An IVP with singularity having continuous solutionsAn IVP with 'dummy' singularityQuestion on a derivation...
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An IVP with singularity having continuous solutions
An IVP with 'dummy' singularityQuestion on a derivation regarding the non-linear ODE $x'' = -U'(x)$, $U$ potentialAbout a Property of maximal solutions of separable ODE's $y'=g(x)h(y)$ for locally Lipschitz $h : Utomathbb R$, $U$ openDeterminants and Differential Equations ProblemA question about the period T of a simple pendulumMinimal order of ODE with given set of fundamental solutions.Proof that the equation $3^x=3+x$ has a solution in the interval $(0,1)$Prove that the equation $5 sqrt{sinleft(x^2+frac{1}{2}right)}=frac{19}{4}$ has exactly one solutiongeneral solution of the FODE $frac{1}{sin(x)} frac{dy}{dx} = ysec(x) -2$Differential equation for HamiltonianHas the following IVP unique solution?
$begingroup$
I am TAing undergraduate differential equation course, and recently have encountered the following problem:
Solve the following initial value problem
$$ y' + (tan x) cdot y = cos ^2 x, quad y(0) = 1$$
and give the largest interval $I$ over which the solution is defined. (In this problem, solving the equation means finding all the continuous, differentiable functions that satisfy the given differential equation at each point of the domain.)
The DE gives the general solution $$y = sin x cdot cos x + cos x$$ for $x in (-frac{pi}{2},frac{pi}{2})$.
The problem is that this solution can be extended to the whole real line $mathbb{R}$ so that the extension, which turns out to be $y = sin x cdot cos x + cos x$ for $x in mathbb{R}$, is continuously differentiable in the whole real line.
Some people in the class suggest therefore the answer be $I=mathbb{R}$.
However I think the answer is $I = (-frac{pi}{2},frac{pi}{2})$.
Here's my thought:
According to Picard's theorem (although the name could be wrong), a unique solution to an IVP exists on the interval where the given DE is 'regular'. Thus the solution exists uniquely on the interval $(-frac{pi}{2},frac{pi}{2})$ over which the initial value is given and the DE is regular. One can find so called 'one-parameter family of solutions' $y = sin x cdot cos x + ccdot cos x$ on intervals other than $(-frac{pi}{2},frac{pi}{2})$, but cannot determine the constant $c$ because we don't have any clues on the behavior of the solutions on those intervals. Therefore one cannot say the solution to the IVP exists uniquely on those intervals, and also cannot say the solution is well-defined on those intervals.
Can anyone tell me which part of my thought is wrong, if there is any? I'd be glad to any explanations. Thank you.
analysis differential
asked Mar 14 at 11:22
Hoil LeeHoil Lee
154
$endgroup$
add a comment |
$begingroup$
I am TAing undergraduate differential equation course, and recently have encountered the following problem:
Solve the following initial value problem
$$ y' + (tan x) cdot y = cos ^2 x, quad y(0) = 1$$
and give the largest interval $I$ over which the solution is defined. (In this problem, solving the equation means finding all the continuous, differentiable functions that satisfy the given differential equation at each point of the domain.)
The DE gives the general solution $$y = sin x cdot cos x + cos x$$ for $x in (-frac{pi}{2},frac{pi}{2})$.
The problem is that this solution can be extended to the whole real line $mathbb{R}$ so that the extension, which turns out to be $y = sin x cdot cos x + cos x$ for $x in mathbb{R}$, is continuously differentiable in the whole real line.
Some people in the class suggest therefore the answer be $I=mathbb{R}$.
However I think the answer is $I = (-frac{pi}{2},frac{pi}{2})$.
Here's my thought:
According to Picard's theorem (although the name could be wrong), a unique solution to an IVP exists on the interval where the given DE is 'regular'. Thus the solution exists uniquely on the interval $(-frac{pi}{2},frac{pi}{2})$ over which the initial value is given and the DE is regular. One can find so called 'one-parameter family of solutions' $y = sin x cdot cos x + ccdot cos x$ on intervals other than $(-frac{pi}{2},frac{pi}{2})$, but cannot determine the constant $c$ because we don't have any clues on the behavior of the solutions on those intervals. Therefore one cannot say the solution to the IVP exists uniquely on those intervals, and also cannot say the solution is well-defined on those intervals.
Can anyone tell me which part of my thought is wrong, if there is any? I'd be glad to any explanations. Thank you.
analysis differential
asked Mar 14 at 11:22
Hoil LeeHoil Lee
154
$endgroup$
$begingroup$
I would say that $(-tfracpi2,tfracpi2)$ is the largest interval on which the solution of the given IVP is defined. I don't know the context, but this is an approach in any standard course on ODEs. In fact, when considering ODEs with analytic RHSs, one could admit removable singularities, but that is a different theory.
$endgroup$
– user539887
Mar 16 at 8:30
$begingroup$
@user539887 Thank you so much for your kind explanations. It helped a lot.
$endgroup$
– Hoil Lee
Mar 16 at 11:56
add a comment |
$begingroup$
I am TAing undergraduate differential equation course, and recently have encountered the following problem:
Solve the following initial value problem
$$ y' + (tan x) cdot y = cos ^2 x, quad y(0) = 1$$
and give the largest interval $I$ over which the solution is defined. (In this problem, solving the equation means finding all the continuous, differentiable functions that satisfy the given differential equation at each point of the domain.)
The DE gives the general solution $$y = sin x cdot cos x + cos x$$ for $x in (-frac{pi}{2},frac{pi}{2})$.
The problem is that this solution can be extended to the whole real line $mathbb{R}$ so that the extension, which turns out to be $y = sin x cdot cos x + cos x$ for $x in mathbb{R}$, is continuously differentiable in the whole real line.
Some people in the class suggest therefore the answer be $I=mathbb{R}$.
However I think the answer is $I = (-frac{pi}{2},frac{pi}{2})$.
Here's my thought:
According to Picard's theorem (although the name could be wrong), a unique solution to an IVP exists on the interval where the given DE is 'regular'. Thus the solution exists uniquely on the interval $(-frac{pi}{2},frac{pi}{2})$ over which the initial value is given and the DE is regular. One can find so called 'one-parameter family of solutions' $y = sin x cdot cos x + ccdot cos x$ on intervals other than $(-frac{pi}{2},frac{pi}{2})$, but cannot determine the constant $c$ because we don't have any clues on the behavior of the solutions on those intervals. Therefore one cannot say the solution to the IVP exists uniquely on those intervals, and also cannot say the solution is well-defined on those intervals.
Can anyone tell me which part of my thought is wrong, if there is any? I'd be glad to any explanations. Thank you.
analysis differential
asked Mar 14 at 11:22
Hoil LeeHoil Lee
154
$endgroup$
I am TAing undergraduate differential equation course, and recently have encountered the following problem:
Solve the following initial value problem
$$ y' + (tan x) cdot y = cos ^2 x, quad y(0) = 1$$
and give the largest interval $I$ over which the solution is defined. (In this problem, solving the equation means finding all the continuous, differentiable functions that satisfy the given differential equation at each point of the domain.)
The DE gives the general solution $$y = sin x cdot cos x + cos x$$ for $x in (-frac{pi}{2},frac{pi}{2})$.
The problem is that this solution can be extended to the whole real line $mathbb{R}$ so that the extension, which turns out to be $y = sin x cdot cos x + cos x$ for $x in mathbb{R}$, is continuously differentiable in the whole real line.
Some people in the class suggest therefore the answer be $I=mathbb{R}$.
However I think the answer is $I = (-frac{pi}{2},frac{pi}{2})$.
Here's my thought:
According to Picard's theorem (although the name could be wrong), a unique solution to an IVP exists on the interval where the given DE is 'regular'. Thus the solution exists uniquely on the interval $(-frac{pi}{2},frac{pi}{2})$ over which the initial value is given and the DE is regular. One can find so called 'one-parameter family of solutions' $y = sin x cdot cos x + ccdot cos x$ on intervals other than $(-frac{pi}{2},frac{pi}{2})$, but cannot determine the constant $c$ because we don't have any clues on the behavior of the solutions on those intervals. Therefore one cannot say the solution to the IVP exists uniquely on those intervals, and also cannot say the solution is well-defined on those intervals.
Can anyone tell me which part of my thought is wrong, if there is any? I'd be glad to any explanations. Thank you.
analysis differential
analysis differential
asked Mar 14 at 11:22
Hoil LeeHoil Lee
154
asked Mar 14 at 11:22
Hoil LeeHoil Lee
154
edited Mar 19 at 6:17
Hoil Lee
asked Mar 14 at 11:22
Hoil LeeHoil Lee
154
asked Mar 14 at 11:22
Hoil LeeHoil Lee
154
asked Mar 14 at 11:22
Hoil LeeHoil Lee
154
154
$begingroup$
I would say that $(-tfracpi2,tfracpi2)$ is the largest interval on which the solution of the given IVP is defined. I don't know the context, but this is an approach in any standard course on ODEs. In fact, when considering ODEs with analytic RHSs, one could admit removable singularities, but that is a different theory.
$endgroup$
– user539887
Mar 16 at 8:30
$begingroup$
@user539887 Thank you so much for your kind explanations. It helped a lot.
$endgroup$
– Hoil Lee
Mar 16 at 11:56
add a comment |
$begingroup$
I would say that $(-tfracpi2,tfracpi2)$ is the largest interval on which the solution of the given IVP is defined. I don't know the context, but this is an approach in any standard course on ODEs. In fact, when considering ODEs with analytic RHSs, one could admit removable singularities, but that is a different theory.
$endgroup$
– user539887
Mar 16 at 8:30
$begingroup$
@user539887 Thank you so much for your kind explanations. It helped a lot.
$endgroup$
– Hoil Lee
Mar 16 at 11:56
$begingroup$
I would say that $(-tfracpi2,tfracpi2)$ is the largest interval on which the solution of the given IVP is defined. I don't know the context, but this is an approach in any standard course on ODEs. In fact, when considering ODEs with analytic RHSs, one could admit removable singularities, but that is a different theory.
$endgroup$
– user539887
Mar 16 at 8:30
$begingroup$
I would say that $(-tfracpi2,tfracpi2)$ is the largest interval on which the solution of the given IVP is defined. I don't know the context, but this is an approach in any standard course on ODEs. In fact, when considering ODEs with analytic RHSs, one could admit removable singularities, but that is a different theory.
$endgroup$
– user539887
Mar 16 at 8:30
$begingroup$
@user539887 Thank you so much for your kind explanations. It helped a lot.
$endgroup$
– Hoil Lee
Mar 16 at 11:56
$begingroup$
@user539887 Thank you so much for your kind explanations. It helped a lot.
$endgroup$
– Hoil Lee
Mar 16 at 11:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I would say you're right on track. In fact, the family of solutions may be broader than that! Consider any functions $f,g:Bbb ZtoBbb R,$ and define $$h(f,g,x):=begin{cases}bigl(f(n)+sin xbigr)cos x & ninBbb Z,xinleft(npi-fracpi2,npi+fracpi2right)\g(n) & ninBbb Z,x=npi+fracpi2.end{cases}$$
Here are a few facts about the function $F(x)=h(f,g,x)$ that I leave to you to verify:
$F:Bbb RtoBbb R,$ regardless of our choices of $f$ and $g.$
$F$ is continuous on $Bbb R$ if and only if $g$ is identically zero.
$F$ is differentiable on $Bbb R$ if and only if $f$ is constant and $g$ is identically zero--that is, if and only if $h(f,g,x)=(c+sin x)cos x$ for some constant $c.$
$y=F(x)$ is a solution to the IVP if and only if $f(0)=1.$
Thus, uniqueness fails (rather badly) outside the interval $left(-fracpi2, fracpi2right),$ but we certainly have existence. The only solution that is everywhere differentiable on the whole real line is $y=(1+sin x)cos x.$ Every solution is identical to this on the interval $left(-fracpi2,fracpi2right),$ but (unless we impose more restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:25
Cameron BuieCameron Buie
86.2k772161
$endgroup$
$begingroup$
Deeply appreciate your answer. It helped a lot.
$endgroup$
– Hoil Lee
Mar 18 at 8:11
$begingroup$
You're very welcome! If you feel that an answer is useful, you can "upvote" it with the arrow buttons. If an answer fully addresses your question, you can "accept" it by clicking the check mark. This signals to other users that the question has been adequately addressed, and gains you reputation points. Welcome to Math.SE!
$endgroup$
– Cameron Buie
Mar 18 at 11:02
$begingroup$
I see. I did upvoted and also accepted. Thanks!
$endgroup$
– Hoil Lee
Mar 19 at 6:15
add a comment |
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$begingroup$
I would say you're right on track. In fact, the family of solutions may be broader than that! Consider any functions $f,g:Bbb ZtoBbb R,$ and define $$h(f,g,x):=begin{cases}bigl(f(n)+sin xbigr)cos x & ninBbb Z,xinleft(npi-fracpi2,npi+fracpi2right)\g(n) & ninBbb Z,x=npi+fracpi2.end{cases}$$
Here are a few facts about the function $F(x)=h(f,g,x)$ that I leave to you to verify:
$F:Bbb RtoBbb R,$ regardless of our choices of $f$ and $g.$
$F$ is continuous on $Bbb R$ if and only if $g$ is identically zero.
$F$ is differentiable on $Bbb R$ if and only if $f$ is constant and $g$ is identically zero--that is, if and only if $h(f,g,x)=(c+sin x)cos x$ for some constant $c.$
$y=F(x)$ is a solution to the IVP if and only if $f(0)=1.$
Thus, uniqueness fails (rather badly) outside the interval $left(-fracpi2, fracpi2right),$ but we certainly have existence. The only solution that is everywhere differentiable on the whole real line is $y=(1+sin x)cos x.$ Every solution is identical to this on the interval $left(-fracpi2,fracpi2right),$ but (unless we impose more restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:25
Cameron BuieCameron Buie
86.2k772161
$endgroup$
$begingroup$
Deeply appreciate your answer. It helped a lot.
$endgroup$
– Hoil Lee
Mar 18 at 8:11
$begingroup$
You're very welcome! If you feel that an answer is useful, you can "upvote" it with the arrow buttons. If an answer fully addresses your question, you can "accept" it by clicking the check mark. This signals to other users that the question has been adequately addressed, and gains you reputation points. Welcome to Math.SE!
$endgroup$
– Cameron Buie
Mar 18 at 11:02
$begingroup$
I see. I did upvoted and also accepted. Thanks!
$endgroup$
– Hoil Lee
Mar 19 at 6:15
add a comment |
$begingroup$
I would say you're right on track. In fact, the family of solutions may be broader than that! Consider any functions $f,g:Bbb ZtoBbb R,$ and define $$h(f,g,x):=begin{cases}bigl(f(n)+sin xbigr)cos x & ninBbb Z,xinleft(npi-fracpi2,npi+fracpi2right)\g(n) & ninBbb Z,x=npi+fracpi2.end{cases}$$
Here are a few facts about the function $F(x)=h(f,g,x)$ that I leave to you to verify:
$F:Bbb RtoBbb R,$ regardless of our choices of $f$ and $g.$
$F$ is continuous on $Bbb R$ if and only if $g$ is identically zero.
$F$ is differentiable on $Bbb R$ if and only if $f$ is constant and $g$ is identically zero--that is, if and only if $h(f,g,x)=(c+sin x)cos x$ for some constant $c.$
$y=F(x)$ is a solution to the IVP if and only if $f(0)=1.$
Thus, uniqueness fails (rather badly) outside the interval $left(-fracpi2, fracpi2right),$ but we certainly have existence. The only solution that is everywhere differentiable on the whole real line is $y=(1+sin x)cos x.$ Every solution is identical to this on the interval $left(-fracpi2,fracpi2right),$ but (unless we impose more restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:25
Cameron BuieCameron Buie
86.2k772161
$endgroup$
$begingroup$
Deeply appreciate your answer. It helped a lot.
$endgroup$
– Hoil Lee
Mar 18 at 8:11
$begingroup$
You're very welcome! If you feel that an answer is useful, you can "upvote" it with the arrow buttons. If an answer fully addresses your question, you can "accept" it by clicking the check mark. This signals to other users that the question has been adequately addressed, and gains you reputation points. Welcome to Math.SE!
$endgroup$
– Cameron Buie
Mar 18 at 11:02
$begingroup$
I see. I did upvoted and also accepted. Thanks!
$endgroup$
– Hoil Lee
Mar 19 at 6:15
add a comment |
$begingroup$
I would say you're right on track. In fact, the family of solutions may be broader than that! Consider any functions $f,g:Bbb ZtoBbb R,$ and define $$h(f,g,x):=begin{cases}bigl(f(n)+sin xbigr)cos x & ninBbb Z,xinleft(npi-fracpi2,npi+fracpi2right)\g(n) & ninBbb Z,x=npi+fracpi2.end{cases}$$
Here are a few facts about the function $F(x)=h(f,g,x)$ that I leave to you to verify:
$F:Bbb RtoBbb R,$ regardless of our choices of $f$ and $g.$
$F$ is continuous on $Bbb R$ if and only if $g$ is identically zero.
$F$ is differentiable on $Bbb R$ if and only if $f$ is constant and $g$ is identically zero--that is, if and only if $h(f,g,x)=(c+sin x)cos x$ for some constant $c.$
$y=F(x)$ is a solution to the IVP if and only if $f(0)=1.$
Thus, uniqueness fails (rather badly) outside the interval $left(-fracpi2, fracpi2right),$ but we certainly have existence. The only solution that is everywhere differentiable on the whole real line is $y=(1+sin x)cos x.$ Every solution is identical to this on the interval $left(-fracpi2,fracpi2right),$ but (unless we impose more restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:25
Cameron BuieCameron Buie
86.2k772161
$endgroup$
I would say you're right on track. In fact, the family of solutions may be broader than that! Consider any functions $f,g:Bbb ZtoBbb R,$ and define $$h(f,g,x):=begin{cases}bigl(f(n)+sin xbigr)cos x & ninBbb Z,xinleft(npi-fracpi2,npi+fracpi2right)\g(n) & ninBbb Z,x=npi+fracpi2.end{cases}$$
Here are a few facts about the function $F(x)=h(f,g,x)$ that I leave to you to verify:
$F:Bbb RtoBbb R,$ regardless of our choices of $f$ and $g.$
$F$ is continuous on $Bbb R$ if and only if $g$ is identically zero.
$F$ is differentiable on $Bbb R$ if and only if $f$ is constant and $g$ is identically zero--that is, if and only if $h(f,g,x)=(c+sin x)cos x$ for some constant $c.$
$y=F(x)$ is a solution to the IVP if and only if $f(0)=1.$
Thus, uniqueness fails (rather badly) outside the interval $left(-fracpi2, fracpi2right),$ but we certainly have existence. The only solution that is everywhere differentiable on the whole real line is $y=(1+sin x)cos x.$ Every solution is identical to this on the interval $left(-fracpi2,fracpi2right),$ but (unless we impose more restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:25
Cameron BuieCameron Buie
86.2k772161
answered Mar 16 at 16:25
Cameron BuieCameron Buie
86.2k772161
answered Mar 16 at 16:25
Cameron BuieCameron Buie
86.2k772161
answered Mar 16 at 16:25
Cameron BuieCameron Buie
86.2k772161
86.2k772161
$begingroup$
Deeply appreciate your answer. It helped a lot.
$endgroup$
– Hoil Lee
Mar 18 at 8:11
$begingroup$
You're very welcome! If you feel that an answer is useful, you can "upvote" it with the arrow buttons. If an answer fully addresses your question, you can "accept" it by clicking the check mark. This signals to other users that the question has been adequately addressed, and gains you reputation points. Welcome to Math.SE!
$endgroup$
– Cameron Buie
Mar 18 at 11:02
$begingroup$
I see. I did upvoted and also accepted. Thanks!
$endgroup$
– Hoil Lee
Mar 19 at 6:15
add a comment |
$begingroup$
Deeply appreciate your answer. It helped a lot.
$endgroup$
– Hoil Lee
Mar 18 at 8:11
$begingroup$
You're very welcome! If you feel that an answer is useful, you can "upvote" it with the arrow buttons. If an answer fully addresses your question, you can "accept" it by clicking the check mark. This signals to other users that the question has been adequately addressed, and gains you reputation points. Welcome to Math.SE!
$endgroup$
– Cameron Buie
Mar 18 at 11:02
$begingroup$
I see. I did upvoted and also accepted. Thanks!
$endgroup$
– Hoil Lee
Mar 19 at 6:15
$begingroup$
Deeply appreciate your answer. It helped a lot.
$endgroup$
– Hoil Lee
Mar 18 at 8:11
$begingroup$
Deeply appreciate your answer. It helped a lot.
$endgroup$
– Hoil Lee
Mar 18 at 8:11
$begingroup$
You're very welcome! If you feel that an answer is useful, you can "upvote" it with the arrow buttons. If an answer fully addresses your question, you can "accept" it by clicking the check mark. This signals to other users that the question has been adequately addressed, and gains you reputation points. Welcome to Math.SE!
$endgroup$
– Cameron Buie
Mar 18 at 11:02
$begingroup$
You're very welcome! If you feel that an answer is useful, you can "upvote" it with the arrow buttons. If an answer fully addresses your question, you can "accept" it by clicking the check mark. This signals to other users that the question has been adequately addressed, and gains you reputation points. Welcome to Math.SE!
$endgroup$
– Cameron Buie
Mar 18 at 11:02
$begingroup$
I see. I did upvoted and also accepted. Thanks!
$endgroup$
– Hoil Lee
Mar 19 at 6:15
$begingroup$
I see. I did upvoted and also accepted. Thanks!
$endgroup$
– Hoil Lee
Mar 19 at 6:15
add a comment |
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I would say that $(-tfracpi2,tfracpi2)$ is the largest interval on which the solution of the given IVP is defined. I don't know the context, but this is an approach in any standard course on ODEs. In fact, when considering ODEs with analytic RHSs, one could admit removable singularities, but that is a different theory.
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– user539887
Mar 16 at 8:30
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@user539887 Thank you so much for your kind explanations. It helped a lot.
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– Hoil Lee
Mar 16 at 11:56