Mixing
Solving nonlinear 1D advection pde with MoC
$begingroup$
I would like to solve the 1D nonlinear advection equation with the method of characteristics. Here is my notation:
begin{equation}
begin{cases}
rho_t + (1+rho)rho_x = 0\
rho = rho(x,t); quad rho(x,0) = frac{1}{1+x^2}
end{cases}
end{equation}
What I have been up to is the following, using the parameter s:
begin{align}
&frac{drho}{dt} = frac{dx}{ds}frac{partial rho}{partial x} + frac{dt}{ds}frac{partial rho}{partial t} = 0\
Longrightarrow ; & frac{dt}{ds} = 1 ;quad t(0)=0 ; Longrightarrow ; t=s \
Longrightarrow ; & frac{drho}{ds} = 0 ;quad rho(0)=rho_0 ; Longrightarrow ; rho = rho_0 \
Longrightarrow ; & frac{dx}{ds}=1+rho= 1+rho_0 ;quad x(0) = f(rho_0) ; Longrightarrow ; x = (1+rho_0)s + f(rho_0)
end{align}
So that I end up with
begin{equation}
f(rho) = x - (1+rho)t ; Longrightarrow ; rho = F(x - [1+rho]t )
end{equation}
I tried to apply several techniques I found to find the solution of the above Riemann problem using the method of characteristics. I have the general form of the solution as F but I would like to have the analytic solution for this case so I can plot for several times and see the shockwave.
Note: This problem comes from this online document, chapter 11.
pde nonlinear-system characteristics
asked Jan 31 at 8:57
DashDash
111
$endgroup$
add a comment |
$begingroup$
I would like to solve the 1D nonlinear advection equation with the method of characteristics. Here is my notation:
begin{equation}
begin{cases}
rho_t + (1+rho)rho_x = 0\
rho = rho(x,t); quad rho(x,0) = frac{1}{1+x^2}
end{cases}
end{equation}
What I have been up to is the following, using the parameter s:
begin{align}
&frac{drho}{dt} = frac{dx}{ds}frac{partial rho}{partial x} + frac{dt}{ds}frac{partial rho}{partial t} = 0\
Longrightarrow ; & frac{dt}{ds} = 1 ;quad t(0)=0 ; Longrightarrow ; t=s \
Longrightarrow ; & frac{drho}{ds} = 0 ;quad rho(0)=rho_0 ; Longrightarrow ; rho = rho_0 \
Longrightarrow ; & frac{dx}{ds}=1+rho= 1+rho_0 ;quad x(0) = f(rho_0) ; Longrightarrow ; x = (1+rho_0)s + f(rho_0)
end{align}
So that I end up with
begin{equation}
f(rho) = x - (1+rho)t ; Longrightarrow ; rho = F(x - [1+rho]t )
end{equation}
I tried to apply several techniques I found to find the solution of the above Riemann problem using the method of characteristics. I have the general form of the solution as F but I would like to have the analytic solution for this case so I can plot for several times and see the shockwave.
Note: This problem comes from this online document, chapter 11.
pde nonlinear-system characteristics
asked Jan 31 at 8:57
DashDash
111
$endgroup$
$begingroup$
You must provide a condition on the "inflow" boundary, for instance $rho(0,t)=1$. If you just want to visualize the solution I can provide two lines of Wolfram code that will do the trick.
$endgroup$
– PierreCarre
Jan 31 at 9:43
$begingroup$
I think you haven't used the initial condition on $rho$ yet, which yields $rho_0 = frac{1}{1+f(rho_0)^2}$?
$endgroup$
– Christoph
Jan 31 at 9:54
$begingroup$
As J.Jacquelin did below, using the unitial condition on $rho$ leads to a cubic equation that I have no idea how to solve.
$endgroup$
– Dash
Jan 31 at 14:48
add a comment |
$begingroup$
I would like to solve the 1D nonlinear advection equation with the method of characteristics. Here is my notation:
begin{equation}
begin{cases}
rho_t + (1+rho)rho_x = 0\
rho = rho(x,t); quad rho(x,0) = frac{1}{1+x^2}
end{cases}
end{equation}
What I have been up to is the following, using the parameter s:
begin{align}
&frac{drho}{dt} = frac{dx}{ds}frac{partial rho}{partial x} + frac{dt}{ds}frac{partial rho}{partial t} = 0\
Longrightarrow ; & frac{dt}{ds} = 1 ;quad t(0)=0 ; Longrightarrow ; t=s \
Longrightarrow ; & frac{drho}{ds} = 0 ;quad rho(0)=rho_0 ; Longrightarrow ; rho = rho_0 \
Longrightarrow ; & frac{dx}{ds}=1+rho= 1+rho_0 ;quad x(0) = f(rho_0) ; Longrightarrow ; x = (1+rho_0)s + f(rho_0)
end{align}
So that I end up with
begin{equation}
f(rho) = x - (1+rho)t ; Longrightarrow ; rho = F(x - [1+rho]t )
end{equation}
I tried to apply several techniques I found to find the solution of the above Riemann problem using the method of characteristics. I have the general form of the solution as F but I would like to have the analytic solution for this case so I can plot for several times and see the shockwave.
Note: This problem comes from this online document, chapter 11.
pde nonlinear-system characteristics
asked Jan 31 at 8:57
DashDash
111
$endgroup$
I would like to solve the 1D nonlinear advection equation with the method of characteristics. Here is my notation:
begin{equation}
begin{cases}
rho_t + (1+rho)rho_x = 0\
rho = rho(x,t); quad rho(x,0) = frac{1}{1+x^2}
end{cases}
end{equation}
What I have been up to is the following, using the parameter s:
begin{align}
&frac{drho}{dt} = frac{dx}{ds}frac{partial rho}{partial x} + frac{dt}{ds}frac{partial rho}{partial t} = 0\
Longrightarrow ; & frac{dt}{ds} = 1 ;quad t(0)=0 ; Longrightarrow ; t=s \
Longrightarrow ; & frac{drho}{ds} = 0 ;quad rho(0)=rho_0 ; Longrightarrow ; rho = rho_0 \
Longrightarrow ; & frac{dx}{ds}=1+rho= 1+rho_0 ;quad x(0) = f(rho_0) ; Longrightarrow ; x = (1+rho_0)s + f(rho_0)
end{align}
So that I end up with
begin{equation}
f(rho) = x - (1+rho)t ; Longrightarrow ; rho = F(x - [1+rho]t )
end{equation}
I tried to apply several techniques I found to find the solution of the above Riemann problem using the method of characteristics. I have the general form of the solution as F but I would like to have the analytic solution for this case so I can plot for several times and see the shockwave.
Note: This problem comes from this online document, chapter 11.
pde nonlinear-system characteristics
pde nonlinear-system characteristics
asked Jan 31 at 8:57
DashDash
111
asked Jan 31 at 8:57
DashDash
111
asked Jan 31 at 8:57
DashDash
111
asked Jan 31 at 8:57
DashDash
111
asked Jan 31 at 8:57
DashDash
111
111
$begingroup$
You must provide a condition on the "inflow" boundary, for instance $rho(0,t)=1$. If you just want to visualize the solution I can provide two lines of Wolfram code that will do the trick.
$endgroup$
– PierreCarre
Jan 31 at 9:43
$begingroup$
I think you haven't used the initial condition on $rho$ yet, which yields $rho_0 = frac{1}{1+f(rho_0)^2}$?
$endgroup$
– Christoph
Jan 31 at 9:54
$begingroup$
As J.Jacquelin did below, using the unitial condition on $rho$ leads to a cubic equation that I have no idea how to solve.
$endgroup$
– Dash
Jan 31 at 14:48
add a comment |
$begingroup$
You must provide a condition on the "inflow" boundary, for instance $rho(0,t)=1$. If you just want to visualize the solution I can provide two lines of Wolfram code that will do the trick.
$endgroup$
– PierreCarre
Jan 31 at 9:43
$begingroup$
I think you haven't used the initial condition on $rho$ yet, which yields $rho_0 = frac{1}{1+f(rho_0)^2}$?
$endgroup$
– Christoph
Jan 31 at 9:54
$begingroup$
As J.Jacquelin did below, using the unitial condition on $rho$ leads to a cubic equation that I have no idea how to solve.
$endgroup$
– Dash
Jan 31 at 14:48
$begingroup$
You must provide a condition on the "inflow" boundary, for instance $rho(0,t)=1$. If you just want to visualize the solution I can provide two lines of Wolfram code that will do the trick.
$endgroup$
– PierreCarre
Jan 31 at 9:43
$begingroup$
You must provide a condition on the "inflow" boundary, for instance $rho(0,t)=1$. If you just want to visualize the solution I can provide two lines of Wolfram code that will do the trick.
$endgroup$
– PierreCarre
Jan 31 at 9:43
$begingroup$
I think you haven't used the initial condition on $rho$ yet, which yields $rho_0 = frac{1}{1+f(rho_0)^2}$?
$endgroup$
– Christoph
Jan 31 at 9:54
$begingroup$
I think you haven't used the initial condition on $rho$ yet, which yields $rho_0 = frac{1}{1+f(rho_0)^2}$?
$endgroup$
– Christoph
Jan 31 at 9:54
$begingroup$
As J.Jacquelin did below, using the unitial condition on $rho$ leads to a cubic equation that I have no idea how to solve.
$endgroup$
– Dash
Jan 31 at 14:48
$begingroup$
As J.Jacquelin did below, using the unitial condition on $rho$ leads to a cubic equation that I have no idea how to solve.
$endgroup$
– Dash
Jan 31 at 14:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your calculus is correct. You found the general solution :
$$rho=Fleft(x-(1+rho)tright)$$
Condition :
$$rho(x,0)=frac{1}{1+x^2}=Fleft(x-(1+rho)0right)=Fleft(xright)$$
The function $F$ is determined :
$$F(X)=frac{1}{1+X^2}$$
We put this function into the general solution where $X=x-(1+rho)t$
$$rho=frac{1}{1+(x-(1+rho)t)^2}$$
This is the solution on implicit form.
In order to obtain the explicit solution, solve the cubic equation for $rho$ :
$$(1+(x-(1+rho)t)^2)rho-1=0$$
ADITION after the discussion in comments.
There is no difficulty to plot the figures that you saw in the document :
https://courses.physics.ucsd.edu/2011/Spring/physics221a/LECTURES/CH11_SHOCKS.pdf
For exemple, to plot the curves corresponding tu figure 11.1 :
$$x=(1+rho)tpmsqrt{frac{1}{rho}-1}$$
Plot the two branches with signs $+$ and $-$.
Inverse the axis if you want it in the standard position.
answered Jan 31 at 11:29
JJacquelinJJacquelin
45.4k21856
$endgroup$
$begingroup$
Thamks for answering. I actually came to the same conclusion a few minutes ago. My next problem is when I'm computing the $delta = b^2 - 4ac$ of this equation, which sign is undetermined because, well, in (x,t) parameters are supposed to be both varying parameters and though $delta$ is not a scalar but a space of solutions.
$endgroup$
– Dash
Jan 31 at 14:26
$begingroup$
Apologies, I had a slightly different in mind indeed.. Still you are right, the formula cannot be written explicitly. Why I would like to have a explicit formula is to plot the solution for several time step to compare with what is shown in the book I mentionned. Another contributor kindly proposed a Wolfram code to do that but I wanted to do it on gnuplot or python. You know, the old way..
$endgroup$
– Dash
Jan 31 at 15:08
$begingroup$
There is no need for an explicit formula to plot $rho(x,t)$ as a function of $t$ for given $x$. Compute the formula of $t$ as a function of $rho$ which is an explicit equation. Plot $t$ as a function of $rho$ with $t$ on horizontal axis and $rho$ on vertical axis.
$endgroup$
– JJacquelin
Jan 31 at 15:25
$begingroup$
If I rewrite the equation as begin{equation} frac{1}{rho}-1=[x−(1+ρ)t]^2 end{equation} Then I cannot isolate t from ρ because of the square
$endgroup$
– Dash
Jan 31 at 17:02
$begingroup$
Here the advection velocity depends on $rho$ itself, which is the problem when I look for an explicit analytic solution. Perhaps I should quit looking for a beautiful formula, but do you have an idea of how they did in fig. 11 of the book to plot the solution ?
$endgroup$
– Dash
Jan 31 at 18:55
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
Your calculus is correct. You found the general solution :
$$rho=Fleft(x-(1+rho)tright)$$
Condition :
$$rho(x,0)=frac{1}{1+x^2}=Fleft(x-(1+rho)0right)=Fleft(xright)$$
The function $F$ is determined :
$$F(X)=frac{1}{1+X^2}$$
We put this function into the general solution where $X=x-(1+rho)t$
$$rho=frac{1}{1+(x-(1+rho)t)^2}$$
This is the solution on implicit form.
In order to obtain the explicit solution, solve the cubic equation for $rho$ :
$$(1+(x-(1+rho)t)^2)rho-1=0$$
ADITION after the discussion in comments.
There is no difficulty to plot the figures that you saw in the document :
https://courses.physics.ucsd.edu/2011/Spring/physics221a/LECTURES/CH11_SHOCKS.pdf
For exemple, to plot the curves corresponding tu figure 11.1 :
$$x=(1+rho)tpmsqrt{frac{1}{rho}-1}$$
Plot the two branches with signs $+$ and $-$.
Inverse the axis if you want it in the standard position.
answered Jan 31 at 11:29
JJacquelinJJacquelin
45.4k21856
$endgroup$
$begingroup$
Thamks for answering. I actually came to the same conclusion a few minutes ago. My next problem is when I'm computing the $delta = b^2 - 4ac$ of this equation, which sign is undetermined because, well, in (x,t) parameters are supposed to be both varying parameters and though $delta$ is not a scalar but a space of solutions.
$endgroup$
– Dash
Jan 31 at 14:26
$begingroup$
Apologies, I had a slightly different in mind indeed.. Still you are right, the formula cannot be written explicitly. Why I would like to have a explicit formula is to plot the solution for several time step to compare with what is shown in the book I mentionned. Another contributor kindly proposed a Wolfram code to do that but I wanted to do it on gnuplot or python. You know, the old way..
$endgroup$
– Dash
Jan 31 at 15:08
$begingroup$
There is no need for an explicit formula to plot $rho(x,t)$ as a function of $t$ for given $x$. Compute the formula of $t$ as a function of $rho$ which is an explicit equation. Plot $t$ as a function of $rho$ with $t$ on horizontal axis and $rho$ on vertical axis.
$endgroup$
– JJacquelin
Jan 31 at 15:25
$begingroup$
If I rewrite the equation as begin{equation} frac{1}{rho}-1=[x−(1+ρ)t]^2 end{equation} Then I cannot isolate t from ρ because of the square
$endgroup$
– Dash
Jan 31 at 17:02
$begingroup$
Here the advection velocity depends on $rho$ itself, which is the problem when I look for an explicit analytic solution. Perhaps I should quit looking for a beautiful formula, but do you have an idea of how they did in fig. 11 of the book to plot the solution ?
$endgroup$
– Dash
Jan 31 at 18:55
|
show 2 more comments
$begingroup$
Your calculus is correct. You found the general solution :
$$rho=Fleft(x-(1+rho)tright)$$
Condition :
$$rho(x,0)=frac{1}{1+x^2}=Fleft(x-(1+rho)0right)=Fleft(xright)$$
The function $F$ is determined :
$$F(X)=frac{1}{1+X^2}$$
We put this function into the general solution where $X=x-(1+rho)t$
$$rho=frac{1}{1+(x-(1+rho)t)^2}$$
This is the solution on implicit form.
In order to obtain the explicit solution, solve the cubic equation for $rho$ :
$$(1+(x-(1+rho)t)^2)rho-1=0$$
ADITION after the discussion in comments.
There is no difficulty to plot the figures that you saw in the document :
https://courses.physics.ucsd.edu/2011/Spring/physics221a/LECTURES/CH11_SHOCKS.pdf
For exemple, to plot the curves corresponding tu figure 11.1 :
$$x=(1+rho)tpmsqrt{frac{1}{rho}-1}$$
Plot the two branches with signs $+$ and $-$.
Inverse the axis if you want it in the standard position.
answered Jan 31 at 11:29
JJacquelinJJacquelin
45.4k21856
$endgroup$
$begingroup$
Thamks for answering. I actually came to the same conclusion a few minutes ago. My next problem is when I'm computing the $delta = b^2 - 4ac$ of this equation, which sign is undetermined because, well, in (x,t) parameters are supposed to be both varying parameters and though $delta$ is not a scalar but a space of solutions.
$endgroup$
– Dash
Jan 31 at 14:26
$begingroup$
Apologies, I had a slightly different in mind indeed.. Still you are right, the formula cannot be written explicitly. Why I would like to have a explicit formula is to plot the solution for several time step to compare with what is shown in the book I mentionned. Another contributor kindly proposed a Wolfram code to do that but I wanted to do it on gnuplot or python. You know, the old way..
$endgroup$
– Dash
Jan 31 at 15:08
$begingroup$
There is no need for an explicit formula to plot $rho(x,t)$ as a function of $t$ for given $x$. Compute the formula of $t$ as a function of $rho$ which is an explicit equation. Plot $t$ as a function of $rho$ with $t$ on horizontal axis and $rho$ on vertical axis.
$endgroup$
– JJacquelin
Jan 31 at 15:25
$begingroup$
If I rewrite the equation as begin{equation} frac{1}{rho}-1=[x−(1+ρ)t]^2 end{equation} Then I cannot isolate t from ρ because of the square
$endgroup$
– Dash
Jan 31 at 17:02
$begingroup$
Here the advection velocity depends on $rho$ itself, which is the problem when I look for an explicit analytic solution. Perhaps I should quit looking for a beautiful formula, but do you have an idea of how they did in fig. 11 of the book to plot the solution ?
$endgroup$
– Dash
Jan 31 at 18:55
|
show 2 more comments
$begingroup$
Your calculus is correct. You found the general solution :
$$rho=Fleft(x-(1+rho)tright)$$
Condition :
$$rho(x,0)=frac{1}{1+x^2}=Fleft(x-(1+rho)0right)=Fleft(xright)$$
The function $F$ is determined :
$$F(X)=frac{1}{1+X^2}$$
We put this function into the general solution where $X=x-(1+rho)t$
$$rho=frac{1}{1+(x-(1+rho)t)^2}$$
This is the solution on implicit form.
In order to obtain the explicit solution, solve the cubic equation for $rho$ :
$$(1+(x-(1+rho)t)^2)rho-1=0$$
ADITION after the discussion in comments.
There is no difficulty to plot the figures that you saw in the document :
https://courses.physics.ucsd.edu/2011/Spring/physics221a/LECTURES/CH11_SHOCKS.pdf
For exemple, to plot the curves corresponding tu figure 11.1 :
$$x=(1+rho)tpmsqrt{frac{1}{rho}-1}$$
Plot the two branches with signs $+$ and $-$.
Inverse the axis if you want it in the standard position.
answered Jan 31 at 11:29
JJacquelinJJacquelin
45.4k21856
$endgroup$
Your calculus is correct. You found the general solution :
$$rho=Fleft(x-(1+rho)tright)$$
Condition :
$$rho(x,0)=frac{1}{1+x^2}=Fleft(x-(1+rho)0right)=Fleft(xright)$$
The function $F$ is determined :
$$F(X)=frac{1}{1+X^2}$$
We put this function into the general solution where $X=x-(1+rho)t$
$$rho=frac{1}{1+(x-(1+rho)t)^2}$$
This is the solution on implicit form.
In order to obtain the explicit solution, solve the cubic equation for $rho$ :
$$(1+(x-(1+rho)t)^2)rho-1=0$$
ADITION after the discussion in comments.
There is no difficulty to plot the figures that you saw in the document :
https://courses.physics.ucsd.edu/2011/Spring/physics221a/LECTURES/CH11_SHOCKS.pdf
For exemple, to plot the curves corresponding tu figure 11.1 :
$$x=(1+rho)tpmsqrt{frac{1}{rho}-1}$$
Plot the two branches with signs $+$ and $-$.
Inverse the axis if you want it in the standard position.
answered Jan 31 at 11:29
JJacquelinJJacquelin
45.4k21856
edited Feb 1 at 16:33
answered Jan 31 at 11:29
JJacquelinJJacquelin
45.4k21856
answered Jan 31 at 11:29
JJacquelinJJacquelin
45.4k21856
answered Jan 31 at 11:29
JJacquelinJJacquelin
45.4k21856
45.4k21856
$begingroup$
Thamks for answering. I actually came to the same conclusion a few minutes ago. My next problem is when I'm computing the $delta = b^2 - 4ac$ of this equation, which sign is undetermined because, well, in (x,t) parameters are supposed to be both varying parameters and though $delta$ is not a scalar but a space of solutions.
$endgroup$
– Dash
Jan 31 at 14:26
$begingroup$
Apologies, I had a slightly different in mind indeed.. Still you are right, the formula cannot be written explicitly. Why I would like to have a explicit formula is to plot the solution for several time step to compare with what is shown in the book I mentionned. Another contributor kindly proposed a Wolfram code to do that but I wanted to do it on gnuplot or python. You know, the old way..
$endgroup$
– Dash
Jan 31 at 15:08
$begingroup$
There is no need for an explicit formula to plot $rho(x,t)$ as a function of $t$ for given $x$. Compute the formula of $t$ as a function of $rho$ which is an explicit equation. Plot $t$ as a function of $rho$ with $t$ on horizontal axis and $rho$ on vertical axis.
$endgroup$
– JJacquelin
Jan 31 at 15:25
$begingroup$
If I rewrite the equation as begin{equation} frac{1}{rho}-1=[x−(1+ρ)t]^2 end{equation} Then I cannot isolate t from ρ because of the square
$endgroup$
– Dash
Jan 31 at 17:02
$begingroup$
Here the advection velocity depends on $rho$ itself, which is the problem when I look for an explicit analytic solution. Perhaps I should quit looking for a beautiful formula, but do you have an idea of how they did in fig. 11 of the book to plot the solution ?
$endgroup$
– Dash
Jan 31 at 18:55
|
show 2 more comments
$begingroup$
Thamks for answering. I actually came to the same conclusion a few minutes ago. My next problem is when I'm computing the $delta = b^2 - 4ac$ of this equation, which sign is undetermined because, well, in (x,t) parameters are supposed to be both varying parameters and though $delta$ is not a scalar but a space of solutions.
$endgroup$
– Dash
Jan 31 at 14:26
$begingroup$
Apologies, I had a slightly different in mind indeed.. Still you are right, the formula cannot be written explicitly. Why I would like to have a explicit formula is to plot the solution for several time step to compare with what is shown in the book I mentionned. Another contributor kindly proposed a Wolfram code to do that but I wanted to do it on gnuplot or python. You know, the old way..
$endgroup$
– Dash
Jan 31 at 15:08
$begingroup$
There is no need for an explicit formula to plot $rho(x,t)$ as a function of $t$ for given $x$. Compute the formula of $t$ as a function of $rho$ which is an explicit equation. Plot $t$ as a function of $rho$ with $t$ on horizontal axis and $rho$ on vertical axis.
$endgroup$
– JJacquelin
Jan 31 at 15:25
$begingroup$
If I rewrite the equation as begin{equation} frac{1}{rho}-1=[x−(1+ρ)t]^2 end{equation} Then I cannot isolate t from ρ because of the square
$endgroup$
– Dash
Jan 31 at 17:02
$begingroup$
Here the advection velocity depends on $rho$ itself, which is the problem when I look for an explicit analytic solution. Perhaps I should quit looking for a beautiful formula, but do you have an idea of how they did in fig. 11 of the book to plot the solution ?
$endgroup$
– Dash
Jan 31 at 18:55
$begingroup$
Thamks for answering. I actually came to the same conclusion a few minutes ago. My next problem is when I'm computing the $delta = b^2 - 4ac$ of this equation, which sign is undetermined because, well, in (x,t) parameters are supposed to be both varying parameters and though $delta$ is not a scalar but a space of solutions.
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– Dash
Jan 31 at 14:26
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Thamks for answering. I actually came to the same conclusion a few minutes ago. My next problem is when I'm computing the $delta = b^2 - 4ac$ of this equation, which sign is undetermined because, well, in (x,t) parameters are supposed to be both varying parameters and though $delta$ is not a scalar but a space of solutions.
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– Dash
Jan 31 at 14:26
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Apologies, I had a slightly different in mind indeed.. Still you are right, the formula cannot be written explicitly. Why I would like to have a explicit formula is to plot the solution for several time step to compare with what is shown in the book I mentionned. Another contributor kindly proposed a Wolfram code to do that but I wanted to do it on gnuplot or python. You know, the old way..
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– Dash
Jan 31 at 15:08
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Apologies, I had a slightly different in mind indeed.. Still you are right, the formula cannot be written explicitly. Why I would like to have a explicit formula is to plot the solution for several time step to compare with what is shown in the book I mentionned. Another contributor kindly proposed a Wolfram code to do that but I wanted to do it on gnuplot or python. You know, the old way..
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– Dash
Jan 31 at 15:08
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There is no need for an explicit formula to plot $rho(x,t)$ as a function of $t$ for given $x$. Compute the formula of $t$ as a function of $rho$ which is an explicit equation. Plot $t$ as a function of $rho$ with $t$ on horizontal axis and $rho$ on vertical axis.
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– JJacquelin
Jan 31 at 15:25
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There is no need for an explicit formula to plot $rho(x,t)$ as a function of $t$ for given $x$. Compute the formula of $t$ as a function of $rho$ which is an explicit equation. Plot $t$ as a function of $rho$ with $t$ on horizontal axis and $rho$ on vertical axis.
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– JJacquelin
Jan 31 at 15:25
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If I rewrite the equation as begin{equation} frac{1}{rho}-1=[x−(1+ρ)t]^2 end{equation} Then I cannot isolate t from ρ because of the square
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– Dash
Jan 31 at 17:02
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If I rewrite the equation as begin{equation} frac{1}{rho}-1=[x−(1+ρ)t]^2 end{equation} Then I cannot isolate t from ρ because of the square
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– Dash
Jan 31 at 17:02
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Here the advection velocity depends on $rho$ itself, which is the problem when I look for an explicit analytic solution. Perhaps I should quit looking for a beautiful formula, but do you have an idea of how they did in fig. 11 of the book to plot the solution ?
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– Dash
Jan 31 at 18:55
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Here the advection velocity depends on $rho$ itself, which is the problem when I look for an explicit analytic solution. Perhaps I should quit looking for a beautiful formula, but do you have an idea of how they did in fig. 11 of the book to plot the solution ?
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– Dash
Jan 31 at 18:55
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You must provide a condition on the "inflow" boundary, for instance $rho(0,t)=1$. If you just want to visualize the solution I can provide two lines of Wolfram code that will do the trick.
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– PierreCarre
Jan 31 at 9:43
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I think you haven't used the initial condition on $rho$ yet, which yields $rho_0 = frac{1}{1+f(rho_0)^2}$?
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– Christoph
Jan 31 at 9:54
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As J.Jacquelin did below, using the unitial condition on $rho$ leads to a cubic equation that I have no idea how to solve.
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– Dash
Jan 31 at 14:48