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Green's Function for Parabolic PDEs

A Green's function is the response of a linear PDE to a point source -- a delta function impulse at a single point in space and time. For parabolic equations, the Green's function encodes the complete solution operator: once you know how the system responds to a point source, you can build the solution for arbitrary data by superposition. In finance, the Green's function is the transition density of the underlying stochastic process.

Intuition

Imagine injecting a unit of heat at a single point \(y\) at time \(s\). The Green's function \(G(t, x; s, y)\) describes how this heat spreads to point \(x\) at a later time \(t > s\).

  • At the injection moment: \(G\) is a delta function concentrated at \(y\)
  • For \(t > s\): the heat diffuses outward, forming a bell curve that flattens and broadens
  • As \(t \to \infty\): the heat dissipates (on unbounded domains) or reaches equilibrium (on bounded domains)

The solution for arbitrary initial data \(f(x)\) is obtained by superposition:

\[ u(t, x) = \int G(t, x; 0, y)\,f(y)\,dy \]

Each point \(y\) contributes its share \(f(y)\) of heat, weighted by the Green's function.

Definition

The Parabolic Operator

Consider a general parabolic operator:

\[ \mathcal{P}u = \frac{\partial u}{\partial t} - \mathcal{L}u \]

where \(\mathcal{L}\) is a second-order elliptic operator:

\[ \mathcal{L} = \mu(x)\frac{\partial}{\partial x} + \frac{1}{2}\sigma^2(x)\frac{\partial^2}{\partial x^2} \]

Green's Function

The Green's function \(G(t, x; s, y)\) is the fundamental solution of \(\mathcal{P}\):

\[ \boxed{ \mathcal{P}_x G(t, x; s, y) = \frac{\partial G}{\partial t} - \mathcal{L}_x G = 0, \quad t > s } \]

with the initial condition:

\[ \lim_{t \downarrow s} G(t, x; s, y) = \delta(x - y) \]

Here \(\mathcal{L}_x\) denotes differentiation with respect to \(x\).

Interpretation: \(G(t, x; s, y)\) is the solution at \((t, x)\) due to a unit impulse at \((s, y)\).

Construction for the Heat Equation

Free-Space Green's Function

For the standard heat equation \(u_t = \frac{1}{2}u_{xx}\) on \(\mathbb{R}\), the Green's function is the heat kernel:

\[ \boxed{ G(t, x; s, y) = \frac{1}{\sqrt{2\pi(t-s)}} \exp\left(-\frac{(x-y)^2}{2(t-s)}\right), \quad t > s } \]

Verification:

  1. Satisfies the heat equation in \((t, x)\) for \(t > s\) -- direct computation (see Fundamental Solution)
  2. Initial condition: As \(t \downarrow s\), \(G(t, x; s, y) \to \delta(x - y)\) (the Gaussian concentrates at \(y\))

Translation Invariance

For constant-coefficient equations, the Green's function depends only on differences:

\[ G(t, x; s, y) = G(t - s, x - y; 0, 0) = \Gamma(t - s, x - y) \]

where \(\Gamma(\tau, z) = \frac{1}{\sqrt{2\pi\tau}}\,e^{-z^2/2\tau}\) is the fundamental solution.

Solution by Superposition

Initial Value Problem

For \(u_t = \frac{1}{2}u_{xx}\) with \(u(0, x) = f(x)\):

\[ \boxed{ u(t, x) = \int_{-\infty}^{\infty} G(t, x; 0, y)\,f(y)\,dy } \]

This is a convolution: \(u(t, \cdot) = G(t, \cdot; 0, 0) * f\).

Source Problem

For \(u_t - \frac{1}{2}u_{xx} = h(t, x)\) with \(u(0, x) = 0\):

\[ u(t, x) = \int_0^t \int_{-\infty}^{\infty} G(t, x; s, y)\,h(s, y)\,dy\,ds \]

This is the Duhamel principle: the response to a distributed source is the integral of responses to point sources.

Combined Problem

For \(u_t - \frac{1}{2}u_{xx} = h(t,x)\) with \(u(0,x) = f(x)\):

\[ u(t, x) = \int G(t, x; 0, y)\,f(y)\,dy + \int_0^t\!\int G(t, x; s, y)\,h(s, y)\,dy\,ds \]

Properties of the Parabolic Green's Function

1. Positivity

\[ G(t, x; s, y) > 0 \quad \text{for all } t > s \]

Heat always flows from source to surroundings -- the Green's function is everywhere positive. This is the PDE manifestation of the fact that transition densities are non-negative.

2. Normalization

\[ \int_{-\infty}^{\infty} G(t, x; s, y)\,dx = 1 \quad \text{for all } t > s \]

The total amount of "heat" (or probability) is conserved. The Green's function is a probability density in \(x\).

3. Semigroup Property (Chapman-Kolmogorov)

For \(s < r < t\):

\[ \boxed{ G(t, x; s, y) = \int_{-\infty}^{\infty} G(t, x; r, z)\,G(r, z; s, y)\,dz } \]

Interpretation: The transition from \(y\) at time \(s\) to \(x\) at time \(t\) can be decomposed into a transition from \(y\) to some intermediate point \(z\) at time \(r\), followed by a transition from \(z\) to \(x\). Summing over all possible intermediate states gives the direct transition.

Probabilistic meaning: This is the Chapman-Kolmogorov equation for the transition density of a Markov process.

4. Symmetry (for Self-Adjoint Operators)

When \(\mathcal{L}\) is self-adjoint (\(\mathcal{L} = \mathcal{L}^*\), which holds for the heat equation but not in general), the Green's function satisfies:

\[ G(t, x; s, y) = G(t, y; s, x) \]

For non-self-adjoint operators (e.g., with drift), this symmetry fails, but a modified symmetry relates \(G\) to the Green's function of the adjoint operator.

5. Smoothing

For \(t > s\), the map \(f \mapsto \int G(t, x; s, y)\,f(y)\,dy\) sends bounded measurable functions to \(C^\infty\) functions. This is the instantaneous regularization property of parabolic equations.

Construction for Variable Coefficients

Parametrix Method

For the general operator \(\mathcal{L} = \mu(x)\partial_x + \frac{1}{2}\sigma^2(x)\partial_{xx}\), the Green's function is constructed by the parametrix method:

  1. Freeze coefficients at the source point \(y\): define \(\mathcal{L}_y = \mu(y)\partial_x + \frac{1}{2}\sigma^2(y)\partial_{xx}\), which has the explicit Gaussian Green's function:
\[ G_0(t, x; s, y) = \frac{1}{\sigma(y)\sqrt{2\pi(t-s)}} \exp\left(-\frac{(x - y - \mu(y)(t-s))^2}{2\sigma^2(y)(t-s)}\right) \]

  1. Correct iteratively: Write \(G = G_0 + G_1 + G_2 + \cdots\) where each correction \(G_n\) accounts for the error from the frozen-coefficient approximation.

  2. Convergence: Under Holder continuity of the coefficients, the series converges and yields a smooth Green's function.

Levi's Method

This iterative construction is called Levi's parametrix method. It is the standard technique for proving existence and deriving short-time asymptotics of the Green's function for variable-coefficient parabolic operators.

Short-Time Asymptotics

For small \(t - s\), the Green's function is approximately Gaussian:

\[ G(t, x; s, y) \approx \frac{1}{\sigma(y)\sqrt{2\pi(t-s)}} \exp\left(-\frac{(x - y - \mu(y)(t-s))^2}{2\sigma^2(y)(t-s)}\right) \]

The corrections are of order \(O((t-s)^{1/2})\) relative to the leading term.

Green's Function and the Adjoint Equation

The Green's function satisfies two PDEs:

Equation Variables PDE
Forward (in \(t, x\)) Source \((s, y)\) fixed \(\partial_t G = \mathcal{L}_x G\)
Backward (in \(s, y\)) Observation \((t, x)\) fixed \(-\partial_s G = \mathcal{L}_y^* G\)

where \(\mathcal{L}^*\) is the formal adjoint:

\[ \mathcal{L}^* p = -\frac{\partial}{\partial y}[\mu(y)p] + \frac{1}{2}\frac{\partial^2}{\partial y^2}[\sigma^2(y)p] \]

Financial Interpretation

  • Forward equation: Fix where the process starts; the Green's function (transition density) evolves via Fokker-Planck as you vary the destination
  • Backward equation: Fix where you observe; the Green's function solves Kolmogorov backward as you vary the origin

Example: Brownian Motion with Drift

For \(dX_t = \mu\,dt + \sigma\,dW_t\) (constant coefficients), the Green's function is:

\[ G(t, x; s, y) = \frac{1}{\sigma\sqrt{2\pi(t-s)}} \exp\left(-\frac{(x - y - \mu(t-s))^2}{2\sigma^2(t-s)}\right) \]

This is the density of \(X_t \mid X_s = y\), i.e., a normal distribution with:

  • Mean: \(y + \mu(t-s)\)
  • Variance: \(\sigma^2(t-s)\)

Verification of the semigroup property: The convolution of two Gaussians is Gaussian with added means and variances -- consistent with the independent increments of Brownian motion with drift.

Example: Ornstein-Uhlenbeck Process

For \(dX_t = -\kappa X_t\,dt + \sigma\,dW_t\), the Green's function is:

\[ G(t, x; s, y) = \frac{1}{\sqrt{2\pi\,v(t-s)}} \exp\left(-\frac{(x - ye^{-\kappa(t-s)})^2}{2\,v(t-s)}\right) \]

where the conditional variance is:

\[ v(\tau) = \frac{\sigma^2}{2\kappa}\left(1 - e^{-2\kappa\tau}\right) \]

Key features:

  • The conditional mean \(ye^{-\kappa\tau}\) reverts toward zero at rate \(\kappa\)
  • The conditional variance \(v(\tau) \to \frac{\sigma^2}{2\kappa}\) as \(\tau \to \infty\) -- the stationary variance
  • The Green's function converges to the stationary density \(N(0, \sigma^2/2\kappa)\) regardless of the starting point

Summary

\[ \boxed{ u(t, x) = \int G(t, x; s, y)\,f(y)\,dy \quad \text{(superposition principle)} } \]
Property Statement
Definition \(\partial_t G = \mathcal{L}_x G\) with \(G(s^+, x; s, y) = \delta(x-y)\)
Positivity \(G > 0\) for \(t > s\)
Normalization \(\int G\,dx = 1\)
Semigroup \(G(t,x;s,y) = \int G(t,x;r,z)\,G(r,z;s,y)\,dz\)
Smoothing Maps \(L^\infty\) to \(C^\infty\)

The Green's function is the fundamental building block for parabolic PDEs: it solves the equation for a point source, and arbitrary solutions are obtained by superposition. In probability, it is the transition density; in finance, it is the state-price density.

See Also

Exercises

Exercise 1. For the heat equation \(\partial_t u = \frac{1}{2}\partial_{xx}u\) on \(\mathbb{R}\), verify that the Green's function \(G(t,x;0,y) = (2\pi t)^{-1/2}\exp(-(x-y)^2/(2t))\) satisfies the delta function initial condition: show that \(\lim_{t \to 0^+} \int_{-\infty}^{\infty} G(t,x;0,y)f(y)\,dy = f(x)\) for continuous \(f\).

Solution to Exercise 1

We must show that \(\lim_{t \to 0^+} \int_{-\infty}^{\infty} G(t,x;0,y)\,f(y)\,dy = f(x)\) for continuous \(f\). Write

\[ \int_{-\infty}^{\infty} G(t,x;0,y)\,f(y)\,dy = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp\!\left(-\frac{(x-y)^2}{2t}\right) f(y)\,dy \]

Substitute \(z = (y - x)/\sqrt{t}\), so \(y = x + z\sqrt{t}\) and \(dy = \sqrt{t}\,dz\):

\[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}\,f(x + z\sqrt{t})\,dz \]

As \(t \to 0^+\), the argument \(x + z\sqrt{t} \to x\) for every fixed \(z\). Since \(f\) is continuous, \(f(x + z\sqrt{t}) \to f(x)\). By the dominated convergence theorem (assuming \(f\) is bounded or has at most polynomial growth, which is sufficient since the Gaussian provides exponential decay), we can pass the limit inside the integral:

\[ \lim_{t \to 0^+} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}\,f(x + z\sqrt{t})\,dz = f(x)\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}\,dz = f(x) \cdot 1 = f(x) \]

This confirms the delta-function initial condition: \(G(t, x; 0, y)\) acts as an approximate identity that selects the value \(f(x)\) in the limit \(t \to 0^+\).

Exercise 2. The superposition principle states \(u(t,x) = \int G(t,x;0,y)\,f(y)\,dy\). For \(f(y) = e^{-y^2}\) and the free-space heat kernel, evaluate this integral explicitly using the convolution of Gaussians.

Solution to Exercise 2

We need to evaluate

\[ u(t, x) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp\!\left(-\frac{(x-y)^2}{2t}\right) e^{-y^2}\,dy \]

Combine the exponents:

\[ -\frac{(x-y)^2}{2t} - y^2 = -\frac{(x-y)^2 + 2ty^2}{2t} = -\frac{x^2 - 2xy + y^2 + 2ty^2}{2t} = -\frac{(1+2t)y^2 - 2xy + x^2}{2t} \]

Complete the square in \(y\):

\[ (1+2t)y^2 - 2xy = (1+2t)\!\left(y - \frac{x}{1+2t}\right)^2 - \frac{x^2}{1+2t} \]

Therefore

\[ -\frac{(1+2t)y^2 - 2xy + x^2}{2t} = -\frac{(1+2t)}{2t}\!\left(y - \frac{x}{1+2t}\right)^2 - \frac{x^2}{2t} + \frac{x^2}{2t(1+2t)} \]

Simplifying the \(x^2\) terms: \(-\frac{x^2}{2t} + \frac{x^2}{2t(1+2t)} = -\frac{x^2}{1+2t}\).

The Gaussian integral in \(y\) evaluates to \(\sqrt{2\pi t/(1+2t)}\), giving

\[ u(t, x) = \frac{1}{\sqrt{2\pi t}} \cdot \sqrt{\frac{2\pi t}{1+2t}} \cdot \exp\!\left(-\frac{x^2}{1+2t}\right) = \frac{1}{\sqrt{1+2t}} \exp\!\left(-\frac{x^2}{1+2t}\right) \]

This is a Gaussian that broadens over time: the initial width \(1\) increases to \(\sqrt{1+2t}\), which is the convolution of two Gaussians with variances \(1\) (from \(f\)) and \(t\) (from the heat kernel) giving total variance \((1+2t)/2\).

Exercise 3. Explain why the Green's function \(G(t,x;s,y)\) for a parabolic PDE satisfies the semigroup property \(G(t,x;s,y) = \int G(t,x;r,z)\,G(r,z;s,y)\,dz\) for \(s < r < t\). What is the probabilistic interpretation via the Chapman-Kolmogorov equation?

Solution to Exercise 3

The semigroup property states that for \(s < r < t\):

\[ G(t, x; s, y) = \int_{-\infty}^{\infty} G(t, x; r, z)\,G(r, z; s, y)\,dz \]

PDE argument: Consider the initial value problem \(\partial_t u = \mathcal{L}_x u\) with \(u(s, x) = \delta(x - y)\). The solution at time \(t\) is \(u(t, x) = G(t, x; s, y)\). We can equivalently solve in two stages: first evolve from \(s\) to \(r\) to get \(u(r, z) = G(r, z; s, y)\), then use this as initial data and evolve from \(r\) to \(t\):

\[ u(t, x) = \int G(t, x; r, z)\,u(r, z)\,dz = \int G(t, x; r, z)\,G(r, z; s, y)\,dz \]

By uniqueness of the PDE solution, this must equal \(G(t, x; s, y)\).

Probabilistic interpretation: This is the Chapman-Kolmogorov equation. For a Markov process \(X_t\):

\[ p(t, x \mid s, y) = \int p(t, x \mid r, z)\,p(r, z \mid s, y)\,dz \]

This says: the probability of going from \(y\) at time \(s\) to \(x\) at time \(t\) equals the sum over all intermediate states \(z\) at time \(r\) of the probability of going \(y \to z\) then \(z \to x\). This follows from the Markov property (the future depends on the past only through the present) and the law of total probability.

Exercise 4. For the operator \(\mathcal{L} = \mu(x)\partial_x + \frac{1}{2}\sigma^2(x)\partial_{xx}\), the Green's function satisfies \(\partial_t G = \mathcal{L}_x G\) as a function of \((t,x)\) and \(\partial_s G = -\mathcal{L}_y^* G\) as a function of \((s,y)\). Identify the adjoint operator \(\mathcal{L}^*\) and explain why it involves the forward (Fokker-Planck) equation.

Solution to Exercise 4

Given \(\mathcal{L} = \mu(x)\partial_x + \frac{1}{2}\sigma^2(x)\partial_{xx}\), the Green's function satisfies the forward equation \(\partial_t G = \mathcal{L}_x G\) in the observation variables \((t, x)\).

The formal adjoint \(\mathcal{L}^*\) is obtained by integration by parts. For any smooth functions \(u, v\) vanishing at infinity:

\[ \int (\mathcal{L}u)\,v\,dx = \int u\,(\mathcal{L}^* v)\,dx \]

Computing term by term:

  • \(\int \mu(x)\,u_x\,v\,dx = -\int u\,\frac{\partial}{\partial x}[\mu(x)\,v]\,dx\) (integration by parts once)
  • \(\int \frac{1}{2}\sigma^2(x)\,u_{xx}\,v\,dx = \int u\,\frac{1}{2}\frac{\partial^2}{\partial x^2}[\sigma^2(x)\,v]\,dx\) (integration by parts twice)

Therefore

\[ \mathcal{L}^* v = -\frac{\partial}{\partial x}[\mu(x)\,v] + \frac{1}{2}\frac{\partial^2}{\partial x^2}[\sigma^2(x)\,v] \]

The backward equation is \(-\partial_s G = \mathcal{L}_y^* G\), which involves \(\mathcal{L}^*\) acting on the source variables.

The adjoint equation \(\partial_t p = \mathcal{L}^* p\) is exactly the Fokker-Planck (forward) equation: it describes how the probability density \(p\) evolves over time. The terms \(-\partial_x[\mu\,p]\) represent the advective flux (drift carries probability) and \(\frac{1}{2}\partial_{xx}[\sigma^2 p]\) represents the diffusive flux (noise spreads probability). The adjoint structure ensures that probability is conserved: \(\frac{d}{dt}\int p\,dx = 0\).

Exercise 5. For geometric Brownian motion \(dS_t = rS_t\,dt + \sigma S_t\,dW_t\), the log-transformation \(X_t = \ln S_t\) gives \(dX_t = (r - \sigma^2/2)\,dt + \sigma\,dW_t\). Write the Green's function for \(X_t\) and explain how it is related to the lognormal transition density of \(S_t\).

Solution to Exercise 5

Under the log-transformation \(X_t = \ln S_t\), Ito's formula gives

\[ dX_t = \left(r - \frac{\sigma^2}{2}\right)dt + \sigma\,dW_t \]

This is Brownian motion with drift \(\mu = r - \sigma^2/2\) and diffusion coefficient \(\sigma\). The transition density of \(X_t\) given \(X_s = \ln S\) is Gaussian:

\[ p_X(t, x \mid s, \ln S) = \frac{1}{\sigma\sqrt{2\pi(t-s)}} \exp\!\left(-\frac{(x - \ln S - (r - \sigma^2/2)(t-s))^2}{2\sigma^2(t-s)}\right) \]

This is the Green's function \(G(t, x; s, \ln S)\) for the PDE \(\partial_t u = (r - \sigma^2/2)\partial_x u + \frac{1}{2}\sigma^2\partial_{xx}u\).

To obtain the transition density of \(S_t\) in the original variable, use the change of variables \(S_T = e^x\), so \(dS_T = e^x\,dx\) and \(p_S(t, S_T \mid s, S) = p_X(t, \ln S_T \mid s, \ln S)/S_T\):

\[ p_S(t, S_T \mid s, S) = \frac{1}{S_T \sigma\sqrt{2\pi(t-s)}} \exp\!\left(-\frac{(\ln(S_T/S) - (r - \sigma^2/2)(t-s))^2}{2\sigma^2(t-s)}\right) \]

This is the lognormal transition density. The log-space Green's function and the lognormal density are related by the Jacobian factor \(1/S_T\) from the exponential change of variables.

Exercise 6. Explain the role of Green's functions in option pricing: the price of a European derivative with payoff \(g(S_T)\) can be written as \(V(t,S) = e^{-r(T-t)}\int G(T,y;t,\ln S)\,g(e^y)\,dy\). What is the financial interpretation of the Green's function as a "state price density"?

Solution to Exercise 6

The European derivative price under risk-neutral pricing is

\[ V(t, S) = e^{-r(T-t)}\,\mathbb{E}^{\mathbb{Q}}[g(S_T) \mid S_t = S] = e^{-r(T-t)}\int_0^{\infty} g(S_T)\,p^{\mathbb{Q}}(T, S_T \mid t, S)\,dS_T \]

In log-space (\(y = \ln S_T\)):

\[ V(t, S) = e^{-r(T-t)}\int_{-\infty}^{\infty} g(e^y)\,G(T, y; t, \ln S)\,dy \]

The Green's function \(G(T, y; t, \ln S)\) plays the role of a state-price density (Arrow-Debreu price). For each terminal state \(y\), the quantity \(e^{-r(T-t)}G(T, y; t, \ln S)\,dy\) is the price today of a security that pays \(\$1\) if the log-price at maturity falls in the interval \([y, y + dy]\) and zero otherwise.

The Green's function encodes all information needed for pricing because:

  • It captures the dynamics of the underlying process (drift, volatility) through the PDE it satisfies.
  • It incorporates the risk-neutral measure through the drift adjustment (\(r\) replaces \(\mu\)).
  • Any European payoff \(g\) can be priced by integrating against \(G\) -- different payoffs simply change the weighting function, not the kernel.
  • The discount factor \(e^{-r(T-t)}\) accounts for the time value of money.

Thus, knowing the Green's function is equivalent to knowing all European option prices simultaneously.

Exercise 7. Consider the generator \(\mathcal{L} = \frac{1}{2}\partial_{xx} - \frac{1}{2}\partial_x\) (Brownian motion with drift \(-1/2\)). Compute the Green's function by completing the square in the exponent, starting from the Gaussian kernel and incorporating the drift shift.

Solution to Exercise 7

The generator is \(\mathcal{L} = \frac{1}{2}\partial_{xx} - \frac{1}{2}\partial_x\), corresponding to the SDE \(dX_t = -\frac{1}{2}\,dt + dW_t\) (Brownian motion with drift \(\mu = -1/2\)).

The Green's function for drift-diffusion \(dX_t = \mu\,dt + \sigma\,dW_t\) with \(\mu = -1/2\) and \(\sigma = 1\) is

\[ G(t, x; s, y) = \frac{1}{\sqrt{2\pi(t-s)}} \exp\!\left(-\frac{(x - y + \frac{1}{2}(t-s))^2}{2(t-s)}\right) \]

Derivation by completing the square: Start with the Gaussian kernel and incorporate the drift. The PDE is \(\partial_t u = \frac{1}{2}\partial_{xx}u - \frac{1}{2}\partial_x u\). Make the substitution \(u(t, x) = e^{\alpha x + \beta t}\,v(t, x)\) to eliminate the first-order term. Substituting:

\[ e^{\alpha x + \beta t}(\beta v + v_t) = \frac{1}{2}e^{\alpha x + \beta t}(\alpha^2 v + 2\alpha v_x + v_{xx}) - \frac{1}{2}e^{\alpha x + \beta t}(\alpha v + v_x) \]

Choosing \(\alpha = 1/2\) eliminates the \(v_x\) terms (since \(2\alpha \cdot \frac{1}{2} - \frac{1}{2} = 0\)), and then \(\beta = -\alpha^2/2 + \alpha/2 = -1/8 + 1/4 = 1/8\). This gives \(v_t = \frac{1}{2}v_{xx}\), the standard heat equation.

The Green's function of the heat equation is \(\Gamma(t, x-y) = (2\pi t)^{-1/2}e^{-(x-y)^2/(2t)}\). Reverting:

\[ G(t, x; 0, y) = e^{-\frac{1}{2}(x-y) - \frac{1}{8}t} \cdot \frac{1}{\sqrt{2\pi t}}\,e^{-(x-y)^2/(2t)} \]

Combining the exponentials:

\[ -\frac{(x-y)^2}{2t} - \frac{x-y}{2} - \frac{t}{8} = -\frac{(x-y)^2 + t(x-y) + t^2/4}{2t} = -\frac{(x - y + t/2)^2}{2t} \]

Therefore

\[ G(t, x; 0, y) = \frac{1}{\sqrt{2\pi t}} \exp\!\left(-\frac{(x - y + t/2)^2}{2t}\right) \]

This is a Gaussian centered at \(y - t/2\) (the source point shifted by the drift \(-1/2\) over time \(t\)), confirming that the drift simply translates the center of the heat kernel.