Given that $\mu_V$, $\lambda_V$, and $\sigma_V$ are constants, we can solve the partial differential equation (PDE):

$\frac{\partial q(V, \tau)}{\partial \tau} = (\mu_V - \lambda_V) \frac{\partial q(V, \tau)}{\partial V} + \frac{1}{2} \sigma_V^2 V^2 \frac{\partial^2 q(V, \tau)}{\partial V^2}$

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

Step-by-Step Solution

1. Transform the PDE:

   First, consider a transformation of the solution $q(V, \tau)$ to a simpler form. Let's define:

   $q(V, \tau) = \frac{1}{V} u(V, \tau)$

   Substituting into the original PDE gives:

   $\frac{\partial}{\partial \tau} \left(\frac{1}{V} u(V, \tau)\right) = (\mu_V - \lambda_V) \frac{\partial}{\partial V} \left(\frac{1}{V} u(V, \tau)\right) + \frac{1}{2} \sigma_V^2 V^2 \frac{\partial^2}{\partial V^2} \left(\frac{1}{V} u(V, \tau)\right)$

   Simplifying this, we obtain a new PDE for $u(V, \tau)$:

   $\frac{\partial u(V, \tau)}{\partial \tau} = (\mu_V - \lambda_V) \frac{\partial u(V, \tau)}{\partial V} + \frac{1}{2} \sigma_V^2 V^2 \frac{\partial^2 u(V, \tau)}{\partial V^2}$

   The boundary condition transforms to $u(V, 0) = 1$.

2. Solve the Transformed PDE:

   The transformed PDE is now in a form that is easier to solve. Notice that this is a linear PDE of the heat equation type, often solved using methods such as separation of variables, the method of characteristics, or by assuming a particular form of solution.

   Since $u(V, \tau)$ doesn't depend on $\tau$ in the initial condition, we assume a solution of the form:

   $u(V, \tau) = \exp\left( A(\tau) V + B(\tau) \right)$

   Substituting into the PDE gives us a system of ODEs for $A(\tau)$ and $B(\tau)$. After some calculations, we obtain:

   $A(\tau) = 0 \quad \text{and} \quad B(\tau) = \text{constant}.$

   Thus, the solution for $u(V, \tau)$ is:

   $u(V, \tau) = \exp\left(B(\tau)\right)$

   Given the boundary condition $u(V, 0) = 1$, we find that $B(\tau) = 0$, and thus:

   $u(V, \tau) = 1$

3. Substitute Back to Find $q(V, \tau)$:

   Since $q(V, \tau) = \frac{1}{V} u(V, \tau)$ and $u(V, \tau) = 1$, we have:

   $q(V, \tau) = \frac{1}{V}$

Final Solution

The solution to the PDE with the given boundary condition is:

$q(V, \tau) = \frac{1}{V}$

This solution indicates that $q(V, \tau)$ is independent of $\tau$, remaining constant at $\frac{1}{V}$.

Expanding Knowledge

1. What are common boundary conditions used in PDEs, and how do they affect the solution?
2. How does the assumption of constant coefficients simplify solving PDEs?
3. What are the typical characteristics of heat equation solutions in one dimension?
4. Can this PDE be generalized to non-constant coefficients, and how would that change the solution approach?
5. How would numerical methods like finite difference or finite element handle solving this PDE?

Tip: When faced with complex PDEs, try simplifying the problem by looking for invariant solutions or using symmetry arguments to reduce the number of variables.