Mixing
Why do we take only intervals for solution of differential equation [closed]
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Suppose the solution to my equation blows up at 0 and is well defined everywhere else. Then why can't we just take R-{0} as the domain of the function? Also can someone give an example of a nonzero function whose first order derivative is zero?
Edit: sorry, nonzero function which is not constant.
Suppose the solution to a differential equation is 1/x then why do we have to take an interval and not R-{0}
integration ordinary-differential-equations interval-arithmetic
asked Jan 28 at 6:10
KorraKorra
185
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closed as off-topic by Dylan, RRL, José Carlos Santos, mrtaurho, Ali Caglayan Jan 28 at 13:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dylan, RRL, José Carlos Santos, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose the solution to my equation blows up at 0 and is well defined everywhere else. Then why can't we just take R-{0} as the domain of the function? Also can someone give an example of a nonzero function whose first order derivative is zero?
Edit: sorry, nonzero function which is not constant.
Suppose the solution to a differential equation is 1/x then why do we have to take an interval and not R-{0}
integration ordinary-differential-equations interval-arithmetic
asked Jan 28 at 6:10
KorraKorra
185
$endgroup$
closed as off-topic by Dylan, RRL, José Carlos Santos, mrtaurho, Ali Caglayan Jan 28 at 13:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dylan, RRL, José Carlos Santos, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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"Also can someone give an example of a nonzero function whose first order derivative is zero?" $f(x)=3$?
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– Arthur
Jan 28 at 6:55
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Intervals of solution are based on the singular points of the differential equation, not the solution itself. Without specific examples I can't say more.
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– Dylan
Jan 28 at 7:04
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@Arthur I edited my question. Can you give one now? Please?
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– Korra
Jan 28 at 7:41
add a comment |
$begingroup$
Suppose the solution to my equation blows up at 0 and is well defined everywhere else. Then why can't we just take R-{0} as the domain of the function? Also can someone give an example of a nonzero function whose first order derivative is zero?
Edit: sorry, nonzero function which is not constant.
Suppose the solution to a differential equation is 1/x then why do we have to take an interval and not R-{0}
integration ordinary-differential-equations interval-arithmetic
asked Jan 28 at 6:10
KorraKorra
185
$endgroup$
Suppose the solution to my equation blows up at 0 and is well defined everywhere else. Then why can't we just take R-{0} as the domain of the function? Also can someone give an example of a nonzero function whose first order derivative is zero?
Edit: sorry, nonzero function which is not constant.
Suppose the solution to a differential equation is 1/x then why do we have to take an interval and not R-{0}
integration ordinary-differential-equations interval-arithmetic
integration ordinary-differential-equations interval-arithmetic
asked Jan 28 at 6:10
KorraKorra
185
asked Jan 28 at 6:10
KorraKorra
185
edited Jan 28 at 7:36
Korra
asked Jan 28 at 6:10
KorraKorra
185
asked Jan 28 at 6:10
KorraKorra
185
asked Jan 28 at 6:10
KorraKorra
185
185
closed as off-topic by Dylan, RRL, José Carlos Santos, mrtaurho, Ali Caglayan Jan 28 at 13:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dylan, RRL, José Carlos Santos, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Dylan, RRL, José Carlos Santos, mrtaurho, Ali Caglayan Jan 28 at 13:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dylan, RRL, José Carlos Santos, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
"Also can someone give an example of a nonzero function whose first order derivative is zero?" $f(x)=3$?
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– Arthur
Jan 28 at 6:55
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Intervals of solution are based on the singular points of the differential equation, not the solution itself. Without specific examples I can't say more.
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– Dylan
Jan 28 at 7:04
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@Arthur I edited my question. Can you give one now? Please?
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– Korra
Jan 28 at 7:41
add a comment |
1
$begingroup$
"Also can someone give an example of a nonzero function whose first order derivative is zero?" $f(x)=3$?
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– Arthur
Jan 28 at 6:55
$begingroup$
Intervals of solution are based on the singular points of the differential equation, not the solution itself. Without specific examples I can't say more.
$endgroup$
– Dylan
Jan 28 at 7:04
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@Arthur I edited my question. Can you give one now? Please?
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– Korra
Jan 28 at 7:41
1
1
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"Also can someone give an example of a nonzero function whose first order derivative is zero?" $f(x)=3$?
$endgroup$
– Arthur
Jan 28 at 6:55
$begingroup$
"Also can someone give an example of a nonzero function whose first order derivative is zero?" $f(x)=3$?
$endgroup$
– Arthur
Jan 28 at 6:55
$begingroup$
Intervals of solution are based on the singular points of the differential equation, not the solution itself. Without specific examples I can't say more.
$endgroup$
– Dylan
Jan 28 at 7:04
$begingroup$
Intervals of solution are based on the singular points of the differential equation, not the solution itself. Without specific examples I can't say more.
$endgroup$
– Dylan
Jan 28 at 7:04
$begingroup$
@Arthur I edited my question. Can you give one now? Please?
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– Korra
Jan 28 at 7:41
$begingroup$
@Arthur I edited my question. Can you give one now? Please?
$endgroup$
– Korra
Jan 28 at 7:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose your equation is
$$
y'(x)=-frac 1{x^2}\
y(1)=1
$$
just to keep it simple. Then clearly, for positive $x$, the solution is $frac1x$. However, the equation itself says nothing about what the value of the function is for negative numbers. Maybe for negative $x$ we have $y=frac1x +1000$? There is just no way to use the equation to extend a solution uniquely past a point of singularity like that.
As for a function with a zero derivative, such a function must be constant on any connected component of its domain. Maybe that can give you some hint on how to find one.
answered Jan 28 at 7:58
ArthurArthur
120k7120204
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Yep, exactly. This is what I was trying to get at, too.
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– goblin
Jan 28 at 8:22
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So basically all I have to do is change the interval from (a,b) to a one which is disconnected? @Arthur
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– Korra
Jan 28 at 11:53
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@Korra If you want a non-constant function with zero derivative, yes. You need a disconnected domain.
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– Arthur
Jan 28 at 15:22
add a comment |
$begingroup$
In full generality, you can look at solutions on any open set $U$, and the theory will work just fine. In particular: given a differential equation $E$, you get a corresponding sheaf $tilde{E}$ by defining $tilde{E}(U)$ to be the set of all solutions to the equation on the open set $U$.
However, the benefit of restricting attention to situations where $U$ is connected is that it ensures uniqueness of solutions to IVPs.
For example, if we know that $frac{dy}{dx} = 1/x$ and that $y(x := 1) = 0$, then we can conclude that $y = mathrm{log}_e(x)$, but only if we're implicitly working on an open set $U$ that contains $1$, does not contain $0$, and happens to be connected. Note that $(0,infty)$ is the largest such set. If we included some extra points in our set, say $U = (-infty,0) cup (0,infty)$, then our IVP would have more than one solution. In particular, for all $A in mathbb{R}$, the function
$$y = begin{cases} mathrm{log}_e(-x)+A, x < 0 \ mathrm{log}_w(x), x > 0 end{cases}$$
solves the aforementioned IVP on the domain $U = (-infty,0) cup (0,infty).$
answered Jan 28 at 7:54
goblingoblin
37.1k1159193
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose your equation is
$$
y'(x)=-frac 1{x^2}\
y(1)=1
$$
just to keep it simple. Then clearly, for positive $x$, the solution is $frac1x$. However, the equation itself says nothing about what the value of the function is for negative numbers. Maybe for negative $x$ we have $y=frac1x +1000$? There is just no way to use the equation to extend a solution uniquely past a point of singularity like that.
As for a function with a zero derivative, such a function must be constant on any connected component of its domain. Maybe that can give you some hint on how to find one.
answered Jan 28 at 7:58
ArthurArthur
120k7120204
$endgroup$
$begingroup$
Yep, exactly. This is what I was trying to get at, too.
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– goblin
Jan 28 at 8:22
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So basically all I have to do is change the interval from (a,b) to a one which is disconnected? @Arthur
$endgroup$
– Korra
Jan 28 at 11:53
$begingroup$
@Korra If you want a non-constant function with zero derivative, yes. You need a disconnected domain.
$endgroup$
– Arthur
Jan 28 at 15:22
add a comment |
$begingroup$
Suppose your equation is
$$
y'(x)=-frac 1{x^2}\
y(1)=1
$$
just to keep it simple. Then clearly, for positive $x$, the solution is $frac1x$. However, the equation itself says nothing about what the value of the function is for negative numbers. Maybe for negative $x$ we have $y=frac1x +1000$? There is just no way to use the equation to extend a solution uniquely past a point of singularity like that.
As for a function with a zero derivative, such a function must be constant on any connected component of its domain. Maybe that can give you some hint on how to find one.
answered Jan 28 at 7:58
ArthurArthur
120k7120204
$endgroup$
$begingroup$
Yep, exactly. This is what I was trying to get at, too.
$endgroup$
– goblin
Jan 28 at 8:22
$begingroup$
So basically all I have to do is change the interval from (a,b) to a one which is disconnected? @Arthur
$endgroup$
– Korra
Jan 28 at 11:53
$begingroup$
@Korra If you want a non-constant function with zero derivative, yes. You need a disconnected domain.
$endgroup$
– Arthur
Jan 28 at 15:22
add a comment |
$begingroup$
Suppose your equation is
$$
y'(x)=-frac 1{x^2}\
y(1)=1
$$
just to keep it simple. Then clearly, for positive $x$, the solution is $frac1x$. However, the equation itself says nothing about what the value of the function is for negative numbers. Maybe for negative $x$ we have $y=frac1x +1000$? There is just no way to use the equation to extend a solution uniquely past a point of singularity like that.
As for a function with a zero derivative, such a function must be constant on any connected component of its domain. Maybe that can give you some hint on how to find one.
answered Jan 28 at 7:58
ArthurArthur
120k7120204
$endgroup$
Suppose your equation is
$$
y'(x)=-frac 1{x^2}\
y(1)=1
$$
just to keep it simple. Then clearly, for positive $x$, the solution is $frac1x$. However, the equation itself says nothing about what the value of the function is for negative numbers. Maybe for negative $x$ we have $y=frac1x +1000$? There is just no way to use the equation to extend a solution uniquely past a point of singularity like that.
As for a function with a zero derivative, such a function must be constant on any connected component of its domain. Maybe that can give you some hint on how to find one.
answered Jan 28 at 7:58
ArthurArthur
120k7120204
answered Jan 28 at 7:58
ArthurArthur
120k7120204
answered Jan 28 at 7:58
ArthurArthur
120k7120204
answered Jan 28 at 7:58
ArthurArthur
120k7120204
120k7120204
$begingroup$
Yep, exactly. This is what I was trying to get at, too.
$endgroup$
– goblin
Jan 28 at 8:22
$begingroup$
So basically all I have to do is change the interval from (a,b) to a one which is disconnected? @Arthur
$endgroup$
– Korra
Jan 28 at 11:53
$begingroup$
@Korra If you want a non-constant function with zero derivative, yes. You need a disconnected domain.
$endgroup$
– Arthur
Jan 28 at 15:22
add a comment |
$begingroup$
Yep, exactly. This is what I was trying to get at, too.
$endgroup$
– goblin
Jan 28 at 8:22
$begingroup$
So basically all I have to do is change the interval from (a,b) to a one which is disconnected? @Arthur
$endgroup$
– Korra
Jan 28 at 11:53
$begingroup$
@Korra If you want a non-constant function with zero derivative, yes. You need a disconnected domain.
$endgroup$
– Arthur
Jan 28 at 15:22
$begingroup$
Yep, exactly. This is what I was trying to get at, too.
$endgroup$
– goblin
Jan 28 at 8:22
$begingroup$
Yep, exactly. This is what I was trying to get at, too.
$endgroup$
– goblin
Jan 28 at 8:22
$begingroup$
So basically all I have to do is change the interval from (a,b) to a one which is disconnected? @Arthur
$endgroup$
– Korra
Jan 28 at 11:53
$begingroup$
So basically all I have to do is change the interval from (a,b) to a one which is disconnected? @Arthur
$endgroup$
– Korra
Jan 28 at 11:53
$begingroup$
@Korra If you want a non-constant function with zero derivative, yes. You need a disconnected domain.
$endgroup$
– Arthur
Jan 28 at 15:22
$begingroup$
@Korra If you want a non-constant function with zero derivative, yes. You need a disconnected domain.
$endgroup$
– Arthur
Jan 28 at 15:22
add a comment |
$begingroup$
In full generality, you can look at solutions on any open set $U$, and the theory will work just fine. In particular: given a differential equation $E$, you get a corresponding sheaf $tilde{E}$ by defining $tilde{E}(U)$ to be the set of all solutions to the equation on the open set $U$.
However, the benefit of restricting attention to situations where $U$ is connected is that it ensures uniqueness of solutions to IVPs.
For example, if we know that $frac{dy}{dx} = 1/x$ and that $y(x := 1) = 0$, then we can conclude that $y = mathrm{log}_e(x)$, but only if we're implicitly working on an open set $U$ that contains $1$, does not contain $0$, and happens to be connected. Note that $(0,infty)$ is the largest such set. If we included some extra points in our set, say $U = (-infty,0) cup (0,infty)$, then our IVP would have more than one solution. In particular, for all $A in mathbb{R}$, the function
$$y = begin{cases} mathrm{log}_e(-x)+A, x < 0 \ mathrm{log}_w(x), x > 0 end{cases}$$
solves the aforementioned IVP on the domain $U = (-infty,0) cup (0,infty).$
answered Jan 28 at 7:54
goblingoblin
37.1k1159193
$endgroup$
add a comment |
$begingroup$
In full generality, you can look at solutions on any open set $U$, and the theory will work just fine. In particular: given a differential equation $E$, you get a corresponding sheaf $tilde{E}$ by defining $tilde{E}(U)$ to be the set of all solutions to the equation on the open set $U$.
However, the benefit of restricting attention to situations where $U$ is connected is that it ensures uniqueness of solutions to IVPs.
For example, if we know that $frac{dy}{dx} = 1/x$ and that $y(x := 1) = 0$, then we can conclude that $y = mathrm{log}_e(x)$, but only if we're implicitly working on an open set $U$ that contains $1$, does not contain $0$, and happens to be connected. Note that $(0,infty)$ is the largest such set. If we included some extra points in our set, say $U = (-infty,0) cup (0,infty)$, then our IVP would have more than one solution. In particular, for all $A in mathbb{R}$, the function
$$y = begin{cases} mathrm{log}_e(-x)+A, x < 0 \ mathrm{log}_w(x), x > 0 end{cases}$$
solves the aforementioned IVP on the domain $U = (-infty,0) cup (0,infty).$
answered Jan 28 at 7:54
goblingoblin
37.1k1159193
$endgroup$
add a comment |
$begingroup$
In full generality, you can look at solutions on any open set $U$, and the theory will work just fine. In particular: given a differential equation $E$, you get a corresponding sheaf $tilde{E}$ by defining $tilde{E}(U)$ to be the set of all solutions to the equation on the open set $U$.
However, the benefit of restricting attention to situations where $U$ is connected is that it ensures uniqueness of solutions to IVPs.
For example, if we know that $frac{dy}{dx} = 1/x$ and that $y(x := 1) = 0$, then we can conclude that $y = mathrm{log}_e(x)$, but only if we're implicitly working on an open set $U$ that contains $1$, does not contain $0$, and happens to be connected. Note that $(0,infty)$ is the largest such set. If we included some extra points in our set, say $U = (-infty,0) cup (0,infty)$, then our IVP would have more than one solution. In particular, for all $A in mathbb{R}$, the function
$$y = begin{cases} mathrm{log}_e(-x)+A, x < 0 \ mathrm{log}_w(x), x > 0 end{cases}$$
solves the aforementioned IVP on the domain $U = (-infty,0) cup (0,infty).$
answered Jan 28 at 7:54
goblingoblin
37.1k1159193
$endgroup$
In full generality, you can look at solutions on any open set $U$, and the theory will work just fine. In particular: given a differential equation $E$, you get a corresponding sheaf $tilde{E}$ by defining $tilde{E}(U)$ to be the set of all solutions to the equation on the open set $U$.
However, the benefit of restricting attention to situations where $U$ is connected is that it ensures uniqueness of solutions to IVPs.
For example, if we know that $frac{dy}{dx} = 1/x$ and that $y(x := 1) = 0$, then we can conclude that $y = mathrm{log}_e(x)$, but only if we're implicitly working on an open set $U$ that contains $1$, does not contain $0$, and happens to be connected. Note that $(0,infty)$ is the largest such set. If we included some extra points in our set, say $U = (-infty,0) cup (0,infty)$, then our IVP would have more than one solution. In particular, for all $A in mathbb{R}$, the function
$$y = begin{cases} mathrm{log}_e(-x)+A, x < 0 \ mathrm{log}_w(x), x > 0 end{cases}$$
solves the aforementioned IVP on the domain $U = (-infty,0) cup (0,infty).$
answered Jan 28 at 7:54
goblingoblin
37.1k1159193
edited Jan 28 at 8:21
answered Jan 28 at 7:54
goblingoblin
37.1k1159193
answered Jan 28 at 7:54
goblingoblin
37.1k1159193
answered Jan 28 at 7:54
goblingoblin
37.1k1159193
37.1k1159193
add a comment |
add a comment |
1
$begingroup$
"Also can someone give an example of a nonzero function whose first order derivative is zero?" $f(x)=3$?
$endgroup$
– Arthur
Jan 28 at 6:55
$begingroup$
Intervals of solution are based on the singular points of the differential equation, not the solution itself. Without specific examples I can't say more.
$endgroup$
– Dylan
Jan 28 at 7:04
$begingroup$
@Arthur I edited my question. Can you give one now? Please?
$endgroup$
– Korra
Jan 28 at 7:41