Non-linear IVP with $f(u_1,u_2):=sqrt{1+sqrt{u_1^2+u_2^2}}begin{pmatrix}u_1+u_2\3 u_2-u_1end{pmatrix}$
$begingroup$
Does the following initial value problem
begin{equation*}
begin{cases}
u'(t) = f(u(t)), \ u(0) = u_0
end{cases}
quad text{with} quad
f(u_1,u_2):=sqrt{1+sqrt{u_1^2+u_2^2}}begin{pmatrix}u_1+u_2\3 u_2-u_1end{pmatrix}
quad text{and} quad
u_0 := begin{pmatrix} 1 \ 0 end{pmatrix}
end{equation*}
have a global solution on $[0, infty)$, where $f: mathbb{R}^2 to mathbb{R}^2$?
Since the system is non-linear, we can't transform this system into a matrix and find eigenvalues etc, so I couldn't come up with a sensible answer.
In our course (ODE) we discussed absolute continuity and Caratheodory solutions to equations but I couldn't think of how this could help.
ordinary-differential-equations nonlinear-system initial-value-problems
asked Jan 16 at 17:17
Viktor GlombikViktor Glombik
1,3122628
$endgroup$
add a comment |
$begingroup$
Does the following initial value problem
begin{equation*}
begin{cases}
u'(t) = f(u(t)), \ u(0) = u_0
end{cases}
quad text{with} quad
f(u_1,u_2):=sqrt{1+sqrt{u_1^2+u_2^2}}begin{pmatrix}u_1+u_2\3 u_2-u_1end{pmatrix}
quad text{and} quad
u_0 := begin{pmatrix} 1 \ 0 end{pmatrix}
end{equation*}
have a global solution on $[0, infty)$, where $f: mathbb{R}^2 to mathbb{R}^2$?
Since the system is non-linear, we can't transform this system into a matrix and find eigenvalues etc, so I couldn't come up with a sensible answer.
In our course (ODE) we discussed absolute continuity and Caratheodory solutions to equations but I couldn't think of how this could help.
ordinary-differential-equations nonlinear-system initial-value-problems
asked Jan 16 at 17:17
Viktor GlombikViktor Glombik
1,3122628
$endgroup$
add a comment |
$begingroup$
Does the following initial value problem
begin{equation*}
begin{cases}
u'(t) = f(u(t)), \ u(0) = u_0
end{cases}
quad text{with} quad
f(u_1,u_2):=sqrt{1+sqrt{u_1^2+u_2^2}}begin{pmatrix}u_1+u_2\3 u_2-u_1end{pmatrix}
quad text{and} quad
u_0 := begin{pmatrix} 1 \ 0 end{pmatrix}
end{equation*}
have a global solution on $[0, infty)$, where $f: mathbb{R}^2 to mathbb{R}^2$?
Since the system is non-linear, we can't transform this system into a matrix and find eigenvalues etc, so I couldn't come up with a sensible answer.
In our course (ODE) we discussed absolute continuity and Caratheodory solutions to equations but I couldn't think of how this could help.
ordinary-differential-equations nonlinear-system initial-value-problems
asked Jan 16 at 17:17
Viktor GlombikViktor Glombik
1,3122628
$endgroup$
Does the following initial value problem
begin{equation*}
begin{cases}
u'(t) = f(u(t)), \ u(0) = u_0
end{cases}
quad text{with} quad
f(u_1,u_2):=sqrt{1+sqrt{u_1^2+u_2^2}}begin{pmatrix}u_1+u_2\3 u_2-u_1end{pmatrix}
quad text{and} quad
u_0 := begin{pmatrix} 1 \ 0 end{pmatrix}
end{equation*}
have a global solution on $[0, infty)$, where $f: mathbb{R}^2 to mathbb{R}^2$?
Since the system is non-linear, we can't transform this system into a matrix and find eigenvalues etc, so I couldn't come up with a sensible answer.
In our course (ODE) we discussed absolute continuity and Caratheodory solutions to equations but I couldn't think of how this could help.
ordinary-differential-equations nonlinear-system initial-value-problems
ordinary-differential-equations nonlinear-system initial-value-problems
asked Jan 16 at 17:17
Viktor GlombikViktor Glombik
1,3122628
asked Jan 16 at 17:17
Viktor GlombikViktor Glombik
1,3122628
asked Jan 16 at 17:17
Viktor GlombikViktor Glombik
1,3122628
asked Jan 16 at 17:17
Viktor GlombikViktor Glombik
1,3122628
asked Jan 16 at 17:17
Viktor GlombikViktor Glombik
1,3122628
1,3122628
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The IVP does indeed have a global solution because the function $f$ is globally Lipschitz. Therefore the intervals of existence obtained by Picard's Theorem are all of $mathbb{R}$. To see this, consider the linear and nonlinear parts separately. for the matrix part, we have $max |lambda|=2$ and for the nonlinear part, we have $sup left|nabla left(sqrt{1+sqrt{u_1^2+u_2^2}}right)right|=1/2$. Therefore $f$ has is globally Lipschitz with constant 1, so the global solution exists.
answered Jan 16 at 17:31
whpowell96whpowell96
49819
$endgroup$
$begingroup$
Why do you consider $nabla$ of the non-linear term? And what is the intuition behind ''being allowed'' to consider both parts separately?
$endgroup$
– Viktor Glombik
Jan 16 at 18:46
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The IVP does indeed have a global solution because the function $f$ is globally Lipschitz. Therefore the intervals of existence obtained by Picard's Theorem are all of $mathbb{R}$. To see this, consider the linear and nonlinear parts separately. for the matrix part, we have $max |lambda|=2$ and for the nonlinear part, we have $sup left|nabla left(sqrt{1+sqrt{u_1^2+u_2^2}}right)right|=1/2$. Therefore $f$ has is globally Lipschitz with constant 1, so the global solution exists.
answered Jan 16 at 17:31
whpowell96whpowell96
49819
$endgroup$
$begingroup$
Why do you consider $nabla$ of the non-linear term? And what is the intuition behind ''being allowed'' to consider both parts separately?
$endgroup$
– Viktor Glombik
Jan 16 at 18:46
add a comment |
$begingroup$
The IVP does indeed have a global solution because the function $f$ is globally Lipschitz. Therefore the intervals of existence obtained by Picard's Theorem are all of $mathbb{R}$. To see this, consider the linear and nonlinear parts separately. for the matrix part, we have $max |lambda|=2$ and for the nonlinear part, we have $sup left|nabla left(sqrt{1+sqrt{u_1^2+u_2^2}}right)right|=1/2$. Therefore $f$ has is globally Lipschitz with constant 1, so the global solution exists.
answered Jan 16 at 17:31
whpowell96whpowell96
49819
$endgroup$
$begingroup$
Why do you consider $nabla$ of the non-linear term? And what is the intuition behind ''being allowed'' to consider both parts separately?
$endgroup$
– Viktor Glombik
Jan 16 at 18:46
add a comment |
$begingroup$
The IVP does indeed have a global solution because the function $f$ is globally Lipschitz. Therefore the intervals of existence obtained by Picard's Theorem are all of $mathbb{R}$. To see this, consider the linear and nonlinear parts separately. for the matrix part, we have $max |lambda|=2$ and for the nonlinear part, we have $sup left|nabla left(sqrt{1+sqrt{u_1^2+u_2^2}}right)right|=1/2$. Therefore $f$ has is globally Lipschitz with constant 1, so the global solution exists.
answered Jan 16 at 17:31
whpowell96whpowell96
49819
$endgroup$
The IVP does indeed have a global solution because the function $f$ is globally Lipschitz. Therefore the intervals of existence obtained by Picard's Theorem are all of $mathbb{R}$. To see this, consider the linear and nonlinear parts separately. for the matrix part, we have $max |lambda|=2$ and for the nonlinear part, we have $sup left|nabla left(sqrt{1+sqrt{u_1^2+u_2^2}}right)right|=1/2$. Therefore $f$ has is globally Lipschitz with constant 1, so the global solution exists.
answered Jan 16 at 17:31
whpowell96whpowell96
49819
answered Jan 16 at 17:31
whpowell96whpowell96
49819
answered Jan 16 at 17:31
whpowell96whpowell96
49819
answered Jan 16 at 17:31
whpowell96whpowell96
49819
49819
$begingroup$
Why do you consider $nabla$ of the non-linear term? And what is the intuition behind ''being allowed'' to consider both parts separately?
$endgroup$
– Viktor Glombik
Jan 16 at 18:46
add a comment |
$begingroup$
Why do you consider $nabla$ of the non-linear term? And what is the intuition behind ''being allowed'' to consider both parts separately?
$endgroup$
– Viktor Glombik
Jan 16 at 18:46
$begingroup$
Why do you consider $nabla$ of the non-linear term? And what is the intuition behind ''being allowed'' to consider both parts separately?
$endgroup$
– Viktor Glombik
Jan 16 at 18:46
$begingroup$
Why do you consider $nabla$ of the non-linear term? And what is the intuition behind ''being allowed'' to consider both parts separately?
$endgroup$
– Viktor Glombik
Jan 16 at 18:46
add a comment |
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