Mixing
Existence and uniqueness of solution of $y'=(x-y)^{2/3}$ such that $y(5)=5$
$begingroup$
Assume the Initial Value Problem:
$$
y'(x)=[x-y(x)]^{frac23}equiv f(x,y(x)), quad y(5)=5
$$
Existence:
Since $f: mathbb{R^2} longrightarrow mathbb{R}$ and $(x_0,y_0)=(5,5) in mathbb{R^2}$, Peano's Theorem ensures that there exists a local solution $z: I longrightarrow mathbb{R}$, where $I$ is a neighborhood of $x_0=5$.
Uniqueness:
I'm not sure whether we can deduce anything about uniqueness. We could take:
$$
f_y=frac{partial}{partial y} f(x,y)=-frac{2}{3(x-y)^{frac13}}, quad forall (x,y):xneq y
$$
Here's the part which confuses me. Does the fact that $f_y$ not bounded for any neighborhood of $x_0=5$ prevents us from proving uniqueness?
ordinary-differential-equations dynamical-systems lipschitz-functions initial-value-problems
edited Jan 26 at 16:59
Did
248k23226465
asked Jan 26 at 16:14
LoneBoneLoneBone
968
$endgroup$
add a comment |
$begingroup$
Assume the Initial Value Problem:
$$
y'(x)=[x-y(x)]^{frac23}equiv f(x,y(x)), quad y(5)=5
$$
Existence:
Since $f: mathbb{R^2} longrightarrow mathbb{R}$ and $(x_0,y_0)=(5,5) in mathbb{R^2}$, Peano's Theorem ensures that there exists a local solution $z: I longrightarrow mathbb{R}$, where $I$ is a neighborhood of $x_0=5$.
Uniqueness:
I'm not sure whether we can deduce anything about uniqueness. We could take:
$$
f_y=frac{partial}{partial y} f(x,y)=-frac{2}{3(x-y)^{frac13}}, quad forall (x,y):xneq y
$$
Here's the part which confuses me. Does the fact that $f_y$ not bounded for any neighborhood of $x_0=5$ prevents us from proving uniqueness?
ordinary-differential-equations dynamical-systems lipschitz-functions initial-value-problems
edited Jan 26 at 16:59
Did
248k23226465
asked Jan 26 at 16:14
LoneBoneLoneBone
968
$endgroup$
add a comment |
$begingroup$
Assume the Initial Value Problem:
$$
y'(x)=[x-y(x)]^{frac23}equiv f(x,y(x)), quad y(5)=5
$$
Existence:
Since $f: mathbb{R^2} longrightarrow mathbb{R}$ and $(x_0,y_0)=(5,5) in mathbb{R^2}$, Peano's Theorem ensures that there exists a local solution $z: I longrightarrow mathbb{R}$, where $I$ is a neighborhood of $x_0=5$.
Uniqueness:
I'm not sure whether we can deduce anything about uniqueness. We could take:
$$
f_y=frac{partial}{partial y} f(x,y)=-frac{2}{3(x-y)^{frac13}}, quad forall (x,y):xneq y
$$
Here's the part which confuses me. Does the fact that $f_y$ not bounded for any neighborhood of $x_0=5$ prevents us from proving uniqueness?
ordinary-differential-equations dynamical-systems lipschitz-functions initial-value-problems
edited Jan 26 at 16:59
Did
248k23226465
asked Jan 26 at 16:14
LoneBoneLoneBone
968
$endgroup$
Assume the Initial Value Problem:
$$
y'(x)=[x-y(x)]^{frac23}equiv f(x,y(x)), quad y(5)=5
$$
Existence:
Since $f: mathbb{R^2} longrightarrow mathbb{R}$ and $(x_0,y_0)=(5,5) in mathbb{R^2}$, Peano's Theorem ensures that there exists a local solution $z: I longrightarrow mathbb{R}$, where $I$ is a neighborhood of $x_0=5$.
Uniqueness:
I'm not sure whether we can deduce anything about uniqueness. We could take:
$$
f_y=frac{partial}{partial y} f(x,y)=-frac{2}{3(x-y)^{frac13}}, quad forall (x,y):xneq y
$$
Here's the part which confuses me. Does the fact that $f_y$ not bounded for any neighborhood of $x_0=5$ prevents us from proving uniqueness?
ordinary-differential-equations dynamical-systems lipschitz-functions initial-value-problems
ordinary-differential-equations dynamical-systems lipschitz-functions initial-value-problems
edited Jan 26 at 16:59
Did
248k23226465
asked Jan 26 at 16:14
LoneBoneLoneBone
968
edited Jan 26 at 16:59
Did
248k23226465
asked Jan 26 at 16:14
LoneBoneLoneBone
968
edited Jan 26 at 16:59
Did
248k23226465
edited Jan 26 at 16:59
Did
248k23226465
edited Jan 26 at 16:59
Did
248k23226465
248k23226465
asked Jan 26 at 16:14
LoneBoneLoneBone
968
asked Jan 26 at 16:14
LoneBoneLoneBone
968
asked Jan 26 at 16:14
LoneBoneLoneBone
968
968
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you have already seen, the function $f$ is not locally Lipschitz continuous around $y=x$, hence you cannot deduce uniqueness using the standard Cauchy-Lipschitz theory.
Nevertheless, the Cauchy problem admits a unique solution.
You can get an intuition of this fact observing that any solution hits the "bad" set ${(x,y): x=y}$ only for $x=5$, since $y'(5) = 0$.
For a formal proof, you can observe that $y(x)$ is a solution of your Cauchy problem if and only if the function $z(x) := x - y(x)$ is a solution to
$$
z' = 1-z^{2/3}, quad, z(5) = 0,
$$
and this Cauchy problem admits unique solution.
answered Jan 26 at 16:55
RigelRigel
11.3k11320
$endgroup$
add a comment |
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$begingroup$
As you have already seen, the function $f$ is not locally Lipschitz continuous around $y=x$, hence you cannot deduce uniqueness using the standard Cauchy-Lipschitz theory.
Nevertheless, the Cauchy problem admits a unique solution.
You can get an intuition of this fact observing that any solution hits the "bad" set ${(x,y): x=y}$ only for $x=5$, since $y'(5) = 0$.
For a formal proof, you can observe that $y(x)$ is a solution of your Cauchy problem if and only if the function $z(x) := x - y(x)$ is a solution to
$$
z' = 1-z^{2/3}, quad, z(5) = 0,
$$
and this Cauchy problem admits unique solution.
answered Jan 26 at 16:55
RigelRigel
11.3k11320
$endgroup$
add a comment |
$begingroup$
As you have already seen, the function $f$ is not locally Lipschitz continuous around $y=x$, hence you cannot deduce uniqueness using the standard Cauchy-Lipschitz theory.
Nevertheless, the Cauchy problem admits a unique solution.
You can get an intuition of this fact observing that any solution hits the "bad" set ${(x,y): x=y}$ only for $x=5$, since $y'(5) = 0$.
For a formal proof, you can observe that $y(x)$ is a solution of your Cauchy problem if and only if the function $z(x) := x - y(x)$ is a solution to
$$
z' = 1-z^{2/3}, quad, z(5) = 0,
$$
and this Cauchy problem admits unique solution.
answered Jan 26 at 16:55
RigelRigel
11.3k11320
$endgroup$
add a comment |
$begingroup$
As you have already seen, the function $f$ is not locally Lipschitz continuous around $y=x$, hence you cannot deduce uniqueness using the standard Cauchy-Lipschitz theory.
Nevertheless, the Cauchy problem admits a unique solution.
You can get an intuition of this fact observing that any solution hits the "bad" set ${(x,y): x=y}$ only for $x=5$, since $y'(5) = 0$.
For a formal proof, you can observe that $y(x)$ is a solution of your Cauchy problem if and only if the function $z(x) := x - y(x)$ is a solution to
$$
z' = 1-z^{2/3}, quad, z(5) = 0,
$$
and this Cauchy problem admits unique solution.
answered Jan 26 at 16:55
RigelRigel
11.3k11320
$endgroup$
As you have already seen, the function $f$ is not locally Lipschitz continuous around $y=x$, hence you cannot deduce uniqueness using the standard Cauchy-Lipschitz theory.
Nevertheless, the Cauchy problem admits a unique solution.
You can get an intuition of this fact observing that any solution hits the "bad" set ${(x,y): x=y}$ only for $x=5$, since $y'(5) = 0$.
For a formal proof, you can observe that $y(x)$ is a solution of your Cauchy problem if and only if the function $z(x) := x - y(x)$ is a solution to
$$
z' = 1-z^{2/3}, quad, z(5) = 0,
$$
and this Cauchy problem admits unique solution.
answered Jan 26 at 16:55
RigelRigel
11.3k11320
answered Jan 26 at 16:55
RigelRigel
11.3k11320
answered Jan 26 at 16:55
RigelRigel
11.3k11320
answered Jan 26 at 16:55
RigelRigel
11.3k11320
11.3k11320
add a comment |
add a comment |
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