Solution to equations involving Plasma dispersion function
I am trying to solve an equation involving a complex argument for the plasma dispersion function as:
$z = x + iota y$,
$ x = omega / sqrt2 k v_{Ti} $
$ y = nu_i /sqrt{2} k v_{Ti} $
$S[z] = -iota sqrt{pi} Exp[-z^2] Erfc[-iota z]$;
I have reduced the equation to a system of equations corresponding to real part and imaginary part as:
$(x - beta)(c_1^2 + c_2^2) + nu_e(d_2c_2 - d_1c_1)=0$,
$gamma(c_1^2 + c_2^2) + nu_e(d_2c_1 - d_1c_2)=0$
I am not able to find how should i solve this system for omega and gamma to obtain plots for $omega$ v/s k and $gamma$ v/s k? $c_1, c_2, d_1, d_2$ are all functions of S(z). $beta, nu_e, nu_i, v_{Ti}$ are the parameters whose values are known.
complex-analysis numerical-methods special-functions
edited Jun 28 '16 at 15:12
J. M. is not a mathematician
60.8k5150289
asked Jan 29 '13 at 21:45
user60149user60149
62
add a comment |
I am trying to solve an equation involving a complex argument for the plasma dispersion function as:
$z = x + iota y$,
$ x = omega / sqrt2 k v_{Ti} $
$ y = nu_i /sqrt{2} k v_{Ti} $
$S[z] = -iota sqrt{pi} Exp[-z^2] Erfc[-iota z]$;
I have reduced the equation to a system of equations corresponding to real part and imaginary part as:
$(x - beta)(c_1^2 + c_2^2) + nu_e(d_2c_2 - d_1c_1)=0$,
$gamma(c_1^2 + c_2^2) + nu_e(d_2c_1 - d_1c_2)=0$
I am not able to find how should i solve this system for omega and gamma to obtain plots for $omega$ v/s k and $gamma$ v/s k? $c_1, c_2, d_1, d_2$ are all functions of S(z). $beta, nu_e, nu_i, v_{Ti}$ are the parameters whose values are known.
complex-analysis numerical-methods special-functions
edited Jun 28 '16 at 15:12
J. M. is not a mathematician
60.8k5150289
asked Jan 29 '13 at 21:45
user60149user60149
62
2
Welcome to MSE! Can you please format your question using MathJax. It improves readability and may get more responses. Regards
– Amzoti
Jan 29 '13 at 21:46
That looks a lot better, but there is still no $omega$ in your reduced equations?
– mrf
Jan 29 '13 at 23:22
add a comment |
I am trying to solve an equation involving a complex argument for the plasma dispersion function as:
$z = x + iota y$,
$ x = omega / sqrt2 k v_{Ti} $
$ y = nu_i /sqrt{2} k v_{Ti} $
$S[z] = -iota sqrt{pi} Exp[-z^2] Erfc[-iota z]$;
I have reduced the equation to a system of equations corresponding to real part and imaginary part as:
$(x - beta)(c_1^2 + c_2^2) + nu_e(d_2c_2 - d_1c_1)=0$,
$gamma(c_1^2 + c_2^2) + nu_e(d_2c_1 - d_1c_2)=0$
I am not able to find how should i solve this system for omega and gamma to obtain plots for $omega$ v/s k and $gamma$ v/s k? $c_1, c_2, d_1, d_2$ are all functions of S(z). $beta, nu_e, nu_i, v_{Ti}$ are the parameters whose values are known.
complex-analysis numerical-methods special-functions
edited Jun 28 '16 at 15:12
J. M. is not a mathematician
60.8k5150289
asked Jan 29 '13 at 21:45
user60149user60149
62
I am trying to solve an equation involving a complex argument for the plasma dispersion function as:
$z = x + iota y$,
$ x = omega / sqrt2 k v_{Ti} $
$ y = nu_i /sqrt{2} k v_{Ti} $
$S[z] = -iota sqrt{pi} Exp[-z^2] Erfc[-iota z]$;
I have reduced the equation to a system of equations corresponding to real part and imaginary part as:
$(x - beta)(c_1^2 + c_2^2) + nu_e(d_2c_2 - d_1c_1)=0$,
$gamma(c_1^2 + c_2^2) + nu_e(d_2c_1 - d_1c_2)=0$
I am not able to find how should i solve this system for omega and gamma to obtain plots for $omega$ v/s k and $gamma$ v/s k? $c_1, c_2, d_1, d_2$ are all functions of S(z). $beta, nu_e, nu_i, v_{Ti}$ are the parameters whose values are known.
complex-analysis numerical-methods special-functions
complex-analysis numerical-methods special-functions
edited Jun 28 '16 at 15:12
J. M. is not a mathematician
60.8k5150289
asked Jan 29 '13 at 21:45
user60149user60149
62
edited Jun 28 '16 at 15:12
J. M. is not a mathematician
60.8k5150289
asked Jan 29 '13 at 21:45
user60149user60149
62
edited Jun 28 '16 at 15:12
J. M. is not a mathematician
60.8k5150289
edited Jun 28 '16 at 15:12
J. M. is not a mathematician
60.8k5150289
edited Jun 28 '16 at 15:12
J. M. is not a mathematician
60.8k5150289
60.8k5150289
asked Jan 29 '13 at 21:45
user60149user60149
62
asked Jan 29 '13 at 21:45
user60149user60149
62
asked Jan 29 '13 at 21:45
user60149user60149
62
62
2
Welcome to MSE! Can you please format your question using MathJax. It improves readability and may get more responses. Regards
– Amzoti
Jan 29 '13 at 21:46
That looks a lot better, but there is still no $omega$ in your reduced equations?
– mrf
Jan 29 '13 at 23:22
add a comment |
2
Welcome to MSE! Can you please format your question using MathJax. It improves readability and may get more responses. Regards
– Amzoti
Jan 29 '13 at 21:46
That looks a lot better, but there is still no $omega$ in your reduced equations?
– mrf
Jan 29 '13 at 23:22
2
2
Welcome to MSE! Can you please format your question using MathJax. It improves readability and may get more responses. Regards
– Amzoti
Jan 29 '13 at 21:46
Welcome to MSE! Can you please format your question using MathJax. It improves readability and may get more responses. Regards
– Amzoti
Jan 29 '13 at 21:46
That looks a lot better, but there is still no $omega$ in your reduced equations?
– mrf
Jan 29 '13 at 23:22
That looks a lot better, but there is still no $omega$ in your reduced equations?
– mrf
Jan 29 '13 at 23:22
add a comment |
1 Answer
1
active
oldest
votes
The most simple and usually best way to solve for this is via the Newton-Rhapson Method, which works even for complex variables; so, you do not need to separate the real and imaginary part.
For example, suppose you want to solve for the roots of a complex function $W(x,y)$, given by
$$W(x,y) = -y^2 - x + frac x{ysqrtbeta} S(xi),,$$
where $xi = (x-1)/y sqrtbeta$, $beta$ and $y$ are real numbers and $x$ is the complex variable.
Then, for $y=y_0$, we will look for $x=x(y_0)$. Choose $x=z_n$ as an initial seed for the Newton's method.
$$z_{n+1} = -frac{ W(z_n, y_0) }{ W'(z_n,y_0) },,$$
where $W'(x,y) = partial W/partial x$. Iterate until $n=N$ where the last equation converges, so that $x(y_0) = z_{N}$.
Now, choose $y=y_1$ and suppose $x=z_n=x(y_0)$ is a good seed. Through the Newton Method you now have $x=x(y_1)$.
Iterate this procedure until some $y_M$.
In the software Mathematica, I have implemented this code:
rama[s_,ys_]:=
(
data={};
xn=s;
Table[
xn=x/.FindRoot[W[x,yn],{x,xn+0. I}];
AppendTo[data,{yn, Re[xn],Im[xn]}];
,{yn,ys,0.02,-0.001}];
Export["data.dat", data]
)
You need to choose the first seed $x=s$ and $y=y_s$. Then, the code will find the solutions $x=x(y)$ for $0.02<y<y_s$ in steps of $-0.001$, and the results will be saved in a file with columns $y$ $Re(x)$ $Im(x)$. FindRoot is a function of Mathematica which uses (I hope) the Newton Method.
P.D.: $W(x,y)=0$ represents the dispersion relation of electromagnetic waves propagating along a background magnetic field in a plasma of electrons and protons, where $beta$ is the beta-parameter, $y$ is the normalized wave-number and $x=x(y)$ is the normalized complex frequency.
answered Aug 24 '13 at 5:32
vagobertovagoberto
1344
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The most simple and usually best way to solve for this is via the Newton-Rhapson Method, which works even for complex variables; so, you do not need to separate the real and imaginary part.
For example, suppose you want to solve for the roots of a complex function $W(x,y)$, given by
$$W(x,y) = -y^2 - x + frac x{ysqrtbeta} S(xi),,$$
where $xi = (x-1)/y sqrtbeta$, $beta$ and $y$ are real numbers and $x$ is the complex variable.
Then, for $y=y_0$, we will look for $x=x(y_0)$. Choose $x=z_n$ as an initial seed for the Newton's method.
$$z_{n+1} = -frac{ W(z_n, y_0) }{ W'(z_n,y_0) },,$$
where $W'(x,y) = partial W/partial x$. Iterate until $n=N$ where the last equation converges, so that $x(y_0) = z_{N}$.
Now, choose $y=y_1$ and suppose $x=z_n=x(y_0)$ is a good seed. Through the Newton Method you now have $x=x(y_1)$.
Iterate this procedure until some $y_M$.
In the software Mathematica, I have implemented this code:
rama[s_,ys_]:=
(
data={};
xn=s;
Table[
xn=x/.FindRoot[W[x,yn],{x,xn+0. I}];
AppendTo[data,{yn, Re[xn],Im[xn]}];
,{yn,ys,0.02,-0.001}];
Export["data.dat", data]
)
You need to choose the first seed $x=s$ and $y=y_s$. Then, the code will find the solutions $x=x(y)$ for $0.02<y<y_s$ in steps of $-0.001$, and the results will be saved in a file with columns $y$ $Re(x)$ $Im(x)$. FindRoot is a function of Mathematica which uses (I hope) the Newton Method.
P.D.: $W(x,y)=0$ represents the dispersion relation of electromagnetic waves propagating along a background magnetic field in a plasma of electrons and protons, where $beta$ is the beta-parameter, $y$ is the normalized wave-number and $x=x(y)$ is the normalized complex frequency.
answered Aug 24 '13 at 5:32
vagobertovagoberto
1344
add a comment |
The most simple and usually best way to solve for this is via the Newton-Rhapson Method, which works even for complex variables; so, you do not need to separate the real and imaginary part.
For example, suppose you want to solve for the roots of a complex function $W(x,y)$, given by
$$W(x,y) = -y^2 - x + frac x{ysqrtbeta} S(xi),,$$
where $xi = (x-1)/y sqrtbeta$, $beta$ and $y$ are real numbers and $x$ is the complex variable.
Then, for $y=y_0$, we will look for $x=x(y_0)$. Choose $x=z_n$ as an initial seed for the Newton's method.
$$z_{n+1} = -frac{ W(z_n, y_0) }{ W'(z_n,y_0) },,$$
where $W'(x,y) = partial W/partial x$. Iterate until $n=N$ where the last equation converges, so that $x(y_0) = z_{N}$.
Now, choose $y=y_1$ and suppose $x=z_n=x(y_0)$ is a good seed. Through the Newton Method you now have $x=x(y_1)$.
Iterate this procedure until some $y_M$.
In the software Mathematica, I have implemented this code:
rama[s_,ys_]:=
(
data={};
xn=s;
Table[
xn=x/.FindRoot[W[x,yn],{x,xn+0. I}];
AppendTo[data,{yn, Re[xn],Im[xn]}];
,{yn,ys,0.02,-0.001}];
Export["data.dat", data]
)
You need to choose the first seed $x=s$ and $y=y_s$. Then, the code will find the solutions $x=x(y)$ for $0.02<y<y_s$ in steps of $-0.001$, and the results will be saved in a file with columns $y$ $Re(x)$ $Im(x)$. FindRoot is a function of Mathematica which uses (I hope) the Newton Method.
P.D.: $W(x,y)=0$ represents the dispersion relation of electromagnetic waves propagating along a background magnetic field in a plasma of electrons and protons, where $beta$ is the beta-parameter, $y$ is the normalized wave-number and $x=x(y)$ is the normalized complex frequency.
answered Aug 24 '13 at 5:32
vagobertovagoberto
1344
add a comment |
The most simple and usually best way to solve for this is via the Newton-Rhapson Method, which works even for complex variables; so, you do not need to separate the real and imaginary part.
For example, suppose you want to solve for the roots of a complex function $W(x,y)$, given by
$$W(x,y) = -y^2 - x + frac x{ysqrtbeta} S(xi),,$$
where $xi = (x-1)/y sqrtbeta$, $beta$ and $y$ are real numbers and $x$ is the complex variable.
Then, for $y=y_0$, we will look for $x=x(y_0)$. Choose $x=z_n$ as an initial seed for the Newton's method.
$$z_{n+1} = -frac{ W(z_n, y_0) }{ W'(z_n,y_0) },,$$
where $W'(x,y) = partial W/partial x$. Iterate until $n=N$ where the last equation converges, so that $x(y_0) = z_{N}$.
Now, choose $y=y_1$ and suppose $x=z_n=x(y_0)$ is a good seed. Through the Newton Method you now have $x=x(y_1)$.
Iterate this procedure until some $y_M$.
In the software Mathematica, I have implemented this code:
rama[s_,ys_]:=
(
data={};
xn=s;
Table[
xn=x/.FindRoot[W[x,yn],{x,xn+0. I}];
AppendTo[data,{yn, Re[xn],Im[xn]}];
,{yn,ys,0.02,-0.001}];
Export["data.dat", data]
)
You need to choose the first seed $x=s$ and $y=y_s$. Then, the code will find the solutions $x=x(y)$ for $0.02<y<y_s$ in steps of $-0.001$, and the results will be saved in a file with columns $y$ $Re(x)$ $Im(x)$. FindRoot is a function of Mathematica which uses (I hope) the Newton Method.
P.D.: $W(x,y)=0$ represents the dispersion relation of electromagnetic waves propagating along a background magnetic field in a plasma of electrons and protons, where $beta$ is the beta-parameter, $y$ is the normalized wave-number and $x=x(y)$ is the normalized complex frequency.
answered Aug 24 '13 at 5:32
vagobertovagoberto
1344
The most simple and usually best way to solve for this is via the Newton-Rhapson Method, which works even for complex variables; so, you do not need to separate the real and imaginary part.
For example, suppose you want to solve for the roots of a complex function $W(x,y)$, given by
$$W(x,y) = -y^2 - x + frac x{ysqrtbeta} S(xi),,$$
where $xi = (x-1)/y sqrtbeta$, $beta$ and $y$ are real numbers and $x$ is the complex variable.
Then, for $y=y_0$, we will look for $x=x(y_0)$. Choose $x=z_n$ as an initial seed for the Newton's method.
$$z_{n+1} = -frac{ W(z_n, y_0) }{ W'(z_n,y_0) },,$$
where $W'(x,y) = partial W/partial x$. Iterate until $n=N$ where the last equation converges, so that $x(y_0) = z_{N}$.
Now, choose $y=y_1$ and suppose $x=z_n=x(y_0)$ is a good seed. Through the Newton Method you now have $x=x(y_1)$.
Iterate this procedure until some $y_M$.
In the software Mathematica, I have implemented this code:
rama[s_,ys_]:=
(
data={};
xn=s;
Table[
xn=x/.FindRoot[W[x,yn],{x,xn+0. I}];
AppendTo[data,{yn, Re[xn],Im[xn]}];
,{yn,ys,0.02,-0.001}];
Export["data.dat", data]
)
You need to choose the first seed $x=s$ and $y=y_s$. Then, the code will find the solutions $x=x(y)$ for $0.02<y<y_s$ in steps of $-0.001$, and the results will be saved in a file with columns $y$ $Re(x)$ $Im(x)$. FindRoot is a function of Mathematica which uses (I hope) the Newton Method.
P.D.: $W(x,y)=0$ represents the dispersion relation of electromagnetic waves propagating along a background magnetic field in a plasma of electrons and protons, where $beta$ is the beta-parameter, $y$ is the normalized wave-number and $x=x(y)$ is the normalized complex frequency.
answered Aug 24 '13 at 5:32
vagobertovagoberto
1344
answered Aug 24 '13 at 5:32
vagobertovagoberto
1344
answered Aug 24 '13 at 5:32
vagobertovagoberto
1344
answered Aug 24 '13 at 5:32
vagobertovagoberto
1344
1344
add a comment |
add a comment |
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2
Welcome to MSE! Can you please format your question using MathJax. It improves readability and may get more responses. Regards
– Amzoti
Jan 29 '13 at 21:46
That looks a lot better, but there is still no $omega$ in your reduced equations?
– mrf
Jan 29 '13 at 23:22