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,
$$\int z^2e^{xz}v(z)dz-x\int e^{xz}v(z)dz=0,$$ 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:
$$v(z)\frac{\left(z^2-x\right)e^{xz}}{z}-\int 2v(z)e^{xz}+v'(z)\frac{\left(z^2-x\right)e^{xz}}{z}dz=0.$$
At this point, I tried to differentiate this expression to reach the form $v'+z^2v=0$:
$$\frac{d}{dz}\left(zv(z)e^{xz}-\frac{xv(z)e^{xz}}{z}-\int 2v(z)e^{xz}dz-\int zv'(z)e^{xz}dz+\int\frac{xv'(z)e^{xz}}{z}dz\right)=0,$$
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:
$$v(z)\left(z^2+xz^3-x-x^2z-2z^2\right)+v'(z)\left(z^3-xz-z^3+xz\right)=0,$$
which leaves:
$$\left(xz^3-z^2-x^2z-x\right)v(z)=0.$$
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.
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:
$$\frac{dv}{dz}=e^{xz} \implies v=\frac{1}{\color{red}{x}}e^{xz}$$