Mixing
PDE Method of Characteristics with 3 independent variables. Any Idea?
$begingroup$
would you please help me with solving the 1st order PDE below?
$$ u_x + u_y + zu_z = u^3 $$
where
$$u(x, y, 1) = h(x, y)$$
using characteristic curves.
As far as I have studied, the characteristic lines are as follow: (am I right?)
$$ frac{dx}{1} = frac{dy}{1} = frac{dz}{z} = frac{du}{u^3} $$
I am trying to figure out how to write formula here. please excuse me for not being expert on this website yet.
thank you
pde
edited Jan 28 at 7:46
Dylan
14.1k31127
asked Jan 27 at 22:36
ShayShay
363
$endgroup$
add a comment |
$begingroup$
would you please help me with solving the 1st order PDE below?
$$ u_x + u_y + zu_z = u^3 $$
where
$$u(x, y, 1) = h(x, y)$$
using characteristic curves.
As far as I have studied, the characteristic lines are as follow: (am I right?)
$$ frac{dx}{1} = frac{dy}{1} = frac{dz}{z} = frac{du}{u^3} $$
I am trying to figure out how to write formula here. please excuse me for not being expert on this website yet.
thank you
pde
edited Jan 28 at 7:46
Dylan
14.1k31127
asked Jan 27 at 22:36
ShayShay
363
$endgroup$
$begingroup$
Welcome to StackExchange! Could you please be so kind and edit your question so that it easier to understand? We like to help, make it easy to help.
$endgroup$
– Michael Paris
Jan 27 at 22:43
$begingroup$
Can you write down the ordinary differential equations which give you the characteristic curves?
$endgroup$
– Christoph
Jan 28 at 0:50
add a comment |
$begingroup$
would you please help me with solving the 1st order PDE below?
$$ u_x + u_y + zu_z = u^3 $$
where
$$u(x, y, 1) = h(x, y)$$
using characteristic curves.
As far as I have studied, the characteristic lines are as follow: (am I right?)
$$ frac{dx}{1} = frac{dy}{1} = frac{dz}{z} = frac{du}{u^3} $$
I am trying to figure out how to write formula here. please excuse me for not being expert on this website yet.
thank you
pde
edited Jan 28 at 7:46
Dylan
14.1k31127
asked Jan 27 at 22:36
ShayShay
363
$endgroup$
would you please help me with solving the 1st order PDE below?
$$ u_x + u_y + zu_z = u^3 $$
where
$$u(x, y, 1) = h(x, y)$$
using characteristic curves.
As far as I have studied, the characteristic lines are as follow: (am I right?)
$$ frac{dx}{1} = frac{dy}{1} = frac{dz}{z} = frac{du}{u^3} $$
I am trying to figure out how to write formula here. please excuse me for not being expert on this website yet.
thank you
pde
pde
edited Jan 28 at 7:46
Dylan
14.1k31127
asked Jan 27 at 22:36
ShayShay
363
edited Jan 28 at 7:46
Dylan
14.1k31127
asked Jan 27 at 22:36
ShayShay
363
edited Jan 28 at 7:46
Dylan
14.1k31127
edited Jan 28 at 7:46
Dylan
14.1k31127
edited Jan 28 at 7:46
Dylan
14.1k31127
14.1k31127
asked Jan 27 at 22:36
ShayShay
363
asked Jan 27 at 22:36
ShayShay
363
asked Jan 27 at 22:36
ShayShay
363
363
$begingroup$
Welcome to StackExchange! Could you please be so kind and edit your question so that it easier to understand? We like to help, make it easy to help.
$endgroup$
– Michael Paris
Jan 27 at 22:43
$begingroup$
Can you write down the ordinary differential equations which give you the characteristic curves?
$endgroup$
– Christoph
Jan 28 at 0:50
add a comment |
$begingroup$
Welcome to StackExchange! Could you please be so kind and edit your question so that it easier to understand? We like to help, make it easy to help.
$endgroup$
– Michael Paris
Jan 27 at 22:43
$begingroup$
Can you write down the ordinary differential equations which give you the characteristic curves?
$endgroup$
– Christoph
Jan 28 at 0:50
$begingroup$
Welcome to StackExchange! Could you please be so kind and edit your question so that it easier to understand? We like to help, make it easy to help.
$endgroup$
– Michael Paris
Jan 27 at 22:43
$begingroup$
Welcome to StackExchange! Could you please be so kind and edit your question so that it easier to understand? We like to help, make it easy to help.
$endgroup$
– Michael Paris
Jan 27 at 22:43
$begingroup$
Can you write down the ordinary differential equations which give you the characteristic curves?
$endgroup$
– Christoph
Jan 28 at 0:50
$begingroup$
Can you write down the ordinary differential equations which give you the characteristic curves?
$endgroup$
– Christoph
Jan 28 at 0:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your characteristic equations are correct. You may solve them as
$dx = frac{dz}{z}$: $x = x_0 + ln(z)$,
$dy = frac{dz}{z}$: $y = y_0 + ln(z)$,
$frac{du}{u^3} = frac{dz}{z}$: $u = left(u_0^{-2} -2 ln(z)right)^{-1/2}$,
where the initial values $(x_0,y_0,u_0)$ are all given at $z=1$ ($ln(z) = 0$). From the initial condition we now obtain
begin{equation}
u_0 = u(x_0,y_0,1) = h(x_0,y_0) stackrel{1., 2.}{=} h(x-ln(z),y-ln(z)),
end{equation}
and therefore
begin{equation}
u(x,y,z) stackrel{3.}{=} left(hleft(x-ln(z),y-ln(z)right)^{-2} -2 ln(z)right)^{-1/2}.
end{equation}
You can now verify that this function $u$ satisfies indeed both the PDE (if $h$ is differentiable) and the initial condition.
answered Jan 28 at 5:53
ChristophChristoph
59616
$endgroup$
$begingroup$
Thank you very much. It is always reassuring to have other opinions for me.
$endgroup$
– Shay
Feb 4 at 13:58
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your characteristic equations are correct. You may solve them as
$dx = frac{dz}{z}$: $x = x_0 + ln(z)$,
$dy = frac{dz}{z}$: $y = y_0 + ln(z)$,
$frac{du}{u^3} = frac{dz}{z}$: $u = left(u_0^{-2} -2 ln(z)right)^{-1/2}$,
where the initial values $(x_0,y_0,u_0)$ are all given at $z=1$ ($ln(z) = 0$). From the initial condition we now obtain
begin{equation}
u_0 = u(x_0,y_0,1) = h(x_0,y_0) stackrel{1., 2.}{=} h(x-ln(z),y-ln(z)),
end{equation}
and therefore
begin{equation}
u(x,y,z) stackrel{3.}{=} left(hleft(x-ln(z),y-ln(z)right)^{-2} -2 ln(z)right)^{-1/2}.
end{equation}
You can now verify that this function $u$ satisfies indeed both the PDE (if $h$ is differentiable) and the initial condition.
answered Jan 28 at 5:53
ChristophChristoph
59616
$endgroup$
$begingroup$
Thank you very much. It is always reassuring to have other opinions for me.
$endgroup$
– Shay
Feb 4 at 13:58
add a comment |
$begingroup$
Your characteristic equations are correct. You may solve them as
$dx = frac{dz}{z}$: $x = x_0 + ln(z)$,
$dy = frac{dz}{z}$: $y = y_0 + ln(z)$,
$frac{du}{u^3} = frac{dz}{z}$: $u = left(u_0^{-2} -2 ln(z)right)^{-1/2}$,
where the initial values $(x_0,y_0,u_0)$ are all given at $z=1$ ($ln(z) = 0$). From the initial condition we now obtain
begin{equation}
u_0 = u(x_0,y_0,1) = h(x_0,y_0) stackrel{1., 2.}{=} h(x-ln(z),y-ln(z)),
end{equation}
and therefore
begin{equation}
u(x,y,z) stackrel{3.}{=} left(hleft(x-ln(z),y-ln(z)right)^{-2} -2 ln(z)right)^{-1/2}.
end{equation}
You can now verify that this function $u$ satisfies indeed both the PDE (if $h$ is differentiable) and the initial condition.
answered Jan 28 at 5:53
ChristophChristoph
59616
$endgroup$
$begingroup$
Thank you very much. It is always reassuring to have other opinions for me.
$endgroup$
– Shay
Feb 4 at 13:58
add a comment |
$begingroup$
Your characteristic equations are correct. You may solve them as
$dx = frac{dz}{z}$: $x = x_0 + ln(z)$,
$dy = frac{dz}{z}$: $y = y_0 + ln(z)$,
$frac{du}{u^3} = frac{dz}{z}$: $u = left(u_0^{-2} -2 ln(z)right)^{-1/2}$,
where the initial values $(x_0,y_0,u_0)$ are all given at $z=1$ ($ln(z) = 0$). From the initial condition we now obtain
begin{equation}
u_0 = u(x_0,y_0,1) = h(x_0,y_0) stackrel{1., 2.}{=} h(x-ln(z),y-ln(z)),
end{equation}
and therefore
begin{equation}
u(x,y,z) stackrel{3.}{=} left(hleft(x-ln(z),y-ln(z)right)^{-2} -2 ln(z)right)^{-1/2}.
end{equation}
You can now verify that this function $u$ satisfies indeed both the PDE (if $h$ is differentiable) and the initial condition.
answered Jan 28 at 5:53
ChristophChristoph
59616
$endgroup$
Your characteristic equations are correct. You may solve them as
$dx = frac{dz}{z}$: $x = x_0 + ln(z)$,
$dy = frac{dz}{z}$: $y = y_0 + ln(z)$,
$frac{du}{u^3} = frac{dz}{z}$: $u = left(u_0^{-2} -2 ln(z)right)^{-1/2}$,
where the initial values $(x_0,y_0,u_0)$ are all given at $z=1$ ($ln(z) = 0$). From the initial condition we now obtain
begin{equation}
u_0 = u(x_0,y_0,1) = h(x_0,y_0) stackrel{1., 2.}{=} h(x-ln(z),y-ln(z)),
end{equation}
and therefore
begin{equation}
u(x,y,z) stackrel{3.}{=} left(hleft(x-ln(z),y-ln(z)right)^{-2} -2 ln(z)right)^{-1/2}.
end{equation}
You can now verify that this function $u$ satisfies indeed both the PDE (if $h$ is differentiable) and the initial condition.
answered Jan 28 at 5:53
ChristophChristoph
59616
edited Jan 28 at 10:36
answered Jan 28 at 5:53
ChristophChristoph
59616
answered Jan 28 at 5:53
ChristophChristoph
59616
answered Jan 28 at 5:53
ChristophChristoph
59616
59616
$begingroup$
Thank you very much. It is always reassuring to have other opinions for me.
$endgroup$
– Shay
Feb 4 at 13:58
add a comment |
$begingroup$
Thank you very much. It is always reassuring to have other opinions for me.
$endgroup$
– Shay
Feb 4 at 13:58
$begingroup$
Thank you very much. It is always reassuring to have other opinions for me.
$endgroup$
– Shay
Feb 4 at 13:58
$begingroup$
Thank you very much. It is always reassuring to have other opinions for me.
$endgroup$
– Shay
Feb 4 at 13:58
add a comment |
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$begingroup$
Welcome to StackExchange! Could you please be so kind and edit your question so that it easier to understand? We like to help, make it easy to help.
$endgroup$
– Michael Paris
Jan 27 at 22:43
$begingroup$
Can you write down the ordinary differential equations which give you the characteristic curves?
$endgroup$
– Christoph
Jan 28 at 0:50