Following the solution provided by *Timothy Falcon Crack - Heard on the Street, Quantitative Questions from Wall Street Job Interviews:
For $S \geq \underline{S} \equiv \frac{\lambda_2 K}{\lambda_2 - 1}$, the value of the perpetual American put option is determined as: $$ V(S) = (K - \underline{S}) \left( \frac{S}{\underline{S}} \right)^{\lambda_2} = \left( \frac{K}{1 - \lambda_2} \right) \left( \frac{S (\lambda_2 - 1)}{\lambda_2 K} \right)^{\lambda_2}, $$ where: $$ \lambda_2 = \frac{-\left(r - q - \frac{1}{2} \sigma^2\right) - \sqrt{\left(r - q - \frac{1}{2} \sigma^2\right)^2 + 2 \sigma^2 r}}{\sigma^2}. $$
Conversely, for a perpetual American call option, within the range $0 \leq S \leq \frac{\lambda_1 K}{\lambda_1 - 1} \equiv \bar{S}$, its value is given by: $$ V(S) = \left( \frac{K}{\lambda_1 - 1} \right) \left( \frac{S (\lambda_1 - 1)}{\lambda_1 K} \right)^{\lambda_1}, $$ where: $$ \lambda_1 = \frac{-\left(r - q - \frac{1}{2} \sigma^2\right) + \sqrt{\left(r - q - \frac{1}{2} \sigma^2\right)^2 + 2 \sigma^2 r}}{\sigma^2}. $$
I understand (though not in depth, as I am not fully confident solving ODEs and PDEs) how the solution for the perpetual American put option is derived. However, I have doubts regarding the solution provided for the perpetual American call option.
The boundary conditions for the call should alter the delta of the option and the payoff: $$ \frac{\partial V}{\partial S} \bigg|_{S = \bar{S}} = 1, \quad \text{and} \quad V(S = \bar{S}) = \bar{S} - K. $$
The representation of the ODE’s solutions remains the same, yielding $\lambda_1$ and $\lambda_2$ as in the put case. These solutions are represented as a linear combination of two linearly independent solutions: $$ V(S) = A_1 S^{\lambda_1} + A_2 S^{\lambda_2}, $$ where $A_1$ and $A_2$ are constants. From the ODE, the degree of the derivatives of $S$ in the coefficients suggests solutions are powers of $S$: $$ V^1 = S^{\lambda_1}, \quad V^2 = S^{\lambda_2}. $$
The solutions for $\lambda_i$ satisfy: $$ \lambda_1 > 0 \quad \text{if } r > 0, \quad \text{and} \quad \lambda_2 < 0 \quad \text{if } r > 0. $$
If $|A_1| > 0$, we observe that: $$ \lim_{S \to \infty} \left( A_1 S^{\lambda_1} + A_2 S^{\lambda_2} \right) = \pm \infty, $$ which forces the solution for the call option to be: $$ V(\bar{S}) = A_2 \bar{S}^{\lambda_2} = \bar{S} - K. $$
Question:
From what I can deduce, the value of the perpetual American call option appears to follow a similar form to the put option: $V(\bar{S}) = A_2 \bar{S}^{\lambda_2} = \bar{S} - K$. How does the author derive the call option value in terms of $\lambda_1$, as provided in the solution?
