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I'm trying to obtain the Laplace transform of Call option price with repect to time to maturity under the CEV process.

The well known Black scholes PDE is given by $$ \frac{1}{2}\sigma(x)^2x^2\frac{\partial^2}{\partial x^2}C(x,\tau)+\mu x\frac{\partial}{\partial x}C(x,\tau)-rC(x,\tau)-\frac{\partial}{\partial \tau}C(x,\tau)=0. $$ where the initial condition $C(x,0)=max(x-K,0)$ and $\sigma(x)=\delta x^\beta$.

Taking a Lapalce transform with respect to $\tau$, we obtain the following ODE : $$ \frac{1}{2}\delta x^{2\beta+2}\frac{\partial^2}{\partial x^2}\hat{C}(x,\lambda)+\mu x\frac{\partial}{\partial x}\hat{C}(x,\lambda)-(\lambda+r)\hat{C}(x,\lambda)=-max(x-K,0). $$ where $\hat{C}(x,\lambda)=\int_0^\infty e^{-\lambda \tau}C(x,\tau)d\tau$.

and the initial condition is transformed to $$ \hat{C}(x,\lambda)=\int_0^\infty e^{-\lambda \tau}C(x,0) d\tau=max(x-K,0)/\lambda $$(is this right??? it seems wrong..)

. Then, $\hat{C}(x,\lambda)$ can be analytically formulated by the case $x>K$ and $x\leq K$.

How to get explicit formula for $\hat{C}(x,\tau)$?. I can't proceed from this stage.

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Firstly, note that by taking Laplace transforms time has been eliminated and as such there are no initial conditions for your ODE anymore. There are only boundary conditions which in this case read $C(\pm \infty, \tau) = 0$. By the way it is customary to denote the price of the underlying as $S$ whereas you denote it as $x$..

If you look at the ODE you see that it is very similar to the Euler equation. Thus it is useful to substitute for $y := \log(x)$. Having done that we end up with : \begin{equation} (i) \frac{\delta}{2} \frac{d^2}{d y^2} \tilde{C}(y,\lambda) + \left(\mu - \frac{\delta}{2} e^{2 \beta y}\right) \frac{d}{d y} \tilde{C}(y,\lambda) - (\lambda + r) \tilde{C}(y,\lambda) = - max(e^y - K,0) \end{equation}

This is allready a much simpler equation yet since the coefficient at the first derivative is not constant we still have to transform it. We solve firstly the homogenuous equation. Here let us try the a following ansatz: \begin{equation} (ii) \tilde{C}(y,\lambda) := \exp(2 A \beta y) \tilde{\tilde{C}}(\exp(2 \beta y)) \end{equation} Inserting this ansatz to the above equation we will obtain another second order equation for the function $\tilde{\tilde{C}}$. The later equation will be grossly simplified if we require the constant term in the coefficient at the zeroth derivative to be zero. This gives: $-2 A^2 b^2 \delta - 2 \mu A b + r + \lambda =0$ which gives $A = (-\mu + \sqrt{\mu^2 + 2 \delta (\lambda+r)})/(2 \beta \delta)$ . Taking all this into account we get a following ODE: \begin{equation} (iii) A \tilde{\tilde{C}}(u,\lambda) + (B_0 - B_1 u )d_u \tilde{\tilde{C}}(u,\lambda) + C u d^2_{u^2} \tilde{\tilde{C}}(u,\lambda) = 0 \end{equation}

where $A = (\mu - \sqrt{\mu^2+2\delta(\lambda+r)})$ and $B_0 = 2 \beta (2\beta \delta + 2 \sqrt{\mu^2 + 2 \delta(\lambda+r)}$, $B_1 = 2\beta\delta$ and $C = (2 \beta)^2 \delta$ and $u := \exp(2 \beta y)$. We substitute for $B_1/C u$ and we recognize the generalized Laguerre differential equation, an equation this is easily solved through a power series method. Now, combining (iii) with (ii) we obtain the fundamental solutions of the homogenous ODE (i) . Having done this we solve the inhomogenuous ODE (i) by means of the Green's function method, for example.

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