Mixing
Uniqueness of the solution of a PDE…2
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How can I prove the uniqueness in $[0, +infty)times (0,1)$ of the solution of a PDE as the following:
$frac{partial}{partial t}v(t,x)=frac{partial}{partial x}Big((x-frac{1}{2})v(t,x)Big)$
$v(0,x)=g(x)$,
where $gin C^infty (mathbb R)$ with support in the interval $(0,1)$.
real-analysis analysis pde
asked Feb 3 at 7:46
user268193user268193
718
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add a comment |
$begingroup$
How can I prove the uniqueness in $[0, +infty)times (0,1)$ of the solution of a PDE as the following:
$frac{partial}{partial t}v(t,x)=frac{partial}{partial x}Big((x-frac{1}{2})v(t,x)Big)$
$v(0,x)=g(x)$,
where $gin C^infty (mathbb R)$ with support in the interval $(0,1)$.
real-analysis analysis pde
asked Feb 3 at 7:46
user268193user268193
718
$endgroup$
$begingroup$
Any boundary condition on $x=0$ and $x=1$?
$endgroup$
– GReyes
Feb 3 at 8:06
$begingroup$
$g(0)=0$,and $g(1)=0$
$endgroup$
– user268193
Feb 3 at 12:52
add a comment |
$begingroup$
How can I prove the uniqueness in $[0, +infty)times (0,1)$ of the solution of a PDE as the following:
$frac{partial}{partial t}v(t,x)=frac{partial}{partial x}Big((x-frac{1}{2})v(t,x)Big)$
$v(0,x)=g(x)$,
where $gin C^infty (mathbb R)$ with support in the interval $(0,1)$.
real-analysis analysis pde
asked Feb 3 at 7:46
user268193user268193
718
$endgroup$
How can I prove the uniqueness in $[0, +infty)times (0,1)$ of the solution of a PDE as the following:
$frac{partial}{partial t}v(t,x)=frac{partial}{partial x}Big((x-frac{1}{2})v(t,x)Big)$
$v(0,x)=g(x)$,
where $gin C^infty (mathbb R)$ with support in the interval $(0,1)$.
real-analysis analysis pde
real-analysis analysis pde
asked Feb 3 at 7:46
user268193user268193
718
asked Feb 3 at 7:46
user268193user268193
718
asked Feb 3 at 7:46
user268193user268193
718
asked Feb 3 at 7:46
user268193user268193
718
asked Feb 3 at 7:46
user268193user268193
718
718
$begingroup$
Any boundary condition on $x=0$ and $x=1$?
$endgroup$
– GReyes
Feb 3 at 8:06
$begingroup$
$g(0)=0$,and $g(1)=0$
$endgroup$
– user268193
Feb 3 at 12:52
add a comment |
$begingroup$
Any boundary condition on $x=0$ and $x=1$?
$endgroup$
– GReyes
Feb 3 at 8:06
$begingroup$
$g(0)=0$,and $g(1)=0$
$endgroup$
– user268193
Feb 3 at 12:52
$begingroup$
Any boundary condition on $x=0$ and $x=1$?
$endgroup$
– GReyes
Feb 3 at 8:06
$begingroup$
Any boundary condition on $x=0$ and $x=1$?
$endgroup$
– GReyes
Feb 3 at 8:06
$begingroup$
$g(0)=0$,and $g(1)=0$
$endgroup$
– user268193
Feb 3 at 12:52
$begingroup$
$g(0)=0$,and $g(1)=0$
$endgroup$
– user268193
Feb 3 at 12:52
add a comment |
1 Answer
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oldest
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@user268193 Uniqueness follows from the fact that the characteristics do not intersect and are in a one-to-one correspondence with the points of the parabolic boundary. The characteristics of your equation satisfy the ODE
$$
dt=dx/(1/2-x)
$$
which are exponential curves approaching $x=1/2$ as $ttoinfty$ and also $x=1/2$ itself. These curves do not intersect and, given any point $(t,x)in[(0,infty)times(0,1)$, you can connect it along a characteristic to a unique point on either part of your boundary $(t,x)in{0}times[0,1]$ or $(0,infty)times{0}$ or $(0,infty)times{1}$. By uniqueness for ODEs along characteristics, the values of $u(t,x)$ are thus uniquely determined by the initial and boundary data.
answered Feb 6 at 7:17
GReyesGReyes
2,52815
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
@user268193 Uniqueness follows from the fact that the characteristics do not intersect and are in a one-to-one correspondence with the points of the parabolic boundary. The characteristics of your equation satisfy the ODE
$$
dt=dx/(1/2-x)
$$
which are exponential curves approaching $x=1/2$ as $ttoinfty$ and also $x=1/2$ itself. These curves do not intersect and, given any point $(t,x)in[(0,infty)times(0,1)$, you can connect it along a characteristic to a unique point on either part of your boundary $(t,x)in{0}times[0,1]$ or $(0,infty)times{0}$ or $(0,infty)times{1}$. By uniqueness for ODEs along characteristics, the values of $u(t,x)$ are thus uniquely determined by the initial and boundary data.
answered Feb 6 at 7:17
GReyesGReyes
2,52815
$endgroup$
add a comment |
$begingroup$
@user268193 Uniqueness follows from the fact that the characteristics do not intersect and are in a one-to-one correspondence with the points of the parabolic boundary. The characteristics of your equation satisfy the ODE
$$
dt=dx/(1/2-x)
$$
which are exponential curves approaching $x=1/2$ as $ttoinfty$ and also $x=1/2$ itself. These curves do not intersect and, given any point $(t,x)in[(0,infty)times(0,1)$, you can connect it along a characteristic to a unique point on either part of your boundary $(t,x)in{0}times[0,1]$ or $(0,infty)times{0}$ or $(0,infty)times{1}$. By uniqueness for ODEs along characteristics, the values of $u(t,x)$ are thus uniquely determined by the initial and boundary data.
answered Feb 6 at 7:17
GReyesGReyes
2,52815
$endgroup$
add a comment |
$begingroup$
@user268193 Uniqueness follows from the fact that the characteristics do not intersect and are in a one-to-one correspondence with the points of the parabolic boundary. The characteristics of your equation satisfy the ODE
$$
dt=dx/(1/2-x)
$$
which are exponential curves approaching $x=1/2$ as $ttoinfty$ and also $x=1/2$ itself. These curves do not intersect and, given any point $(t,x)in[(0,infty)times(0,1)$, you can connect it along a characteristic to a unique point on either part of your boundary $(t,x)in{0}times[0,1]$ or $(0,infty)times{0}$ or $(0,infty)times{1}$. By uniqueness for ODEs along characteristics, the values of $u(t,x)$ are thus uniquely determined by the initial and boundary data.
answered Feb 6 at 7:17
GReyesGReyes
2,52815
$endgroup$
@user268193 Uniqueness follows from the fact that the characteristics do not intersect and are in a one-to-one correspondence with the points of the parabolic boundary. The characteristics of your equation satisfy the ODE
$$
dt=dx/(1/2-x)
$$
which are exponential curves approaching $x=1/2$ as $ttoinfty$ and also $x=1/2$ itself. These curves do not intersect and, given any point $(t,x)in[(0,infty)times(0,1)$, you can connect it along a characteristic to a unique point on either part of your boundary $(t,x)in{0}times[0,1]$ or $(0,infty)times{0}$ or $(0,infty)times{1}$. By uniqueness for ODEs along characteristics, the values of $u(t,x)$ are thus uniquely determined by the initial and boundary data.
answered Feb 6 at 7:17
GReyesGReyes
2,52815
answered Feb 6 at 7:17
GReyesGReyes
2,52815
answered Feb 6 at 7:17
GReyesGReyes
2,52815
answered Feb 6 at 7:17
GReyesGReyes
2,52815
2,52815
add a comment |
add a comment |
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$begingroup$
Any boundary condition on $x=0$ and $x=1$?
$endgroup$
– GReyes
Feb 3 at 8:06
$begingroup$
$g(0)=0$,and $g(1)=0$
$endgroup$
– user268193
Feb 3 at 12:52