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How to find all solutions of the ODE $x'=3x^frac23, x(0)=0$
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Problem: Find all the solutions of the IVP $$x'=3x^frac23, x(0)=0$$ for $tgeq 0$.
Here $3x^frac23$ is not $C^1$, so the existence and uniqueness theorem does not apply here.
My guess the solutions is $$x=left{beginmatrix
0 & textif 0leq t< t_0 \
(t-t_0)^3 & text if tgeq t_0
endmatrixright.$$
$t_0inmathbbR^+$ or $t_0rightarrow+infty$.
But my professor told me that there are a lot more!
My main question is: How can I find all the solutions, and then prove that they are all the solutions, rigorously?
real-analysis differential-equations analysis
asked Jul 25 at 16:43
Eric Yewen Sun
427216
 |Â
show 9 more comments
up vote
2
down vote
favorite
Problem: Find all the solutions of the IVP $$x'=3x^frac23, x(0)=0$$ for $tgeq 0$.
Here $3x^frac23$ is not $C^1$, so the existence and uniqueness theorem does not apply here.
My guess the solutions is $$x=left{beginmatrix
0 & textif 0leq t< t_0 \
(t-t_0)^3 & text if tgeq t_0
endmatrixright.$$
$t_0inmathbbR^+$ or $t_0rightarrow+infty$.
But my professor told me that there are a lot more!
My main question is: How can I find all the solutions, and then prove that they are all the solutions, rigorously?
real-analysis differential-equations analysis
asked Jul 25 at 16:43
Eric Yewen Sun
427216
@mfl I know. I want all the solutions. When $t_0=+infty$, $x=0$ is the solution in my expression.
– Eric Yewen Sun
Jul 25 at 16:49
2
@EricSun $+inftynotinBbb R^+$
– Holo
Jul 25 at 16:49
@Holo In some general sense, and it is not the point. Trivial solution does no help to my question.
– Eric Yewen Sun
Jul 25 at 16:50
@Vasya Yes it can be, and my expression exactly means that, but my question is: are these all the solutions and how to prove it?
– Eric Yewen Sun
Jul 25 at 16:56
1
Wit these conventions about two-thirds powers, a proof that there is no other solution is not that difficult, actually... First show that $x(t)geqslant0$ for every $tgeqslant0$ and every solution. Next, assume that $x(t_1)=x_1$ with $x_1>0$ for some $t_1>0$, and apply Cauchy-Lipschitz to deduce the value of $x(t)$ for every $t$ in some interval $[t_2,+infty)$ depending on $(t_1,x_1)$, with $x(t_2)=0$. Finally show that $x(t)=0$ for every $t$ in $[0,t_2]$. End of the proof.
– Did
Jul 25 at 19:08
 |Â
show 9 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem: Find all the solutions of the IVP $$x'=3x^frac23, x(0)=0$$ for $tgeq 0$.
Here $3x^frac23$ is not $C^1$, so the existence and uniqueness theorem does not apply here.
My guess the solutions is $$x=left{beginmatrix
0 & textif 0leq t< t_0 \
(t-t_0)^3 & text if tgeq t_0
endmatrixright.$$
$t_0inmathbbR^+$ or $t_0rightarrow+infty$.
But my professor told me that there are a lot more!
My main question is: How can I find all the solutions, and then prove that they are all the solutions, rigorously?
real-analysis differential-equations analysis
asked Jul 25 at 16:43
Eric Yewen Sun
427216
Problem: Find all the solutions of the IVP $$x'=3x^frac23, x(0)=0$$ for $tgeq 0$.
Here $3x^frac23$ is not $C^1$, so the existence and uniqueness theorem does not apply here.
My guess the solutions is $$x=left{beginmatrix
0 & textif 0leq t< t_0 \
(t-t_0)^3 & text if tgeq t_0
endmatrixright.$$
$t_0inmathbbR^+$ or $t_0rightarrow+infty$.
But my professor told me that there are a lot more!
My main question is: How can I find all the solutions, and then prove that they are all the solutions, rigorously?
real-analysis differential-equations analysis
asked Jul 25 at 16:43
Eric Yewen Sun
427216
edited Jul 25 at 17:02
asked Jul 25 at 16:43
Eric Yewen Sun
427216
asked Jul 25 at 16:43
Eric Yewen Sun
427216
asked Jul 25 at 16:43
Eric Yewen Sun
427216
427216
@mfl I know. I want all the solutions. When $t_0=+infty$, $x=0$ is the solution in my expression.
– Eric Yewen Sun
Jul 25 at 16:49
2
@EricSun $+inftynotinBbb R^+$
– Holo
Jul 25 at 16:49
@Holo In some general sense, and it is not the point. Trivial solution does no help to my question.
– Eric Yewen Sun
Jul 25 at 16:50
@Vasya Yes it can be, and my expression exactly means that, but my question is: are these all the solutions and how to prove it?
– Eric Yewen Sun
Jul 25 at 16:56
1
Wit these conventions about two-thirds powers, a proof that there is no other solution is not that difficult, actually... First show that $x(t)geqslant0$ for every $tgeqslant0$ and every solution. Next, assume that $x(t_1)=x_1$ with $x_1>0$ for some $t_1>0$, and apply Cauchy-Lipschitz to deduce the value of $x(t)$ for every $t$ in some interval $[t_2,+infty)$ depending on $(t_1,x_1)$, with $x(t_2)=0$. Finally show that $x(t)=0$ for every $t$ in $[0,t_2]$. End of the proof.
– Did
Jul 25 at 19:08
 |Â
show 9 more comments
@mfl I know. I want all the solutions. When $t_0=+infty$, $x=0$ is the solution in my expression.
– Eric Yewen Sun
Jul 25 at 16:49
2
@EricSun $+inftynotinBbb R^+$
– Holo
Jul 25 at 16:49
@Holo In some general sense, and it is not the point. Trivial solution does no help to my question.
– Eric Yewen Sun
Jul 25 at 16:50
@Vasya Yes it can be, and my expression exactly means that, but my question is: are these all the solutions and how to prove it?
– Eric Yewen Sun
Jul 25 at 16:56
1
Wit these conventions about two-thirds powers, a proof that there is no other solution is not that difficult, actually... First show that $x(t)geqslant0$ for every $tgeqslant0$ and every solution. Next, assume that $x(t_1)=x_1$ with $x_1>0$ for some $t_1>0$, and apply Cauchy-Lipschitz to deduce the value of $x(t)$ for every $t$ in some interval $[t_2,+infty)$ depending on $(t_1,x_1)$, with $x(t_2)=0$. Finally show that $x(t)=0$ for every $t$ in $[0,t_2]$. End of the proof.
– Did
Jul 25 at 19:08
@mfl I know. I want all the solutions. When $t_0=+infty$, $x=0$ is the solution in my expression.
– Eric Yewen Sun
Jul 25 at 16:49
@mfl I know. I want all the solutions. When $t_0=+infty$, $x=0$ is the solution in my expression.
– Eric Yewen Sun
Jul 25 at 16:49
2
2
@EricSun $+inftynotinBbb R^+$
– Holo
Jul 25 at 16:49
@EricSun $+inftynotinBbb R^+$
– Holo
Jul 25 at 16:49
@Holo In some general sense, and it is not the point. Trivial solution does no help to my question.
– Eric Yewen Sun
Jul 25 at 16:50
@Holo In some general sense, and it is not the point. Trivial solution does no help to my question.
– Eric Yewen Sun
Jul 25 at 16:50
@Vasya Yes it can be, and my expression exactly means that, but my question is: are these all the solutions and how to prove it?
– Eric Yewen Sun
Jul 25 at 16:56
@Vasya Yes it can be, and my expression exactly means that, but my question is: are these all the solutions and how to prove it?
– Eric Yewen Sun
Jul 25 at 16:56
1
1
Wit these conventions about two-thirds powers, a proof that there is no other solution is not that difficult, actually... First show that $x(t)geqslant0$ for every $tgeqslant0$ and every solution. Next, assume that $x(t_1)=x_1$ with $x_1>0$ for some $t_1>0$, and apply Cauchy-Lipschitz to deduce the value of $x(t)$ for every $t$ in some interval $[t_2,+infty)$ depending on $(t_1,x_1)$, with $x(t_2)=0$. Finally show that $x(t)=0$ for every $t$ in $[0,t_2]$. End of the proof.
– Did
Jul 25 at 19:08
Wit these conventions about two-thirds powers, a proof that there is no other solution is not that difficult, actually... First show that $x(t)geqslant0$ for every $tgeqslant0$ and every solution. Next, assume that $x(t_1)=x_1$ with $x_1>0$ for some $t_1>0$, and apply Cauchy-Lipschitz to deduce the value of $x(t)$ for every $t$ in some interval $[t_2,+infty)$ depending on $(t_1,x_1)$, with $x(t_2)=0$. Finally show that $x(t)=0$ for every $t$ in $[0,t_2]$. End of the proof.
– Did
Jul 25 at 19:08
 |Â
show 9 more comments
1 Answer
1
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1
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First of all, we can build $x'$, so $x$ is continuous. Since $x^2/3=(x^1/3)^2$, we have $x'ge 0$, so $x$ is increasing. Then $x$ is either constant, zero, on $J:=[0,infty)$, or there exist a maximal $t_0=sup x^-1( 0 )=max x^-1( 0 )$ in $J$ so that $x(t_0)=0$. We will only consider this last case, $t_0<infty$.
Then we have $x'>0$ on $(t_0,infty)$. Now we forget about the condition in $0$ for a while, solve the given differential equation in $(t_0,infty)$, knowing $x>0$. On $(0,infty)$, the function $tto t^1/3$ is of class $C^1$, so we can build as compositum the differentiable function
$$
y = x^1/3 ,
$$
which satisfies by the chain rule
$$
y'=
(x^1/3)'
=frac 13x^-2/3cdot x'
=frac 13x^-2/3cdot 3 x^2/3=1 .
$$
From $y'=1$ on $(t_0,infty)$ we find a constant $C$ with $y(t)=t+C$ on $(t_0,infty)$. So far we know about $x$ that it is $0$ on $[0,t_0]$, it is $tto y^3(t)=(t+C)^3$, and it is continuous in $t_0$. The only matching constant is $C=-t_0$. We get the solutions and only the solutions from the OP. (It is clear that these are of class $C^1$ and satisfy the given differential equation.)
answered Jul 25 at 19:08
dan_fulea
4,1421211
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First of all, we can build $x'$, so $x$ is continuous. Since $x^2/3=(x^1/3)^2$, we have $x'ge 0$, so $x$ is increasing. Then $x$ is either constant, zero, on $J:=[0,infty)$, or there exist a maximal $t_0=sup x^-1( 0 )=max x^-1( 0 )$ in $J$ so that $x(t_0)=0$. We will only consider this last case, $t_0<infty$.
Then we have $x'>0$ on $(t_0,infty)$. Now we forget about the condition in $0$ for a while, solve the given differential equation in $(t_0,infty)$, knowing $x>0$. On $(0,infty)$, the function $tto t^1/3$ is of class $C^1$, so we can build as compositum the differentiable function
$$
y = x^1/3 ,
$$
which satisfies by the chain rule
$$
y'=
(x^1/3)'
=frac 13x^-2/3cdot x'
=frac 13x^-2/3cdot 3 x^2/3=1 .
$$
From $y'=1$ on $(t_0,infty)$ we find a constant $C$ with $y(t)=t+C$ on $(t_0,infty)$. So far we know about $x$ that it is $0$ on $[0,t_0]$, it is $tto y^3(t)=(t+C)^3$, and it is continuous in $t_0$. The only matching constant is $C=-t_0$. We get the solutions and only the solutions from the OP. (It is clear that these are of class $C^1$ and satisfy the given differential equation.)
answered Jul 25 at 19:08
dan_fulea
4,1421211
add a comment |Â
up vote
1
down vote
accepted
First of all, we can build $x'$, so $x$ is continuous. Since $x^2/3=(x^1/3)^2$, we have $x'ge 0$, so $x$ is increasing. Then $x$ is either constant, zero, on $J:=[0,infty)$, or there exist a maximal $t_0=sup x^-1( 0 )=max x^-1( 0 )$ in $J$ so that $x(t_0)=0$. We will only consider this last case, $t_0<infty$.
Then we have $x'>0$ on $(t_0,infty)$. Now we forget about the condition in $0$ for a while, solve the given differential equation in $(t_0,infty)$, knowing $x>0$. On $(0,infty)$, the function $tto t^1/3$ is of class $C^1$, so we can build as compositum the differentiable function
$$
y = x^1/3 ,
$$
which satisfies by the chain rule
$$
y'=
(x^1/3)'
=frac 13x^-2/3cdot x'
=frac 13x^-2/3cdot 3 x^2/3=1 .
$$
From $y'=1$ on $(t_0,infty)$ we find a constant $C$ with $y(t)=t+C$ on $(t_0,infty)$. So far we know about $x$ that it is $0$ on $[0,t_0]$, it is $tto y^3(t)=(t+C)^3$, and it is continuous in $t_0$. The only matching constant is $C=-t_0$. We get the solutions and only the solutions from the OP. (It is clear that these are of class $C^1$ and satisfy the given differential equation.)
answered Jul 25 at 19:08
dan_fulea
4,1421211
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First of all, we can build $x'$, so $x$ is continuous. Since $x^2/3=(x^1/3)^2$, we have $x'ge 0$, so $x$ is increasing. Then $x$ is either constant, zero, on $J:=[0,infty)$, or there exist a maximal $t_0=sup x^-1( 0 )=max x^-1( 0 )$ in $J$ so that $x(t_0)=0$. We will only consider this last case, $t_0<infty$.
Then we have $x'>0$ on $(t_0,infty)$. Now we forget about the condition in $0$ for a while, solve the given differential equation in $(t_0,infty)$, knowing $x>0$. On $(0,infty)$, the function $tto t^1/3$ is of class $C^1$, so we can build as compositum the differentiable function
$$
y = x^1/3 ,
$$
which satisfies by the chain rule
$$
y'=
(x^1/3)'
=frac 13x^-2/3cdot x'
=frac 13x^-2/3cdot 3 x^2/3=1 .
$$
From $y'=1$ on $(t_0,infty)$ we find a constant $C$ with $y(t)=t+C$ on $(t_0,infty)$. So far we know about $x$ that it is $0$ on $[0,t_0]$, it is $tto y^3(t)=(t+C)^3$, and it is continuous in $t_0$. The only matching constant is $C=-t_0$. We get the solutions and only the solutions from the OP. (It is clear that these are of class $C^1$ and satisfy the given differential equation.)
answered Jul 25 at 19:08
dan_fulea
4,1421211
First of all, we can build $x'$, so $x$ is continuous. Since $x^2/3=(x^1/3)^2$, we have $x'ge 0$, so $x$ is increasing. Then $x$ is either constant, zero, on $J:=[0,infty)$, or there exist a maximal $t_0=sup x^-1( 0 )=max x^-1( 0 )$ in $J$ so that $x(t_0)=0$. We will only consider this last case, $t_0<infty$.
Then we have $x'>0$ on $(t_0,infty)$. Now we forget about the condition in $0$ for a while, solve the given differential equation in $(t_0,infty)$, knowing $x>0$. On $(0,infty)$, the function $tto t^1/3$ is of class $C^1$, so we can build as compositum the differentiable function
$$
y = x^1/3 ,
$$
which satisfies by the chain rule
$$
y'=
(x^1/3)'
=frac 13x^-2/3cdot x'
=frac 13x^-2/3cdot 3 x^2/3=1 .
$$
From $y'=1$ on $(t_0,infty)$ we find a constant $C$ with $y(t)=t+C$ on $(t_0,infty)$. So far we know about $x$ that it is $0$ on $[0,t_0]$, it is $tto y^3(t)=(t+C)^3$, and it is continuous in $t_0$. The only matching constant is $C=-t_0$. We get the solutions and only the solutions from the OP. (It is clear that these are of class $C^1$ and satisfy the given differential equation.)
answered Jul 25 at 19:08
dan_fulea
4,1421211
answered Jul 25 at 19:08
dan_fulea
4,1421211
answered Jul 25 at 19:08
dan_fulea
4,1421211
answered Jul 25 at 19:08
dan_fulea
4,1421211
4,1421211
add a comment |Â
add a comment |Â
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@mfl I know. I want all the solutions. When $t_0=+infty$, $x=0$ is the solution in my expression.
– Eric Yewen Sun
Jul 25 at 16:49
2
@EricSun $+inftynotinBbb R^+$
– Holo
Jul 25 at 16:49
@Holo In some general sense, and it is not the point. Trivial solution does no help to my question.
– Eric Yewen Sun
Jul 25 at 16:50
@Vasya Yes it can be, and my expression exactly means that, but my question is: are these all the solutions and how to prove it?
– Eric Yewen Sun
Jul 25 at 16:56
1
Wit these conventions about two-thirds powers, a proof that there is no other solution is not that difficult, actually... First show that $x(t)geqslant0$ for every $tgeqslant0$ and every solution. Next, assume that $x(t_1)=x_1$ with $x_1>0$ for some $t_1>0$, and apply Cauchy-Lipschitz to deduce the value of $x(t)$ for every $t$ in some interval $[t_2,+infty)$ depending on $(t_1,x_1)$, with $x(t_2)=0$. Finally show that $x(t)=0$ for every $t$ in $[0,t_2]$. End of the proof.
– Did
Jul 25 at 19:08