Why Smooth Pasting Can Set the Second Derivative to Zero

Summary

The question concerns a boundary in an investment problem with proportional transaction costs. At the boundary, it is optimal to pay a cost and move into another region of the value function. The paper reportedly imposes a first-derivative condition there, and the question asks why optimality would also imply that the second derivative vanishes.

The answer offers a local intuition: if the marginal value, represented by the first derivative, is constant throughout a neighborhood on one side of the boundary, then its rate of change at the boundary is zero. A first-order Taylor expansion of the derivative expresses that rate of change through the second derivative, yielding the stated zero-curvature condition. This is a brief explanation of the algebra under the answer’s neighborhood assumption, not a full derivation of the optimal-control boundary conditions. The source question uses a cost parameter while the answer switches notation, and it does not establish that the first derivative is constant in every transaction-cost model; the conclusion depends on that premise.

Key ideas

• The boundary condition for the first derivative represents a specified marginal value at the intervention point.
• If the first derivative stays constant nearby on one side, its local rate of change is zero.
• A Taylor expansion relates that rate of change to the second derivative.
• The explanation relies on a local constancy assumption and is not a general derivation for every model.

Tags

# Smooth pasting conditions for optimal investment with transactions costs


# Smooth pasting conditions for optimal investment with transactions costs












I'm reading this paper relating to optimal investment with transaction costs where some value function $F(x)$ is optimized. At some boundary $x=u$ it will be optimal to pay a proportional cost $C$ which gives the boundary condition

\begin{equation} F'(u) = -C \end{equation}

The author argues that optimality also implies a boundary condition for the second derivative

\begin{equation} F''(u) = 0 \end{equation}

but I'm struggling to understand why this is the case. Any hints that will help me understand the intuition behind this condition?

## Answer by ir7 (score 1)

https://quant.stackexchange.com/a/55481

I think that they are saying that, at special point $u$, we have:

$$ F'(u) = -\rho. $$

Also, that in a neighborhood from the left, we have:

$$ F'(u- dU) = -\rho $$

for any small positive $dU$.

Then, with Taylor on the left:

$$ F'(u) - F''(u)dU = - \rho $$

Hence: $$ F''(u) = 0 $$

Basically, if the first derivative of a function is constant in a neighborhood of a point, then its second derivative must be null at that point.

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