Boundary Value Problem and FullSimplify
3
$begingroup$
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
345111
$endgroup$
add a comment |
3
$begingroup$
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
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
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 |
3
3
3
$begingroup$
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
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 Mar 21 at 15:41
MarcoB
37.9k556114
asked Mar 21 at 14:57
dpholmesdpholmes
345111
edited Mar 21 at 15:41
MarcoB
37.9k556114
asked Mar 21 at 14:57
dpholmesdpholmes
345111
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
345111
asked Mar 21 at 14:57
dpholmesdpholmes
345111
asked Mar 21 at 14:57
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
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
5
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193714%2fboundary-value-problem-and-fullsimplify%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
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 |
5
$begingroup$
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 |
5
5
5
$begingroup$
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 |
draft saved
draft discarded
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193714%2fboundary-value-problem-and-fullsimplify%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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