Solving a Piecewise Cauchy–Euler Bond Pricing Equation

Summary

The document examines a perpetual coupon defaultable bond model described by a Cauchy–Euler differential equation, with the discount rate changing at a threshold in the issuer’s asset value. For a constant rate, the stated solution combines a constant term with powers of the asset value, and its coefficients depend on boundary conditions. The question is whether solutions for the two rate regions can be joined at the threshold.

The answer gives a piecewise solution for the special case where the asset drift equals the rate in the higher-value region. It uses a separate constant-plus-power form on each side of the threshold, then determines coefficients by substituting into the equation and applying boundary conditions and smooth pasting. This is a limited result rather than a general derivation for arbitrary parameters. The document supplies no numerical example or discussion of boundary-condition choices, so the setup and matching conditions must be adapted to the specific bond model.

Key ideas

• A threshold-dependent discount rate divides the pricing equation into two regions.
• For a constant rate, the equation has a solution formed from constant and power terms.
• The answer constructs separate solution forms on either side of the threshold for a special parameter case.
• Coefficients are determined using the differential equation, boundary conditions, and smooth pasting.

Tags

# Cauchy-Euler ODE with indicator function in coefficient


# Cauchy-Euler ODE with indicator function in coefficient












Consider the following Cauchy-Euler ODE, which is in particular the asset pricing equation for a (perpetual coupon defaultable) bond:

$$\frac12 \sigma^2 V^2 F_{vv}(V,t) + \mu V F_{v}(V,t) - r F(V,t) + C = 0$$

where $k \in \mathbb{R}_+$ and $$r = \begin{cases} r_1 \; \text{ if } \; V > k \\ r_2 \; \text{otherwise} \end{cases}$$

If $r$ was constant, the general solution has the form: $$F(V) = A_0 + A_1 V + A_2 V^{-x}$$ where $x \equiv \frac{m + \sqrt{m^2 + 2 r \sigma^2}}{\sigma^2}; \; \; m \equiv \mu - \frac{\sigma^2}{2}$ and the coefficients have to be determined using boundary conditions.

Is it possible to derive a similar general solution for the case of $r$ above? I'm hoping that one could solve separately the ODE over the two ranges separately and then paste the two solutions with a condition on the first derivative. Is this approach correct?

## Answer by Pollo Gi (score 1, accepted)

https://quant.stackexchange.com/a/63730

I solved it for the case $\mu = r_1$, the solution in $\mathbb{C}^1$ takes the guessed form $$F(V) = \begin{cases} A_0 + A_1 V + A_2 V^{-x} \; \text{ if } \; V>k \\ B_0 + B_1 V + B_2 V^{-y} \; \text{ else } \end{cases}$$ where the constants can be found by plugging the guess into the ODE and the remaining ones by imposing boundary conditions and smooth pasting at the discontinuity.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.