Method of characteristics for system of linear transport equations
$begingroup$
If I have a system of pde
$$begin{cases}
u_t+v_x=0\
v_t+u_x=0\
u(x,0)=u_0(x), v(x,0)=v_0(x)end{cases}$$
how to extend the idea of method of characteristics to this situation? How do I parametrize the initial data and write the explicit solution to the problem for all $x$ and $t$ in terms of the functions $u_0$ and $v_0$?
pde systems-of-equations hyperbolic-equations linear-pde
edited Dec 9 '18 at 12:56
Harry49
6,43731132
asked Nov 15 '18 at 2:27
dxdydzdxdydz
33110
$endgroup$
add a comment |
$begingroup$
If I have a system of pde
$$begin{cases}
u_t+v_x=0\
v_t+u_x=0\
u(x,0)=u_0(x), v(x,0)=v_0(x)end{cases}$$
how to extend the idea of method of characteristics to this situation? How do I parametrize the initial data and write the explicit solution to the problem for all $x$ and $t$ in terms of the functions $u_0$ and $v_0$?
pde systems-of-equations hyperbolic-equations linear-pde
edited Dec 9 '18 at 12:56
Harry49
6,43731132
asked Nov 15 '18 at 2:27
dxdydzdxdydz
33110
$endgroup$
1
$begingroup$
A great solution to method of characteristics for a system of PDEs is given here.
$endgroup$
– Mattos
Nov 15 '18 at 3:47
add a comment |
$begingroup$
If I have a system of pde
$$begin{cases}
u_t+v_x=0\
v_t+u_x=0\
u(x,0)=u_0(x), v(x,0)=v_0(x)end{cases}$$
how to extend the idea of method of characteristics to this situation? How do I parametrize the initial data and write the explicit solution to the problem for all $x$ and $t$ in terms of the functions $u_0$ and $v_0$?
pde systems-of-equations hyperbolic-equations linear-pde
edited Dec 9 '18 at 12:56
Harry49
6,43731132
asked Nov 15 '18 at 2:27
dxdydzdxdydz
33110
$endgroup$
If I have a system of pde
$$begin{cases}
u_t+v_x=0\
v_t+u_x=0\
u(x,0)=u_0(x), v(x,0)=v_0(x)end{cases}$$
how to extend the idea of method of characteristics to this situation? How do I parametrize the initial data and write the explicit solution to the problem for all $x$ and $t$ in terms of the functions $u_0$ and $v_0$?
pde systems-of-equations hyperbolic-equations linear-pde
pde systems-of-equations hyperbolic-equations linear-pde
edited Dec 9 '18 at 12:56
Harry49
6,43731132
asked Nov 15 '18 at 2:27
dxdydzdxdydz
33110
edited Dec 9 '18 at 12:56
Harry49
6,43731132
asked Nov 15 '18 at 2:27
dxdydzdxdydz
33110
edited Dec 9 '18 at 12:56
Harry49
6,43731132
edited Dec 9 '18 at 12:56
Harry49
6,43731132
edited Dec 9 '18 at 12:56
Harry49
6,43731132
6,43731132
asked Nov 15 '18 at 2:27
dxdydzdxdydz
33110
asked Nov 15 '18 at 2:27
dxdydzdxdydz
33110
asked Nov 15 '18 at 2:27
dxdydzdxdydz
33110
33110
1
$begingroup$
A great solution to method of characteristics for a system of PDEs is given here.
$endgroup$
– Mattos
Nov 15 '18 at 3:47
add a comment |
1
$begingroup$
A great solution to method of characteristics for a system of PDEs is given here.
$endgroup$
– Mattos
Nov 15 '18 at 3:47
1
1
$begingroup$
A great solution to method of characteristics for a system of PDEs is given here.
$endgroup$
– Mattos
Nov 15 '18 at 3:47
$begingroup$
A great solution to method of characteristics for a system of PDEs is given here.
$endgroup$
– Mattos
Nov 15 '18 at 3:47
add a comment |
1 Answer
1
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oldest
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The system rewrites as ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$ with ${bf u} = (u,v)^top$. We diagonalize the matrix as ${bf M} = {bf S}, {bf J}, {bf S}^{-1}$ where $bf J$ is diagonal. Setting ${bf v} = {bf S}^{-1}{bf u}$, one obtains a diagonal system ${bf v}_t + {bf J}, {bf v}_x = {bf 0}$ which rows can be solved independently by using the method of characteristics. Then, $bf u$ is deduced from ${bf u} = {bf S},{bf v}$. Here, we find
begin{aligned}
u(x,t)&= frac{1}{2}big(u_0(x-t)+u_0(x+t)big)+frac{1}{2}big(v_0(x-t)-v_0(x+t)big)\
v(x,t)&= frac{1}{2}big(v_0(x-t)+v_0(x+t)big)+frac{1}{2}big(u_0(x-t)-u_0(x+t)big)
end{aligned}
This method works only for linear first-order systems ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$, which matrix $bf M$ can be diagonalized in $Bbb R$ (linear hyperbolic systems).
answered Dec 9 '18 at 12:55
Harry49Harry49
6,43731132
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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 system rewrites as ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$ with ${bf u} = (u,v)^top$. We diagonalize the matrix as ${bf M} = {bf S}, {bf J}, {bf S}^{-1}$ where $bf J$ is diagonal. Setting ${bf v} = {bf S}^{-1}{bf u}$, one obtains a diagonal system ${bf v}_t + {bf J}, {bf v}_x = {bf 0}$ which rows can be solved independently by using the method of characteristics. Then, $bf u$ is deduced from ${bf u} = {bf S},{bf v}$. Here, we find
begin{aligned}
u(x,t)&= frac{1}{2}big(u_0(x-t)+u_0(x+t)big)+frac{1}{2}big(v_0(x-t)-v_0(x+t)big)\
v(x,t)&= frac{1}{2}big(v_0(x-t)+v_0(x+t)big)+frac{1}{2}big(u_0(x-t)-u_0(x+t)big)
end{aligned}
This method works only for linear first-order systems ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$, which matrix $bf M$ can be diagonalized in $Bbb R$ (linear hyperbolic systems).
answered Dec 9 '18 at 12:55
Harry49Harry49
6,43731132
$endgroup$
add a comment |
$begingroup$
The system rewrites as ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$ with ${bf u} = (u,v)^top$. We diagonalize the matrix as ${bf M} = {bf S}, {bf J}, {bf S}^{-1}$ where $bf J$ is diagonal. Setting ${bf v} = {bf S}^{-1}{bf u}$, one obtains a diagonal system ${bf v}_t + {bf J}, {bf v}_x = {bf 0}$ which rows can be solved independently by using the method of characteristics. Then, $bf u$ is deduced from ${bf u} = {bf S},{bf v}$. Here, we find
begin{aligned}
u(x,t)&= frac{1}{2}big(u_0(x-t)+u_0(x+t)big)+frac{1}{2}big(v_0(x-t)-v_0(x+t)big)\
v(x,t)&= frac{1}{2}big(v_0(x-t)+v_0(x+t)big)+frac{1}{2}big(u_0(x-t)-u_0(x+t)big)
end{aligned}
This method works only for linear first-order systems ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$, which matrix $bf M$ can be diagonalized in $Bbb R$ (linear hyperbolic systems).
answered Dec 9 '18 at 12:55
Harry49Harry49
6,43731132
$endgroup$
add a comment |
$begingroup$
The system rewrites as ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$ with ${bf u} = (u,v)^top$. We diagonalize the matrix as ${bf M} = {bf S}, {bf J}, {bf S}^{-1}$ where $bf J$ is diagonal. Setting ${bf v} = {bf S}^{-1}{bf u}$, one obtains a diagonal system ${bf v}_t + {bf J}, {bf v}_x = {bf 0}$ which rows can be solved independently by using the method of characteristics. Then, $bf u$ is deduced from ${bf u} = {bf S},{bf v}$. Here, we find
begin{aligned}
u(x,t)&= frac{1}{2}big(u_0(x-t)+u_0(x+t)big)+frac{1}{2}big(v_0(x-t)-v_0(x+t)big)\
v(x,t)&= frac{1}{2}big(v_0(x-t)+v_0(x+t)big)+frac{1}{2}big(u_0(x-t)-u_0(x+t)big)
end{aligned}
This method works only for linear first-order systems ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$, which matrix $bf M$ can be diagonalized in $Bbb R$ (linear hyperbolic systems).
answered Dec 9 '18 at 12:55
Harry49Harry49
6,43731132
$endgroup$
The system rewrites as ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$ with ${bf u} = (u,v)^top$. We diagonalize the matrix as ${bf M} = {bf S}, {bf J}, {bf S}^{-1}$ where $bf J$ is diagonal. Setting ${bf v} = {bf S}^{-1}{bf u}$, one obtains a diagonal system ${bf v}_t + {bf J}, {bf v}_x = {bf 0}$ which rows can be solved independently by using the method of characteristics. Then, $bf u$ is deduced from ${bf u} = {bf S},{bf v}$. Here, we find
begin{aligned}
u(x,t)&= frac{1}{2}big(u_0(x-t)+u_0(x+t)big)+frac{1}{2}big(v_0(x-t)-v_0(x+t)big)\
v(x,t)&= frac{1}{2}big(v_0(x-t)+v_0(x+t)big)+frac{1}{2}big(u_0(x-t)-u_0(x+t)big)
end{aligned}
This method works only for linear first-order systems ${bf u}_t + {bf M}, {bf u}_x = {bf 0}$, which matrix $bf M$ can be diagonalized in $Bbb R$ (linear hyperbolic systems).
answered Dec 9 '18 at 12:55
Harry49Harry49
6,43731132
answered Dec 9 '18 at 12:55
Harry49Harry49
6,43731132
answered Dec 9 '18 at 12:55
Harry49Harry49
6,43731132
answered Dec 9 '18 at 12:55
Harry49Harry49
6,43731132
6,43731132
add a comment |
add a comment |
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$begingroup$
A great solution to method of characteristics for a system of PDEs is given here.
$endgroup$
– Mattos
Nov 15 '18 at 3:47