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Are characteristics the only solution to the advection equation in 1+1D?
Clash Royale CLAN TAG#URR8PPP
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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 $v$:
$$ fracpartial upartial t + v fracpartial upartial x = 0tag1$$
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
$$ fracpartial upartial t = f(x) fracpartial gpartial t$$
$$ fracpartial upartial x = g(t) fracpartial fpartial x
$$
Inserting into the advection equation and restructuring a little, we get
$$frac1g fracpartial gpartial t = frac1ffracpartial fpartial x = -lambda $$
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
edited 37 mins ago
Qmechanic♦
99.1k121781096
asked 6 hours ago
lemdan
877
add a comment |
up vote
3
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 $v$:
$$ fracpartial upartial t + v fracpartial upartial x = 0tag1$$
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
$$ fracpartial upartial t = f(x) fracpartial gpartial t$$
$$ fracpartial upartial x = g(t) fracpartial fpartial x
$$
Inserting into the advection equation and restructuring a little, we get
$$frac1g fracpartial gpartial t = frac1ffracpartial fpartial x = -lambda $$
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
edited 37 mins ago
Qmechanic♦
99.1k121781096
asked 6 hours ago
lemdan
877
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
1 hour ago
add a comment |
up vote
3
down vote
favorite
up vote
3
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 $v$:
$$ fracpartial upartial t + v fracpartial upartial x = 0tag1$$
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
$$ fracpartial upartial t = f(x) fracpartial gpartial t$$
$$ fracpartial upartial x = g(t) fracpartial fpartial x
$$
Inserting into the advection equation and restructuring a little, we get
$$frac1g fracpartial gpartial t = frac1ffracpartial fpartial x = -lambda $$
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
edited 37 mins ago
Qmechanic♦
99.1k121781096
asked 6 hours ago
lemdan
877
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 $v$:
$$ fracpartial upartial t + v fracpartial upartial x = 0tag1$$
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
$$ fracpartial upartial t = f(x) fracpartial gpartial t$$
$$ fracpartial upartial x = g(t) fracpartial fpartial x
$$
Inserting into the advection equation and restructuring a little, we get
$$frac1g fracpartial gpartial t = frac1ffracpartial fpartial x = -lambda $$
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 37 mins ago
Qmechanic♦
99.1k121781096
asked 6 hours ago
lemdan
877
edited 37 mins ago
Qmechanic♦
99.1k121781096
asked 6 hours ago
lemdan
877
edited 37 mins ago
Qmechanic♦
99.1k121781096
edited 37 mins ago
Qmechanic♦
99.1k121781096
edited 37 mins ago
Qmechanic♦
99.1k121781096
99.1k121781096
asked 6 hours ago
lemdan
877
asked 6 hours ago
lemdan
877
asked 6 hours ago
lemdan
877
877
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
1 hour ago
add a comment |
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
1 hour ago
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
1 hour ago
I voted to migrate this to Mathematics.
– AccidentalFourierTransform
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
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 $fracpartial upartial x^+~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
answered 5 hours ago
Qmechanic♦
99.1k121781096
I see that using $fracpartial upartial x^+ = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
4 hours ago
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
38 mins ago
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.
answered 2 hours ago
Chester Miller
13.4k2623
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
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 $fracpartial upartial x^+~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
answered 5 hours ago
Qmechanic♦
99.1k121781096
I see that using $fracpartial upartial x^+ = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
4 hours ago
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
38 mins ago
add a comment |
up vote
4
down vote
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 $fracpartial upartial x^+~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
answered 5 hours ago
Qmechanic♦
99.1k121781096
I see that using $fracpartial upartial x^+ = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
4 hours ago
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
38 mins ago
add a comment |
up vote
4
down vote
up vote
4
down vote
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 $fracpartial upartial x^+~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
answered 5 hours ago
Qmechanic♦
99.1k121781096
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 $fracpartial upartial x^+~=~0$.
Deduce that $u=u(x^-)$ is a function of $x^-$ only.
answered 5 hours ago
Qmechanic♦
99.1k121781096
edited 37 mins ago
answered 5 hours ago
Qmechanic♦
99.1k121781096
answered 5 hours ago
Qmechanic♦
99.1k121781096
answered 5 hours ago
Qmechanic♦
99.1k121781096
99.1k121781096
I see that using $fracpartial upartial x^+ = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
4 hours ago
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
38 mins ago
add a comment |
I see that using $fracpartial upartial x^+ = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
4 hours ago
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
38 mins ago
I see that using $fracpartial upartial x^+ = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
4 hours ago
I see that using $fracpartial upartial x^+ = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
– lemdan
4 hours ago
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
38 mins ago
There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
– Qmechanic♦
38 mins ago
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.
answered 2 hours ago
Chester Miller
13.4k2623
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.
answered 2 hours ago
Chester Miller
13.4k2623
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.
answered 2 hours ago
Chester Miller
13.4k2623
The equation is linear, and the solution to a linear equation in one unknown is always unique.
answered 2 hours ago
Chester Miller
13.4k2623
answered 2 hours ago
Chester Miller
13.4k2623
answered 2 hours ago
Chester Miller
13.4k2623
answered 2 hours ago
Chester Miller
13.4k2623
13.4k2623
add a comment |
add a comment |
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I voted to migrate this to Mathematics.
– AccidentalFourierTransform
1 hour ago