 && 0 < y < d],
DirichletCondition[{u[x, y] == 0, v[x, y] == 0},
y == 0 || y == d]},
DirichletCondition[p[x, y] == 0, x == L && 0 < y < d]}, {u, v,
p}, {x, y} \[Element] reg1,
Method -> {"FiniteElement",
"InterpolationOrder" -> {u -> 2, v -> 2, p -> 1},
"MeshOptions" -> {"MaxCellMeasure" -> 0.0000005}}];
ux = If[y <= 0, 0, UX[i][x, y]]; vy = If[y <= 0, 0, VY[i][x, y]];
Tfs[i] =
NDSolveValue[{rde[y] ((ux*D[T[x, y], x] + vy*D[T[x, y], y])) -
Inactive[Div][
NeumannValue[q, y == -e],
DirichletCondition[{T[x, y] == 1}, x == 0 && 0 <= y <= d]},
T, {x, y} \[Element] reg2,
Method -> {"FiniteElement", "InterpolationOrder" -> {T -> 2},
"MeshOptions" -> {"MaxCellMeasure" -> 0.0000001}}] //
Quiet;, {i, 1, nn}]; // AbsoluteTiming


My question is:

Why do we need an iterator to reach a steady-state solution?

In other words, the time-dependent code solved for the flow and temperature field at each time-step (but within each time-step there seemed to be no need for further iterations to reach a converged solution). Going by that logic, NDSolve should have been able to directly solve for the steady-state solution (i.e, without the time-dependent terms), but it cannot as I tried below:

{U, V, P} =
NDSolveValue[{{Inactive[
u[x, y], {x, y}]), {x, y}] + D[p[x, y], x] +
u[x, y]*D[u[x, y], x] + v[x, y]*D[u[x, y], y],
Inactive[
v[x, y], {x, y}]), {x, y}] + D[p[x, y], y] +
u[x, y]*D[v[x, y], x] + v[x, y]*D[v[x, y], y],
D[u[x, y], x] + D[v[x, y], y]} == {0, 0, 0} /. \[Mu] ->
nu, {DirichletCondition[{u[x, y] == uavg, v[x, y] == 0},
x == 0 && 0 < y < d],
DirichletCondition[{u[x, y] == 0, v[x, y] == 0},
y == 0 || y == d]},
DirichletCondition[p[x, y] == 0, x == L && 0 < y < d]}, {u, v,
p}, {x, y} \[Element] reg1,
Method -> {"FiniteElement",
"InterpolationOrder" -> {u -> 2, v -> 2, p -> 1},
"MeshOptions" -> {"MaxCellMeasure" -> 0.0000005}


I would like to have a descriptive explanation of this difference between modelling a transient system and a steady-state solution? In essence, why within each time-step in the transient solver there is no need of iteration ? However, a steady-state solution requires one to iterate. I hope I could clearly explain my query.

• Please, pay attention, that in my code with UX[i], VY[i], P[i] we solve on every step the linear system of equations, while in your code with {U, V, P} you try to solve nonlinear system. This is a big difference for NDSolve, since we use FEM. Dec 10, 2022 at 11:52
• Thankyou for your comment Alex. I do understand that in my last code block, I try to directly solve a highly non-linear system with significant convective terms. However, when I modified your original code to solve for the steady-problem (by removing the time terms, i.e., the second code block), it worked fine. Hence, I wanted to understand this basic difference between modelling a steady-state and a time-dependent system. It seems my understanding is flawed. Dec 10, 2022 at 12:55
• In my code the false transient method is implemented, it is why numerical solution converges to steady state in a case of the time term elimination - see explanation on community.wolfram.com/groups/-/m/t/1433064?p_p_auth=4O8xgpur Dec 10, 2022 at 13:59
• @AlexTrounev Thankyou for the answer. I went to the link and found some helpful notes. One of your codes there on square driven lid cavity used the method of false transients, which is helpful. I have one other request. Can you run the first code block (i.e., reciprocating flow) for these parameter values f=1, t0=0.15, tflow=700 (Other parameters remain same). I have seen that whenever I am giving a large tflow, the calculation stops midway with a kernel error, although I am running the calculation on a system with 16GB of memory. Dec 11, 2022 at 11:29
• Yes, you are right, there are some unstably solutions. In your case just put t0=0.3 to compute solution. Regardless your method at $t0\rightarrow \infty$, this is also nice method for some cases. Dec 12, 2022 at 4:54