Derivation of the conduction of heat in mutliple dimensions

Let R be some three dimensional region, and S be the surface area of said region.

Using the conservation law we have that the rate of change of energy in R is the sum of the energy entering the boundary, with the energy generated inside the region, minus the energy leaving the boundary.

We have

\text{Flux} = \vec{\phi}(x,t)

And the energy across the border is the normal component to the boundary with the flux.

\frac{d}{dt}\left(\iiint_R c(x)\rho(x)u(x,t) dx\right) = - \oiint_S(\phi \cdot n) dn + \iiint_R Q(x,t) dx

The divergence theorem states that if \vec{A} is a continously differentiable vector, \vec{A} = (A_x,A_y,A_z). Thus

\iiint_R (\nabla \cdot \vec{A}) dx = \oiint_S (\vec{A} \cdot n)dn

Where

\nabla \cdot \vec{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}

Using the Divergence theorem, we can rewrite our equation for heat as

\iiint_R c(x)\rho(x)\frac{\partial u}{\partial t}(x,t) + \nabla \cdot \phi(x,t) - Q(x,t)\; dx = 0

Therefore

c(x)\rho(x)\frac{\partial u}{\partial t}(x,t) + \nabla \cdot \phi(x,t) - Q(x,t)\; = 0

We can convert flux into tempature using Fouriers law from our one dimensional case,

\bar\phi(\bar x,t) = -k_0(\bar x)\nabla u

Thus we get for our multiple dimensional heat equation

c(x)\rho(x)\frac{\partial u}{\partial t}(x,t) = \nabla \cdot (k_0(\bar x)\nabla u) + Q(x,t)\;

If \rho, k_0 are both constants we achieve

\frac{\partial u}{\partial t}(\bar x,t) = k\nabla \cdot (\nabla u) + \frac{1}{c\rho}Q(\bar x,t)

Where k = k_0/c\rho and is our thermal diffusitivy. We sometimes write

\nabla \cdot (\nabla u) = \nabla^2 u = \Delta u

Where

\Delta u = \nabla(\frac{\partial U_x}{\partial x}, \frac{\partial U_y}{\partial y},\frac{\partial U_z}{\partial z}) = \frac{\partial^2 U_x}{\partial x^2} + \frac{\partial^2 U_y}{\partial y^2} + \frac{\partial^2 U_z}{\partial z^2}

If Q = 0, then the heat equation becomes

\frac{\partial u}{\partial t} = k\Delta u

We now must complement the system with initial conditions and boundary conditions.

Boundary conditions

We have prescribed tempature

u(\bar x,t) = T(\bar x, t) \quad \forall \bar x \in S

Prescribed Flux

\bar \phi (\bar x, t) \cdot \bar n(\bar x) = g(\bar x,t)\quad \forall \bar x \in S

In the case of an insulated surface, you would have \bar \phi \cdot \bar n = 0 = \nabla u \cdot \bar n.

We also have Newtons law of cooling

-k_0(\bar x) \nabla u \cdot \bar n = H(u(\bar x,t)- U(\bar x,t)) \quad \forall \bar x \in S

Where U(\bar x,t) is the tempature of the surrounding material.

Boundary conditions can be different in different parts of the domain.

If the boundary conditions and sources of heat (Q) are independent of time, there will exist a steady state solution such that

u_{eq}(\bar x) \implies \frac{\partial u_{eq}}{\partial t} = 0

If c,\rho,k_0 are constants, then

- k\Delta u = \frac{Q}{c\rho} \iff - \Delta u = \frac{Q}{k_0}

This is called the Poisson equation. In the particular case that Q is zero, then \Delta u = 0. This is called the Laplace equation.

Harmonic functions satisfy the equation

-\Delta u (\bar x) = 0\quad \bar x \in \mathbb{R}

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