Parabolic cylinder functionsParabolic cylinder function as the solution of differential equationParabolic PDEs: Boundary conditionsSolution of a parabolic PDERegularity of a parabolic equationPeriodic solutions to the wav equation with seperated solutionsSolving the parabolic equationSeparable solution to a nonlinear parabolic PDEThe Wronskian of parabolic cylinder functionsPartial differential equation transformationSolution to the parabolic cylinder equation
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Parabolic cylinder functions
Parabolic cylinder function as the solution of differential equationParabolic PDEs: Boundary conditionsSolution of a parabolic PDERegularity of a parabolic equationPeriodic solutions to the wav equation with seperated solutionsSolving the parabolic equationSeparable solution to a nonlinear parabolic PDEThe Wronskian of parabolic cylinder functionsPartial differential equation transformationSolution to the parabolic cylinder equation
$begingroup$
My question is the same as this question but more general. I am dealing with parabolic cylinder functions and misunderstand some moments. As the source, I use this. The authors state that these equations
$$fracd^2 wd z^2+(az^2+bz+c)w=0;quad fracd^2 wdz^2+left(fracz^24-aright)w=0;$$
$$fracd^2 wdz^2-left(fracz^24+aright)=0;quad fracd^2 wdz^2+left(nu+frac12-fracz^24right)w=0$$
can be easily transformed to each other. I start with the first equation and comlete square
$$az^2+bz+c=aleft(z+fracb2aright)^2+left(c-fracb^24aright),$$
then change the variable $zrightarrow z+b/2a$ and denote $lambda=c-b^2/4a$, it gives
$$fracd^2 wd z^2+(az^2+lambda)w=0,$$
where I perfom the change $zrightarrow(4a)^-1/4z$ and $arightarrow-lambda(4a)^-1/2$ and obtain
$$ fracd^2 wdz^2+left(fracz^24-aright)w=0.quad (*)$$
Ok, it's good. From the DLMF, I know that the solutions of $(*)$ equation are the functions $W(a,,pm z)$. Then, from this I know how to connect the solutions $W(a,,pm z)$ and $U$-functions (see previous link). The connection formulas between $W$ and $U$ are very complicated (for me).
Then, I know that the solutions of
$$fracd^2 wdz^2+left(nu+frac12-fracz^24right)w=0quad(**)$$
are $D_nu(z)$ and easily connected with $U$-function. Also, I see that someone just write down solutions of equation with $az^2+bz+c$ in terms of $D_nu(z)$. So, I misunderstand how the complicated connection formula between $W$-function and $U$-function (I need it to connect $D$-function and $W$-function) allows to easily write down the solution of equation with $az^2+bz+c$ in terms of $D$-function.
pde special-functions
asked Feb 18 at 10:52
Artem AlexandrovArtem Alexandrov
306
$endgroup$
add a comment |
$begingroup$
My question is the same as this question but more general. I am dealing with parabolic cylinder functions and misunderstand some moments. As the source, I use this. The authors state that these equations
$$fracd^2 wd z^2+(az^2+bz+c)w=0;quad fracd^2 wdz^2+left(fracz^24-aright)w=0;$$
$$fracd^2 wdz^2-left(fracz^24+aright)=0;quad fracd^2 wdz^2+left(nu+frac12-fracz^24right)w=0$$
can be easily transformed to each other. I start with the first equation and comlete square
$$az^2+bz+c=aleft(z+fracb2aright)^2+left(c-fracb^24aright),$$
then change the variable $zrightarrow z+b/2a$ and denote $lambda=c-b^2/4a$, it gives
$$fracd^2 wd z^2+(az^2+lambda)w=0,$$
where I perfom the change $zrightarrow(4a)^-1/4z$ and $arightarrow-lambda(4a)^-1/2$ and obtain
$$ fracd^2 wdz^2+left(fracz^24-aright)w=0.quad (*)$$
Ok, it's good. From the DLMF, I know that the solutions of $(*)$ equation are the functions $W(a,,pm z)$. Then, from this I know how to connect the solutions $W(a,,pm z)$ and $U$-functions (see previous link). The connection formulas between $W$ and $U$ are very complicated (for me).
Then, I know that the solutions of
$$fracd^2 wdz^2+left(nu+frac12-fracz^24right)w=0quad(**)$$
are $D_nu(z)$ and easily connected with $U$-function. Also, I see that someone just write down solutions of equation with $az^2+bz+c$ in terms of $D_nu(z)$. So, I misunderstand how the complicated connection formula between $W$-function and $U$-function (I need it to connect $D$-function and $W$-function) allows to easily write down the solution of equation with $az^2+bz+c$ in terms of $D$-function.
pde special-functions
asked Feb 18 at 10:52
Artem AlexandrovArtem Alexandrov
306
$endgroup$
add a comment |
$begingroup$
My question is the same as this question but more general. I am dealing with parabolic cylinder functions and misunderstand some moments. As the source, I use this. The authors state that these equations
$$fracd^2 wd z^2+(az^2+bz+c)w=0;quad fracd^2 wdz^2+left(fracz^24-aright)w=0;$$
$$fracd^2 wdz^2-left(fracz^24+aright)=0;quad fracd^2 wdz^2+left(nu+frac12-fracz^24right)w=0$$
can be easily transformed to each other. I start with the first equation and comlete square
$$az^2+bz+c=aleft(z+fracb2aright)^2+left(c-fracb^24aright),$$
then change the variable $zrightarrow z+b/2a$ and denote $lambda=c-b^2/4a$, it gives
$$fracd^2 wd z^2+(az^2+lambda)w=0,$$
where I perfom the change $zrightarrow(4a)^-1/4z$ and $arightarrow-lambda(4a)^-1/2$ and obtain
$$ fracd^2 wdz^2+left(fracz^24-aright)w=0.quad (*)$$
Ok, it's good. From the DLMF, I know that the solutions of $(*)$ equation are the functions $W(a,,pm z)$. Then, from this I know how to connect the solutions $W(a,,pm z)$ and $U$-functions (see previous link). The connection formulas between $W$ and $U$ are very complicated (for me).
Then, I know that the solutions of
$$fracd^2 wdz^2+left(nu+frac12-fracz^24right)w=0quad(**)$$
are $D_nu(z)$ and easily connected with $U$-function. Also, I see that someone just write down solutions of equation with $az^2+bz+c$ in terms of $D_nu(z)$. So, I misunderstand how the complicated connection formula between $W$-function and $U$-function (I need it to connect $D$-function and $W$-function) allows to easily write down the solution of equation with $az^2+bz+c$ in terms of $D$-function.
pde special-functions
asked Feb 18 at 10:52
Artem AlexandrovArtem Alexandrov
306
$endgroup$
My question is the same as this question but more general. I am dealing with parabolic cylinder functions and misunderstand some moments. As the source, I use this. The authors state that these equations
$$fracd^2 wd z^2+(az^2+bz+c)w=0;quad fracd^2 wdz^2+left(fracz^24-aright)w=0;$$
$$fracd^2 wdz^2-left(fracz^24+aright)=0;quad fracd^2 wdz^2+left(nu+frac12-fracz^24right)w=0$$
can be easily transformed to each other. I start with the first equation and comlete square
$$az^2+bz+c=aleft(z+fracb2aright)^2+left(c-fracb^24aright),$$
then change the variable $zrightarrow z+b/2a$ and denote $lambda=c-b^2/4a$, it gives
$$fracd^2 wd z^2+(az^2+lambda)w=0,$$
where I perfom the change $zrightarrow(4a)^-1/4z$ and $arightarrow-lambda(4a)^-1/2$ and obtain
$$ fracd^2 wdz^2+left(fracz^24-aright)w=0.quad (*)$$
Ok, it's good. From the DLMF, I know that the solutions of $(*)$ equation are the functions $W(a,,pm z)$. Then, from this I know how to connect the solutions $W(a,,pm z)$ and $U$-functions (see previous link). The connection formulas between $W$ and $U$ are very complicated (for me).
Then, I know that the solutions of
$$fracd^2 wdz^2+left(nu+frac12-fracz^24right)w=0quad(**)$$
are $D_nu(z)$ and easily connected with $U$-function. Also, I see that someone just write down solutions of equation with $az^2+bz+c$ in terms of $D_nu(z)$. So, I misunderstand how the complicated connection formula between $W$-function and $U$-function (I need it to connect $D$-function and $W$-function) allows to easily write down the solution of equation with $az^2+bz+c$ in terms of $D$-function.
pde special-functions
pde special-functions
asked Feb 18 at 10:52
Artem AlexandrovArtem Alexandrov
306
asked Feb 18 at 10:52
Artem AlexandrovArtem Alexandrov
306
asked Feb 18 at 10:52
Artem AlexandrovArtem Alexandrov
306
asked Feb 18 at 10:52
Artem AlexandrovArtem Alexandrov
306
asked Feb 18 at 10:52
Artem AlexandrovArtem Alexandrov
306
306
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let me try to present the answer. One can start from the following equation:
$$fracd^2wdz^2+(az^2+lambda)=0,$$
then substitute $zrightarrow (4a)^-1/4z$ and $lambdarightarrow(4a)^-1/2lambda$ and obtain:
$$fracd^2wdz^2+left(fracz^24+lambdaright)=0.$$
In this equation we also introduce new variables as $zrightarrow xe^ipi/4$ and $rhorightarrow -ilambda$. It gives
$$boxedfracd^2wdx^2-left(fracz^24+lambdaright)=0.$$
This equation has the solutions $U(rho,,z)$ and the connection formula is $U(rho,,z)=D_-rho-1/2(z)$.
answered Feb 20 at 9:48
Artem AlexandrovArtem Alexandrov
306
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let me try to present the answer. One can start from the following equation:
$$fracd^2wdz^2+(az^2+lambda)=0,$$
then substitute $zrightarrow (4a)^-1/4z$ and $lambdarightarrow(4a)^-1/2lambda$ and obtain:
$$fracd^2wdz^2+left(fracz^24+lambdaright)=0.$$
In this equation we also introduce new variables as $zrightarrow xe^ipi/4$ and $rhorightarrow -ilambda$. It gives
$$boxedfracd^2wdx^2-left(fracz^24+lambdaright)=0.$$
This equation has the solutions $U(rho,,z)$ and the connection formula is $U(rho,,z)=D_-rho-1/2(z)$.
answered Feb 20 at 9:48
Artem AlexandrovArtem Alexandrov
306
$endgroup$
add a comment |
$begingroup$
Let me try to present the answer. One can start from the following equation:
$$fracd^2wdz^2+(az^2+lambda)=0,$$
then substitute $zrightarrow (4a)^-1/4z$ and $lambdarightarrow(4a)^-1/2lambda$ and obtain:
$$fracd^2wdz^2+left(fracz^24+lambdaright)=0.$$
In this equation we also introduce new variables as $zrightarrow xe^ipi/4$ and $rhorightarrow -ilambda$. It gives
$$boxedfracd^2wdx^2-left(fracz^24+lambdaright)=0.$$
This equation has the solutions $U(rho,,z)$ and the connection formula is $U(rho,,z)=D_-rho-1/2(z)$.
answered Feb 20 at 9:48
Artem AlexandrovArtem Alexandrov
306
$endgroup$
add a comment |
$begingroup$
Let me try to present the answer. One can start from the following equation:
$$fracd^2wdz^2+(az^2+lambda)=0,$$
then substitute $zrightarrow (4a)^-1/4z$ and $lambdarightarrow(4a)^-1/2lambda$ and obtain:
$$fracd^2wdz^2+left(fracz^24+lambdaright)=0.$$
In this equation we also introduce new variables as $zrightarrow xe^ipi/4$ and $rhorightarrow -ilambda$. It gives
$$boxedfracd^2wdx^2-left(fracz^24+lambdaright)=0.$$
This equation has the solutions $U(rho,,z)$ and the connection formula is $U(rho,,z)=D_-rho-1/2(z)$.
answered Feb 20 at 9:48
Artem AlexandrovArtem Alexandrov
306
$endgroup$
Let me try to present the answer. One can start from the following equation:
$$fracd^2wdz^2+(az^2+lambda)=0,$$
then substitute $zrightarrow (4a)^-1/4z$ and $lambdarightarrow(4a)^-1/2lambda$ and obtain:
$$fracd^2wdz^2+left(fracz^24+lambdaright)=0.$$
In this equation we also introduce new variables as $zrightarrow xe^ipi/4$ and $rhorightarrow -ilambda$. It gives
$$boxedfracd^2wdx^2-left(fracz^24+lambdaright)=0.$$
This equation has the solutions $U(rho,,z)$ and the connection formula is $U(rho,,z)=D_-rho-1/2(z)$.
answered Feb 20 at 9:48
Artem AlexandrovArtem Alexandrov
306
edited Mar 17 at 20:32
answered Feb 20 at 9:48
Artem AlexandrovArtem Alexandrov
306
answered Feb 20 at 9:48
Artem AlexandrovArtem Alexandrov
306
answered Feb 20 at 9:48
Artem AlexandrovArtem Alexandrov
306
306
add a comment |
add a comment |
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