Proof ODE do not have solution in Real numbers domain
$begingroup$
I need to prove that the following ODE do not have solution in Real numbers domain:
$$
begin{cases}
big(y'big)^2=y-lvert xrvert\[1.2ex]
y(x=0)=0
end{cases}
$$
I found the complex solution but have trouble in explaining why there is no real solution.
ordinary-differential-equations
edited Dec 20 '18 at 16:59
user10354138
7,5472925
asked Dec 20 '18 at 16:22
Amit PerelmanAmit Perelman
1
$endgroup$
add a comment |
$begingroup$
I need to prove that the following ODE do not have solution in Real numbers domain:
$$
begin{cases}
big(y'big)^2=y-lvert xrvert\[1.2ex]
y(x=0)=0
end{cases}
$$
I found the complex solution but have trouble in explaining why there is no real solution.
ordinary-differential-equations
edited Dec 20 '18 at 16:59
user10354138
7,5472925
asked Dec 20 '18 at 16:22
Amit PerelmanAmit Perelman
1
$endgroup$
add a comment |
$begingroup$
I need to prove that the following ODE do not have solution in Real numbers domain:
$$
begin{cases}
big(y'big)^2=y-lvert xrvert\[1.2ex]
y(x=0)=0
end{cases}
$$
I found the complex solution but have trouble in explaining why there is no real solution.
ordinary-differential-equations
edited Dec 20 '18 at 16:59
user10354138
7,5472925
asked Dec 20 '18 at 16:22
Amit PerelmanAmit Perelman
1
$endgroup$
I need to prove that the following ODE do not have solution in Real numbers domain:
$$
begin{cases}
big(y'big)^2=y-lvert xrvert\[1.2ex]
y(x=0)=0
end{cases}
$$
I found the complex solution but have trouble in explaining why there is no real solution.
ordinary-differential-equations
ordinary-differential-equations
edited Dec 20 '18 at 16:59
user10354138
7,5472925
asked Dec 20 '18 at 16:22
Amit PerelmanAmit Perelman
1
edited Dec 20 '18 at 16:59
user10354138
7,5472925
asked Dec 20 '18 at 16:22
Amit PerelmanAmit Perelman
1
edited Dec 20 '18 at 16:59
user10354138
7,5472925
edited Dec 20 '18 at 16:59
user10354138
7,5472925
edited Dec 20 '18 at 16:59
user10354138
7,5472925
7,5472925
asked Dec 20 '18 at 16:22
Amit PerelmanAmit Perelman
1
asked Dec 20 '18 at 16:22
Amit PerelmanAmit Perelman
1
asked Dec 20 '18 at 16:22
Amit PerelmanAmit Perelman
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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You want to get the ODE into the form $y'(x) = f(x,y)$, then you want to analyse the existence and uniqueness theorem for a 1st Non-Linear Order ODE. (The absolute value should ring alarm bells I think).
This should help: https://faculty.math.illinois.edu/~tyson/existence.pdf
or this: https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
answered Dec 20 '18 at 16:27
Joel BiffinJoel Biffin
1018
$endgroup$
add a comment |
$begingroup$
From the differential equation, $y'(0) = 0$, while you said $y(0)=0$. From the definition of derivative,
$lim_{x to 0} y(x)/x = 0$. In particular $y(x) - |x| < 0$ for $x$ sufficiently close to, but not equal to, $0$, and that can't be $(y'(x))^2$ if $y'(x)$ is real.
answered Dec 20 '18 at 16:47
Robert IsraelRobert Israel
331k23220475
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
You want to get the ODE into the form $y'(x) = f(x,y)$, then you want to analyse the existence and uniqueness theorem for a 1st Non-Linear Order ODE. (The absolute value should ring alarm bells I think).
This should help: https://faculty.math.illinois.edu/~tyson/existence.pdf
or this: https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
answered Dec 20 '18 at 16:27
Joel BiffinJoel Biffin
1018
$endgroup$
add a comment |
$begingroup$
You want to get the ODE into the form $y'(x) = f(x,y)$, then you want to analyse the existence and uniqueness theorem for a 1st Non-Linear Order ODE. (The absolute value should ring alarm bells I think).
This should help: https://faculty.math.illinois.edu/~tyson/existence.pdf
or this: https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
answered Dec 20 '18 at 16:27
Joel BiffinJoel Biffin
1018
$endgroup$
add a comment |
$begingroup$
You want to get the ODE into the form $y'(x) = f(x,y)$, then you want to analyse the existence and uniqueness theorem for a 1st Non-Linear Order ODE. (The absolute value should ring alarm bells I think).
This should help: https://faculty.math.illinois.edu/~tyson/existence.pdf
or this: https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
answered Dec 20 '18 at 16:27
Joel BiffinJoel Biffin
1018
$endgroup$
You want to get the ODE into the form $y'(x) = f(x,y)$, then you want to analyse the existence and uniqueness theorem for a 1st Non-Linear Order ODE. (The absolute value should ring alarm bells I think).
This should help: https://faculty.math.illinois.edu/~tyson/existence.pdf
or this: https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem
answered Dec 20 '18 at 16:27
Joel BiffinJoel Biffin
1018
answered Dec 20 '18 at 16:27
Joel BiffinJoel Biffin
1018
answered Dec 20 '18 at 16:27
Joel BiffinJoel Biffin
1018
answered Dec 20 '18 at 16:27
Joel BiffinJoel Biffin
1018
1018
add a comment |
add a comment |
$begingroup$
From the differential equation, $y'(0) = 0$, while you said $y(0)=0$. From the definition of derivative,
$lim_{x to 0} y(x)/x = 0$. In particular $y(x) - |x| < 0$ for $x$ sufficiently close to, but not equal to, $0$, and that can't be $(y'(x))^2$ if $y'(x)$ is real.
answered Dec 20 '18 at 16:47
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
From the differential equation, $y'(0) = 0$, while you said $y(0)=0$. From the definition of derivative,
$lim_{x to 0} y(x)/x = 0$. In particular $y(x) - |x| < 0$ for $x$ sufficiently close to, but not equal to, $0$, and that can't be $(y'(x))^2$ if $y'(x)$ is real.
answered Dec 20 '18 at 16:47
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
From the differential equation, $y'(0) = 0$, while you said $y(0)=0$. From the definition of derivative,
$lim_{x to 0} y(x)/x = 0$. In particular $y(x) - |x| < 0$ for $x$ sufficiently close to, but not equal to, $0$, and that can't be $(y'(x))^2$ if $y'(x)$ is real.
answered Dec 20 '18 at 16:47
Robert IsraelRobert Israel
331k23220475
$endgroup$
From the differential equation, $y'(0) = 0$, while you said $y(0)=0$. From the definition of derivative,
$lim_{x to 0} y(x)/x = 0$. In particular $y(x) - |x| < 0$ for $x$ sufficiently close to, but not equal to, $0$, and that can't be $(y'(x))^2$ if $y'(x)$ is real.
answered Dec 20 '18 at 16:47
Robert IsraelRobert Israel
331k23220475
answered Dec 20 '18 at 16:47
Robert IsraelRobert Israel
331k23220475
answered Dec 20 '18 at 16:47
Robert IsraelRobert Israel
331k23220475
answered Dec 20 '18 at 16:47
Robert IsraelRobert Israel
331k23220475
331k23220475
add a comment |
add a comment |
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