Mixing
Does $y'=|y|^a$ have any global solutions?
up vote
2
down vote
favorite
Assume the differential equation
$$
y'=|y|^a
$$
My intuition tells me that since it involves an absolute value, there might not be any solutions defined everywhere, except for the case $a=0$, where $y(x)=x+c$.
To show this, let
$$
f(y)=|y|^a$$
$bullet,$For $a<0$:
$f$ is not defined for $y=0$ plus it's not bounded
$bullet,$ For $a=0$:
$$y'=1 iff y(x)=x+c, quad x in mathbb{R}$$
$bullet,$ For $a>0$:
$f$ is defined $forall y in mathbb{R}$, but it's not bounded
Can we thus conclude that the only global solution of $y'=|y|^a$ is $y(x)=x+c$?
differential-equations dynamical-systems stability-in-odes initial-value-problems
asked yesterday
Jevaut
5199
|
show 2 more comments
up vote
2
down vote
favorite
Assume the differential equation
$$
y'=|y|^a
$$
My intuition tells me that since it involves an absolute value, there might not be any solutions defined everywhere, except for the case $a=0$, where $y(x)=x+c$.
To show this, let
$$
f(y)=|y|^a$$
$bullet,$For $a<0$:
$f$ is not defined for $y=0$ plus it's not bounded
$bullet,$ For $a=0$:
$$y'=1 iff y(x)=x+c, quad x in mathbb{R}$$
$bullet,$ For $a>0$:
$f$ is defined $forall y in mathbb{R}$, but it's not bounded
Can we thus conclude that the only global solution of $y'=|y|^a$ is $y(x)=x+c$?
differential-equations dynamical-systems stability-in-odes initial-value-problems
asked yesterday
Jevaut
5199
What do you mean by a global solution? Should $y(x)$ be defined on $mathbb{R}$?
– Paul Frost
yesterday
@PaulFrost Exactly
– Jevaut
yesterday
For $a=1$ you can take $y = e^{x+c}$.
– Paul Frost
yesterday
You consider $y' = f(y)$ with $f(y) = lvert y rvert^a$. The problem with $f$ is that $f(0)$ is undefined for $a le 0$ (recall that not only $1/0$, but also $0^0$ is undefined). For $a=0$ you may of course interpret your definition as $f(y) equiv 1$ (in which case you get your solution $y(x) = x + c$), but note $f(y) equiv 1$ does not really agree with the original definition. So for $a le 0$ solutions $y(x)$ must satisfy $y(x) ne 0$ for all $x$.
– Paul Frost
yesterday
So, technically there are no global solutions for $aleq 0$?
– Jevaut
yesterday
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Assume the differential equation
$$
y'=|y|^a
$$
My intuition tells me that since it involves an absolute value, there might not be any solutions defined everywhere, except for the case $a=0$, where $y(x)=x+c$.
To show this, let
$$
f(y)=|y|^a$$
$bullet,$For $a<0$:
$f$ is not defined for $y=0$ plus it's not bounded
$bullet,$ For $a=0$:
$$y'=1 iff y(x)=x+c, quad x in mathbb{R}$$
$bullet,$ For $a>0$:
$f$ is defined $forall y in mathbb{R}$, but it's not bounded
Can we thus conclude that the only global solution of $y'=|y|^a$ is $y(x)=x+c$?
differential-equations dynamical-systems stability-in-odes initial-value-problems
asked yesterday
Jevaut
5199
Assume the differential equation
$$
y'=|y|^a
$$
My intuition tells me that since it involves an absolute value, there might not be any solutions defined everywhere, except for the case $a=0$, where $y(x)=x+c$.
To show this, let
$$
f(y)=|y|^a$$
$bullet,$For $a<0$:
$f$ is not defined for $y=0$ plus it's not bounded
$bullet,$ For $a=0$:
$$y'=1 iff y(x)=x+c, quad x in mathbb{R}$$
$bullet,$ For $a>0$:
$f$ is defined $forall y in mathbb{R}$, but it's not bounded
Can we thus conclude that the only global solution of $y'=|y|^a$ is $y(x)=x+c$?
differential-equations dynamical-systems stability-in-odes initial-value-problems
differential-equations dynamical-systems stability-in-odes initial-value-problems
asked yesterday
Jevaut
5199
asked yesterday
Jevaut
5199
asked yesterday
Jevaut
5199
asked yesterday
Jevaut
5199
asked yesterday
Jevaut
5199
5199
What do you mean by a global solution? Should $y(x)$ be defined on $mathbb{R}$?
– Paul Frost
yesterday
@PaulFrost Exactly
– Jevaut
yesterday
For $a=1$ you can take $y = e^{x+c}$.
– Paul Frost
yesterday
You consider $y' = f(y)$ with $f(y) = lvert y rvert^a$. The problem with $f$ is that $f(0)$ is undefined for $a le 0$ (recall that not only $1/0$, but also $0^0$ is undefined). For $a=0$ you may of course interpret your definition as $f(y) equiv 1$ (in which case you get your solution $y(x) = x + c$), but note $f(y) equiv 1$ does not really agree with the original definition. So for $a le 0$ solutions $y(x)$ must satisfy $y(x) ne 0$ for all $x$.
– Paul Frost
yesterday
So, technically there are no global solutions for $aleq 0$?
– Jevaut
yesterday
|
show 2 more comments
What do you mean by a global solution? Should $y(x)$ be defined on $mathbb{R}$?
– Paul Frost
yesterday
@PaulFrost Exactly
– Jevaut
yesterday
For $a=1$ you can take $y = e^{x+c}$.
– Paul Frost
yesterday
You consider $y' = f(y)$ with $f(y) = lvert y rvert^a$. The problem with $f$ is that $f(0)$ is undefined for $a le 0$ (recall that not only $1/0$, but also $0^0$ is undefined). For $a=0$ you may of course interpret your definition as $f(y) equiv 1$ (in which case you get your solution $y(x) = x + c$), but note $f(y) equiv 1$ does not really agree with the original definition. So for $a le 0$ solutions $y(x)$ must satisfy $y(x) ne 0$ for all $x$.
– Paul Frost
yesterday
So, technically there are no global solutions for $aleq 0$?
– Jevaut
yesterday
What do you mean by a global solution? Should $y(x)$ be defined on $mathbb{R}$?
– Paul Frost
yesterday
What do you mean by a global solution? Should $y(x)$ be defined on $mathbb{R}$?
– Paul Frost
yesterday
@PaulFrost Exactly
– Jevaut
yesterday
@PaulFrost Exactly
– Jevaut
yesterday
For $a=1$ you can take $y = e^{x+c}$.
– Paul Frost
yesterday
For $a=1$ you can take $y = e^{x+c}$.
– Paul Frost
yesterday
You consider $y' = f(y)$ with $f(y) = lvert y rvert^a$. The problem with $f$ is that $f(0)$ is undefined for $a le 0$ (recall that not only $1/0$, but also $0^0$ is undefined). For $a=0$ you may of course interpret your definition as $f(y) equiv 1$ (in which case you get your solution $y(x) = x + c$), but note $f(y) equiv 1$ does not really agree with the original definition. So for $a le 0$ solutions $y(x)$ must satisfy $y(x) ne 0$ for all $x$.
– Paul Frost
yesterday
You consider $y' = f(y)$ with $f(y) = lvert y rvert^a$. The problem with $f$ is that $f(0)$ is undefined for $a le 0$ (recall that not only $1/0$, but also $0^0$ is undefined). For $a=0$ you may of course interpret your definition as $f(y) equiv 1$ (in which case you get your solution $y(x) = x + c$), but note $f(y) equiv 1$ does not really agree with the original definition. So for $a le 0$ solutions $y(x)$ must satisfy $y(x) ne 0$ for all $x$.
– Paul Frost
yesterday
So, technically there are no global solutions for $aleq 0$?
– Jevaut
yesterday
So, technically there are no global solutions for $aleq 0$?
– Jevaut
yesterday
|
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
I have summarised some of the information that has already been stated and made an attempt to complete the answer.
I believe that your question may be rephrased as follows: given $a in mathbb{R}$ does there exist $y:mathbb{R}longrightarrowmathbb{R}$ such that $(Dy)(t) = |y(t)|^a$ (1) for all $t in mathbb{R}$?
Remark: it is assumed that $0^0$ is not defined.
Consider four cases (the first two cases are not necessary to answer your question, but it is always useful to exhibit closed-form analytic solutions when they exist):
Case I: $a = 1$. If $c in mathbb{R}$, then $y(t)=ce^{mathsf{sgn}(c) t}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case II: $0 < a < 1$. Then $y(t)=sgn(t)(1-a)^{frac{1}{1-a}}(sgn(t)t)^{frac{1}{1-a}}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case III: $a > 1$. Then $y(t) = 0$ is a solution of (1) on $mathbb{R}$ (of course, $y(t) = 0$ is also a solution for $0 < a leq 1$).
Case IV: $a leq 0$. Suppose that $y(t)$ is a solution of (1) on $mathbb{R}$. Then, $y(t)neq 0$ for all $t in mathbb{R}$. Given that $(Dy)(t)=|y(t)|^a$, $(Dy)(t) > 0$ for all $t in mathbb{R}$. However, then $y$ must be an increasing function. Therefore, by differentiability of $y$, either $y(t) < 0$ for all $t in mathbb{R}$ or $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) < 0$ for all $t in mathbb{R}$. Then $(Dy)(t)=(-y(t))^a$ for all $t in mathbb{R}$. Thus, the second derivative exists and is given by $(D^2 y)(t)=-a(-y(t))^{2a-1}$ for all $t in mathbb{R}$. Under assumptions $a leq 0$ and $y(t) < 0$, $ 0 leq (D^2 y)(t)$ for all $t in mathbb{R}$. Therefore, $y(t)$ is convex on $mathbb{R}$. Thus, $y(t)$ is constant (link). However, $y(t)$ is also an increasing function. Thus, a contradiction is reached and, therefore, $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) > 0$ for all $t in mathbb{R}$. Then, consider a function $q(t)$ such that $q(t)=y(-t)$ for all $t in mathbb{R}$. Thus, $q(t)>0$, Also, $(Dq)(t)=-q(t)^a=-(y(-t))^a$ and $(D^2q)(t)=aq(t)^{2a-1}$ for all $t in mathbb{R}$. Thus, $(Dq)(t) < 0$ and $(D^2q)(t) leq 0$ for all $t in mathbb{R}$. Hence, $q$ is decreasing and concave on $mathbb{R}$. However, in this case, it cannot be bounded below (link). Thus, by contradiction, if $a leq 0$, then (1) does not have any global solutions on $mathbb{R}$.
In conclusion, if $a > 0$, then there always exists at least one global solution of (1). Otherwise, (1) does not have any global solutions.
answered yesterday
xanonec
1586
@Jevaut I provided an update for Case IV. Hopefully, now, the solution is more transparent than it was before.
– xanonec
14 hours ago
add a comment |
up vote
0
down vote
There is no requirement that $f$ be bounded. When $a=1/2$ you can use $y(x)=(x^2/4) mathrm{sgn,} x$, for example and you can also translate $xmapsto x+c$ to get other solutions. Also don't forget Cauchy-Lipschitz fails, so $yequiv0$ is a solution too.
answered yesterday
Richard Martin
1,3438
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I have summarised some of the information that has already been stated and made an attempt to complete the answer.
I believe that your question may be rephrased as follows: given $a in mathbb{R}$ does there exist $y:mathbb{R}longrightarrowmathbb{R}$ such that $(Dy)(t) = |y(t)|^a$ (1) for all $t in mathbb{R}$?
Remark: it is assumed that $0^0$ is not defined.
Consider four cases (the first two cases are not necessary to answer your question, but it is always useful to exhibit closed-form analytic solutions when they exist):
Case I: $a = 1$. If $c in mathbb{R}$, then $y(t)=ce^{mathsf{sgn}(c) t}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case II: $0 < a < 1$. Then $y(t)=sgn(t)(1-a)^{frac{1}{1-a}}(sgn(t)t)^{frac{1}{1-a}}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case III: $a > 1$. Then $y(t) = 0$ is a solution of (1) on $mathbb{R}$ (of course, $y(t) = 0$ is also a solution for $0 < a leq 1$).
Case IV: $a leq 0$. Suppose that $y(t)$ is a solution of (1) on $mathbb{R}$. Then, $y(t)neq 0$ for all $t in mathbb{R}$. Given that $(Dy)(t)=|y(t)|^a$, $(Dy)(t) > 0$ for all $t in mathbb{R}$. However, then $y$ must be an increasing function. Therefore, by differentiability of $y$, either $y(t) < 0$ for all $t in mathbb{R}$ or $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) < 0$ for all $t in mathbb{R}$. Then $(Dy)(t)=(-y(t))^a$ for all $t in mathbb{R}$. Thus, the second derivative exists and is given by $(D^2 y)(t)=-a(-y(t))^{2a-1}$ for all $t in mathbb{R}$. Under assumptions $a leq 0$ and $y(t) < 0$, $ 0 leq (D^2 y)(t)$ for all $t in mathbb{R}$. Therefore, $y(t)$ is convex on $mathbb{R}$. Thus, $y(t)$ is constant (link). However, $y(t)$ is also an increasing function. Thus, a contradiction is reached and, therefore, $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) > 0$ for all $t in mathbb{R}$. Then, consider a function $q(t)$ such that $q(t)=y(-t)$ for all $t in mathbb{R}$. Thus, $q(t)>0$, Also, $(Dq)(t)=-q(t)^a=-(y(-t))^a$ and $(D^2q)(t)=aq(t)^{2a-1}$ for all $t in mathbb{R}$. Thus, $(Dq)(t) < 0$ and $(D^2q)(t) leq 0$ for all $t in mathbb{R}$. Hence, $q$ is decreasing and concave on $mathbb{R}$. However, in this case, it cannot be bounded below (link). Thus, by contradiction, if $a leq 0$, then (1) does not have any global solutions on $mathbb{R}$.
In conclusion, if $a > 0$, then there always exists at least one global solution of (1). Otherwise, (1) does not have any global solutions.
answered yesterday
xanonec
1586
@Jevaut I provided an update for Case IV. Hopefully, now, the solution is more transparent than it was before.
– xanonec
14 hours ago
add a comment |
up vote
2
down vote
accepted
I have summarised some of the information that has already been stated and made an attempt to complete the answer.
I believe that your question may be rephrased as follows: given $a in mathbb{R}$ does there exist $y:mathbb{R}longrightarrowmathbb{R}$ such that $(Dy)(t) = |y(t)|^a$ (1) for all $t in mathbb{R}$?
Remark: it is assumed that $0^0$ is not defined.
Consider four cases (the first two cases are not necessary to answer your question, but it is always useful to exhibit closed-form analytic solutions when they exist):
Case I: $a = 1$. If $c in mathbb{R}$, then $y(t)=ce^{mathsf{sgn}(c) t}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case II: $0 < a < 1$. Then $y(t)=sgn(t)(1-a)^{frac{1}{1-a}}(sgn(t)t)^{frac{1}{1-a}}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case III: $a > 1$. Then $y(t) = 0$ is a solution of (1) on $mathbb{R}$ (of course, $y(t) = 0$ is also a solution for $0 < a leq 1$).
Case IV: $a leq 0$. Suppose that $y(t)$ is a solution of (1) on $mathbb{R}$. Then, $y(t)neq 0$ for all $t in mathbb{R}$. Given that $(Dy)(t)=|y(t)|^a$, $(Dy)(t) > 0$ for all $t in mathbb{R}$. However, then $y$ must be an increasing function. Therefore, by differentiability of $y$, either $y(t) < 0$ for all $t in mathbb{R}$ or $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) < 0$ for all $t in mathbb{R}$. Then $(Dy)(t)=(-y(t))^a$ for all $t in mathbb{R}$. Thus, the second derivative exists and is given by $(D^2 y)(t)=-a(-y(t))^{2a-1}$ for all $t in mathbb{R}$. Under assumptions $a leq 0$ and $y(t) < 0$, $ 0 leq (D^2 y)(t)$ for all $t in mathbb{R}$. Therefore, $y(t)$ is convex on $mathbb{R}$. Thus, $y(t)$ is constant (link). However, $y(t)$ is also an increasing function. Thus, a contradiction is reached and, therefore, $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) > 0$ for all $t in mathbb{R}$. Then, consider a function $q(t)$ such that $q(t)=y(-t)$ for all $t in mathbb{R}$. Thus, $q(t)>0$, Also, $(Dq)(t)=-q(t)^a=-(y(-t))^a$ and $(D^2q)(t)=aq(t)^{2a-1}$ for all $t in mathbb{R}$. Thus, $(Dq)(t) < 0$ and $(D^2q)(t) leq 0$ for all $t in mathbb{R}$. Hence, $q$ is decreasing and concave on $mathbb{R}$. However, in this case, it cannot be bounded below (link). Thus, by contradiction, if $a leq 0$, then (1) does not have any global solutions on $mathbb{R}$.
In conclusion, if $a > 0$, then there always exists at least one global solution of (1). Otherwise, (1) does not have any global solutions.
answered yesterday
xanonec
1586
@Jevaut I provided an update for Case IV. Hopefully, now, the solution is more transparent than it was before.
– xanonec
14 hours ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I have summarised some of the information that has already been stated and made an attempt to complete the answer.
I believe that your question may be rephrased as follows: given $a in mathbb{R}$ does there exist $y:mathbb{R}longrightarrowmathbb{R}$ such that $(Dy)(t) = |y(t)|^a$ (1) for all $t in mathbb{R}$?
Remark: it is assumed that $0^0$ is not defined.
Consider four cases (the first two cases are not necessary to answer your question, but it is always useful to exhibit closed-form analytic solutions when they exist):
Case I: $a = 1$. If $c in mathbb{R}$, then $y(t)=ce^{mathsf{sgn}(c) t}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case II: $0 < a < 1$. Then $y(t)=sgn(t)(1-a)^{frac{1}{1-a}}(sgn(t)t)^{frac{1}{1-a}}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case III: $a > 1$. Then $y(t) = 0$ is a solution of (1) on $mathbb{R}$ (of course, $y(t) = 0$ is also a solution for $0 < a leq 1$).
Case IV: $a leq 0$. Suppose that $y(t)$ is a solution of (1) on $mathbb{R}$. Then, $y(t)neq 0$ for all $t in mathbb{R}$. Given that $(Dy)(t)=|y(t)|^a$, $(Dy)(t) > 0$ for all $t in mathbb{R}$. However, then $y$ must be an increasing function. Therefore, by differentiability of $y$, either $y(t) < 0$ for all $t in mathbb{R}$ or $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) < 0$ for all $t in mathbb{R}$. Then $(Dy)(t)=(-y(t))^a$ for all $t in mathbb{R}$. Thus, the second derivative exists and is given by $(D^2 y)(t)=-a(-y(t))^{2a-1}$ for all $t in mathbb{R}$. Under assumptions $a leq 0$ and $y(t) < 0$, $ 0 leq (D^2 y)(t)$ for all $t in mathbb{R}$. Therefore, $y(t)$ is convex on $mathbb{R}$. Thus, $y(t)$ is constant (link). However, $y(t)$ is also an increasing function. Thus, a contradiction is reached and, therefore, $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) > 0$ for all $t in mathbb{R}$. Then, consider a function $q(t)$ such that $q(t)=y(-t)$ for all $t in mathbb{R}$. Thus, $q(t)>0$, Also, $(Dq)(t)=-q(t)^a=-(y(-t))^a$ and $(D^2q)(t)=aq(t)^{2a-1}$ for all $t in mathbb{R}$. Thus, $(Dq)(t) < 0$ and $(D^2q)(t) leq 0$ for all $t in mathbb{R}$. Hence, $q$ is decreasing and concave on $mathbb{R}$. However, in this case, it cannot be bounded below (link). Thus, by contradiction, if $a leq 0$, then (1) does not have any global solutions on $mathbb{R}$.
In conclusion, if $a > 0$, then there always exists at least one global solution of (1). Otherwise, (1) does not have any global solutions.
answered yesterday
xanonec
1586
I have summarised some of the information that has already been stated and made an attempt to complete the answer.
I believe that your question may be rephrased as follows: given $a in mathbb{R}$ does there exist $y:mathbb{R}longrightarrowmathbb{R}$ such that $(Dy)(t) = |y(t)|^a$ (1) for all $t in mathbb{R}$?
Remark: it is assumed that $0^0$ is not defined.
Consider four cases (the first two cases are not necessary to answer your question, but it is always useful to exhibit closed-form analytic solutions when they exist):
Case I: $a = 1$. If $c in mathbb{R}$, then $y(t)=ce^{mathsf{sgn}(c) t}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case II: $0 < a < 1$. Then $y(t)=sgn(t)(1-a)^{frac{1}{1-a}}(sgn(t)t)^{frac{1}{1-a}}$ is a solution of (1) on $mathbb{R}$ (this case is not necessary: see Case III).
Case III: $a > 1$. Then $y(t) = 0$ is a solution of (1) on $mathbb{R}$ (of course, $y(t) = 0$ is also a solution for $0 < a leq 1$).
Case IV: $a leq 0$. Suppose that $y(t)$ is a solution of (1) on $mathbb{R}$. Then, $y(t)neq 0$ for all $t in mathbb{R}$. Given that $(Dy)(t)=|y(t)|^a$, $(Dy)(t) > 0$ for all $t in mathbb{R}$. However, then $y$ must be an increasing function. Therefore, by differentiability of $y$, either $y(t) < 0$ for all $t in mathbb{R}$ or $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) < 0$ for all $t in mathbb{R}$. Then $(Dy)(t)=(-y(t))^a$ for all $t in mathbb{R}$. Thus, the second derivative exists and is given by $(D^2 y)(t)=-a(-y(t))^{2a-1}$ for all $t in mathbb{R}$. Under assumptions $a leq 0$ and $y(t) < 0$, $ 0 leq (D^2 y)(t)$ for all $t in mathbb{R}$. Therefore, $y(t)$ is convex on $mathbb{R}$. Thus, $y(t)$ is constant (link). However, $y(t)$ is also an increasing function. Thus, a contradiction is reached and, therefore, $y(t) > 0$ for all $t in mathbb{R}$.
Suppose $y(t) > 0$ for all $t in mathbb{R}$. Then, consider a function $q(t)$ such that $q(t)=y(-t)$ for all $t in mathbb{R}$. Thus, $q(t)>0$, Also, $(Dq)(t)=-q(t)^a=-(y(-t))^a$ and $(D^2q)(t)=aq(t)^{2a-1}$ for all $t in mathbb{R}$. Thus, $(Dq)(t) < 0$ and $(D^2q)(t) leq 0$ for all $t in mathbb{R}$. Hence, $q$ is decreasing and concave on $mathbb{R}$. However, in this case, it cannot be bounded below (link). Thus, by contradiction, if $a leq 0$, then (1) does not have any global solutions on $mathbb{R}$.
In conclusion, if $a > 0$, then there always exists at least one global solution of (1). Otherwise, (1) does not have any global solutions.
answered yesterday
xanonec
1586
edited 17 hours ago
answered yesterday
xanonec
1586
answered yesterday
xanonec
1586
answered yesterday
xanonec
1586
1586
@Jevaut I provided an update for Case IV. Hopefully, now, the solution is more transparent than it was before.
– xanonec
14 hours ago
add a comment |
@Jevaut I provided an update for Case IV. Hopefully, now, the solution is more transparent than it was before.
– xanonec
14 hours ago
@Jevaut I provided an update for Case IV. Hopefully, now, the solution is more transparent than it was before.
– xanonec
14 hours ago
@Jevaut I provided an update for Case IV. Hopefully, now, the solution is more transparent than it was before.
– xanonec
14 hours ago
add a comment |
up vote
0
down vote
There is no requirement that $f$ be bounded. When $a=1/2$ you can use $y(x)=(x^2/4) mathrm{sgn,} x$, for example and you can also translate $xmapsto x+c$ to get other solutions. Also don't forget Cauchy-Lipschitz fails, so $yequiv0$ is a solution too.
answered yesterday
Richard Martin
1,3438
add a comment |
up vote
0
down vote
There is no requirement that $f$ be bounded. When $a=1/2$ you can use $y(x)=(x^2/4) mathrm{sgn,} x$, for example and you can also translate $xmapsto x+c$ to get other solutions. Also don't forget Cauchy-Lipschitz fails, so $yequiv0$ is a solution too.
answered yesterday
Richard Martin
1,3438
add a comment |
up vote
0
down vote
up vote
0
down vote
There is no requirement that $f$ be bounded. When $a=1/2$ you can use $y(x)=(x^2/4) mathrm{sgn,} x$, for example and you can also translate $xmapsto x+c$ to get other solutions. Also don't forget Cauchy-Lipschitz fails, so $yequiv0$ is a solution too.
answered yesterday
Richard Martin
1,3438
There is no requirement that $f$ be bounded. When $a=1/2$ you can use $y(x)=(x^2/4) mathrm{sgn,} x$, for example and you can also translate $xmapsto x+c$ to get other solutions. Also don't forget Cauchy-Lipschitz fails, so $yequiv0$ is a solution too.
answered yesterday
Richard Martin
1,3438
edited yesterday
answered yesterday
Richard Martin
1,3438
answered yesterday
Richard Martin
1,3438
answered yesterday
Richard Martin
1,3438
1,3438
add a comment |
add a comment |
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%2fmath.stackexchange.com%2fquestions%2f3005128%2fdoes-y-ya-have-any-global-solutions%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
What do you mean by a global solution? Should $y(x)$ be defined on $mathbb{R}$?
– Paul Frost
yesterday
@PaulFrost Exactly
– Jevaut
yesterday
For $a=1$ you can take $y = e^{x+c}$.
– Paul Frost
yesterday
You consider $y' = f(y)$ with $f(y) = lvert y rvert^a$. The problem with $f$ is that $f(0)$ is undefined for $a le 0$ (recall that not only $1/0$, but also $0^0$ is undefined). For $a=0$ you may of course interpret your definition as $f(y) equiv 1$ (in which case you get your solution $y(x) = x + c$), but note $f(y) equiv 1$ does not really agree with the original definition. So for $a le 0$ solutions $y(x)$ must satisfy $y(x) ne 0$ for all $x$.
– Paul Frost
yesterday
So, technically there are no global solutions for $aleq 0$?
– Jevaut
yesterday