The Laplace Operator

Let us reason about a problem such as the heating of a rectangular box. There is some source inside it that heats up the whole structure. Perhaps it is an ordinary potbelly stove. Picture it? Such a rusty contraption, inside which logs are burning and keeping us warm. We want to compute the thermal field of this whole structure.

What is temperature? In essence it is a quantity proportional to the square of the average speed of the particles. It turns out that the kinetic energy of a substance's particles is proportional to the product of temperature and mass. And so we arrive at the well-known formula from thermodynamics that relates the amount of heat ΔQ1\Delta Q_1 required to change the temperature by ΔT\Delta T

ΔQ1\displaystyle \Delta Q_1 =c⋅ρ⋅V⋅ΔT\displaystyle = c \cdot \rho \cdot V \cdot \Delta T
(1)

where cc — the heat capacity, ρ\rho — the density, VV — the volume.

But this is all static, whereas we would like to add some dynamics to the formula. To do this, let us consider the inflow and outflow of energy over a time interval Δt\Delta t. Energy arrives from the burning of the wood, and the burning power can be denoted by f(x,y,z,t)f(x,y,z, t). Then the amount of energy from the burning can be denoted by the following formula

ΔQ2\displaystyle \Delta Q_2 =c⋅ρ⋅f(x,y,z,t)⋅Δt\displaystyle = c \cdot \rho \cdot f(x,y,z, t) \cdot \Delta t
(2)

Where is the energy spent? Obviously, it is spent on heating the surrounding environment through the faces of the box. Let us analyze the temperature TT at some geometric-mean point of the box. Since the box is symmetric, let us consider the energy along one coordinate xx and then generalize to the other two. Clearly, the energy flux through a face will be proportional to the area of the face Δy⋅Δz\Delta y \cdot \Delta z and inversely proportional to the length of the face Δx\Delta x. Then, up to a proportionality coefficient kk, the following formula holds

ΔQ3\displaystyle \Delta Q_3 =k⋅Δy⋅Δz⋅Tx−TΔxΔt\displaystyle = k \cdot \Delta y \cdot \Delta z \cdot \dfrac{T_{x} - T}{\Delta x} \Delta t +k⋅Δy⋅Δz⋅Tx+Δx−TΔxΔt\displaystyle + k \cdot \Delta y \cdot \Delta z \cdot \dfrac{T_{x + \Delta x} - T}{\Delta x} \Delta t =k⋅V⋅Tx+Δx−2⋅T+TxΔx⋅ΔxΔt\displaystyle = k \cdot V \cdot \dfrac{T_{x + \Delta x} - 2 \cdot T + T_{x}}{\Delta x \cdot \Delta x} \Delta t
(3)

From the law of conservation of energy it follows that

ΔQ1\displaystyle \Delta Q_1 =ΔQ2\displaystyle = \Delta Q_2 +ΔQ3\displaystyle + \Delta Q_3
(4)

After substituting, adding the other two faces to the formula for ΔQ3\Delta Q_3, and performing obvious transformations, we obtain

ΔTΔt\displaystyle \dfrac{\Delta T}{\Delta t} =kcρ⋅(Tx+Δx−2⋅T+TxΔx2+Ty+Δy−2⋅T+TyΔy2+Tz+Δz−2⋅T+TzΔz2)\displaystyle = \dfrac{k}{c \rho} \cdot \left( \dfrac{T_{x + \Delta x} - 2 \cdot T + T_{x}}{\Delta x^2} + \dfrac{T_{y + \Delta y} - 2 \cdot T + T_{y}}{\Delta y^2} + \dfrac{T_{z + \Delta z} - 2 \cdot T + T_{z}}{\Delta z^2} \right) +f(x,y,z,t)\displaystyle + f(x,y,z, t)
(5)

Letting the time interval and the dimensions of the box tend to zero, we obtain a partial differential equation

dTdt\displaystyle \dfrac{dT}{dt} =a2⋅(∂2T∂x2+∂2T∂y2+∂2T∂z2)\displaystyle = a^2 \cdot \left( \dfrac{\partial^ 2 T}{\partial x^2} + \dfrac{\partial^ 2 T}{\partial y^2} + \dfrac{\partial^ 2 T}{\partial z^2} \right) +f(x,y,z,t)\displaystyle + f(x,y,z, t)
(6)

where aa — the proportionality coefficient, called the thermal diffusivity coefficient.

We are interested not so much in the essence of the heat equation as in the quantity that appeared on the right-hand side. In essence it is a double partial derivative, which can be represented as the divergence of the gradient. That is, a measure that determines, at each point, how quickly energy flows away.

ΔT\displaystyle \Delta T =div grad T\displaystyle = div \space grad \space T
(7)

where Δ\Delta — the so-called Laplace operator.

Sometimes it is convenient to denote the Laplace operator as the dot product of the nabla operators, that is

Δ\displaystyle \Delta =∇⋅∇\displaystyle = \nabla \cdot \nabla
(8)

Taking all of the above into account, the heat equation can be rewritten in the following form

dTdt\displaystyle \dfrac{dT}{dt} =a2⋅ΔT\displaystyle = a^2 \cdot \Delta T +f(x,y,z,t)\displaystyle + f(x,y,z, t)
(9)