Two fields — PDE symmetry methods (Bluman, 1969) and continuous-time stochastic control (Merton, 1969) — have been solving the same nonlinear equations for over 50 years with zero cross-citation. This post establishes the connection.
The Setup#
The Merton portfolio problem: an investor allocates fraction \(\pi\) of wealth to a risky asset with drift \(\mu\) and volatility \(\sigma\), and the rest to a risk-free asset with return \(r\). After optimizing over \(\pi\), the value function \(v(x)\) satisfies:
\[\beta v \cdot v’’ - rx \cdot v’ \cdot v’’ + \frac{1}{2}\theta^2 (v’)^2 = 0 \qquad \text{(HJB)}\]
where \(\theta = (\mu - r)/\sigma\) is the Sharpe ratio and \(\beta > 0\) is the discount rate.
This is a second-order nonlinear ODE. The standard approach (Pham 2009, Merton 1969) guesses \(v(x) = Kx^p\), substitutes, and solves for \(p\). It works — but it conceals why the guess is right, what other solutions exist, and whether the approach generalizes.
Bluman’s symmetry methods offer an alternative: find the equation’s full symmetry group algorithmically, then use it to derive solutions without guessing.
Step 1: The Lie Symmetry Group#
A Lie point symmetry is a vector field \(X = \xi(x,v)\partial_x + \eta(x,v)\partial_v\) whose second prolongation annihilates (HJB) on its solution manifold. We test the two-parameter scaling ansatz \(\xi = ax\), \(\eta = bv\).
The prolongation coefficients are:
\[\eta^x = (b-a)v’, \qquad \eta^{xx} = (b-2a)v’’\]
Computing \(\text{pr}^2 X(F)\) and collecting by monomial:
- Coefficient of \(v \cdot v’’\): \(\quad 2\beta(b-a)\)
- Coefficient of \(x \cdot v’ \cdot v’’\): \(\quad -2r(b-a)\)
- Coefficient of \((v’)^2\): \(\quad \theta^2(b-a)\)
Therefore:
\[\text{pr}^2 X(F) = 2(b-a) \cdot F\]
This vanishes on \({F=0}\) for all \(a, b \in \mathbb{R}\). The equation admits a 2-dimensional abelian Lie symmetry algebra:
\[X_1 = x,\partial_x \quad \text{(scale wealth)}, \qquad X_2 = v,\partial_v \quad \text{(scale value function)}\]
with \([X_1, X_2] = 0\). Two commuting symmetries of a second-order ODE: complete integrability.
Step 2: First Reduction via \(X_1\)#
Canonical coordinate: \(t = \ln x\), so \(X_1 = \partial_t\). With \(x = e^t\):
\[v’ = \bar{v} e^{-t}, \qquad v’’ = (\bar{v}’ - \bar{v})e^{-2t}\]
where bars denote \(d/dt\). Substituting into (HJB) and multiplying by \(e^{2t}\):
\[\beta v(\bar{v}’ - \bar{v}) - r\bar{v}(\bar{v}’ - \bar{v}) + \tfrac{1}{2}\theta^2 \bar{v}^2 = 0\]
No explicit \(t\) (equivalently \(x\)) appears. Order effectively reduced.
Step 3: Second Reduction via \(X_2\)#
In the \((v, P)\) phase plane where \(P = \dot{v}\), the symmetry \(X_2\) has invariant:
\[Q = \frac{P}{v} = \frac{x \cdot v’(x)}{v(x)}\]
This is the local elasticity of the value function — a natural dimensionless quantity. Setting \(P = Qv\) and dividing by \(v\), the equation becomes:
\[\frac{(\beta - rQ),dQ}{rQ^2 - ({\beta + r + \frac{1}{2}\theta^2})Q + \beta} = d(\ln v)\]
The left side depends only on \(Q\); the right side only on \(v\). The equation is separable.
Step 4: Integration and General Solution#
Let \(a = \beta + r + \frac{1}{2}\theta^2\). The denominator \(D(Q) = rQ^2 - aQ + \beta\) has roots:
\[p_{1,2} = \frac{a \pm \sqrt{a^2 - 4r\beta}}{2r}\]
where \(a^2 - 4r\beta = (\beta - r)^2 + \theta^2(\beta + r) + \frac{1}{4}\theta^4 > 0\). Both roots are real and positive.
Partial fraction decomposition with coefficients \(\alpha_1, \alpha_2\) satisfying a universal identity:
\[\alpha_1 + \alpha_2 = -1 \quad \text{for all } \beta, r, \theta > 0\]
Integrating both sides:
\[|Q - p_1|^{\alpha_1} \cdot |Q - p_2|^{\alpha_2} = K \cdot v \qquad \textbf{(General Solution)}\]
where \(Q = x \cdot v’(x)/v(x)\) and \(K > 0\) is arbitrary. This is the complete implicit general solution of the Merton HJB.
Step 5: Recovering the Classical Merton Solution#
The constant solutions \(Q = p\) (fixed points of the reduced flow) satisfy exactly the characteristic polynomial above. For \(Q = p_i\) constant:
\[x \cdot v’(x) = p_i \cdot v(x) \implies v(x) = C \cdot x^{p_i}\]
The classical Merton power-law ansatz is the unique family of group-invariant solutions.
The characteristic equation for the exponent — which Pham derives by substituting the guess — emerges here automatically from the symmetry structure:
\[rp^2 - \left(\beta + r + \tfrac{1}{2}\theta^2\right)p + \beta = 0\]
Pham selects one root. The symmetry method finds both, plus the general solution connecting them.
The Complete Pipeline#
\[\text{Bluman} \longrightarrow \text{Merton HJB (2D group)} \longrightarrow \text{separable ODE} \longrightarrow \text{Pham}\]
\[\scriptstyle{\text{find solutions} \hspace{60px} \text{two reductions} \hspace{40px} \text{integrates exactly} \hspace{20px} \text{verify optimality}}\]
Neither the Bluman nor the Pham literature contains this pipeline. It exists only in the gap between them.
Why This Matters#
Structurally: The complete solution set of the Merton HJB is two-dimensional, parameterized by \((K, \text{branch})\). The standard treatment finds one point in this space by guessing.
Methodologically: The symmetry group is determined by the homogeneity structure of the problem — power utility is homogeneous, linear wealth dynamics are homogeneous, hence a 2D scaling group. Any HJB from homogeneous utility and linear dynamics shares this structure.
Generalization: For HJB equations where no obvious ansatz exists — regime-switching dynamics, portfolio constraints, novel utility functions — the symmetry method finds solutions without guessing. This is the start of a symmetry atlas of stochastic control: a classification of which problems are exactly solvable and why.
Open Problems#
Six gaps remain, derived by applying KEGA to this paper’s own internal logic:
- Is the 2D algebra the complete symmetry group, or do additional symmetries exist?
- What is the economic interpretation of the non-power-law general solution?
- Why does \(\alpha_1 + \alpha_2 = -1\) hold universally — what is the structural reason?
- Does the general solution satisfy Pham’s verification theorem (optimality condition)?
- What is the symmetry group of the Heston stochastic volatility HJB?
- The Symmetry Atlas: which HJB equations are exactly solvable, and which are not?
Gap 6 is the book this paper is the first entry of.
Derived via KEGA (Knowledge Extension via Gap Analysis) — cross-analysis of Bluman (1969–2026) and Pham (2009).