Uniqueness of solution based on characteristic curves
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I have a pde
$$begin{cases} u_t − xu_x = 2u & xinmathbb{R}, t>0\
u(x, 0) = frac{1}{1+x^2}
end{cases}$$
I've solved it using method of characteristics ($u=frac{1}{1+x^2e^{2t}}e^{2t})$ and plotted charactersitic curves.
Consider the upper half-space since $t>0$.
How to argue using the drawing whether or not it is the unique solution? Thank you.
pde characteristics
edited Dec 9 '18 at 12:28
Harry49
6,17331132
asked Nov 18 '18 at 22:38
dxdydzdxdydz
1789
$endgroup$
add a comment |
$begingroup$
I have a pde
$$begin{cases} u_t − xu_x = 2u & xinmathbb{R}, t>0\
u(x, 0) = frac{1}{1+x^2}
end{cases}$$
I've solved it using method of characteristics ($u=frac{1}{1+x^2e^{2t}}e^{2t})$ and plotted charactersitic curves.
Consider the upper half-space since $t>0$.
How to argue using the drawing whether or not it is the unique solution? Thank you.
pde characteristics
edited Dec 9 '18 at 12:28
Harry49
6,17331132
asked Nov 18 '18 at 22:38
dxdydzdxdydz
1789
$endgroup$
$begingroup$
Are you plotting in the $(t, u)$ plane? $(x, u)$? $(t, x)$?
$endgroup$
– Mattos
Nov 19 '18 at 1:46
$begingroup$
@Mattos This is the projection on the $(x,t)$-plane
$endgroup$
– dxdydz
Nov 19 '18 at 2:01
add a comment |
$begingroup$
I have a pde
$$begin{cases} u_t − xu_x = 2u & xinmathbb{R}, t>0\
u(x, 0) = frac{1}{1+x^2}
end{cases}$$
I've solved it using method of characteristics ($u=frac{1}{1+x^2e^{2t}}e^{2t})$ and plotted charactersitic curves.
Consider the upper half-space since $t>0$.
How to argue using the drawing whether or not it is the unique solution? Thank you.
pde characteristics
edited Dec 9 '18 at 12:28
Harry49
6,17331132
asked Nov 18 '18 at 22:38
dxdydzdxdydz
1789
$endgroup$
I have a pde
$$begin{cases} u_t − xu_x = 2u & xinmathbb{R}, t>0\
u(x, 0) = frac{1}{1+x^2}
end{cases}$$
I've solved it using method of characteristics ($u=frac{1}{1+x^2e^{2t}}e^{2t})$ and plotted charactersitic curves.
Consider the upper half-space since $t>0$.
How to argue using the drawing whether or not it is the unique solution? Thank you.
pde characteristics
pde characteristics
edited Dec 9 '18 at 12:28
Harry49
6,17331132
asked Nov 18 '18 at 22:38
dxdydzdxdydz
1789
edited Dec 9 '18 at 12:28
Harry49
6,17331132
asked Nov 18 '18 at 22:38
dxdydzdxdydz
1789
edited Dec 9 '18 at 12:28
Harry49
6,17331132
edited Dec 9 '18 at 12:28
Harry49
6,17331132
edited Dec 9 '18 at 12:28
Harry49
6,17331132
6,17331132
asked Nov 18 '18 at 22:38
dxdydzdxdydz
1789
asked Nov 18 '18 at 22:38
dxdydzdxdydz
1789
asked Nov 18 '18 at 22:38
dxdydzdxdydz
1789
1789
$begingroup$
Are you plotting in the $(t, u)$ plane? $(x, u)$? $(t, x)$?
$endgroup$
– Mattos
Nov 19 '18 at 1:46
$begingroup$
@Mattos This is the projection on the $(x,t)$-plane
$endgroup$
– dxdydz
Nov 19 '18 at 2:01
add a comment |
$begingroup$
Are you plotting in the $(t, u)$ plane? $(x, u)$? $(t, x)$?
$endgroup$
– Mattos
Nov 19 '18 at 1:46
$begingroup$
@Mattos This is the projection on the $(x,t)$-plane
$endgroup$
– dxdydz
Nov 19 '18 at 2:01
$begingroup$
Are you plotting in the $(t, u)$ plane? $(x, u)$? $(t, x)$?
$endgroup$
– Mattos
Nov 19 '18 at 1:46
$begingroup$
Are you plotting in the $(t, u)$ plane? $(x, u)$? $(t, x)$?
$endgroup$
– Mattos
Nov 19 '18 at 1:46
$begingroup$
@Mattos This is the projection on the $(x,t)$-plane
$endgroup$
– dxdydz
Nov 19 '18 at 2:01
$begingroup$
@Mattos This is the projection on the $(x,t)$-plane
$endgroup$
– dxdydz
Nov 19 '18 at 2:01
add a comment |
1 Answer
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The method of characteristics transforms the PDE into an ODE system. Therefore, existence and uniqueness is guaranteed under the assumptions of the Picard-Lindelöf theorem.
answered Dec 9 '18 at 12:27
Harry49Harry49
6,17331132
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add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
The method of characteristics transforms the PDE into an ODE system. Therefore, existence and uniqueness is guaranteed under the assumptions of the Picard-Lindelöf theorem.
answered Dec 9 '18 at 12:27
Harry49Harry49
6,17331132
$endgroup$
add a comment |
$begingroup$
The method of characteristics transforms the PDE into an ODE system. Therefore, existence and uniqueness is guaranteed under the assumptions of the Picard-Lindelöf theorem.
answered Dec 9 '18 at 12:27
Harry49Harry49
6,17331132
$endgroup$
add a comment |
$begingroup$
The method of characteristics transforms the PDE into an ODE system. Therefore, existence and uniqueness is guaranteed under the assumptions of the Picard-Lindelöf theorem.
answered Dec 9 '18 at 12:27
Harry49Harry49
6,17331132
$endgroup$
The method of characteristics transforms the PDE into an ODE system. Therefore, existence and uniqueness is guaranteed under the assumptions of the Picard-Lindelöf theorem.
answered Dec 9 '18 at 12:27
Harry49Harry49
6,17331132
answered Dec 9 '18 at 12:27
Harry49Harry49
6,17331132
answered Dec 9 '18 at 12:27
Harry49Harry49
6,17331132
answered Dec 9 '18 at 12:27
Harry49Harry49
6,17331132
6,17331132
add a comment |
add a comment |
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$begingroup$
Are you plotting in the $(t, u)$ plane? $(x, u)$? $(t, x)$?
$endgroup$
– Mattos
Nov 19 '18 at 1:46
$begingroup$
@Mattos This is the projection on the $(x,t)$-plane
$endgroup$
– dxdydz
Nov 19 '18 at 2:01