$(-v+u)u_ξ = u_{ξξ}$ $rightarrow$ $frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0$
$(-v+u)u_ξ = u_{ξξ}$ $rightarrow$ $frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0 $ where u=u(ξ,$tau$) , $t=tau$ , $x-vt = epsilon ξ$ and $frac{partial u}{partial t} + ufrac{partial u}{partial x} = epsilon frac{partial^2 u}{partial x^2}$
$frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0$
gives me $u_{ξξ} - v_ξu -vu_ξ + uu_ξ=0$
I cant seem to prove the identity
pde partial-derivative
edited Dec 10 at 9:45
Harry49
5,99121031
asked Dec 9 at 13:16
pablo_mathscobar
836
add a comment |
$(-v+u)u_ξ = u_{ξξ}$ $rightarrow$ $frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0 $ where u=u(ξ,$tau$) , $t=tau$ , $x-vt = epsilon ξ$ and $frac{partial u}{partial t} + ufrac{partial u}{partial x} = epsilon frac{partial^2 u}{partial x^2}$
$frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0$
gives me $u_{ξξ} - v_ξu -vu_ξ + uu_ξ=0$
I cant seem to prove the identity
pde partial-derivative
edited Dec 10 at 9:45
Harry49
5,99121031
asked Dec 9 at 13:16
pablo_mathscobar
836
Actually, $$(-v+u)u_ξ = u_{ξξ}$$ is equivalent to $$frac{partial}{partial ξ}(u_ξ + vu - tfrac{1}{2} u^2) =0$$ not what you write. To go further, you may want to 1. check the statement, 2. realize that $v$ is a constant here, 3. replace everypartial_ξbypartial ξ, 4. conclude.
– Did
Dec 9 at 14:23
@Did how do you know that v is constant
– pablo_mathscobar
Dec 9 at 16:22
'Cause otherwise the result is wrong. But say, isn't it your job to know the statement of the homework you were given?
– Did
Dec 9 at 16:24
@Did i am aware.. but there was nothing in the question that explicitly said v is constant. Since $v = x/t - epsilon * xi/t$ i thought you could take the partial wrt to $xi$
– pablo_mathscobar
Dec 9 at 16:28
add a comment |
$(-v+u)u_ξ = u_{ξξ}$ $rightarrow$ $frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0 $ where u=u(ξ,$tau$) , $t=tau$ , $x-vt = epsilon ξ$ and $frac{partial u}{partial t} + ufrac{partial u}{partial x} = epsilon frac{partial^2 u}{partial x^2}$
$frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0$
gives me $u_{ξξ} - v_ξu -vu_ξ + uu_ξ=0$
I cant seem to prove the identity
pde partial-derivative
edited Dec 10 at 9:45
Harry49
5,99121031
asked Dec 9 at 13:16
pablo_mathscobar
836
$(-v+u)u_ξ = u_{ξξ}$ $rightarrow$ $frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0 $ where u=u(ξ,$tau$) , $t=tau$ , $x-vt = epsilon ξ$ and $frac{partial u}{partial t} + ufrac{partial u}{partial x} = epsilon frac{partial^2 u}{partial x^2}$
$frac{partial}{partial_ξ}(u_ξ - vu + frac{1}{2} u^2) =0$
gives me $u_{ξξ} - v_ξu -vu_ξ + uu_ξ=0$
I cant seem to prove the identity
pde partial-derivative
pde partial-derivative
edited Dec 10 at 9:45
Harry49
5,99121031
asked Dec 9 at 13:16
pablo_mathscobar
836
edited Dec 10 at 9:45
Harry49
5,99121031
asked Dec 9 at 13:16
pablo_mathscobar
836
edited Dec 10 at 9:45
Harry49
5,99121031
edited Dec 10 at 9:45
Harry49
5,99121031
edited Dec 10 at 9:45
Harry49
5,99121031
5,99121031
asked Dec 9 at 13:16
pablo_mathscobar
836
asked Dec 9 at 13:16
pablo_mathscobar
836
asked Dec 9 at 13:16
pablo_mathscobar
836
836
Actually, $$(-v+u)u_ξ = u_{ξξ}$$ is equivalent to $$frac{partial}{partial ξ}(u_ξ + vu - tfrac{1}{2} u^2) =0$$ not what you write. To go further, you may want to 1. check the statement, 2. realize that $v$ is a constant here, 3. replace everypartial_ξbypartial ξ, 4. conclude.
– Did
Dec 9 at 14:23
@Did how do you know that v is constant
– pablo_mathscobar
Dec 9 at 16:22
'Cause otherwise the result is wrong. But say, isn't it your job to know the statement of the homework you were given?
– Did
Dec 9 at 16:24
@Did i am aware.. but there was nothing in the question that explicitly said v is constant. Since $v = x/t - epsilon * xi/t$ i thought you could take the partial wrt to $xi$
– pablo_mathscobar
Dec 9 at 16:28
add a comment |
Actually, $$(-v+u)u_ξ = u_{ξξ}$$ is equivalent to $$frac{partial}{partial ξ}(u_ξ + vu - tfrac{1}{2} u^2) =0$$ not what you write. To go further, you may want to 1. check the statement, 2. realize that $v$ is a constant here, 3. replace everypartial_ξbypartial ξ, 4. conclude.
– Did
Dec 9 at 14:23
@Did how do you know that v is constant
– pablo_mathscobar
Dec 9 at 16:22
'Cause otherwise the result is wrong. But say, isn't it your job to know the statement of the homework you were given?
– Did
Dec 9 at 16:24
@Did i am aware.. but there was nothing in the question that explicitly said v is constant. Since $v = x/t - epsilon * xi/t$ i thought you could take the partial wrt to $xi$
– pablo_mathscobar
Dec 9 at 16:28
Actually, $$(-v+u)u_ξ = u_{ξξ}$$ is equivalent to $$frac{partial}{partial ξ}(u_ξ + vu - tfrac{1}{2} u^2) =0$$ not what you write. To go further, you may want to 1. check the statement, 2. realize that $v$ is a constant here, 3. replace every
partial_ξ by partial ξ, 4. conclude.– Did
Dec 9 at 14:23
Actually, $$(-v+u)u_ξ = u_{ξξ}$$ is equivalent to $$frac{partial}{partial ξ}(u_ξ + vu - tfrac{1}{2} u^2) =0$$ not what you write. To go further, you may want to 1. check the statement, 2. realize that $v$ is a constant here, 3. replace every
partial_ξ by partial ξ, 4. conclude.– Did
Dec 9 at 14:23
@Did how do you know that v is constant
– pablo_mathscobar
Dec 9 at 16:22
@Did how do you know that v is constant
– pablo_mathscobar
Dec 9 at 16:22
'Cause otherwise the result is wrong. But say, isn't it your job to know the statement of the homework you were given?
– Did
Dec 9 at 16:24
'Cause otherwise the result is wrong. But say, isn't it your job to know the statement of the homework you were given?
– Did
Dec 9 at 16:24
@Did i am aware.. but there was nothing in the question that explicitly said v is constant. Since $v = x/t - epsilon * xi/t$ i thought you could take the partial wrt to $xi$
– pablo_mathscobar
Dec 9 at 16:28
@Did i am aware.. but there was nothing in the question that explicitly said v is constant. Since $v = x/t - epsilon * xi/t$ i thought you could take the partial wrt to $xi$
– pablo_mathscobar
Dec 9 at 16:28
add a comment |
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Actually, $$(-v+u)u_ξ = u_{ξξ}$$ is equivalent to $$frac{partial}{partial ξ}(u_ξ + vu - tfrac{1}{2} u^2) =0$$ not what you write. To go further, you may want to 1. check the statement, 2. realize that $v$ is a constant here, 3. replace every
partial_ξbypartial ξ, 4. conclude.– Did
Dec 9 at 14:23
@Did how do you know that v is constant
– pablo_mathscobar
Dec 9 at 16:22
'Cause otherwise the result is wrong. But say, isn't it your job to know the statement of the homework you were given?
– Did
Dec 9 at 16:24
@Did i am aware.. but there was nothing in the question that explicitly said v is constant. Since $v = x/t - epsilon * xi/t$ i thought you could take the partial wrt to $xi$
– pablo_mathscobar
Dec 9 at 16:28