Solving a equation set
$begingroup$
For any positive integer $m$, let $y=y(z)$ be a complex function, such that
begin{align*}
left{ {begin{array}{*{20}{c}} frac{dy}{dz}+y^m=1,
\ \
frac{dlog(y)}{dz}=frac{m^2e^{-mz}}{1-e^{-m^2z}}, \
end{array} } right.
end{align*}
with the initial conditions $y(0)=0$ and $limlimits_{zrightarrow infty}y(z)=1$. Does this function $y(z)$ exist?
For example, if $m=1$, we have $y(z)=1-e^{-z}$. If $m=2$, we have $y(z)=tanh(z)$.
ordinary-differential-equations
asked Dec 19 '18 at 14:02
xuce1234xuce1234
486
$endgroup$
add a comment |
$begingroup$
For any positive integer $m$, let $y=y(z)$ be a complex function, such that
begin{align*}
left{ {begin{array}{*{20}{c}} frac{dy}{dz}+y^m=1,
\ \
frac{dlog(y)}{dz}=frac{m^2e^{-mz}}{1-e^{-m^2z}}, \
end{array} } right.
end{align*}
with the initial conditions $y(0)=0$ and $limlimits_{zrightarrow infty}y(z)=1$. Does this function $y(z)$ exist?
For example, if $m=1$, we have $y(z)=1-e^{-z}$. If $m=2$, we have $y(z)=tanh(z)$.
ordinary-differential-equations
asked Dec 19 '18 at 14:02
xuce1234xuce1234
486
$endgroup$
$begingroup$
The function $1-e^{-z}$ has an essential singularity at $infty$, and $lim_{xtoinfty}(1-e^{-z})$ does not exist. May be you mean $lim_{xto+infty, xinBbb R}$?
$endgroup$
– Julián Aguirre
Dec 19 '18 at 14:13
$begingroup$
Using $$frac{d log y}{dz}= frac{1}{y} frac{dy}{dz}$$ in the second equation leads to $$frac{dy}{dz} = frac{m^2 e^{-mz}}{1-e^{-m^2z}} y.$$ Then, using the first equation, we obtain that $y$ is defined implicitly by the relation $$frac{m^2 e^{-mz}}{1-e^{-m^2z}} y=1-y^m,$$ which agrees with your result for $m=1$.
$endgroup$
– rafa11111
Dec 19 '18 at 14:15
add a comment |
$begingroup$
For any positive integer $m$, let $y=y(z)$ be a complex function, such that
begin{align*}
left{ {begin{array}{*{20}{c}} frac{dy}{dz}+y^m=1,
\ \
frac{dlog(y)}{dz}=frac{m^2e^{-mz}}{1-e^{-m^2z}}, \
end{array} } right.
end{align*}
with the initial conditions $y(0)=0$ and $limlimits_{zrightarrow infty}y(z)=1$. Does this function $y(z)$ exist?
For example, if $m=1$, we have $y(z)=1-e^{-z}$. If $m=2$, we have $y(z)=tanh(z)$.
ordinary-differential-equations
asked Dec 19 '18 at 14:02
xuce1234xuce1234
486
$endgroup$
For any positive integer $m$, let $y=y(z)$ be a complex function, such that
begin{align*}
left{ {begin{array}{*{20}{c}} frac{dy}{dz}+y^m=1,
\ \
frac{dlog(y)}{dz}=frac{m^2e^{-mz}}{1-e^{-m^2z}}, \
end{array} } right.
end{align*}
with the initial conditions $y(0)=0$ and $limlimits_{zrightarrow infty}y(z)=1$. Does this function $y(z)$ exist?
For example, if $m=1$, we have $y(z)=1-e^{-z}$. If $m=2$, we have $y(z)=tanh(z)$.
ordinary-differential-equations
ordinary-differential-equations
asked Dec 19 '18 at 14:02
xuce1234xuce1234
486
asked Dec 19 '18 at 14:02
xuce1234xuce1234
486
asked Dec 19 '18 at 14:02
xuce1234xuce1234
486
asked Dec 19 '18 at 14:02
xuce1234xuce1234
486
asked Dec 19 '18 at 14:02
xuce1234xuce1234
486
486
$begingroup$
The function $1-e^{-z}$ has an essential singularity at $infty$, and $lim_{xtoinfty}(1-e^{-z})$ does not exist. May be you mean $lim_{xto+infty, xinBbb R}$?
$endgroup$
– Julián Aguirre
Dec 19 '18 at 14:13
$begingroup$
Using $$frac{d log y}{dz}= frac{1}{y} frac{dy}{dz}$$ in the second equation leads to $$frac{dy}{dz} = frac{m^2 e^{-mz}}{1-e^{-m^2z}} y.$$ Then, using the first equation, we obtain that $y$ is defined implicitly by the relation $$frac{m^2 e^{-mz}}{1-e^{-m^2z}} y=1-y^m,$$ which agrees with your result for $m=1$.
$endgroup$
– rafa11111
Dec 19 '18 at 14:15
add a comment |
$begingroup$
The function $1-e^{-z}$ has an essential singularity at $infty$, and $lim_{xtoinfty}(1-e^{-z})$ does not exist. May be you mean $lim_{xto+infty, xinBbb R}$?
$endgroup$
– Julián Aguirre
Dec 19 '18 at 14:13
$begingroup$
Using $$frac{d log y}{dz}= frac{1}{y} frac{dy}{dz}$$ in the second equation leads to $$frac{dy}{dz} = frac{m^2 e^{-mz}}{1-e^{-m^2z}} y.$$ Then, using the first equation, we obtain that $y$ is defined implicitly by the relation $$frac{m^2 e^{-mz}}{1-e^{-m^2z}} y=1-y^m,$$ which agrees with your result for $m=1$.
$endgroup$
– rafa11111
Dec 19 '18 at 14:15
$begingroup$
The function $1-e^{-z}$ has an essential singularity at $infty$, and $lim_{xtoinfty}(1-e^{-z})$ does not exist. May be you mean $lim_{xto+infty, xinBbb R}$?
$endgroup$
– Julián Aguirre
Dec 19 '18 at 14:13
$begingroup$
The function $1-e^{-z}$ has an essential singularity at $infty$, and $lim_{xtoinfty}(1-e^{-z})$ does not exist. May be you mean $lim_{xto+infty, xinBbb R}$?
$endgroup$
– Julián Aguirre
Dec 19 '18 at 14:13
$begingroup$
Using $$frac{d log y}{dz}= frac{1}{y} frac{dy}{dz}$$ in the second equation leads to $$frac{dy}{dz} = frac{m^2 e^{-mz}}{1-e^{-m^2z}} y.$$ Then, using the first equation, we obtain that $y$ is defined implicitly by the relation $$frac{m^2 e^{-mz}}{1-e^{-m^2z}} y=1-y^m,$$ which agrees with your result for $m=1$.
$endgroup$
– rafa11111
Dec 19 '18 at 14:15
$begingroup$
Using $$frac{d log y}{dz}= frac{1}{y} frac{dy}{dz}$$ in the second equation leads to $$frac{dy}{dz} = frac{m^2 e^{-mz}}{1-e^{-m^2z}} y.$$ Then, using the first equation, we obtain that $y$ is defined implicitly by the relation $$frac{m^2 e^{-mz}}{1-e^{-m^2z}} y=1-y^m,$$ which agrees with your result for $m=1$.
$endgroup$
– rafa11111
Dec 19 '18 at 14:15
add a comment |
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$begingroup$
The function $1-e^{-z}$ has an essential singularity at $infty$, and $lim_{xtoinfty}(1-e^{-z})$ does not exist. May be you mean $lim_{xto+infty, xinBbb R}$?
$endgroup$
– Julián Aguirre
Dec 19 '18 at 14:13
$begingroup$
Using $$frac{d log y}{dz}= frac{1}{y} frac{dy}{dz}$$ in the second equation leads to $$frac{dy}{dz} = frac{m^2 e^{-mz}}{1-e^{-m^2z}} y.$$ Then, using the first equation, we obtain that $y$ is defined implicitly by the relation $$frac{m^2 e^{-mz}}{1-e^{-m^2z}} y=1-y^m,$$ which agrees with your result for $m=1$.
$endgroup$
– rafa11111
Dec 19 '18 at 14:15