Are characteristics the only solution to the advection equation in 1+1D?
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I'm currently reading about fluid dynamics and the Riemann problem, and a very commonly used equation to introduce the topic is the 1+1D advection equation with constant coefficient $v$:
$$ frac{partial u}{partial t} + v frac{partial u}{partial x} = 0tag{1}$$
for which a solution is
$$ u(x,t) = u(x-vt, 0) = u_0(x-vt) $$
where $u_0 = u(t=0)$ is some initial condition.
This can be easily derived using the method of separation of variables: Let $u(x,t) = f(x)g(y)$.
Then
$$ frac{partial u}{partial t} = f(x) frac{partial g}{partial t}$$
$$ frac{partial u}{partial x} = g(t) frac{partial f}{partial x}
$$
Inserting into the advection equation and restructuring a little, we get
$$frac{1}{g } frac{partial g}{partial t} = frac{1}{f}frac{partial f}{partial x} = -lambda $$
where $lambda$ is some constant. Solving each equation separately gives us
$$ g = K_1 e^{-lambda v t} $$
$$ f = K_2 e^{lambda x} $$
$$ Rightarrow u(x,t) = fg = K e^{lambda (x - vt)} $$
with $K_1$, $K_2$ and $K=K_1 K_2$ are constants stemming from integration.
With
$$u_0 = u(x,t=0) = K e^{lambda x}$$
one can easily see that the solution can be expressed as
$$u(x,t) = u_0(x-vt)$$
So far, so good. Here's my question: Is that the only solution of the 1+1D advection equation with constant coefficients? Is there a proof that this is the only solution?
fluid-dynamics waves mathematics differential-equations
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up vote
4
down vote
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I'm currently reading about fluid dynamics and the Riemann problem, and a very commonly used equation to introduce the topic is the 1+1D advection equation with constant coefficient $v$:
$$ frac{partial u}{partial t} + v frac{partial u}{partial x} = 0tag{1}$$
for which a solution is
$$ u(x,t) = u(x-vt, 0) = u_0(x-vt) $$
where $u_0 = u(t=0)$ is some initial condition.
This can be easily derived using the method of separation of variables: Let $u(x,t) = f(x)g(y)$.
Then
$$ frac{partial u}{partial t} = f(x) frac{partial g}{partial t}$$
$$ frac{partial u}{partial x} = g(t) frac{partial f}{partial x}
$$
Inserting into the advection equation and restructuring a little, we get
$$frac{1}{g } frac{partial g}{partial t} = frac{1}{f}frac{partial f}{partial x} = -lambda $$
where $lambda$ is some constant. Solving each equation separately gives us
$$ g = K_1 e^{-lambda v t} $$
$$ f = K_2 e^{lambda x} $$
$$ Rightarrow u(x,t) = fg = K e^{lambda (x - vt)} $$
with $K_1$, $K_2$ and $K=K_1 K_2$ are constants stemming from integration.
With
$$u_0 = u(x,t=0) = K e^{lambda x}$$
one can easily see that the solution can be expressed as
$$u(x,t) = u_0(x-vt)$$
So far, so good. Here's my question: Is that the only solution of the 1+1D advection equation with constant coefficients? Is there a proof that this is the only solution?
fluid-dynamics waves mathematics differential-equations
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
Nov 9 at 17:19
Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
– Kyle Kanos
Nov 10 at 18:04
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm currently reading about fluid dynamics and the Riemann problem, and a very commonly used equation to introduce the topic is the 1+1D advection equation with constant coefficient $v$:
$$ frac{partial u}{partial t} + v frac{partial u}{partial x} = 0tag{1}$$
for which a solution is
$$ u(x,t) = u(x-vt, 0) = u_0(x-vt) $$
where $u_0 = u(t=0)$ is some initial condition.
This can be easily derived using the method of separation of variables: Let $u(x,t) = f(x)g(y)$.
Then
$$ frac{partial u}{partial t} = f(x) frac{partial g}{partial t}$$
$$ frac{partial u}{partial x} = g(t) frac{partial f}{partial x}
$$
Inserting into the advection equation and restructuring a little, we get
$$frac{1}{g } frac{partial g}{partial t} = frac{1}{f}frac{partial f}{partial x} = -lambda $$
where $lambda$ is some constant. Solving each equation separately gives us
$$ g = K_1 e^{-lambda v t} $$
$$ f = K_2 e^{lambda x} $$
$$ Rightarrow u(x,t) = fg = K e^{lambda (x - vt)} $$
with $K_1$, $K_2$ and $K=K_1 K_2$ are constants stemming from integration.
With
$$u_0 = u(x,t=0) = K e^{lambda x}$$
one can easily see that the solution can be expressed as
$$u(x,t) = u_0(x-vt)$$
So far, so good. Here's my question: Is that the only solution of the 1+1D advection equation with constant coefficients? Is there a proof that this is the only solution?
fluid-dynamics waves mathematics differential-equations
I'm currently reading about fluid dynamics and the Riemann problem, and a very commonly used equation to introduce the topic is the 1+1D advection equation with constant coefficient $v$:
$$ frac{partial u}{partial t} + v frac{partial u}{partial x} = 0tag{1}$$
for which a solution is
$$ u(x,t) = u(x-vt, 0) = u_0(x-vt) $$
where $u_0 = u(t=0)$ is some initial condition.
This can be easily derived using the method of separation of variables: Let $u(x,t) = f(x)g(y)$.
Then
$$ frac{partial u}{partial t} = f(x) frac{partial g}{partial t}$$
$$ frac{partial u}{partial x} = g(t) frac{partial f}{partial x}
$$
Inserting into the advection equation and restructuring a little, we get
$$frac{1}{g } frac{partial g}{partial t} = frac{1}{f}frac{partial f}{partial x} = -lambda $$
where $lambda$ is some constant. Solving each equation separately gives us
$$ g = K_1 e^{-lambda v t} $$
$$ f = K_2 e^{lambda x} $$
$$ Rightarrow u(x,t) = fg = K e^{lambda (x - vt)} $$
with $K_1$, $K_2$ and $K=K_1 K_2$ are constants stemming from integration.
With
$$u_0 = u(x,t=0) = K e^{lambda x}$$
one can easily see that the solution can be expressed as
$$u(x,t) = u_0(x-vt)$$
So far, so good. Here's my question: Is that the only solution of the 1+1D advection equation with constant coefficients? Is there a proof that this is the only solution?
fluid-dynamics waves mathematics differential-equations
fluid-dynamics waves mathematics differential-equations
edited Nov 9 at 17:45
Qmechanic♦
99.7k121781119
99.7k121781119
asked Nov 9 at 12:20
lemdan
949
949
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
Nov 9 at 17:19
Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
– Kyle Kanos
Nov 10 at 18:04
add a comment |
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
Nov 9 at 17:19
Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
– Kyle Kanos
Nov 10 at 18:04
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
Nov 9 at 17:19
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
Nov 9 at 17:19
Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
– Kyle Kanos
Nov 10 at 18:04
Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
– Kyle Kanos
Nov 10 at 18:04
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Yes, it is the only solution. Hints for proof:
Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.
Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
Nov 9 at 14:12
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
Nov 9 at 17:44
add a comment |
up vote
2
down vote
The equation is linear, and the solution to a linear equation in one unknown is always unique.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Yes, it is the only solution. Hints for proof:
Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.
Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
Nov 9 at 14:12
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
Nov 9 at 17:44
add a comment |
up vote
4
down vote
accepted
Yes, it is the only solution. Hints for proof:
Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.
Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
Nov 9 at 14:12
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
Nov 9 at 17:44
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Yes, it is the only solution. Hints for proof:
Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.
Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
Yes, it is the only solution. Hints for proof:
Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.
Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
edited Nov 9 at 17:45
answered Nov 9 at 12:34
Qmechanic♦
99.7k121781119
99.7k121781119
I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
Nov 9 at 14:12
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
Nov 9 at 17:44
add a comment |
I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
Nov 9 at 14:12
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
Nov 9 at 17:44
I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
Nov 9 at 14:12
I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
Nov 9 at 14:12
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
Nov 9 at 17:44
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
Nov 9 at 17:44
add a comment |
up vote
2
down vote
The equation is linear, and the solution to a linear equation in one unknown is always unique.
add a comment |
up vote
2
down vote
The equation is linear, and the solution to a linear equation in one unknown is always unique.
add a comment |
up vote
2
down vote
up vote
2
down vote
The equation is linear, and the solution to a linear equation in one unknown is always unique.
The equation is linear, and the solution to a linear equation in one unknown is always unique.
answered Nov 9 at 15:24
Chester Miller
13.9k2623
13.9k2623
add a comment |
add a comment |
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I voted to migrate this to Mathematics.
– AccidentalFourierTransform
Nov 9 at 17:19
Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
– Kyle Kanos
Nov 10 at 18:04