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Asymptotic approximation for $epsilon^2 y'' = axy$
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I would like to find the first two terms of the asymptotic approximation for $epsilon^2 y'' = axy$ on $0leq x <
infty$ where $y(infty) = 0$ and $a>0$.
Here is my work so far, with the standard WKB method, we plug in
$$y(x) = expleft[frac1delta sum_n=0^infty delta^n S_n(x)right]$$ and divide by it; our ODE simplifies to
$$epsilon^2left[frac1delta S_0'' + S_1 + cdots + frac1delta^2 S_0'^2 + 2 frac1delta S_0'S_1' + cdots right] -ax = 0 quad quad (star)$$
Our largest term $fracepsilon^2delta^2 S_0'^2 $must match with $-ax$, so we make $delta = epsilon$, and get
$$S_0'^2 sim ax$$
thus
$$S_0 sim pm sqrta frac23 x^3/2.$$
And by matching $O(epsilon)$ terms in $(star)$, we get
$$2S_0'S_1' + S_0'' = 0$$
and we get $S_1 sim - frac14 log(ax)$.
So our approximation
$$y (x)sim C_1 (ax)^-1/4 exp left[frac1epsilon sqrta frac23 x^3/2right] + C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right]$$
And from our condition $y(infty) = 0$, we get $C_1 = 0$.
But the above approximation does not make sense at $x=0$, how can we take care of this problem?
I know boundary layer theory but often it has problems when the coefficient of $y'$ is constant zero. I also worked out the WKB method for a boundary layer problem here, I am not sure how would I apply to this case.
Edit: If I add a boundary layer at $x=0$, we let $xi = x/delta$, then the ODE in terms of delta will become
$$fracepsilon^2delta^2 (y_0'' + epsilon y_1'' + cdots ) = axidelta(y_0 + epsilon y_1, + cdots)$$
which gives
$$fracepsilon^2delta^2 y_0'' + fracepsilon^3delta^2 y_1'' + cdots - delta axi y_0 + cdots = 0$$
If I let $delta = epsilon$, I will get $y_0 = axi + b$, how would I match this with $y(x)$ from WKB method
differential-equations asymptotics perturbation-theory
asked Jul 18 at 3:01
Xiao
4,42311333
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up vote
2
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favorite
I would like to find the first two terms of the asymptotic approximation for $epsilon^2 y'' = axy$ on $0leq x <
infty$ where $y(infty) = 0$ and $a>0$.
Here is my work so far, with the standard WKB method, we plug in
$$y(x) = expleft[frac1delta sum_n=0^infty delta^n S_n(x)right]$$ and divide by it; our ODE simplifies to
$$epsilon^2left[frac1delta S_0'' + S_1 + cdots + frac1delta^2 S_0'^2 + 2 frac1delta S_0'S_1' + cdots right] -ax = 0 quad quad (star)$$
Our largest term $fracepsilon^2delta^2 S_0'^2 $must match with $-ax$, so we make $delta = epsilon$, and get
$$S_0'^2 sim ax$$
thus
$$S_0 sim pm sqrta frac23 x^3/2.$$
And by matching $O(epsilon)$ terms in $(star)$, we get
$$2S_0'S_1' + S_0'' = 0$$
and we get $S_1 sim - frac14 log(ax)$.
So our approximation
$$y (x)sim C_1 (ax)^-1/4 exp left[frac1epsilon sqrta frac23 x^3/2right] + C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right]$$
And from our condition $y(infty) = 0$, we get $C_1 = 0$.
But the above approximation does not make sense at $x=0$, how can we take care of this problem?
I know boundary layer theory but often it has problems when the coefficient of $y'$ is constant zero. I also worked out the WKB method for a boundary layer problem here, I am not sure how would I apply to this case.
Edit: If I add a boundary layer at $x=0$, we let $xi = x/delta$, then the ODE in terms of delta will become
$$fracepsilon^2delta^2 (y_0'' + epsilon y_1'' + cdots ) = axidelta(y_0 + epsilon y_1, + cdots)$$
which gives
$$fracepsilon^2delta^2 y_0'' + fracepsilon^3delta^2 y_1'' + cdots - delta axi y_0 + cdots = 0$$
If I let $delta = epsilon$, I will get $y_0 = axi + b$, how would I match this with $y(x)$ from WKB method
differential-equations asymptotics perturbation-theory
asked Jul 18 at 3:01
Xiao
4,42311333
1
This is looking like the Airy function.
– Claude Leibovici
Jul 18 at 4:50
@ClaudeLeibovici, It is the Airy equation with $epsilon^2$ before the $y''$, could you say a bit more about this please. Thanks!
– Xiao
Jul 19 at 3:01
1
Well, without using any perturbation theory, we have the solution and we can analyze it. I must confess that I am not familiar at all with this kind of analysis using perturbation theory,
– Claude Leibovici
Jul 19 at 3:07
1
Have a look at math.arizona.edu/~meissen/docs/asymptotics.pdf .It could be of some interest for you.
– Claude Leibovici
Jul 19 at 3:31
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I would like to find the first two terms of the asymptotic approximation for $epsilon^2 y'' = axy$ on $0leq x <
infty$ where $y(infty) = 0$ and $a>0$.
Here is my work so far, with the standard WKB method, we plug in
$$y(x) = expleft[frac1delta sum_n=0^infty delta^n S_n(x)right]$$ and divide by it; our ODE simplifies to
$$epsilon^2left[frac1delta S_0'' + S_1 + cdots + frac1delta^2 S_0'^2 + 2 frac1delta S_0'S_1' + cdots right] -ax = 0 quad quad (star)$$
Our largest term $fracepsilon^2delta^2 S_0'^2 $must match with $-ax$, so we make $delta = epsilon$, and get
$$S_0'^2 sim ax$$
thus
$$S_0 sim pm sqrta frac23 x^3/2.$$
And by matching $O(epsilon)$ terms in $(star)$, we get
$$2S_0'S_1' + S_0'' = 0$$
and we get $S_1 sim - frac14 log(ax)$.
So our approximation
$$y (x)sim C_1 (ax)^-1/4 exp left[frac1epsilon sqrta frac23 x^3/2right] + C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right]$$
And from our condition $y(infty) = 0$, we get $C_1 = 0$.
But the above approximation does not make sense at $x=0$, how can we take care of this problem?
I know boundary layer theory but often it has problems when the coefficient of $y'$ is constant zero. I also worked out the WKB method for a boundary layer problem here, I am not sure how would I apply to this case.
Edit: If I add a boundary layer at $x=0$, we let $xi = x/delta$, then the ODE in terms of delta will become
$$fracepsilon^2delta^2 (y_0'' + epsilon y_1'' + cdots ) = axidelta(y_0 + epsilon y_1, + cdots)$$
which gives
$$fracepsilon^2delta^2 y_0'' + fracepsilon^3delta^2 y_1'' + cdots - delta axi y_0 + cdots = 0$$
If I let $delta = epsilon$, I will get $y_0 = axi + b$, how would I match this with $y(x)$ from WKB method
differential-equations asymptotics perturbation-theory
asked Jul 18 at 3:01
Xiao
4,42311333
I would like to find the first two terms of the asymptotic approximation for $epsilon^2 y'' = axy$ on $0leq x <
infty$ where $y(infty) = 0$ and $a>0$.
Here is my work so far, with the standard WKB method, we plug in
$$y(x) = expleft[frac1delta sum_n=0^infty delta^n S_n(x)right]$$ and divide by it; our ODE simplifies to
$$epsilon^2left[frac1delta S_0'' + S_1 + cdots + frac1delta^2 S_0'^2 + 2 frac1delta S_0'S_1' + cdots right] -ax = 0 quad quad (star)$$
Our largest term $fracepsilon^2delta^2 S_0'^2 $must match with $-ax$, so we make $delta = epsilon$, and get
$$S_0'^2 sim ax$$
thus
$$S_0 sim pm sqrta frac23 x^3/2.$$
And by matching $O(epsilon)$ terms in $(star)$, we get
$$2S_0'S_1' + S_0'' = 0$$
and we get $S_1 sim - frac14 log(ax)$.
So our approximation
$$y (x)sim C_1 (ax)^-1/4 exp left[frac1epsilon sqrta frac23 x^3/2right] + C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right]$$
And from our condition $y(infty) = 0$, we get $C_1 = 0$.
But the above approximation does not make sense at $x=0$, how can we take care of this problem?
I know boundary layer theory but often it has problems when the coefficient of $y'$ is constant zero. I also worked out the WKB method for a boundary layer problem here, I am not sure how would I apply to this case.
Edit: If I add a boundary layer at $x=0$, we let $xi = x/delta$, then the ODE in terms of delta will become
$$fracepsilon^2delta^2 (y_0'' + epsilon y_1'' + cdots ) = axidelta(y_0 + epsilon y_1, + cdots)$$
which gives
$$fracepsilon^2delta^2 y_0'' + fracepsilon^3delta^2 y_1'' + cdots - delta axi y_0 + cdots = 0$$
If I let $delta = epsilon$, I will get $y_0 = axi + b$, how would I match this with $y(x)$ from WKB method
differential-equations asymptotics perturbation-theory
asked Jul 18 at 3:01
Xiao
4,42311333
edited Aug 2 at 15:43
asked Jul 18 at 3:01
Xiao
4,42311333
asked Jul 18 at 3:01
Xiao
4,42311333
asked Jul 18 at 3:01
Xiao
4,42311333
4,42311333
1
This is looking like the Airy function.
– Claude Leibovici
Jul 18 at 4:50
@ClaudeLeibovici, It is the Airy equation with $epsilon^2$ before the $y''$, could you say a bit more about this please. Thanks!
– Xiao
Jul 19 at 3:01
1
Well, without using any perturbation theory, we have the solution and we can analyze it. I must confess that I am not familiar at all with this kind of analysis using perturbation theory,
– Claude Leibovici
Jul 19 at 3:07
1
Have a look at math.arizona.edu/~meissen/docs/asymptotics.pdf .It could be of some interest for you.
– Claude Leibovici
Jul 19 at 3:31
add a comment |Â
1
This is looking like the Airy function.
– Claude Leibovici
Jul 18 at 4:50
@ClaudeLeibovici, It is the Airy equation with $epsilon^2$ before the $y''$, could you say a bit more about this please. Thanks!
– Xiao
Jul 19 at 3:01
1
Well, without using any perturbation theory, we have the solution and we can analyze it. I must confess that I am not familiar at all with this kind of analysis using perturbation theory,
– Claude Leibovici
Jul 19 at 3:07
1
Have a look at math.arizona.edu/~meissen/docs/asymptotics.pdf .It could be of some interest for you.
– Claude Leibovici
Jul 19 at 3:31
1
1
This is looking like the Airy function.
– Claude Leibovici
Jul 18 at 4:50
This is looking like the Airy function.
– Claude Leibovici
Jul 18 at 4:50
@ClaudeLeibovici, It is the Airy equation with $epsilon^2$ before the $y''$, could you say a bit more about this please. Thanks!
– Xiao
Jul 19 at 3:01
@ClaudeLeibovici, It is the Airy equation with $epsilon^2$ before the $y''$, could you say a bit more about this please. Thanks!
– Xiao
Jul 19 at 3:01
1
1
Well, without using any perturbation theory, we have the solution and we can analyze it. I must confess that I am not familiar at all with this kind of analysis using perturbation theory,
– Claude Leibovici
Jul 19 at 3:07
Well, without using any perturbation theory, we have the solution and we can analyze it. I must confess that I am not familiar at all with this kind of analysis using perturbation theory,
– Claude Leibovici
Jul 19 at 3:07
1
1
Have a look at math.arizona.edu/~meissen/docs/asymptotics.pdf .It could be of some interest for you.
– Claude Leibovici
Jul 19 at 3:31
Have a look at math.arizona.edu/~meissen/docs/asymptotics.pdf .It could be of some interest for you.
– Claude Leibovici
Jul 19 at 3:31
add a comment |Â
2 Answers
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up vote
4
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Your asymptotic approximation is correct :
$$y (x)sim C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right] tag 1$$
according to the boundary condition $y(infty)=0$.
Note : A second order ODE requires at least two boundary conditions to expect having a unique solution. Since the wording of the question specifies only the condition $y(infty)=0$, the solution remains undetermined : The coefficient $C_2$ cannot be computed.
Another key point is that an asymptotic formula is valid only for $xtoinfty$, not for $xto 0$.So, don't expect Eq.$(1)$ makes sens at $x=0$. It is normal that the asymptotic approximation be far to be correct for small values of $x$.
If you want an approximate around $x=0$, expand $y(x)$ in power series.
$$y(x)simeq c_0+c_1x+c_2x^2+...$$
Of course, this formula is not valid for large $x$, so the boundary condition at infinity cannot be fulfilled. Moreover, two boundary conditions should be required close to $x=0$ in order to determine the coefficients $c_0$ , $c_1$, $c_2$ , …
In fact, as Claude Leibovici quite rightly wrote in comments, the exact solving involves the Airy functions.
$$y(x)=c_1textAileft(sqrt[3]fracaepsilon^2:xright)+c_2textBileft(sqrt[3]fracaepsilon^2:xright)$$
Ai and Bi are the Airy functions http://mathworld.wolfram.com/AiryFunctions.html
The condition $y(infty)=0$ is satisfied any $c_1$ and $c_2$, which shows that the condition $y(infty)=0$ is not sufficient to fully determine a solution.
On physicists viewpoint, defining the properties of a boundary layer is certainly a way to introduce some boundary conditions in the range around $x=0$. Supposing that $y(x)simeq y_0+y'_0 x$ in a layer of a given thickness $delta$ allows to compute $c_1$ and $c_2$ as a function of $y_0$ , $y'_0$ and $delta$, thanks to the series expansion of the Airy functions. Then the asymptotic expansion of the Airy functions will leads to the coefficient $C_2$ in the asymptotic approximation $(1)$.
http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/06/
http://functions.wolfram.com/Bessel-TypeFunctions/AiryBi/06/
answered Jul 20 at 8:17
JJacquelin
40k21649
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I found the example from my text book. I will add the details here:
- First we see that in WKB approximation above
$$y(x) sim expleftfrac1epsilonS_0 + S_1right = C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right], quad epsilon rightarrow 0^+,$$
we need to have
$epsilon S_2 << 1$ because generally we have $y(x) sim expleftfrac1epsilonS_0 + S_1 + epsilon S_2 + cdots right$.
Solving for $S_2$ explicitly, we get $S_2 sim C x^-3/2$, thus
$epsilon S_2 <<1 Rightarrow epsilon^2/3 << x,$ and we have
$$colorbluey_one(x) sim C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right] quad epsilon rightarrow 0^+ text on x>> epsilon^2/3.$$ - Next, when $x$ is near zero, we let $t = (a/epsilon^2)^1/3x$, so our ODE $epsilon^2 y''(x) = axy(x)$ can be rewritten as $y''(t) = ty(t)$, and this is the Airy equation. And we know the solution is
$$y(t) = D_1 textAi(t) + D_2 textBi(t)$$
thus
$$colororangey_two(x) = D_1 textAi((a/epsilon^2)^1/3x) + D_2 textBi((a/epsilon^2)^1/3x).$$ - Finally would like to match them on $epsilon^2/3 <<x<<1$. First we see $y_one$ and $y_two$ are very different, so we will approximate $y_two$. From a irregular singular point (ISP) analysis of the Airy equation at infinity, we will get
$$textAi(t) sim frac12sqrtpi t^1/4 expleft[-frac23t^3/2right], quad t rightarrow infty $$
$$textBi(t) sim frac1sqrtpi t^1/4 expleft[frac23t^3/2right], quad t rightarrow infty,$$
(the two constants $frac12sqrtpi$ and $frac1sqrtpi$ are a bit more delicate and can not be obtained from the ISP analysis, but we will assume we have this fact.) So
$$colororangey_two(x) sim fracD_12sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[-frac23((a/epsilon^2)^1/3x)^3/2right] + \
fracD_2sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[frac23((a/epsilon^2)^1/3x)^3/2right] text when x>>epsilon^2/3 text and epsilon rightarrow 0^+. $$
After matching with $y_one$, we get $D_2 = 0$, $D_1 = 2sqrtpi(aepsilon)^-1/6 C_2$.
Now getting a uniform formula is non trivial, it follows from Langer's formula we have
$$y_uniform (x) sim C_2 2sqrtpi(aepsilon)^-1/6 textAi((a/epsilon^2)^1/3x).$$
answered Aug 2 at 21:25
Xiao
4,42311333
I think the Langer's formula is the exact solution in this case. Where is the asymptotic approximation?
– SimonChan
Aug 11 at 18:59
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Your asymptotic approximation is correct :
$$y (x)sim C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right] tag 1$$
according to the boundary condition $y(infty)=0$.
Note : A second order ODE requires at least two boundary conditions to expect having a unique solution. Since the wording of the question specifies only the condition $y(infty)=0$, the solution remains undetermined : The coefficient $C_2$ cannot be computed.
Another key point is that an asymptotic formula is valid only for $xtoinfty$, not for $xto 0$.So, don't expect Eq.$(1)$ makes sens at $x=0$. It is normal that the asymptotic approximation be far to be correct for small values of $x$.
If you want an approximate around $x=0$, expand $y(x)$ in power series.
$$y(x)simeq c_0+c_1x+c_2x^2+...$$
Of course, this formula is not valid for large $x$, so the boundary condition at infinity cannot be fulfilled. Moreover, two boundary conditions should be required close to $x=0$ in order to determine the coefficients $c_0$ , $c_1$, $c_2$ , …
In fact, as Claude Leibovici quite rightly wrote in comments, the exact solving involves the Airy functions.
$$y(x)=c_1textAileft(sqrt[3]fracaepsilon^2:xright)+c_2textBileft(sqrt[3]fracaepsilon^2:xright)$$
Ai and Bi are the Airy functions http://mathworld.wolfram.com/AiryFunctions.html
The condition $y(infty)=0$ is satisfied any $c_1$ and $c_2$, which shows that the condition $y(infty)=0$ is not sufficient to fully determine a solution.
On physicists viewpoint, defining the properties of a boundary layer is certainly a way to introduce some boundary conditions in the range around $x=0$. Supposing that $y(x)simeq y_0+y'_0 x$ in a layer of a given thickness $delta$ allows to compute $c_1$ and $c_2$ as a function of $y_0$ , $y'_0$ and $delta$, thanks to the series expansion of the Airy functions. Then the asymptotic expansion of the Airy functions will leads to the coefficient $C_2$ in the asymptotic approximation $(1)$.
http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/06/
http://functions.wolfram.com/Bessel-TypeFunctions/AiryBi/06/
answered Jul 20 at 8:17
JJacquelin
40k21649
add a comment |Â
up vote
4
down vote
Your asymptotic approximation is correct :
$$y (x)sim C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right] tag 1$$
according to the boundary condition $y(infty)=0$.
Note : A second order ODE requires at least two boundary conditions to expect having a unique solution. Since the wording of the question specifies only the condition $y(infty)=0$, the solution remains undetermined : The coefficient $C_2$ cannot be computed.
Another key point is that an asymptotic formula is valid only for $xtoinfty$, not for $xto 0$.So, don't expect Eq.$(1)$ makes sens at $x=0$. It is normal that the asymptotic approximation be far to be correct for small values of $x$.
If you want an approximate around $x=0$, expand $y(x)$ in power series.
$$y(x)simeq c_0+c_1x+c_2x^2+...$$
Of course, this formula is not valid for large $x$, so the boundary condition at infinity cannot be fulfilled. Moreover, two boundary conditions should be required close to $x=0$ in order to determine the coefficients $c_0$ , $c_1$, $c_2$ , …
In fact, as Claude Leibovici quite rightly wrote in comments, the exact solving involves the Airy functions.
$$y(x)=c_1textAileft(sqrt[3]fracaepsilon^2:xright)+c_2textBileft(sqrt[3]fracaepsilon^2:xright)$$
Ai and Bi are the Airy functions http://mathworld.wolfram.com/AiryFunctions.html
The condition $y(infty)=0$ is satisfied any $c_1$ and $c_2$, which shows that the condition $y(infty)=0$ is not sufficient to fully determine a solution.
On physicists viewpoint, defining the properties of a boundary layer is certainly a way to introduce some boundary conditions in the range around $x=0$. Supposing that $y(x)simeq y_0+y'_0 x$ in a layer of a given thickness $delta$ allows to compute $c_1$ and $c_2$ as a function of $y_0$ , $y'_0$ and $delta$, thanks to the series expansion of the Airy functions. Then the asymptotic expansion of the Airy functions will leads to the coefficient $C_2$ in the asymptotic approximation $(1)$.
http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/06/
http://functions.wolfram.com/Bessel-TypeFunctions/AiryBi/06/
answered Jul 20 at 8:17
JJacquelin
40k21649
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Your asymptotic approximation is correct :
$$y (x)sim C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right] tag 1$$
according to the boundary condition $y(infty)=0$.
Note : A second order ODE requires at least two boundary conditions to expect having a unique solution. Since the wording of the question specifies only the condition $y(infty)=0$, the solution remains undetermined : The coefficient $C_2$ cannot be computed.
Another key point is that an asymptotic formula is valid only for $xtoinfty$, not for $xto 0$.So, don't expect Eq.$(1)$ makes sens at $x=0$. It is normal that the asymptotic approximation be far to be correct for small values of $x$.
If you want an approximate around $x=0$, expand $y(x)$ in power series.
$$y(x)simeq c_0+c_1x+c_2x^2+...$$
Of course, this formula is not valid for large $x$, so the boundary condition at infinity cannot be fulfilled. Moreover, two boundary conditions should be required close to $x=0$ in order to determine the coefficients $c_0$ , $c_1$, $c_2$ , …
In fact, as Claude Leibovici quite rightly wrote in comments, the exact solving involves the Airy functions.
$$y(x)=c_1textAileft(sqrt[3]fracaepsilon^2:xright)+c_2textBileft(sqrt[3]fracaepsilon^2:xright)$$
Ai and Bi are the Airy functions http://mathworld.wolfram.com/AiryFunctions.html
The condition $y(infty)=0$ is satisfied any $c_1$ and $c_2$, which shows that the condition $y(infty)=0$ is not sufficient to fully determine a solution.
On physicists viewpoint, defining the properties of a boundary layer is certainly a way to introduce some boundary conditions in the range around $x=0$. Supposing that $y(x)simeq y_0+y'_0 x$ in a layer of a given thickness $delta$ allows to compute $c_1$ and $c_2$ as a function of $y_0$ , $y'_0$ and $delta$, thanks to the series expansion of the Airy functions. Then the asymptotic expansion of the Airy functions will leads to the coefficient $C_2$ in the asymptotic approximation $(1)$.
http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/06/
http://functions.wolfram.com/Bessel-TypeFunctions/AiryBi/06/
answered Jul 20 at 8:17
JJacquelin
40k21649
Your asymptotic approximation is correct :
$$y (x)sim C_2 (ax)^-1/4 exp left[-frac1epsilon sqrta frac23 x^3/2right] tag 1$$
according to the boundary condition $y(infty)=0$.
Note : A second order ODE requires at least two boundary conditions to expect having a unique solution. Since the wording of the question specifies only the condition $y(infty)=0$, the solution remains undetermined : The coefficient $C_2$ cannot be computed.
Another key point is that an asymptotic formula is valid only for $xtoinfty$, not for $xto 0$.So, don't expect Eq.$(1)$ makes sens at $x=0$. It is normal that the asymptotic approximation be far to be correct for small values of $x$.
If you want an approximate around $x=0$, expand $y(x)$ in power series.
$$y(x)simeq c_0+c_1x+c_2x^2+...$$
Of course, this formula is not valid for large $x$, so the boundary condition at infinity cannot be fulfilled. Moreover, two boundary conditions should be required close to $x=0$ in order to determine the coefficients $c_0$ , $c_1$, $c_2$ , …
In fact, as Claude Leibovici quite rightly wrote in comments, the exact solving involves the Airy functions.
$$y(x)=c_1textAileft(sqrt[3]fracaepsilon^2:xright)+c_2textBileft(sqrt[3]fracaepsilon^2:xright)$$
Ai and Bi are the Airy functions http://mathworld.wolfram.com/AiryFunctions.html
The condition $y(infty)=0$ is satisfied any $c_1$ and $c_2$, which shows that the condition $y(infty)=0$ is not sufficient to fully determine a solution.
On physicists viewpoint, defining the properties of a boundary layer is certainly a way to introduce some boundary conditions in the range around $x=0$. Supposing that $y(x)simeq y_0+y'_0 x$ in a layer of a given thickness $delta$ allows to compute $c_1$ and $c_2$ as a function of $y_0$ , $y'_0$ and $delta$, thanks to the series expansion of the Airy functions. Then the asymptotic expansion of the Airy functions will leads to the coefficient $C_2$ in the asymptotic approximation $(1)$.
http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/06/
http://functions.wolfram.com/Bessel-TypeFunctions/AiryBi/06/
answered Jul 20 at 8:17
JJacquelin
40k21649
answered Jul 20 at 8:17
JJacquelin
40k21649
answered Jul 20 at 8:17
JJacquelin
40k21649
answered Jul 20 at 8:17
JJacquelin
40k21649
40k21649
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I found the example from my text book. I will add the details here:
- First we see that in WKB approximation above
$$y(x) sim expleftfrac1epsilonS_0 + S_1right = C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right], quad epsilon rightarrow 0^+,$$
we need to have
$epsilon S_2 << 1$ because generally we have $y(x) sim expleftfrac1epsilonS_0 + S_1 + epsilon S_2 + cdots right$.
Solving for $S_2$ explicitly, we get $S_2 sim C x^-3/2$, thus
$epsilon S_2 <<1 Rightarrow epsilon^2/3 << x,$ and we have
$$colorbluey_one(x) sim C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right] quad epsilon rightarrow 0^+ text on x>> epsilon^2/3.$$ - Next, when $x$ is near zero, we let $t = (a/epsilon^2)^1/3x$, so our ODE $epsilon^2 y''(x) = axy(x)$ can be rewritten as $y''(t) = ty(t)$, and this is the Airy equation. And we know the solution is
$$y(t) = D_1 textAi(t) + D_2 textBi(t)$$
thus
$$colororangey_two(x) = D_1 textAi((a/epsilon^2)^1/3x) + D_2 textBi((a/epsilon^2)^1/3x).$$ - Finally would like to match them on $epsilon^2/3 <<x<<1$. First we see $y_one$ and $y_two$ are very different, so we will approximate $y_two$. From a irregular singular point (ISP) analysis of the Airy equation at infinity, we will get
$$textAi(t) sim frac12sqrtpi t^1/4 expleft[-frac23t^3/2right], quad t rightarrow infty $$
$$textBi(t) sim frac1sqrtpi t^1/4 expleft[frac23t^3/2right], quad t rightarrow infty,$$
(the two constants $frac12sqrtpi$ and $frac1sqrtpi$ are a bit more delicate and can not be obtained from the ISP analysis, but we will assume we have this fact.) So
$$colororangey_two(x) sim fracD_12sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[-frac23((a/epsilon^2)^1/3x)^3/2right] + \
fracD_2sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[frac23((a/epsilon^2)^1/3x)^3/2right] text when x>>epsilon^2/3 text and epsilon rightarrow 0^+. $$
After matching with $y_one$, we get $D_2 = 0$, $D_1 = 2sqrtpi(aepsilon)^-1/6 C_2$.
Now getting a uniform formula is non trivial, it follows from Langer's formula we have
$$y_uniform (x) sim C_2 2sqrtpi(aepsilon)^-1/6 textAi((a/epsilon^2)^1/3x).$$
answered Aug 2 at 21:25
Xiao
4,42311333
I think the Langer's formula is the exact solution in this case. Where is the asymptotic approximation?
– SimonChan
Aug 11 at 18:59
add a comment |Â
up vote
0
down vote
I found the example from my text book. I will add the details here:
- First we see that in WKB approximation above
$$y(x) sim expleftfrac1epsilonS_0 + S_1right = C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right], quad epsilon rightarrow 0^+,$$
we need to have
$epsilon S_2 << 1$ because generally we have $y(x) sim expleftfrac1epsilonS_0 + S_1 + epsilon S_2 + cdots right$.
Solving for $S_2$ explicitly, we get $S_2 sim C x^-3/2$, thus
$epsilon S_2 <<1 Rightarrow epsilon^2/3 << x,$ and we have
$$colorbluey_one(x) sim C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right] quad epsilon rightarrow 0^+ text on x>> epsilon^2/3.$$ - Next, when $x$ is near zero, we let $t = (a/epsilon^2)^1/3x$, so our ODE $epsilon^2 y''(x) = axy(x)$ can be rewritten as $y''(t) = ty(t)$, and this is the Airy equation. And we know the solution is
$$y(t) = D_1 textAi(t) + D_2 textBi(t)$$
thus
$$colororangey_two(x) = D_1 textAi((a/epsilon^2)^1/3x) + D_2 textBi((a/epsilon^2)^1/3x).$$ - Finally would like to match them on $epsilon^2/3 <<x<<1$. First we see $y_one$ and $y_two$ are very different, so we will approximate $y_two$. From a irregular singular point (ISP) analysis of the Airy equation at infinity, we will get
$$textAi(t) sim frac12sqrtpi t^1/4 expleft[-frac23t^3/2right], quad t rightarrow infty $$
$$textBi(t) sim frac1sqrtpi t^1/4 expleft[frac23t^3/2right], quad t rightarrow infty,$$
(the two constants $frac12sqrtpi$ and $frac1sqrtpi$ are a bit more delicate and can not be obtained from the ISP analysis, but we will assume we have this fact.) So
$$colororangey_two(x) sim fracD_12sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[-frac23((a/epsilon^2)^1/3x)^3/2right] + \
fracD_2sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[frac23((a/epsilon^2)^1/3x)^3/2right] text when x>>epsilon^2/3 text and epsilon rightarrow 0^+. $$
After matching with $y_one$, we get $D_2 = 0$, $D_1 = 2sqrtpi(aepsilon)^-1/6 C_2$.
Now getting a uniform formula is non trivial, it follows from Langer's formula we have
$$y_uniform (x) sim C_2 2sqrtpi(aepsilon)^-1/6 textAi((a/epsilon^2)^1/3x).$$
answered Aug 2 at 21:25
Xiao
4,42311333
I think the Langer's formula is the exact solution in this case. Where is the asymptotic approximation?
– SimonChan
Aug 11 at 18:59
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I found the example from my text book. I will add the details here:
- First we see that in WKB approximation above
$$y(x) sim expleftfrac1epsilonS_0 + S_1right = C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right], quad epsilon rightarrow 0^+,$$
we need to have
$epsilon S_2 << 1$ because generally we have $y(x) sim expleftfrac1epsilonS_0 + S_1 + epsilon S_2 + cdots right$.
Solving for $S_2$ explicitly, we get $S_2 sim C x^-3/2$, thus
$epsilon S_2 <<1 Rightarrow epsilon^2/3 << x,$ and we have
$$colorbluey_one(x) sim C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right] quad epsilon rightarrow 0^+ text on x>> epsilon^2/3.$$ - Next, when $x$ is near zero, we let $t = (a/epsilon^2)^1/3x$, so our ODE $epsilon^2 y''(x) = axy(x)$ can be rewritten as $y''(t) = ty(t)$, and this is the Airy equation. And we know the solution is
$$y(t) = D_1 textAi(t) + D_2 textBi(t)$$
thus
$$colororangey_two(x) = D_1 textAi((a/epsilon^2)^1/3x) + D_2 textBi((a/epsilon^2)^1/3x).$$ - Finally would like to match them on $epsilon^2/3 <<x<<1$. First we see $y_one$ and $y_two$ are very different, so we will approximate $y_two$. From a irregular singular point (ISP) analysis of the Airy equation at infinity, we will get
$$textAi(t) sim frac12sqrtpi t^1/4 expleft[-frac23t^3/2right], quad t rightarrow infty $$
$$textBi(t) sim frac1sqrtpi t^1/4 expleft[frac23t^3/2right], quad t rightarrow infty,$$
(the two constants $frac12sqrtpi$ and $frac1sqrtpi$ are a bit more delicate and can not be obtained from the ISP analysis, but we will assume we have this fact.) So
$$colororangey_two(x) sim fracD_12sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[-frac23((a/epsilon^2)^1/3x)^3/2right] + \
fracD_2sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[frac23((a/epsilon^2)^1/3x)^3/2right] text when x>>epsilon^2/3 text and epsilon rightarrow 0^+. $$
After matching with $y_one$, we get $D_2 = 0$, $D_1 = 2sqrtpi(aepsilon)^-1/6 C_2$.
Now getting a uniform formula is non trivial, it follows from Langer's formula we have
$$y_uniform (x) sim C_2 2sqrtpi(aepsilon)^-1/6 textAi((a/epsilon^2)^1/3x).$$
answered Aug 2 at 21:25
Xiao
4,42311333
I found the example from my text book. I will add the details here:
- First we see that in WKB approximation above
$$y(x) sim expleftfrac1epsilonS_0 + S_1right = C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right], quad epsilon rightarrow 0^+,$$
we need to have
$epsilon S_2 << 1$ because generally we have $y(x) sim expleftfrac1epsilonS_0 + S_1 + epsilon S_2 + cdots right$.
Solving for $S_2$ explicitly, we get $S_2 sim C x^-3/2$, thus
$epsilon S_2 <<1 Rightarrow epsilon^2/3 << x,$ and we have
$$colorbluey_one(x) sim C_2 (ax)^1/4 expleft[-frac1epsilonsqrta frac23 x^3/2 right] quad epsilon rightarrow 0^+ text on x>> epsilon^2/3.$$ - Next, when $x$ is near zero, we let $t = (a/epsilon^2)^1/3x$, so our ODE $epsilon^2 y''(x) = axy(x)$ can be rewritten as $y''(t) = ty(t)$, and this is the Airy equation. And we know the solution is
$$y(t) = D_1 textAi(t) + D_2 textBi(t)$$
thus
$$colororangey_two(x) = D_1 textAi((a/epsilon^2)^1/3x) + D_2 textBi((a/epsilon^2)^1/3x).$$ - Finally would like to match them on $epsilon^2/3 <<x<<1$. First we see $y_one$ and $y_two$ are very different, so we will approximate $y_two$. From a irregular singular point (ISP) analysis of the Airy equation at infinity, we will get
$$textAi(t) sim frac12sqrtpi t^1/4 expleft[-frac23t^3/2right], quad t rightarrow infty $$
$$textBi(t) sim frac1sqrtpi t^1/4 expleft[frac23t^3/2right], quad t rightarrow infty,$$
(the two constants $frac12sqrtpi$ and $frac1sqrtpi$ are a bit more delicate and can not be obtained from the ISP analysis, but we will assume we have this fact.) So
$$colororangey_two(x) sim fracD_12sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[-frac23((a/epsilon^2)^1/3x)^3/2right] + \
fracD_2sqrtpi ((a/epsilon^2)^1/3x)^1/4 expleft[frac23((a/epsilon^2)^1/3x)^3/2right] text when x>>epsilon^2/3 text and epsilon rightarrow 0^+. $$
After matching with $y_one$, we get $D_2 = 0$, $D_1 = 2sqrtpi(aepsilon)^-1/6 C_2$.
Now getting a uniform formula is non trivial, it follows from Langer's formula we have
$$y_uniform (x) sim C_2 2sqrtpi(aepsilon)^-1/6 textAi((a/epsilon^2)^1/3x).$$
answered Aug 2 at 21:25
Xiao
4,42311333
answered Aug 2 at 21:25
Xiao
4,42311333
answered Aug 2 at 21:25
Xiao
4,42311333
answered Aug 2 at 21:25
Xiao
4,42311333
4,42311333
I think the Langer's formula is the exact solution in this case. Where is the asymptotic approximation?
– SimonChan
Aug 11 at 18:59
add a comment |Â
I think the Langer's formula is the exact solution in this case. Where is the asymptotic approximation?
– SimonChan
Aug 11 at 18:59
I think the Langer's formula is the exact solution in this case. Where is the asymptotic approximation?
– SimonChan
Aug 11 at 18:59
I think the Langer's formula is the exact solution in this case. Where is the asymptotic approximation?
– SimonChan
Aug 11 at 18:59
add a comment |Â
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1
This is looking like the Airy function.
– Claude Leibovici
Jul 18 at 4:50
@ClaudeLeibovici, It is the Airy equation with $epsilon^2$ before the $y''$, could you say a bit more about this please. Thanks!
– Xiao
Jul 19 at 3:01
1
Well, without using any perturbation theory, we have the solution and we can analyze it. I must confess that I am not familiar at all with this kind of analysis using perturbation theory,
– Claude Leibovici
Jul 19 at 3:07
1
Have a look at math.arizona.edu/~meissen/docs/asymptotics.pdf .It could be of some interest for you.
– Claude Leibovici
Jul 19 at 3:31