Connection between harmonic functions and harmonic analyis?

Question

Harmonic analysis is concern with generalization of Fourier series(of any function) while harmonic functions are solutions to the Laplacian.

Are these matters related somehow? I cant really see how, they seams to be rather unrelated subjects.

Answer 1

Fourier series are decompositions of functions $f$ into eigenfunctions of the Laplacian $\nabla^2 g_i = \lambda_i g_i$. We write $f(\mathbf{x}) = \sum_i f_i g_i(\mathbf{x})$. For example, on a rectangular domain with periodic boundary conditions, $\nabla^2 \exp(i \mathbf{k} \cdot \mathbf{x}) = -|\mathbf{k}|^2$ and writing $f(\mathbf{x}) = \sum_\mathbf{k} f_\mathbf{k} \exp(i \mathbf{k} \cdot \mathbf{x})$ is the Fourier decomposition.

Harmonic functions are functions annihilated by the Laplacian (its kernel), i.e. solutions to $\nabla^2 h = 0$. This means they are the zero eigenfunctions, $g_i$ with $\lambda_i = 0$.

So harmonic functions are a special subset of the Fourier basis.

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For a simple example of where this subset might arise, suppose that you study the heat equation $$\partial u/\partial t = \nabla^2 u \qquad \text{with initial conditions} \qquad u(t=0,\mathbf{x}) = \sum u_i(t=0) g_i(\mathbf{x})$$ where we have Fourier-decomposed the initial conditions. Then in Fourier space, the equation is $\dot{u}_i(t) = \lambda_i u_i(t)$ with solution $$u(t,\mathbf{x}) = \sum_i u_i(t=0) \exp(\lambda_i t) g_i(\mathbf{x})$$ Now assuming that the Laplacian has no positive eigenvalues (exercise: find out why for simple boundary conditions), almost all of the exponentials decay away, so at late times we have $$u(t,\mathbf{x}) \to \sum_{i'} u_{i'}(t=0) h_{i'}(\mathbf{x})$$ where we sum only over harmonic functions with $\lambda_i = 0$.

(In fact, in general there's only one harmonic function given suitable boundary conditions. Another thing to check!)