Laplacian Operator

So far we have encountered

$$\nabla\phi = \left(\frac{\partial \phi}{\partial x},\, \frac{\partial \phi} {\partial y},\, \frac{\partial \phi}{\partial z}\right),$$ (1.139)

which is a vector field formed from a scalar field, and

$$\nabla\cdot{\bf A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} +\frac{\partial A_z}{\partial z},$$ (1.140)

which is a scalar field formed from a vector field. There are two ways in which we can combine gradient and divergence. We can either form the vector field $\nabla(\nabla\cdot{\bf A})$ or the scalar field $\nabla\cdot(\nabla\phi)$. The former is not particularly interesting, but the scalar field $\nabla\cdot(\nabla\phi)$ turns up in a great many physical problems, and is, therefore, worthy of discussion.

Let us introduce the heat flow vector ${\bf h}$, which is the rate of flow of heat energy per unit area across a surface perpendicular to the direction of ${\bf h}$. In many substances, heat flows directly down the temperature gradient, so that we can write

$${\bf h} = - \kappa \,\,\nabla T,$$ (1.141)

where $\kappa$ is the thermal conductivity. The net rate of heat flow $\oint_S {\bf h}\cdot d{\bf S}$ out of some closed surface $S$ must be equal to the rate of decrease of heat energy in the volume $V$ enclosed by $S$. Thus, we have

$$\oint_S {\bf h}\cdot d{\bf S} = - \frac{\partial}{\partial t}\left( \int c\, T\,dV\right),$$ (1.142)

where $c$ is the specific heat. It follows from the divergence theorem that

$$\nabla\cdot{\bf h} = -c\,\frac{\partial T}{\partial t}.$$ (1.143)

Taking the divergence of both sides of Equation (A.141), and making use of Equation (A.143), we obtain

$$\nabla\cdot\left(\kappa \,\nabla T\right) = c\,\frac{\partial T}{\partial t}.$$ (1.144)

If $\kappa$ is constant then the previous equation can be written

$$\nabla\cdot (\nabla T) = \frac{c}{\kappa} \frac{ \partial T}{\partial t}.$$ (1.145)

The scalar field $\nabla\cdot (\nabla T)$ takes the form

$$\nabla\cdot(\nabla T) = \frac{\partial}{\partial x}\!\left(\frac{\partial T}{\partial x... ...right)+ \frac{\partial}{\partial z}\!\left(\frac{\partial T}{\partial z}\right)$$

$$=\frac{\partial^{\,2} T}{\partial x^{\,2}}+\frac{\partial^{\,2} T... ...rtial y^{\,2}} + \frac{\partial^{\,2} T}{\partial z^{\,2}} \equiv \nabla^{2} T.$$ (1.146)

Here, the scalar differential operator

$$\nabla^{\,2} \equiv \frac{\partial^{\,2}}{\partial x^{\,2}}+ \frac{\partial^{\,2}}{\partial y^{\,2}}+ \frac{\partial^{\,2}}{\partial z^{\,2}}$$ (1.147)

is called the Laplacian. The Laplacian is a good scalar operator (i.e., it is coordinate independent) because it is formed from a combination of divergence (another good scalar operator) and gradient (a good vector operator).

What is the physical significance of the Laplacian? In one dimension, $\nabla^{\,2} T$ reduces to $\partial^{\,2} T/\partial x^{\,2}$. Now, $\partial^{\,2} T/\partial x^{\,2}$ is positive if $T(x)$ is concave (from above), and negative if it is convex. So, if $T$ is less than the average of $T$ in its surroundings then $\nabla^{\,2} T$ is positive, and vice versa.

In two dimensions,

$$\nabla^{\,2} T = \frac{\partial^{\,2} T}{\partial x^{\,2}}+ \frac{\partial^{\,2} T}{\partial y^{\,2}}.$$ (1.148)

Consider a local minimum of the temperature. At the minimum, the slope of $T$ increases in all directions, so $\nabla^{\,2} T$ is positive. Likewise, $\nabla^{\,2} T$ is negative at a local maximum. Consider, now, a steep-sided valley in $T$. Suppose that the bottom of the valley runs parallel to the $x$-axis. At the bottom of the valley $\partial^{\,2} T/\partial y^{\,2}$ is large and positive, whereas $\partial^{\,2} T/\partial x^{\,2}$ is small and may even be negative. Thus, $\nabla^{\,2} T$ is positive, and this is associated with $T$ being less than the average local value.

Let us now return to the heat conduction problem:

$$\nabla^{\,2} T = \frac{c}{\kappa} \frac{\partial T}{\partial t}.$$ (1.149)

It is clear that if $\nabla^{\,2} T$ is positive in some small region then the value of $T$ there is less than the local average value, so $\partial T/\partial t>0$: that is, the region heats up. Likewise, if $\nabla^{\,2} T$ is negative then the value of $T$ is greater than the local average value, and heat flows out of the region: that is, $\partial T/\partial t<0$. Thus, the previous heat conduction equation makes physical sense.