03 The Laplacian and Laplace’s equation

Reminders of some differential operators

Gradient

Gradient operator in two dimensions

Let \(u:\mathbb{R}^2\to \mathbb{R}\) be a differentiable function. The gradient of \(u\), denoted \(\nabla u\), is the vector field with

\[ (\nabla u)(x,y) = \begin{bmatrix} \displaystyle \frac{\partial u}{\partial x}(x,y) \\ \displaystyle \frac{\partial u}{\partial y}(x,y) \end{bmatrix}. \]

• The fundamental properties of gradient fields are:
  • The dot product \((\nabla u)(x,y) \cdot \mathbf{v}\) is the (approximate) change in \(u\), at the point \((x,y)\), in the direction of a vector \(\mathbf{v}\).
  • The gradient vector at a point \((x,y)\) is orthogonal to the level curve of \(u\) through \((x,y)\).
  • The gradient vector at a point \((x,y)\) points in the direction of maximum increase of \(u\).
• As long as \(\|\mathbf{v}\|\) is small, we have: \[ (\nabla u)(x,y) \cdot \mathbf{v} \approx u(x+v_1, y+v_2) - u(x,y). \]

Divergence

Divergence operator in two dimensions

Let \(\mathbf{F}:\mathbb{R}^2\to \mathbb{R}^2\) be a continuously differentiable vector field with components \(F_1\) and \(F_2\). The divergence of \(\mathbf{F}\), denoted \(\nabla \cdot \mathbf{F}\), is the scalar field with

\[ (\nabla \cdot \mathbf{F})(x,y) = \frac{\partial F_1}{\partial x}(x,y) + \frac{\partial F_2}{\partial y}(x,y). \]

• How should you think of the divergence?

  • Imagine a small circle \(C\) of radius \(r\) centered at \((x_0,y_0)\). Then the integral \[ \int_C \mathbf{F} \cdot \mathbf{n} \, ds \] is the net flux of \(\mathbf{F}\) across the circle. (Think of \(\mathbf{F}\) as a velocity field of some “flow”.)
  • As \(r\to 0\), this integral certainly vanishes since the circle shrinks to a point.
  • In fact, it goes to \(0\) as fast as the area of the circle, and the divergence is the rate: \[ (\nabla \cdot \mathbf{F})(x_0,y_0) = \lim_{r\to 0^+} \frac{1}{\pi r^2} \int_C \mathbf{F} \cdot \mathbf{n} \, ds. \]

The Divergence Theorem

Theorem (Divergence Theorem).

Let \(\mathbf{F}:\mathbb{R}^2\to \mathbb{R}^2\) be a continuously differentiable vector field, and let \(R\) be a region in \(\mathbb{R}^2\) with a piecewise smooth boundary \(\partial R\). Then \[ \int_{\partial R} \mathbf{F} \cdot \mathbf{n} \, ds = \int_R (\nabla \cdot \mathbf{F}) \, dA. \]

The Laplacian and operators in arbitrary dimensions

Laplacian

Laplacian operator in two dimensions

Let \(u:\mathbb{R}^2\to \mathbb{R}\) be a function with continuous second-order partial derivatives. The Laplacian of \(u\), denoted \(\nabla^2 u\), is the scalar field with

\[ (\nabla^2 u)(x,y) = \frac{\partial^2 u}{\partial x^2}(x,y) + \frac{\partial^2 u}{\partial y^2}(x,y). \]

• The motivation for the notation comes from: \[ \nabla^2 u = \nabla \cdot (\nabla u), \] which shows that the Laplacian is the divergence of the gradient of \(u\).

• How should you think of the Laplacian?

  • Imagine a small circle \(C\) of radius \(r\) centered at \((x_0,y_0)\). Then \[ \frac{1}{2\pi r} \int_C u(x,y) \, ds - u(x_0,y_0) \] is the difference between the average value of \(u\) on the circle and the value at the center.
  • As \(r\to 0\), this quantity certainly vanishes since \(u(x,y) \to u(x_0,y_0)\) for all \((x,y)\in C\).
  • In fact, it goes to \(0\) as fast as \(r^2\), and the Laplacian is (proportional to) the rate: \[ (\nabla^2 u)(x_0,y_0) = \lim_{r\to 0^+} \frac{4}{r^2} \left( \frac{1}{2\pi r} \int_C u(x,y) \, ds - u(x_0,y_0)\right). = \]

Differential operators in arbitrary dimensions

• The gradient, divergence, and Laplacian all have natural generalizations to higher dimensions: \[ \nabla u = \begin{bmatrix} \frac{\partial u}{\partial x_1} \\ \vdots \\ \frac{\partial u}{\partial x_n} \end{bmatrix}, \quad \nabla \cdot \mathbf{F} = \sum_{i=1}^n \frac{\partial F_i}{\partial x_i}, \quad \nabla^2 u = \sum_{i=1}^n \frac{\partial^2 u}{\partial x_i^2}. \]

• But the Laplacian has a slightly different interpretation if \(u\) is “time-dependent.”

  • For example, let \(u:\mathbb{R}^3 \to \mathbb{R}\), and write the third variable as \(t\). So, our function is \(u(x,y,t)\).
  • Then the Laplacian only acts on the spatial variables \(x\) and \(y\), not on \(t\): \[ \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}. \]
  • If our function is \(u(x,t)\), then: \[ \nabla^2 u = \frac{\partial^2 u}{\partial x^2}. \]

The heat and wave equations revisited

• This means we can rewrite the heat and wave equations in terms of the Laplacian acting only on the spatial variables!

The heat equation in arbitrary dimensions

Let \(u:\mathbb{R}^{n+1} \to \mathbb{R}\) be function with continuous second-order partial derivatives, and write \[ u = u(x_1,\dots,x_n,t). \] Then the heat equation is \[ \frac{\partial u}{\partial t} = k \nabla^2 u, \] where \(\nabla^2\) is the Laplacian acting only on the first \(n\) spatial variables and \(k\) is a positive constant.

The wave equation in arbitrary dimensions

Let \(u:\mathbb{R}^{n+1} \to \mathbb{R}\) be function with continuous second-order partial derivatives, and write \[ u = u(x_1,\dots,x_n,t). \] Then the wave equation is \[ \frac{\partial^2 u}{\partial t^2} = a^2 \nabla^2 u, \] where \(\nabla^2\) is the Laplacian acting only on the first \(n\) spatial variables and \(a\) is a positive constant.

Laplace’s equation and harmonic functions

Laplace’s equation

Laplace’s equation in arbitrary dimensions

Let \(u:\mathbb{R}^{n} \to \mathbb{R}\) be function with continuous second-order partial derivatives, and write \[ u = u(x_1,\dots,x_n). \] Then Laplace’s equation is \[ \nabla^2 u = 0, \] where \(\nabla^2\) is the Laplacian acting on all \(n\) spatial variables.

• As you may check, \(\nabla^2\) is a linear differential operator: \[ \nabla^2 (u+v) = \nabla^2 u + \nabla^2 v, \quad \nabla^2 (cu) = c \nabla^2 u. \]
• Thus, solutions to Laplace’s equation are functions in the kernel of the Laplacian operator.
• Solutions are called harmonic functions.
• Using our intuition for the Laplacian, we see that harmonic functions are those that are equal to their average values in every small neighborhood.
• As David Griffiths in Introduction to Electrodynamics says: harmonic functions are “as boring as they possibly could be”.

Exercise: Harmonic functions in one dimension

Find all solutions \(u(x)\) to Laplace’s equation \(\nabla^2 u =0\) over the unit interval \([0,1]\), subject to the boundary conditions \[ u(0) = a \quad \text{and} \quad u(1) = b, \] where \(a\) and \(b\) are given constants.

1. Show that no solution has a maximum or minimum in the interior of the interval. (Assume \(a\neq b\).)
2. Let \(u(x)\) be a solution. For each \(x_0\) in the interior of \([0,1]\), show that \(u(x_0)\) is the average of its values on the boundary of the subinterval \([x_0-\epsilon, x_0+\epsilon]\) for all \(\epsilon > 0\) such that \([x_0-\epsilon, x_0+\epsilon] \subset [0,1]\).