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Solving the Airy Equation using Laplace Transform
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I have been trying to read the book Airy Functions and Applications to Physics, Olivier Vallee & Manuel Soares, for research on the Airy Functions, but I am stuck on using the Laplace transform to solve the Airy equation to get the function in the correct form.
The book states (pg 5):
We consider the following homogeneous second order differential equation called the Airy's equation $$y''-xy=0.$$ This differential equation may be solved by the method of Laplace, i.e. in seeking a solution as an integral $$y=int_C e^{xz}v(z)dz,$$ this is equivalent to solve the first order differential equation $$ v'+z^2v=0.$$ We thus obtain the solution to the Airy's equation, except a normalisation constant, $$y=int_C e^{xz-z^3/3}dz.$$
Below is my working:
Using the suggestion to find a solution to the Airy Equation $y''-xy=0$ in the form $y=int_Ce^{xz}v(z)dz$, the first step would be to differentiate the expression for $y$ to find $y'$ & $y''$:
$$frac{dy}{dx}=int ze^{xz}v(z)dz,$$
$$frac{d^2y}{dx^2}=int z^2e^{xz}v(z)dz.$$
Then,
$$int z^2e^{xz}v(z)dz-xint e^{xz}v(z)dz=0,$$ so
$$int e^{xz}v(z)(z^2-x)dz=0.$$
Then I tried to use integration by parts with $u=(z^2-x)v(z)=z^2v(z)-xv(z)$, so $frac{du}{dz}=2zv(z)+z^2v'(z)-xv'(z)=2zv(z)+v'(z)(z^2-x)$, and $frac{dv}{dz}=e^{xz}$, so $v=frac{1}{z}e^{xz}$.
Following through with this gives:
$$v(z)(z^2-x)frac{1}{z}e^{xz}-int(2zv(z)+v'(z)(z^2-x))frac{1}{z}e^{xz}dz=0.$$
This (sort of) simplified to:
$$v(z)frac{left(z^2-xright)e^{xz}}{z}-int 2v(z)e^{xz}+v'(z)frac{left(z^2-xright)e^{xz}}{z}dz=0.$$
At this point, I tried to differentiate this expression to reach the form $v'+z^2v=0$:
$$frac{d}{dz}left(zv(z)e^{xz}-frac{xv(z)e^{xz}}{z}-int 2v(z)e^{xz}dz-int zv'(z)e^{xz}dz+intfrac{xv'(z)e^{xz}}{z}dzright)=0,$$
which gives
$$v(z)e^{xz}+zv'(z)e^{xz}+xzv(z)e^{xz}-frac{xv(z)e^{xz}}{z^2}-frac{xv'(x)e^{xz}}{z}-frac{x^2v(z)e^{xz}}{z}-2v(z)e^{xz}-zv'(z)e^{xz}+frac{xv'(z)e^{xz}}{z}=0.$$
The $e^{xz}$ terms all cancel through, leaving:
$$v(z)+zv'(z)+xzv(z)-frac{xv(z)}{z^2}-frac{xv'(z)}{z}-frac{x^2v(z)}{z}-2v(z)-zv'(z)+frac{xv'(z)}{z}=0.$$
From here, I removed the fractions:
$$z^2v(z)+z^3v'(z)+xz^3v(z)-xv(z)-xzv'(z)-x^2zv(z)-2z^2v(z)-z^3v'(z)+xzv'(z)=0,$$
and collected the $v(z)$ and $v'(z)$ terms:
$$v(z)left(z^2+xz^3-x-x^2z-2z^2right)+v'(z)left(z^3-xz-z^3+xzright)=0,$$
which leaves:
$$left(xz^3-z^2-x^2z-xright)v(z)=0.$$
This clearly does not match the form given from the book ($v'+z^2v=0$), but I cannot see where I have made the mistake.
Is my mistake something obvious I am missing, could it be due to the nature of the complex function (and if so, what), or is my error deeper? Any help would be much appreciated.
ordinary-differential-equations laplace-transform homogeneous-equation airy-functions
asked Jan 14 at 19:07
FatsFats
468
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add a comment |
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I have been trying to read the book Airy Functions and Applications to Physics, Olivier Vallee & Manuel Soares, for research on the Airy Functions, but I am stuck on using the Laplace transform to solve the Airy equation to get the function in the correct form.
The book states (pg 5):
We consider the following homogeneous second order differential equation called the Airy's equation $$y''-xy=0.$$ This differential equation may be solved by the method of Laplace, i.e. in seeking a solution as an integral $$y=int_C e^{xz}v(z)dz,$$ this is equivalent to solve the first order differential equation $$ v'+z^2v=0.$$ We thus obtain the solution to the Airy's equation, except a normalisation constant, $$y=int_C e^{xz-z^3/3}dz.$$
Below is my working:
Using the suggestion to find a solution to the Airy Equation $y''-xy=0$ in the form $y=int_Ce^{xz}v(z)dz$, the first step would be to differentiate the expression for $y$ to find $y'$ & $y''$:
$$frac{dy}{dx}=int ze^{xz}v(z)dz,$$
$$frac{d^2y}{dx^2}=int z^2e^{xz}v(z)dz.$$
Then,
$$int z^2e^{xz}v(z)dz-xint e^{xz}v(z)dz=0,$$ so
$$int e^{xz}v(z)(z^2-x)dz=0.$$
Then I tried to use integration by parts with $u=(z^2-x)v(z)=z^2v(z)-xv(z)$, so $frac{du}{dz}=2zv(z)+z^2v'(z)-xv'(z)=2zv(z)+v'(z)(z^2-x)$, and $frac{dv}{dz}=e^{xz}$, so $v=frac{1}{z}e^{xz}$.
Following through with this gives:
$$v(z)(z^2-x)frac{1}{z}e^{xz}-int(2zv(z)+v'(z)(z^2-x))frac{1}{z}e^{xz}dz=0.$$
This (sort of) simplified to:
$$v(z)frac{left(z^2-xright)e^{xz}}{z}-int 2v(z)e^{xz}+v'(z)frac{left(z^2-xright)e^{xz}}{z}dz=0.$$
At this point, I tried to differentiate this expression to reach the form $v'+z^2v=0$:
$$frac{d}{dz}left(zv(z)e^{xz}-frac{xv(z)e^{xz}}{z}-int 2v(z)e^{xz}dz-int zv'(z)e^{xz}dz+intfrac{xv'(z)e^{xz}}{z}dzright)=0,$$
which gives
$$v(z)e^{xz}+zv'(z)e^{xz}+xzv(z)e^{xz}-frac{xv(z)e^{xz}}{z^2}-frac{xv'(x)e^{xz}}{z}-frac{x^2v(z)e^{xz}}{z}-2v(z)e^{xz}-zv'(z)e^{xz}+frac{xv'(z)e^{xz}}{z}=0.$$
The $e^{xz}$ terms all cancel through, leaving:
$$v(z)+zv'(z)+xzv(z)-frac{xv(z)}{z^2}-frac{xv'(z)}{z}-frac{x^2v(z)}{z}-2v(z)-zv'(z)+frac{xv'(z)}{z}=0.$$
From here, I removed the fractions:
$$z^2v(z)+z^3v'(z)+xz^3v(z)-xv(z)-xzv'(z)-x^2zv(z)-2z^2v(z)-z^3v'(z)+xzv'(z)=0,$$
and collected the $v(z)$ and $v'(z)$ terms:
$$v(z)left(z^2+xz^3-x-x^2z-2z^2right)+v'(z)left(z^3-xz-z^3+xzright)=0,$$
which leaves:
$$left(xz^3-z^2-x^2z-xright)v(z)=0.$$
This clearly does not match the form given from the book ($v'+z^2v=0$), but I cannot see where I have made the mistake.
Is my mistake something obvious I am missing, could it be due to the nature of the complex function (and if so, what), or is my error deeper? Any help would be much appreciated.
ordinary-differential-equations laplace-transform homogeneous-equation airy-functions
asked Jan 14 at 19:07
FatsFats
468
$endgroup$
add a comment |
$begingroup$
I have been trying to read the book Airy Functions and Applications to Physics, Olivier Vallee & Manuel Soares, for research on the Airy Functions, but I am stuck on using the Laplace transform to solve the Airy equation to get the function in the correct form.
The book states (pg 5):
We consider the following homogeneous second order differential equation called the Airy's equation $$y''-xy=0.$$ This differential equation may be solved by the method of Laplace, i.e. in seeking a solution as an integral $$y=int_C e^{xz}v(z)dz,$$ this is equivalent to solve the first order differential equation $$ v'+z^2v=0.$$ We thus obtain the solution to the Airy's equation, except a normalisation constant, $$y=int_C e^{xz-z^3/3}dz.$$
Below is my working:
Using the suggestion to find a solution to the Airy Equation $y''-xy=0$ in the form $y=int_Ce^{xz}v(z)dz$, the first step would be to differentiate the expression for $y$ to find $y'$ & $y''$:
$$frac{dy}{dx}=int ze^{xz}v(z)dz,$$
$$frac{d^2y}{dx^2}=int z^2e^{xz}v(z)dz.$$
Then,
$$int z^2e^{xz}v(z)dz-xint e^{xz}v(z)dz=0,$$ so
$$int e^{xz}v(z)(z^2-x)dz=0.$$
Then I tried to use integration by parts with $u=(z^2-x)v(z)=z^2v(z)-xv(z)$, so $frac{du}{dz}=2zv(z)+z^2v'(z)-xv'(z)=2zv(z)+v'(z)(z^2-x)$, and $frac{dv}{dz}=e^{xz}$, so $v=frac{1}{z}e^{xz}$.
Following through with this gives:
$$v(z)(z^2-x)frac{1}{z}e^{xz}-int(2zv(z)+v'(z)(z^2-x))frac{1}{z}e^{xz}dz=0.$$
This (sort of) simplified to:
$$v(z)frac{left(z^2-xright)e^{xz}}{z}-int 2v(z)e^{xz}+v'(z)frac{left(z^2-xright)e^{xz}}{z}dz=0.$$
At this point, I tried to differentiate this expression to reach the form $v'+z^2v=0$:
$$frac{d}{dz}left(zv(z)e^{xz}-frac{xv(z)e^{xz}}{z}-int 2v(z)e^{xz}dz-int zv'(z)e^{xz}dz+intfrac{xv'(z)e^{xz}}{z}dzright)=0,$$
which gives
$$v(z)e^{xz}+zv'(z)e^{xz}+xzv(z)e^{xz}-frac{xv(z)e^{xz}}{z^2}-frac{xv'(x)e^{xz}}{z}-frac{x^2v(z)e^{xz}}{z}-2v(z)e^{xz}-zv'(z)e^{xz}+frac{xv'(z)e^{xz}}{z}=0.$$
The $e^{xz}$ terms all cancel through, leaving:
$$v(z)+zv'(z)+xzv(z)-frac{xv(z)}{z^2}-frac{xv'(z)}{z}-frac{x^2v(z)}{z}-2v(z)-zv'(z)+frac{xv'(z)}{z}=0.$$
From here, I removed the fractions:
$$z^2v(z)+z^3v'(z)+xz^3v(z)-xv(z)-xzv'(z)-x^2zv(z)-2z^2v(z)-z^3v'(z)+xzv'(z)=0,$$
and collected the $v(z)$ and $v'(z)$ terms:
$$v(z)left(z^2+xz^3-x-x^2z-2z^2right)+v'(z)left(z^3-xz-z^3+xzright)=0,$$
which leaves:
$$left(xz^3-z^2-x^2z-xright)v(z)=0.$$
This clearly does not match the form given from the book ($v'+z^2v=0$), but I cannot see where I have made the mistake.
Is my mistake something obvious I am missing, could it be due to the nature of the complex function (and if so, what), or is my error deeper? Any help would be much appreciated.
ordinary-differential-equations laplace-transform homogeneous-equation airy-functions
asked Jan 14 at 19:07
FatsFats
468
$endgroup$
I have been trying to read the book Airy Functions and Applications to Physics, Olivier Vallee & Manuel Soares, for research on the Airy Functions, but I am stuck on using the Laplace transform to solve the Airy equation to get the function in the correct form.
The book states (pg 5):
We consider the following homogeneous second order differential equation called the Airy's equation $$y''-xy=0.$$ This differential equation may be solved by the method of Laplace, i.e. in seeking a solution as an integral $$y=int_C e^{xz}v(z)dz,$$ this is equivalent to solve the first order differential equation $$ v'+z^2v=0.$$ We thus obtain the solution to the Airy's equation, except a normalisation constant, $$y=int_C e^{xz-z^3/3}dz.$$
Below is my working:
Using the suggestion to find a solution to the Airy Equation $y''-xy=0$ in the form $y=int_Ce^{xz}v(z)dz$, the first step would be to differentiate the expression for $y$ to find $y'$ & $y''$:
$$frac{dy}{dx}=int ze^{xz}v(z)dz,$$
$$frac{d^2y}{dx^2}=int z^2e^{xz}v(z)dz.$$
Then,
$$int z^2e^{xz}v(z)dz-xint e^{xz}v(z)dz=0,$$ so
$$int e^{xz}v(z)(z^2-x)dz=0.$$
Then I tried to use integration by parts with $u=(z^2-x)v(z)=z^2v(z)-xv(z)$, so $frac{du}{dz}=2zv(z)+z^2v'(z)-xv'(z)=2zv(z)+v'(z)(z^2-x)$, and $frac{dv}{dz}=e^{xz}$, so $v=frac{1}{z}e^{xz}$.
Following through with this gives:
$$v(z)(z^2-x)frac{1}{z}e^{xz}-int(2zv(z)+v'(z)(z^2-x))frac{1}{z}e^{xz}dz=0.$$
This (sort of) simplified to:
$$v(z)frac{left(z^2-xright)e^{xz}}{z}-int 2v(z)e^{xz}+v'(z)frac{left(z^2-xright)e^{xz}}{z}dz=0.$$
At this point, I tried to differentiate this expression to reach the form $v'+z^2v=0$:
$$frac{d}{dz}left(zv(z)e^{xz}-frac{xv(z)e^{xz}}{z}-int 2v(z)e^{xz}dz-int zv'(z)e^{xz}dz+intfrac{xv'(z)e^{xz}}{z}dzright)=0,$$
which gives
$$v(z)e^{xz}+zv'(z)e^{xz}+xzv(z)e^{xz}-frac{xv(z)e^{xz}}{z^2}-frac{xv'(x)e^{xz}}{z}-frac{x^2v(z)e^{xz}}{z}-2v(z)e^{xz}-zv'(z)e^{xz}+frac{xv'(z)e^{xz}}{z}=0.$$
The $e^{xz}$ terms all cancel through, leaving:
$$v(z)+zv'(z)+xzv(z)-frac{xv(z)}{z^2}-frac{xv'(z)}{z}-frac{x^2v(z)}{z}-2v(z)-zv'(z)+frac{xv'(z)}{z}=0.$$
From here, I removed the fractions:
$$z^2v(z)+z^3v'(z)+xz^3v(z)-xv(z)-xzv'(z)-x^2zv(z)-2z^2v(z)-z^3v'(z)+xzv'(z)=0,$$
and collected the $v(z)$ and $v'(z)$ terms:
$$v(z)left(z^2+xz^3-x-x^2z-2z^2right)+v'(z)left(z^3-xz-z^3+xzright)=0,$$
which leaves:
$$left(xz^3-z^2-x^2z-xright)v(z)=0.$$
This clearly does not match the form given from the book ($v'+z^2v=0$), but I cannot see where I have made the mistake.
Is my mistake something obvious I am missing, could it be due to the nature of the complex function (and if so, what), or is my error deeper? Any help would be much appreciated.
ordinary-differential-equations laplace-transform homogeneous-equation airy-functions
ordinary-differential-equations laplace-transform homogeneous-equation airy-functions
asked Jan 14 at 19:07
FatsFats
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asked Jan 14 at 19:07
FatsFats
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asked Jan 14 at 19:07
FatsFats
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asked Jan 14 at 19:07
FatsFats
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asked Jan 14 at 19:07
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You are overthinking it. Starting from the beginning
$$ int_C z^2 e^{xz} v(z) dz - int_C xe^{xz} v(z) dz = 0 $$
Integration by parts gives
$$ int_C xe^{xz} v(z) dz = e^{xz}v(z) Big vert_C - int_C e^{xz} v'(z) dz $$
Here $C$ is a vertical line on the complex plane, such that all singularities of $v(z)$ lies on the left of it. If we hand-wave away that the first term goes to $0$, we can simplify
$$ int_C e^{xz}(z^2 v(z) + v'(z)) dz = 0 $$
Notice the LHS is a function of $x$ but the RHS is identically $0$, so it must follow that
$$ v'(z) + z^2 v(z) = 0 $$
Which can be solved by separation of variables
$$ v(z) = e^{-z^3/3} $$
up to a multiplicative constant.
This answer doesn't justify everything, but hopefully it can explain some things.
It should also be noted that this text uses a less accessible form of the inverse Laplace transform which uses complex analysis. A more common definition is the forward transform
$$ v(z) = int_0^{infty} e^{-zx} y(x) dx $$
Applying the forward transform to the original equation, and using integration by parts, we can show that
begin{align}
int_0^infty e^{-zx} y''(x) dx &= -y'(0) + int_0^infty ze^{-zx} y'(x) dx \
&= -y'(0) - zy(0) + int_0^infty z^2e^{-zx}y(x) dx \ &= -zy(0) - y'(0) + z^2v(z) \
int_0^infty e^{-zx} xy dx &= -int_0^infty frac{d}{dz} e^{-zx} y(x) dx \
&= -v'(z)
end{align}
where the second result is justified by differentiation under the integral. Then
$$ v'(z) + z^2v(z) = zy(0) + y'(0) $$
Letting $y(0) = y'(0) = 0$ leads to the given result.
Edit: Your mistake is in the very first integration by parts:
$$frac{dv}{dz}=e^{xz} implies v=frac{1}{color{red}{x}}e^{xz}$$
answered Jan 15 at 6:35
DylanDylan
12.9k31027
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I knew the main bulk of my error would be something trivial, which I couldn't see, so thank you for pointing that out! And, thank you for the second method with the forward transform - I think it's much easier to follow (and less likely for me to get wrong again). The one part I am slightly confused about is how I would know the $C$ is a vertical line on the complex plane, as it wasn't stated in the book.
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– Fats
Jan 16 at 16:04
$begingroup$
I'm not entirely sure why they would do this. Maybe physicists do things differently? In every differential equations class, they will generally teach you the forward transform, give a look-up table to go backwards, and not even mention the inverse transform at all.
$endgroup$
– Dylan
Jan 16 at 17:52
$begingroup$
Another follow up, how then could I use the forward transform and $v(z)=e^{-z^3/3}$ to find $y(x)$?
$endgroup$
– Fats
Jan 17 at 15:20
$begingroup$
The Airy function is non-elementary, so you'll just have to leave it as an integral.
$endgroup$
– Dylan
Jan 17 at 17:33
add a comment |
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$begingroup$
You are overthinking it. Starting from the beginning
$$ int_C z^2 e^{xz} v(z) dz - int_C xe^{xz} v(z) dz = 0 $$
Integration by parts gives
$$ int_C xe^{xz} v(z) dz = e^{xz}v(z) Big vert_C - int_C e^{xz} v'(z) dz $$
Here $C$ is a vertical line on the complex plane, such that all singularities of $v(z)$ lies on the left of it. If we hand-wave away that the first term goes to $0$, we can simplify
$$ int_C e^{xz}(z^2 v(z) + v'(z)) dz = 0 $$
Notice the LHS is a function of $x$ but the RHS is identically $0$, so it must follow that
$$ v'(z) + z^2 v(z) = 0 $$
Which can be solved by separation of variables
$$ v(z) = e^{-z^3/3} $$
up to a multiplicative constant.
This answer doesn't justify everything, but hopefully it can explain some things.
It should also be noted that this text uses a less accessible form of the inverse Laplace transform which uses complex analysis. A more common definition is the forward transform
$$ v(z) = int_0^{infty} e^{-zx} y(x) dx $$
Applying the forward transform to the original equation, and using integration by parts, we can show that
begin{align}
int_0^infty e^{-zx} y''(x) dx &= -y'(0) + int_0^infty ze^{-zx} y'(x) dx \
&= -y'(0) - zy(0) + int_0^infty z^2e^{-zx}y(x) dx \ &= -zy(0) - y'(0) + z^2v(z) \
int_0^infty e^{-zx} xy dx &= -int_0^infty frac{d}{dz} e^{-zx} y(x) dx \
&= -v'(z)
end{align}
where the second result is justified by differentiation under the integral. Then
$$ v'(z) + z^2v(z) = zy(0) + y'(0) $$
Letting $y(0) = y'(0) = 0$ leads to the given result.
Edit: Your mistake is in the very first integration by parts:
$$frac{dv}{dz}=e^{xz} implies v=frac{1}{color{red}{x}}e^{xz}$$
answered Jan 15 at 6:35
DylanDylan
12.9k31027
$endgroup$
$begingroup$
I knew the main bulk of my error would be something trivial, which I couldn't see, so thank you for pointing that out! And, thank you for the second method with the forward transform - I think it's much easier to follow (and less likely for me to get wrong again). The one part I am slightly confused about is how I would know the $C$ is a vertical line on the complex plane, as it wasn't stated in the book.
$endgroup$
– Fats
Jan 16 at 16:04
$begingroup$
I'm not entirely sure why they would do this. Maybe physicists do things differently? In every differential equations class, they will generally teach you the forward transform, give a look-up table to go backwards, and not even mention the inverse transform at all.
$endgroup$
– Dylan
Jan 16 at 17:52
$begingroup$
Another follow up, how then could I use the forward transform and $v(z)=e^{-z^3/3}$ to find $y(x)$?
$endgroup$
– Fats
Jan 17 at 15:20
$begingroup$
The Airy function is non-elementary, so you'll just have to leave it as an integral.
$endgroup$
– Dylan
Jan 17 at 17:33
add a comment |
$begingroup$
You are overthinking it. Starting from the beginning
$$ int_C z^2 e^{xz} v(z) dz - int_C xe^{xz} v(z) dz = 0 $$
Integration by parts gives
$$ int_C xe^{xz} v(z) dz = e^{xz}v(z) Big vert_C - int_C e^{xz} v'(z) dz $$
Here $C$ is a vertical line on the complex plane, such that all singularities of $v(z)$ lies on the left of it. If we hand-wave away that the first term goes to $0$, we can simplify
$$ int_C e^{xz}(z^2 v(z) + v'(z)) dz = 0 $$
Notice the LHS is a function of $x$ but the RHS is identically $0$, so it must follow that
$$ v'(z) + z^2 v(z) = 0 $$
Which can be solved by separation of variables
$$ v(z) = e^{-z^3/3} $$
up to a multiplicative constant.
This answer doesn't justify everything, but hopefully it can explain some things.
It should also be noted that this text uses a less accessible form of the inverse Laplace transform which uses complex analysis. A more common definition is the forward transform
$$ v(z) = int_0^{infty} e^{-zx} y(x) dx $$
Applying the forward transform to the original equation, and using integration by parts, we can show that
begin{align}
int_0^infty e^{-zx} y''(x) dx &= -y'(0) + int_0^infty ze^{-zx} y'(x) dx \
&= -y'(0) - zy(0) + int_0^infty z^2e^{-zx}y(x) dx \ &= -zy(0) - y'(0) + z^2v(z) \
int_0^infty e^{-zx} xy dx &= -int_0^infty frac{d}{dz} e^{-zx} y(x) dx \
&= -v'(z)
end{align}
where the second result is justified by differentiation under the integral. Then
$$ v'(z) + z^2v(z) = zy(0) + y'(0) $$
Letting $y(0) = y'(0) = 0$ leads to the given result.
Edit: Your mistake is in the very first integration by parts:
$$frac{dv}{dz}=e^{xz} implies v=frac{1}{color{red}{x}}e^{xz}$$
answered Jan 15 at 6:35
DylanDylan
12.9k31027
$endgroup$
$begingroup$
I knew the main bulk of my error would be something trivial, which I couldn't see, so thank you for pointing that out! And, thank you for the second method with the forward transform - I think it's much easier to follow (and less likely for me to get wrong again). The one part I am slightly confused about is how I would know the $C$ is a vertical line on the complex plane, as it wasn't stated in the book.
$endgroup$
– Fats
Jan 16 at 16:04
$begingroup$
I'm not entirely sure why they would do this. Maybe physicists do things differently? In every differential equations class, they will generally teach you the forward transform, give a look-up table to go backwards, and not even mention the inverse transform at all.
$endgroup$
– Dylan
Jan 16 at 17:52
$begingroup$
Another follow up, how then could I use the forward transform and $v(z)=e^{-z^3/3}$ to find $y(x)$?
$endgroup$
– Fats
Jan 17 at 15:20
$begingroup$
The Airy function is non-elementary, so you'll just have to leave it as an integral.
$endgroup$
– Dylan
Jan 17 at 17:33
add a comment |
$begingroup$
You are overthinking it. Starting from the beginning
$$ int_C z^2 e^{xz} v(z) dz - int_C xe^{xz} v(z) dz = 0 $$
Integration by parts gives
$$ int_C xe^{xz} v(z) dz = e^{xz}v(z) Big vert_C - int_C e^{xz} v'(z) dz $$
Here $C$ is a vertical line on the complex plane, such that all singularities of $v(z)$ lies on the left of it. If we hand-wave away that the first term goes to $0$, we can simplify
$$ int_C e^{xz}(z^2 v(z) + v'(z)) dz = 0 $$
Notice the LHS is a function of $x$ but the RHS is identically $0$, so it must follow that
$$ v'(z) + z^2 v(z) = 0 $$
Which can be solved by separation of variables
$$ v(z) = e^{-z^3/3} $$
up to a multiplicative constant.
This answer doesn't justify everything, but hopefully it can explain some things.
It should also be noted that this text uses a less accessible form of the inverse Laplace transform which uses complex analysis. A more common definition is the forward transform
$$ v(z) = int_0^{infty} e^{-zx} y(x) dx $$
Applying the forward transform to the original equation, and using integration by parts, we can show that
begin{align}
int_0^infty e^{-zx} y''(x) dx &= -y'(0) + int_0^infty ze^{-zx} y'(x) dx \
&= -y'(0) - zy(0) + int_0^infty z^2e^{-zx}y(x) dx \ &= -zy(0) - y'(0) + z^2v(z) \
int_0^infty e^{-zx} xy dx &= -int_0^infty frac{d}{dz} e^{-zx} y(x) dx \
&= -v'(z)
end{align}
where the second result is justified by differentiation under the integral. Then
$$ v'(z) + z^2v(z) = zy(0) + y'(0) $$
Letting $y(0) = y'(0) = 0$ leads to the given result.
Edit: Your mistake is in the very first integration by parts:
$$frac{dv}{dz}=e^{xz} implies v=frac{1}{color{red}{x}}e^{xz}$$
answered Jan 15 at 6:35
DylanDylan
12.9k31027
$endgroup$
You are overthinking it. Starting from the beginning
$$ int_C z^2 e^{xz} v(z) dz - int_C xe^{xz} v(z) dz = 0 $$
Integration by parts gives
$$ int_C xe^{xz} v(z) dz = e^{xz}v(z) Big vert_C - int_C e^{xz} v'(z) dz $$
Here $C$ is a vertical line on the complex plane, such that all singularities of $v(z)$ lies on the left of it. If we hand-wave away that the first term goes to $0$, we can simplify
$$ int_C e^{xz}(z^2 v(z) + v'(z)) dz = 0 $$
Notice the LHS is a function of $x$ but the RHS is identically $0$, so it must follow that
$$ v'(z) + z^2 v(z) = 0 $$
Which can be solved by separation of variables
$$ v(z) = e^{-z^3/3} $$
up to a multiplicative constant.
This answer doesn't justify everything, but hopefully it can explain some things.
It should also be noted that this text uses a less accessible form of the inverse Laplace transform which uses complex analysis. A more common definition is the forward transform
$$ v(z) = int_0^{infty} e^{-zx} y(x) dx $$
Applying the forward transform to the original equation, and using integration by parts, we can show that
begin{align}
int_0^infty e^{-zx} y''(x) dx &= -y'(0) + int_0^infty ze^{-zx} y'(x) dx \
&= -y'(0) - zy(0) + int_0^infty z^2e^{-zx}y(x) dx \ &= -zy(0) - y'(0) + z^2v(z) \
int_0^infty e^{-zx} xy dx &= -int_0^infty frac{d}{dz} e^{-zx} y(x) dx \
&= -v'(z)
end{align}
where the second result is justified by differentiation under the integral. Then
$$ v'(z) + z^2v(z) = zy(0) + y'(0) $$
Letting $y(0) = y'(0) = 0$ leads to the given result.
Edit: Your mistake is in the very first integration by parts:
$$frac{dv}{dz}=e^{xz} implies v=frac{1}{color{red}{x}}e^{xz}$$
answered Jan 15 at 6:35
DylanDylan
12.9k31027
edited Jan 16 at 17:45
answered Jan 15 at 6:35
DylanDylan
12.9k31027
answered Jan 15 at 6:35
DylanDylan
12.9k31027
answered Jan 15 at 6:35
DylanDylan
12.9k31027
12.9k31027
$begingroup$
I knew the main bulk of my error would be something trivial, which I couldn't see, so thank you for pointing that out! And, thank you for the second method with the forward transform - I think it's much easier to follow (and less likely for me to get wrong again). The one part I am slightly confused about is how I would know the $C$ is a vertical line on the complex plane, as it wasn't stated in the book.
$endgroup$
– Fats
Jan 16 at 16:04
$begingroup$
I'm not entirely sure why they would do this. Maybe physicists do things differently? In every differential equations class, they will generally teach you the forward transform, give a look-up table to go backwards, and not even mention the inverse transform at all.
$endgroup$
– Dylan
Jan 16 at 17:52
$begingroup$
Another follow up, how then could I use the forward transform and $v(z)=e^{-z^3/3}$ to find $y(x)$?
$endgroup$
– Fats
Jan 17 at 15:20
$begingroup$
The Airy function is non-elementary, so you'll just have to leave it as an integral.
$endgroup$
– Dylan
Jan 17 at 17:33
add a comment |
$begingroup$
I knew the main bulk of my error would be something trivial, which I couldn't see, so thank you for pointing that out! And, thank you for the second method with the forward transform - I think it's much easier to follow (and less likely for me to get wrong again). The one part I am slightly confused about is how I would know the $C$ is a vertical line on the complex plane, as it wasn't stated in the book.
$endgroup$
– Fats
Jan 16 at 16:04
$begingroup$
I'm not entirely sure why they would do this. Maybe physicists do things differently? In every differential equations class, they will generally teach you the forward transform, give a look-up table to go backwards, and not even mention the inverse transform at all.
$endgroup$
– Dylan
Jan 16 at 17:52
$begingroup$
Another follow up, how then could I use the forward transform and $v(z)=e^{-z^3/3}$ to find $y(x)$?
$endgroup$
– Fats
Jan 17 at 15:20
$begingroup$
The Airy function is non-elementary, so you'll just have to leave it as an integral.
$endgroup$
– Dylan
Jan 17 at 17:33
$begingroup$
I knew the main bulk of my error would be something trivial, which I couldn't see, so thank you for pointing that out! And, thank you for the second method with the forward transform - I think it's much easier to follow (and less likely for me to get wrong again). The one part I am slightly confused about is how I would know the $C$ is a vertical line on the complex plane, as it wasn't stated in the book.
$endgroup$
– Fats
Jan 16 at 16:04
$begingroup$
I knew the main bulk of my error would be something trivial, which I couldn't see, so thank you for pointing that out! And, thank you for the second method with the forward transform - I think it's much easier to follow (and less likely for me to get wrong again). The one part I am slightly confused about is how I would know the $C$ is a vertical line on the complex plane, as it wasn't stated in the book.
$endgroup$
– Fats
Jan 16 at 16:04
$begingroup$
I'm not entirely sure why they would do this. Maybe physicists do things differently? In every differential equations class, they will generally teach you the forward transform, give a look-up table to go backwards, and not even mention the inverse transform at all.
$endgroup$
– Dylan
Jan 16 at 17:52
$begingroup$
I'm not entirely sure why they would do this. Maybe physicists do things differently? In every differential equations class, they will generally teach you the forward transform, give a look-up table to go backwards, and not even mention the inverse transform at all.
$endgroup$
– Dylan
Jan 16 at 17:52
$begingroup$
Another follow up, how then could I use the forward transform and $v(z)=e^{-z^3/3}$ to find $y(x)$?
$endgroup$
– Fats
Jan 17 at 15:20
$begingroup$
Another follow up, how then could I use the forward transform and $v(z)=e^{-z^3/3}$ to find $y(x)$?
$endgroup$
– Fats
Jan 17 at 15:20
$begingroup$
The Airy function is non-elementary, so you'll just have to leave it as an integral.
$endgroup$
– Dylan
Jan 17 at 17:33
$begingroup$
The Airy function is non-elementary, so you'll just have to leave it as an integral.
$endgroup$
– Dylan
Jan 17 at 17:33
add a comment |
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