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 yesterday
MarcoB
37.9k556114
asked yesterday
dpholmesdpholmes
345111
$endgroup$
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 yesterday
MarcoB
37.9k556114
asked yesterday
dpholmesdpholmes
345111
$endgroup$
$begingroup$
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
yesterday
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
yesterday
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 yesterday
MarcoB
37.9k556114
asked yesterday
dpholmesdpholmes
345111
$endgroup$
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 yesterday
MarcoB
37.9k556114
asked yesterday
dpholmesdpholmes
345111
edited yesterday
MarcoB
37.9k556114
asked yesterday
dpholmesdpholmes
345111
edited yesterday
MarcoB
37.9k556114
edited yesterday
MarcoB
37.9k556114
edited yesterday
MarcoB
37.9k556114
37.9k556114
asked yesterday
dpholmesdpholmes
345111
asked yesterday
dpholmesdpholmes
345111
asked yesterday
dpholmesdpholmes
345111
345111
$begingroup$
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
yesterday
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
yesterday
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
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
yesterday
$begingroup$
@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
yesterday
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 yesterday
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 yesterday
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 yesterday
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 yesterday
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 -> $MachinePrecision
]
answered yesterday
MarcoBMarcoB
37.9k556114
answered yesterday
MarcoBMarcoB
37.9k556114
answered yesterday
MarcoBMarcoB
37.9k556114
answered yesterday
MarcoBMarcoB
37.9k556114
37.9k556114
add a comment |
add a comment |
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$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
yesterday
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
yesterday