The given partial differential equation (PDE) is:

$\frac{\partial q(V, \tau)}{\partial \tau} = (\mu_{V} - \lambda_{V}) \frac{\partial q(V, \tau)}{\partial V} + \frac{1}{2} \sigma_{V}^2 V^2 \frac{\partial^2 q(V, \tau)}{\partial V^2}$

with the boundary condition $q(V, 0) = \frac{1}{V}$.

Solution Method

This PDE is a linear parabolic equation, commonly encountered in financial mathematics, particularly in models like the Cox-Ingersoll-Ross (CIR) or similar interest rate models. Here's a step-by-step outline of how to approach solving this PDE:

1. Transform the PDE: To simplify the PDE, we can often use a change of variables or guess a solution form based on the boundary condition. Let's try the transformation: $q(V, \tau) = \frac{1}{V} u(V, \tau)$ Substituting this into the original PDE, we get: $\frac{\partial u}{\partial \tau} = (\mu_V - \lambda_V)\frac{\partial u}{\partial V} + \frac{1}{2} \sigma_V^2 V^2 \frac{\partial^2 u}{\partial V^2}$ and the boundary condition transforms to $u(V, 0) = 1$.

2. Solve the transformed PDE: The transformed equation is now in a form that is easier to solve. The solution approach typically involves assuming a solution form, using a method like separation of variables, or solving via a Green's function, depending on the complexity of the coefficients.

   Assuming a separable solution $u(V, \tau) = f(V)g(\tau)$, we substitute this into the transformed PDE and separate variables to solve the resulting ODEs for $f(V)$ and $g(\tau)$.

3. Revert to the original function: After finding $u(V, \tau)$, we substitute back to find $q(V, \tau) = \frac{1}{V} u(V, \tau)$.

Specific Solution

Given the linearity of the PDE, the solution will likely involve an exponential or power function depending on the specific form of $\mu_V$, $\lambda_V$, and $\sigma_V$. Assuming the parameters are constants, the general solution will take a form similar to:

$q(V, \tau) = \frac{1}{V} \exp\left( a(\tau) V + b(\tau) \right)$

Where $a(\tau)$ and $b(\tau)$ are functions determined by solving the corresponding ordinary differential equations (ODEs) derived from substituting this assumed solution form into the transformed PDE.

Since the boundary condition is $q(V, 0) = \frac{1}{V}$, this implies specific conditions on $a(0)$ and $b(0)$ which will help solve these ODEs.

Final Expression

Without loss of generality and assuming constant coefficients, the final solution can be complex and involve special functions or integrals if the coefficients are non-constant. A complete analytical solution would require explicit forms of $\mu_V$, $\lambda_V$, and $\sigma_V$.

Would you like a more detailed solution involving the specifics of these functions, or have any questions?

Expanding Knowledge

1. How does the choice of boundary conditions affect the solution of a PDE?
2. What are common applications of this type of PDE in financial mathematics?
3. How does the separation of variables method work for PDEs like this?
4. What are Green's functions, and how are they used in solving PDEs?
5. How can numerical methods be applied if an analytical solution is too complex?

Tip: When solving PDEs, always check if the equation allows for symmetry or a specific transformation to simplify the process.