Determine characteristics for a PDE
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Given the PDE $frac{partialrho}{partial t} + 4 frac{partialrho}{partial x}= 0 $ determine the characteristics.
I understand how to solve this PDE using the method of characteristics as $rho=f(x-4t)$, but I do not understand what is meant by "determine the characteristics".
Which equations are the characteristics?
differential-equations pde
asked Dec 1 at 20:01
bphi
17816
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up vote
1
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favorite
Given the PDE $frac{partialrho}{partial t} + 4 frac{partialrho}{partial x}= 0 $ determine the characteristics.
I understand how to solve this PDE using the method of characteristics as $rho=f(x-4t)$, but I do not understand what is meant by "determine the characteristics".
Which equations are the characteristics?
differential-equations pde
asked Dec 1 at 20:01
bphi
17816
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given the PDE $frac{partialrho}{partial t} + 4 frac{partialrho}{partial x}= 0 $ determine the characteristics.
I understand how to solve this PDE using the method of characteristics as $rho=f(x-4t)$, but I do not understand what is meant by "determine the characteristics".
Which equations are the characteristics?
differential-equations pde
asked Dec 1 at 20:01
bphi
17816
Given the PDE $frac{partialrho}{partial t} + 4 frac{partialrho}{partial x}= 0 $ determine the characteristics.
I understand how to solve this PDE using the method of characteristics as $rho=f(x-4t)$, but I do not understand what is meant by "determine the characteristics".
Which equations are the characteristics?
differential-equations pde
differential-equations pde
asked Dec 1 at 20:01
bphi
17816
asked Dec 1 at 20:01
bphi
17816
asked Dec 1 at 20:01
bphi
17816
asked Dec 1 at 20:01
bphi
17816
asked Dec 1 at 20:01
bphi
17816
17816
add a comment |
add a comment |
1 Answer
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If you're already solving the PDE with the method of characteristics, than you're computing the "characteristics"; the solution is a union of characteristic curves.
They are integral curves of a vector field which the solution is tangent to at every point. In other words, the solution is constant along them and the equation reduces to a ODE along the characteristic curves.
In your case, the curve $x(s), t(s)$ such that
$$
dfrac{dx}{ds} = 4 \
dfrac{dt}{ds} = 1
$$
answered Dec 1 at 20:21
al0
618614
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you're already solving the PDE with the method of characteristics, than you're computing the "characteristics"; the solution is a union of characteristic curves.
They are integral curves of a vector field which the solution is tangent to at every point. In other words, the solution is constant along them and the equation reduces to a ODE along the characteristic curves.
In your case, the curve $x(s), t(s)$ such that
$$
dfrac{dx}{ds} = 4 \
dfrac{dt}{ds} = 1
$$
answered Dec 1 at 20:21
al0
618614
add a comment |
up vote
1
down vote
accepted
If you're already solving the PDE with the method of characteristics, than you're computing the "characteristics"; the solution is a union of characteristic curves.
They are integral curves of a vector field which the solution is tangent to at every point. In other words, the solution is constant along them and the equation reduces to a ODE along the characteristic curves.
In your case, the curve $x(s), t(s)$ such that
$$
dfrac{dx}{ds} = 4 \
dfrac{dt}{ds} = 1
$$
answered Dec 1 at 20:21
al0
618614
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you're already solving the PDE with the method of characteristics, than you're computing the "characteristics"; the solution is a union of characteristic curves.
They are integral curves of a vector field which the solution is tangent to at every point. In other words, the solution is constant along them and the equation reduces to a ODE along the characteristic curves.
In your case, the curve $x(s), t(s)$ such that
$$
dfrac{dx}{ds} = 4 \
dfrac{dt}{ds} = 1
$$
answered Dec 1 at 20:21
al0
618614
If you're already solving the PDE with the method of characteristics, than you're computing the "characteristics"; the solution is a union of characteristic curves.
They are integral curves of a vector field which the solution is tangent to at every point. In other words, the solution is constant along them and the equation reduces to a ODE along the characteristic curves.
In your case, the curve $x(s), t(s)$ such that
$$
dfrac{dx}{ds} = 4 \
dfrac{dt}{ds} = 1
$$
answered Dec 1 at 20:21
al0
618614
answered Dec 1 at 20:21
al0
618614
answered Dec 1 at 20:21
al0
618614
answered Dec 1 at 20:21
al0
618614
618614
add a comment |
add a comment |
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