Can someone find this inverse (and solve Burger's equation)?
up vote
0
down vote
favorite
My question regards the following function:
$f(x) = begin{cases}frac{x}{logleft(frac{x+1}{x-1}right)}&text{ for }x>1,\
0&text{ for x=1}end{cases}$
which arose in circumstances I will describe below. Quite simply, I'd like to know if someone could find an expression for $f^{-1}$ of any kind, be it in closed form or not (e.g., a power series or an integral).
My motivation for this question has got to do with the steady-state viscous Burger's equation
$kappaddot{w}(x) = w(x)dot{w}(x)$ with $xin[-1,1]$ and $w(-1)=-w(1) = 1$ and $kappa>0$.
After some deliberation I gathered that a solution is given by
$w(x) = ccdot frac{e^{-frac{c}{kappa} y} - 1}{e^{-frac{c}{kappa} y}+1}$
where $c$ is chosen so that the boundary conditions hold, which is equivalent to, lo and behold,
$kappa = frac{c}{logleft(frac{c+1}{c-1}right)}$.
Also, references to relevant literature are appreciated. I have mostly relied on Wolfram's articles on the (inverse) hyperbolic tangent:
http://mathworld.wolfram.com/HyperbolicTangent.html
and
http://mathworld.wolfram.com/InverseHyperbolicTangent.html
real-analysis differential-equations
add a comment |
up vote
0
down vote
favorite
My question regards the following function:
$f(x) = begin{cases}frac{x}{logleft(frac{x+1}{x-1}right)}&text{ for }x>1,\
0&text{ for x=1}end{cases}$
which arose in circumstances I will describe below. Quite simply, I'd like to know if someone could find an expression for $f^{-1}$ of any kind, be it in closed form or not (e.g., a power series or an integral).
My motivation for this question has got to do with the steady-state viscous Burger's equation
$kappaddot{w}(x) = w(x)dot{w}(x)$ with $xin[-1,1]$ and $w(-1)=-w(1) = 1$ and $kappa>0$.
After some deliberation I gathered that a solution is given by
$w(x) = ccdot frac{e^{-frac{c}{kappa} y} - 1}{e^{-frac{c}{kappa} y}+1}$
where $c$ is chosen so that the boundary conditions hold, which is equivalent to, lo and behold,
$kappa = frac{c}{logleft(frac{c+1}{c-1}right)}$.
Also, references to relevant literature are appreciated. I have mostly relied on Wolfram's articles on the (inverse) hyperbolic tangent:
http://mathworld.wolfram.com/HyperbolicTangent.html
and
http://mathworld.wolfram.com/InverseHyperbolicTangent.html
real-analysis differential-equations
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My question regards the following function:
$f(x) = begin{cases}frac{x}{logleft(frac{x+1}{x-1}right)}&text{ for }x>1,\
0&text{ for x=1}end{cases}$
which arose in circumstances I will describe below. Quite simply, I'd like to know if someone could find an expression for $f^{-1}$ of any kind, be it in closed form or not (e.g., a power series or an integral).
My motivation for this question has got to do with the steady-state viscous Burger's equation
$kappaddot{w}(x) = w(x)dot{w}(x)$ with $xin[-1,1]$ and $w(-1)=-w(1) = 1$ and $kappa>0$.
After some deliberation I gathered that a solution is given by
$w(x) = ccdot frac{e^{-frac{c}{kappa} y} - 1}{e^{-frac{c}{kappa} y}+1}$
where $c$ is chosen so that the boundary conditions hold, which is equivalent to, lo and behold,
$kappa = frac{c}{logleft(frac{c+1}{c-1}right)}$.
Also, references to relevant literature are appreciated. I have mostly relied on Wolfram's articles on the (inverse) hyperbolic tangent:
http://mathworld.wolfram.com/HyperbolicTangent.html
and
http://mathworld.wolfram.com/InverseHyperbolicTangent.html
real-analysis differential-equations
My question regards the following function:
$f(x) = begin{cases}frac{x}{logleft(frac{x+1}{x-1}right)}&text{ for }x>1,\
0&text{ for x=1}end{cases}$
which arose in circumstances I will describe below. Quite simply, I'd like to know if someone could find an expression for $f^{-1}$ of any kind, be it in closed form or not (e.g., a power series or an integral).
My motivation for this question has got to do with the steady-state viscous Burger's equation
$kappaddot{w}(x) = w(x)dot{w}(x)$ with $xin[-1,1]$ and $w(-1)=-w(1) = 1$ and $kappa>0$.
After some deliberation I gathered that a solution is given by
$w(x) = ccdot frac{e^{-frac{c}{kappa} y} - 1}{e^{-frac{c}{kappa} y}+1}$
where $c$ is chosen so that the boundary conditions hold, which is equivalent to, lo and behold,
$kappa = frac{c}{logleft(frac{c+1}{c-1}right)}$.
Also, references to relevant literature are appreciated. I have mostly relied on Wolfram's articles on the (inverse) hyperbolic tangent:
http://mathworld.wolfram.com/HyperbolicTangent.html
and
http://mathworld.wolfram.com/InverseHyperbolicTangent.html
real-analysis differential-equations
real-analysis differential-equations
edited Nov 15 at 23:02
asked Nov 15 at 16:42
IAnemaet
385
385
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999945%2fcan-someone-find-this-inverse-and-solve-burgers-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown