Mixing
Differential equation $frac{dx}{dt} = kx^2$ prove that I have all the solutions
$begingroup$
In the book Differential Equations and Boundary Value Problems by Edwards and Penney, section 1.1, question 43, a differential equation is provided:
$$
frac{dx}{dt} = kx^2
$$
and a solution:
$$
x(t)=frac{1}{C-kt}.
$$
The question then says "Determine by inspection a solution of the initial value problem $x'=kx^2, x(0)=0.$"
I reason that since the provided solution never equals $0$, I can safely ignore it. Also, if $x=0$, then $x'=0$, which means that only the functions which might be solutions are ones that have a local min or max at the point $(0,0)$.
The answer in the back of the book is given as the identically zero function. This makes me squint, since while that clearly is a solution, I don't see how to prove that it's the only one. At this point, I realize that the question does say a solution, and not every solution, so I guess it works as an answer to the question, but now I'm still left wondering if the identically zero function is in fact the only solution that satisfies that initial condition, and more particularly, how to prove it, if it's true.
I see how to achieve the solution for all $x$ where $xneq 0$, so that part is clear to me. I just don't see how to go from "the point $(t,0)$ must be a local extremum if it solves $x$" to "$x$ is identically $0$", or actually, if that conclusion is even true.
ordinary-differential-equations
edited Jan 16 at 2:49
NoChance
3,67121221
asked Jan 16 at 2:36
Michael LeueMichael Leue
6
$endgroup$
add a comment |
$begingroup$
In the book Differential Equations and Boundary Value Problems by Edwards and Penney, section 1.1, question 43, a differential equation is provided:
$$
frac{dx}{dt} = kx^2
$$
and a solution:
$$
x(t)=frac{1}{C-kt}.
$$
The question then says "Determine by inspection a solution of the initial value problem $x'=kx^2, x(0)=0.$"
I reason that since the provided solution never equals $0$, I can safely ignore it. Also, if $x=0$, then $x'=0$, which means that only the functions which might be solutions are ones that have a local min or max at the point $(0,0)$.
The answer in the back of the book is given as the identically zero function. This makes me squint, since while that clearly is a solution, I don't see how to prove that it's the only one. At this point, I realize that the question does say a solution, and not every solution, so I guess it works as an answer to the question, but now I'm still left wondering if the identically zero function is in fact the only solution that satisfies that initial condition, and more particularly, how to prove it, if it's true.
I see how to achieve the solution for all $x$ where $xneq 0$, so that part is clear to me. I just don't see how to go from "the point $(t,0)$ must be a local extremum if it solves $x$" to "$x$ is identically $0$", or actually, if that conclusion is even true.
ordinary-differential-equations
edited Jan 16 at 2:49
NoChance
3,67121221
asked Jan 16 at 2:36
Michael LeueMichael Leue
6
$endgroup$
$begingroup$
For any autonomous first-order ODE $x' = f(x)$ the constant functions $x(t) = x_0$ where $f(x_0) = 0$, i. e. the zeros of $f$, are solutions. These cannot be obtained by any value of $C$ in the general solution. So indeed with just the general solution $x(t) = (C-kt)^{-1}$ you do not have all the solutions of the ODE.
$endgroup$
– Christoph
Jan 16 at 3:16
add a comment |
$begingroup$
In the book Differential Equations and Boundary Value Problems by Edwards and Penney, section 1.1, question 43, a differential equation is provided:
$$
frac{dx}{dt} = kx^2
$$
and a solution:
$$
x(t)=frac{1}{C-kt}.
$$
The question then says "Determine by inspection a solution of the initial value problem $x'=kx^2, x(0)=0.$"
I reason that since the provided solution never equals $0$, I can safely ignore it. Also, if $x=0$, then $x'=0$, which means that only the functions which might be solutions are ones that have a local min or max at the point $(0,0)$.
The answer in the back of the book is given as the identically zero function. This makes me squint, since while that clearly is a solution, I don't see how to prove that it's the only one. At this point, I realize that the question does say a solution, and not every solution, so I guess it works as an answer to the question, but now I'm still left wondering if the identically zero function is in fact the only solution that satisfies that initial condition, and more particularly, how to prove it, if it's true.
I see how to achieve the solution for all $x$ where $xneq 0$, so that part is clear to me. I just don't see how to go from "the point $(t,0)$ must be a local extremum if it solves $x$" to "$x$ is identically $0$", or actually, if that conclusion is even true.
ordinary-differential-equations
edited Jan 16 at 2:49
NoChance
3,67121221
asked Jan 16 at 2:36
Michael LeueMichael Leue
6
$endgroup$
In the book Differential Equations and Boundary Value Problems by Edwards and Penney, section 1.1, question 43, a differential equation is provided:
$$
frac{dx}{dt} = kx^2
$$
and a solution:
$$
x(t)=frac{1}{C-kt}.
$$
The question then says "Determine by inspection a solution of the initial value problem $x'=kx^2, x(0)=0.$"
I reason that since the provided solution never equals $0$, I can safely ignore it. Also, if $x=0$, then $x'=0$, which means that only the functions which might be solutions are ones that have a local min or max at the point $(0,0)$.
The answer in the back of the book is given as the identically zero function. This makes me squint, since while that clearly is a solution, I don't see how to prove that it's the only one. At this point, I realize that the question does say a solution, and not every solution, so I guess it works as an answer to the question, but now I'm still left wondering if the identically zero function is in fact the only solution that satisfies that initial condition, and more particularly, how to prove it, if it's true.
I see how to achieve the solution for all $x$ where $xneq 0$, so that part is clear to me. I just don't see how to go from "the point $(t,0)$ must be a local extremum if it solves $x$" to "$x$ is identically $0$", or actually, if that conclusion is even true.
ordinary-differential-equations
ordinary-differential-equations
edited Jan 16 at 2:49
NoChance
3,67121221
asked Jan 16 at 2:36
Michael LeueMichael Leue
6
edited Jan 16 at 2:49
NoChance
3,67121221
asked Jan 16 at 2:36
Michael LeueMichael Leue
6
edited Jan 16 at 2:49
NoChance
3,67121221
edited Jan 16 at 2:49
NoChance
3,67121221
edited Jan 16 at 2:49
NoChance
3,67121221
3,67121221
asked Jan 16 at 2:36
Michael LeueMichael Leue
6
asked Jan 16 at 2:36
Michael LeueMichael Leue
6
asked Jan 16 at 2:36
Michael LeueMichael Leue
6
6
$begingroup$
For any autonomous first-order ODE $x' = f(x)$ the constant functions $x(t) = x_0$ where $f(x_0) = 0$, i. e. the zeros of $f$, are solutions. These cannot be obtained by any value of $C$ in the general solution. So indeed with just the general solution $x(t) = (C-kt)^{-1}$ you do not have all the solutions of the ODE.
$endgroup$
– Christoph
Jan 16 at 3:16
add a comment |
$begingroup$
For any autonomous first-order ODE $x' = f(x)$ the constant functions $x(t) = x_0$ where $f(x_0) = 0$, i. e. the zeros of $f$, are solutions. These cannot be obtained by any value of $C$ in the general solution. So indeed with just the general solution $x(t) = (C-kt)^{-1}$ you do not have all the solutions of the ODE.
$endgroup$
– Christoph
Jan 16 at 3:16
$begingroup$
For any autonomous first-order ODE $x' = f(x)$ the constant functions $x(t) = x_0$ where $f(x_0) = 0$, i. e. the zeros of $f$, are solutions. These cannot be obtained by any value of $C$ in the general solution. So indeed with just the general solution $x(t) = (C-kt)^{-1}$ you do not have all the solutions of the ODE.
$endgroup$
– Christoph
Jan 16 at 3:16
$begingroup$
For any autonomous first-order ODE $x' = f(x)$ the constant functions $x(t) = x_0$ where $f(x_0) = 0$, i. e. the zeros of $f$, are solutions. These cannot be obtained by any value of $C$ in the general solution. So indeed with just the general solution $x(t) = (C-kt)^{-1}$ you do not have all the solutions of the ODE.
$endgroup$
– Christoph
Jan 16 at 3:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is a well known theorem giving conditions for existence and uniqueness of first order differential equations $ y' = F(x, y), y(x_0) = y_0 $.
The statement of the theorem is that if $F(x, y)$ be a real-valued function continuous on some rectangle $ R = { (x, y) | |x - x_0| leq a, |y - y_0| leq b}$, and $ partial F / partial y$ is continuous on $ R$, then there exists an interval $ I = [x_0 - h, x_0 + h] $, $h leq a$, such that the initial value problem has a unique solution.
answered Jan 16 at 2:48
Klint QinamiKlint Qinami
1,137410
$endgroup$
$begingroup$
So...the reasoning is that since $x(t)=0$ is always a solution for every $t$, then since that solution is unique, no other function which is a solution can hit the same point?
$endgroup$
– Michael Leue
Jan 16 at 2:51
$begingroup$
Sort of. All this theorem initially gives you is a small interval (-h, h) for which the solution is unique. You can then extend this by repeated application of the theorem to all of R.
$endgroup$
– Klint Qinami
Jan 16 at 2:55
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
There is a well known theorem giving conditions for existence and uniqueness of first order differential equations $ y' = F(x, y), y(x_0) = y_0 $.
The statement of the theorem is that if $F(x, y)$ be a real-valued function continuous on some rectangle $ R = { (x, y) | |x - x_0| leq a, |y - y_0| leq b}$, and $ partial F / partial y$ is continuous on $ R$, then there exists an interval $ I = [x_0 - h, x_0 + h] $, $h leq a$, such that the initial value problem has a unique solution.
answered Jan 16 at 2:48
Klint QinamiKlint Qinami
1,137410
$endgroup$
$begingroup$
So...the reasoning is that since $x(t)=0$ is always a solution for every $t$, then since that solution is unique, no other function which is a solution can hit the same point?
$endgroup$
– Michael Leue
Jan 16 at 2:51
$begingroup$
Sort of. All this theorem initially gives you is a small interval (-h, h) for which the solution is unique. You can then extend this by repeated application of the theorem to all of R.
$endgroup$
– Klint Qinami
Jan 16 at 2:55
add a comment |
$begingroup$
There is a well known theorem giving conditions for existence and uniqueness of first order differential equations $ y' = F(x, y), y(x_0) = y_0 $.
The statement of the theorem is that if $F(x, y)$ be a real-valued function continuous on some rectangle $ R = { (x, y) | |x - x_0| leq a, |y - y_0| leq b}$, and $ partial F / partial y$ is continuous on $ R$, then there exists an interval $ I = [x_0 - h, x_0 + h] $, $h leq a$, such that the initial value problem has a unique solution.
answered Jan 16 at 2:48
Klint QinamiKlint Qinami
1,137410
$endgroup$
$begingroup$
So...the reasoning is that since $x(t)=0$ is always a solution for every $t$, then since that solution is unique, no other function which is a solution can hit the same point?
$endgroup$
– Michael Leue
Jan 16 at 2:51
$begingroup$
Sort of. All this theorem initially gives you is a small interval (-h, h) for which the solution is unique. You can then extend this by repeated application of the theorem to all of R.
$endgroup$
– Klint Qinami
Jan 16 at 2:55
add a comment |
$begingroup$
There is a well known theorem giving conditions for existence and uniqueness of first order differential equations $ y' = F(x, y), y(x_0) = y_0 $.
The statement of the theorem is that if $F(x, y)$ be a real-valued function continuous on some rectangle $ R = { (x, y) | |x - x_0| leq a, |y - y_0| leq b}$, and $ partial F / partial y$ is continuous on $ R$, then there exists an interval $ I = [x_0 - h, x_0 + h] $, $h leq a$, such that the initial value problem has a unique solution.
answered Jan 16 at 2:48
Klint QinamiKlint Qinami
1,137410
$endgroup$
There is a well known theorem giving conditions for existence and uniqueness of first order differential equations $ y' = F(x, y), y(x_0) = y_0 $.
The statement of the theorem is that if $F(x, y)$ be a real-valued function continuous on some rectangle $ R = { (x, y) | |x - x_0| leq a, |y - y_0| leq b}$, and $ partial F / partial y$ is continuous on $ R$, then there exists an interval $ I = [x_0 - h, x_0 + h] $, $h leq a$, such that the initial value problem has a unique solution.
answered Jan 16 at 2:48
Klint QinamiKlint Qinami
1,137410
edited Jan 16 at 2:58
answered Jan 16 at 2:48
Klint QinamiKlint Qinami
1,137410
answered Jan 16 at 2:48
Klint QinamiKlint Qinami
1,137410
answered Jan 16 at 2:48
Klint QinamiKlint Qinami
1,137410
1,137410
$begingroup$
So...the reasoning is that since $x(t)=0$ is always a solution for every $t$, then since that solution is unique, no other function which is a solution can hit the same point?
$endgroup$
– Michael Leue
Jan 16 at 2:51
$begingroup$
Sort of. All this theorem initially gives you is a small interval (-h, h) for which the solution is unique. You can then extend this by repeated application of the theorem to all of R.
$endgroup$
– Klint Qinami
Jan 16 at 2:55
add a comment |
$begingroup$
So...the reasoning is that since $x(t)=0$ is always a solution for every $t$, then since that solution is unique, no other function which is a solution can hit the same point?
$endgroup$
– Michael Leue
Jan 16 at 2:51
$begingroup$
Sort of. All this theorem initially gives you is a small interval (-h, h) for which the solution is unique. You can then extend this by repeated application of the theorem to all of R.
$endgroup$
– Klint Qinami
Jan 16 at 2:55
$begingroup$
So...the reasoning is that since $x(t)=0$ is always a solution for every $t$, then since that solution is unique, no other function which is a solution can hit the same point?
$endgroup$
– Michael Leue
Jan 16 at 2:51
$begingroup$
So...the reasoning is that since $x(t)=0$ is always a solution for every $t$, then since that solution is unique, no other function which is a solution can hit the same point?
$endgroup$
– Michael Leue
Jan 16 at 2:51
$begingroup$
Sort of. All this theorem initially gives you is a small interval (-h, h) for which the solution is unique. You can then extend this by repeated application of the theorem to all of R.
$endgroup$
– Klint Qinami
Jan 16 at 2:55
$begingroup$
Sort of. All this theorem initially gives you is a small interval (-h, h) for which the solution is unique. You can then extend this by repeated application of the theorem to all of R.
$endgroup$
– Klint Qinami
Jan 16 at 2:55
add a comment |
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$begingroup$
For any autonomous first-order ODE $x' = f(x)$ the constant functions $x(t) = x_0$ where $f(x_0) = 0$, i. e. the zeros of $f$, are solutions. These cannot be obtained by any value of $C$ in the general solution. So indeed with just the general solution $x(t) = (C-kt)^{-1}$ you do not have all the solutions of the ODE.
$endgroup$
– Christoph
Jan 16 at 3:16