Boundary Value Problem and FullSimplifyI failed to solve a set of one-dimension fluid mechanics PDEs with NDSolveDSolve gives complex function although the solution is a real oneNumerical solution of coupled ODEs with boundary conditionsNDSolve and strange “nonlinear coefficients problem”Reaction-diffusion PDE with NDSolve: either very slow or very inaccurateSolution of nonlinear system with boundary conditionsDSolve, NDSolve with WhenEvent Give Incorrect Solution for Simple ODEInhomogeneous Neumann boundary conditions for diffusion equationAnalyitic and numerical solutions plots of PDE are different!Not sure how to set up the Laplacian/Poisson Equation
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Boundary Value Problem and FullSimplify
I failed to solve a set of one-dimension fluid mechanics PDEs with NDSolveDSolve gives complex function although the solution is a real oneNumerical solution of coupled ODEs with boundary conditionsNDSolve and strange “nonlinear coefficients problem”Reaction-diffusion PDE with NDSolve: either very slow or very inaccurateSolution of nonlinear system with boundary conditionsDSolve, NDSolve with WhenEvent Give Incorrect Solution for Simple ODEInhomogeneous Neumann boundary conditions for diffusion equationAnalyitic and numerical solutions plots of PDE are different!Not sure how to set up the Laplacian/Poisson Equation
$begingroup$
I'm confused about the output Mathematica is giving me when solving a boundary value problem of the form:
eq = ϵ y''[t] + 2 y'[t] + 2 y[t] == 0;
bc1 = y[0] == 0;
bc2 = y[1] == 1;
aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t][[1]][[1]]
This yields the correct answer, and produces plots like this for ep=1, ep=0.1, and ep=0.01.
Plot[
aSol /. ϵ -> 1,
aSol /. ϵ -> 0.1,
aSol /. ϵ -> 0.01,
t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
So far, so good!
However, if I simply ask Mathematica to FullSimplify[aSol], the resulting solution is no longer correct, and it does not satisfy one of the boundary conditions:
aSolSimpl = FullSimplify[aSol]
Plot[
aSol /. ϵ -> 0.05,
aSolSimpl /. ϵ -> 0.05
, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
What's going wrong here?
differential-equations simplifying-expressions
edited Mar 21 at 15:41
MarcoB
37.9k556114
asked Mar 21 at 14:57
dpholmesdpholmes
350111
$endgroup$
add a comment |
$begingroup$
I'm confused about the output Mathematica is giving me when solving a boundary value problem of the form:
eq = ϵ y''[t] + 2 y'[t] + 2 y[t] == 0;
bc1 = y[0] == 0;
bc2 = y[1] == 1;
aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t][[1]][[1]]
This yields the correct answer, and produces plots like this for ep=1, ep=0.1, and ep=0.01.
Plot[
aSol /. ϵ -> 1,
aSol /. ϵ -> 0.1,
aSol /. ϵ -> 0.01,
t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
So far, so good!
However, if I simply ask Mathematica to FullSimplify[aSol], the resulting solution is no longer correct, and it does not satisfy one of the boundary conditions:
aSolSimpl = FullSimplify[aSol]
Plot[
aSol /. ϵ -> 0.05,
aSolSimpl /. ϵ -> 0.05
, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
What's going wrong here?
differential-equations simplifying-expressions
edited Mar 21 at 15:41
MarcoB
37.9k556114
asked Mar 21 at 14:57
dpholmesdpholmes
350111
$endgroup$
$begingroup$
i think because you assign Epsilon different values. i used Full Simplify and it was fine with me use it as follow to check aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t] /. [Epsilon] -> 1 aSolSimpl = FullSimplify[aSol] /. [Epsilon] -> 1 Plot[aSol, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"] Plot[aSolSimpl, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
$endgroup$
– Alrubaie
Mar 21 at 15:19
1
$begingroup$
@dpholmes PlottingPlot3D[Evaluate[aSol, FullSimplify[aSol, [Epsilon] > 0]], t, 0, 1, [Epsilon], 0, 1]reveals that it might be a precision problem.
$endgroup$
– Henrik Schumacher
Mar 21 at 15:20
add a comment |
$begingroup$
I'm confused about the output Mathematica is giving me when solving a boundary value problem of the form:
eq = ϵ y''[t] + 2 y'[t] + 2 y[t] == 0;
bc1 = y[0] == 0;
bc2 = y[1] == 1;
aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t][[1]][[1]]
This yields the correct answer, and produces plots like this for ep=1, ep=0.1, and ep=0.01.
Plot[
aSol /. ϵ -> 1,
aSol /. ϵ -> 0.1,
aSol /. ϵ -> 0.01,
t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
So far, so good!
However, if I simply ask Mathematica to FullSimplify[aSol], the resulting solution is no longer correct, and it does not satisfy one of the boundary conditions:
aSolSimpl = FullSimplify[aSol]
Plot[
aSol /. ϵ -> 0.05,
aSolSimpl /. ϵ -> 0.05
, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
What's going wrong here?
differential-equations simplifying-expressions
edited Mar 21 at 15:41
MarcoB
37.9k556114
asked Mar 21 at 14:57
dpholmesdpholmes
350111
$endgroup$
I'm confused about the output Mathematica is giving me when solving a boundary value problem of the form:
eq = ϵ y''[t] + 2 y'[t] + 2 y[t] == 0;
bc1 = y[0] == 0;
bc2 = y[1] == 1;
aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t][[1]][[1]]
This yields the correct answer, and produces plots like this for ep=1, ep=0.1, and ep=0.01.
Plot[
aSol /. ϵ -> 1,
aSol /. ϵ -> 0.1,
aSol /. ϵ -> 0.01,
t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
So far, so good!
However, if I simply ask Mathematica to FullSimplify[aSol], the resulting solution is no longer correct, and it does not satisfy one of the boundary conditions:
aSolSimpl = FullSimplify[aSol]
Plot[
aSol /. ϵ -> 0.05,
aSolSimpl /. ϵ -> 0.05
, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
What's going wrong here?
differential-equations simplifying-expressions
differential-equations simplifying-expressions
edited Mar 21 at 15:41
MarcoB
37.9k556114
asked Mar 21 at 14:57
dpholmesdpholmes
350111
edited Mar 21 at 15:41
MarcoB
37.9k556114
asked Mar 21 at 14:57
dpholmesdpholmes
350111
edited Mar 21 at 15:41
MarcoB
37.9k556114
edited Mar 21 at 15:41
MarcoB
37.9k556114
edited Mar 21 at 15:41
MarcoB
37.9k556114
37.9k556114
asked Mar 21 at 14:57
dpholmesdpholmes
350111
asked Mar 21 at 14:57
dpholmesdpholmes
350111
asked Mar 21 at 14:57
dpholmesdpholmes
350111
350111
$begingroup$
i think because you assign Epsilon different values. i used Full Simplify and it was fine with me use it as follow to check aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t] /. [Epsilon] -> 1 aSolSimpl = FullSimplify[aSol] /. [Epsilon] -> 1 Plot[aSol, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"] Plot[aSolSimpl, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
$endgroup$
– Alrubaie
Mar 21 at 15:19
1
$begingroup$
@dpholmes PlottingPlot3D[Evaluate[aSol, FullSimplify[aSol, [Epsilon] > 0]], t, 0, 1, [Epsilon], 0, 1]reveals that it might be a precision problem.
$endgroup$
– Henrik Schumacher
Mar 21 at 15:20
add a comment |
$begingroup$
i think because you assign Epsilon different values. i used Full Simplify and it was fine with me use it as follow to check aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t] /. [Epsilon] -> 1 aSolSimpl = FullSimplify[aSol] /. [Epsilon] -> 1 Plot[aSol, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"] Plot[aSolSimpl, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
$endgroup$
– Alrubaie
Mar 21 at 15:19
1
$begingroup$
@dpholmes PlottingPlot3D[Evaluate[aSol, FullSimplify[aSol, [Epsilon] > 0]], t, 0, 1, [Epsilon], 0, 1]reveals that it might be a precision problem.
$endgroup$
– Henrik Schumacher
Mar 21 at 15:20
$begingroup$
i think because you assign Epsilon different values. i used Full Simplify and it was fine with me use it as follow to check aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t] /. [Epsilon] -> 1 aSolSimpl = FullSimplify[aSol] /. [Epsilon] -> 1 Plot[aSol, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"] Plot[aSolSimpl, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
$endgroup$
– Alrubaie
Mar 21 at 15:19
$begingroup$
i think because you assign Epsilon different values. i used Full Simplify and it was fine with me use it as follow to check aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t] /. [Epsilon] -> 1 aSolSimpl = FullSimplify[aSol] /. [Epsilon] -> 1 Plot[aSol, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"] Plot[aSolSimpl, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
$endgroup$
– Alrubaie
Mar 21 at 15:19
1
1
$begingroup$
@dpholmes Plotting
Plot3D[Evaluate[aSol, FullSimplify[aSol, [Epsilon] > 0]], t, 0, 1, [Epsilon], 0, 1] reveals that it might be a precision problem.$endgroup$
– Henrik Schumacher
Mar 21 at 15:20
$begingroup$
@dpholmes Plotting
Plot3D[Evaluate[aSol, FullSimplify[aSol, [Epsilon] > 0]], t, 0, 1, [Epsilon], 0, 1] reveals that it might be a precision problem.$endgroup$
– Henrik Schumacher
Mar 21 at 15:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This behavior seems due to precision problems, as Henrik suggested in comments:
aSol = DSolveValue[eq, bc1, bc2, y[t], t];
aSolSimpl = FullSimplify[aSol];
Plot[Evaluate[aSol /. ϵ -> 1, 1/10, 1/100], t, 0, 1]
Plot[
Evaluate[aSolSimpl /. ϵ -> 1, 1/10, 1/100], t, 0, 1,
WorkingPrecision -> $MachinePrecision
]
answered Mar 21 at 15:53
MarcoBMarcoB
37.9k556114
$endgroup$
add a comment |
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$begingroup$
This behavior seems due to precision problems, as Henrik suggested in comments:
aSol = DSolveValue[eq, bc1, bc2, y[t], t];
aSolSimpl = FullSimplify[aSol];
Plot[Evaluate[aSol /. ϵ -> 1, 1/10, 1/100], t, 0, 1]
Plot[
Evaluate[aSolSimpl /. ϵ -> 1, 1/10, 1/100], t, 0, 1,
WorkingPrecision -> $MachinePrecision
]
answered Mar 21 at 15:53
MarcoBMarcoB
37.9k556114
$endgroup$
add a comment |
$begingroup$
This behavior seems due to precision problems, as Henrik suggested in comments:
aSol = DSolveValue[eq, bc1, bc2, y[t], t];
aSolSimpl = FullSimplify[aSol];
Plot[Evaluate[aSol /. ϵ -> 1, 1/10, 1/100], t, 0, 1]
Plot[
Evaluate[aSolSimpl /. ϵ -> 1, 1/10, 1/100], t, 0, 1,
WorkingPrecision -> $MachinePrecision
]
answered Mar 21 at 15:53
MarcoBMarcoB
37.9k556114
$endgroup$
add a comment |
$begingroup$
This behavior seems due to precision problems, as Henrik suggested in comments:
aSol = DSolveValue[eq, bc1, bc2, y[t], t];
aSolSimpl = FullSimplify[aSol];
Plot[Evaluate[aSol /. ϵ -> 1, 1/10, 1/100], t, 0, 1]
Plot[
Evaluate[aSolSimpl /. ϵ -> 1, 1/10, 1/100], t, 0, 1,
WorkingPrecision -> $MachinePrecision
]
answered Mar 21 at 15:53
MarcoBMarcoB
37.9k556114
$endgroup$
This behavior seems due to precision problems, as Henrik suggested in comments:
aSol = DSolveValue[eq, bc1, bc2, y[t], t];
aSolSimpl = FullSimplify[aSol];
Plot[Evaluate[aSol /. ϵ -> 1, 1/10, 1/100], t, 0, 1]
Plot[
Evaluate[aSolSimpl /. ϵ -> 1, 1/10, 1/100], t, 0, 1,
WorkingPrecision -> $MachinePrecision
]
answered Mar 21 at 15:53
MarcoBMarcoB
37.9k556114
answered Mar 21 at 15:53
MarcoBMarcoB
37.9k556114
answered Mar 21 at 15:53
MarcoBMarcoB
37.9k556114
answered Mar 21 at 15:53
MarcoBMarcoB
37.9k556114
37.9k556114
add a comment |
add a comment |
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$begingroup$
i think because you assign Epsilon different values. i used Full Simplify and it was fine with me use it as follow to check aSol = y[t] /. DSolve[eq, bc1, bc2, y[t], t] /. [Epsilon] -> 1 aSolSimpl = FullSimplify[aSol] /. [Epsilon] -> 1 Plot[aSol, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"] Plot[aSolSimpl, t, 0, 1, Frame -> True, FrameLabel -> "t", "y(t)"]
$endgroup$
– Alrubaie
Mar 21 at 15:19
1
$begingroup$
@dpholmes Plotting
Plot3D[Evaluate[aSol, FullSimplify[aSol, [Epsilon] > 0]], t, 0, 1, [Epsilon], 0, 1]reveals that it might be a precision problem.$endgroup$
– Henrik Schumacher
Mar 21 at 15:20