Boundary Value Problems/Derivation of Heat Equation

Boundary Value Problems

C - Heat Capacity Joules per degree Kelvin c - specific heat capacity is C per unit mass so j/(gm deg)

A quick derivation of heat equation.

First Principles


 * Heat is measured in calories which is just another form of joules. 1 cal = 1.484 joules.


 * Heat flow. By observation we know that heat(energy) flows from a higher(energy) temperature to a lower(energy) temperature.


 * Let $$ V $$ represent a volume in $$ \mathcal{R}^3 $$ . At each point in the volume $$ V $$ let $$ u(x,y,z,t) $$ represent a scalar field that is the temperature at each point in the  volume. If the $$ \nabla u $$ is not equal to zero at a point $$ P(x_0,y_0,z_0) $$ this indicates that there will be a flow of heat from the higher temperature to the lower temperature in the region about that point. $$ - \nabla u  $$ represents the direction of decreasing temperature and points in the direction of energy flow.

The $$ -\nabla u = v $$ is a vector field the provides a direction and magnitude for the heat flow at any point in the volume.

Let $$ \Delta V $$ be a small volume in $$ \mathbf{V} $$. Choose $$ S_1 \mbox{ and } S_2 $$ represent two parallel end areas of this volume. Each end area has a normal $$ n_1 \mbox{ and } n_2 $$. The net flow through the end area $$ \Delta A $$ is the amount of flow $$ v $$ normal to the surface $$ S_1 $$ times the surface area $$ \Delta A $$ of $$ S_1 $$

$$ \mbox{ net flux is }F = -k \nabla u \cdot \hat{n} \Delta A $$. The constant $$ k $$ is a material property.

For a cubic volume the net flux through each side area $$ S_B $$ will be $$ \lim_{N\to\infty} \sum_{i=1}^{N} - k\nabla u_i \cdot n \Delta A_i = \int \int_{S_B} -k \nabla u_i \cdot n dA$$

If there are any sources or sinks in the block these are included as an additive term $$ E = \int_B f(x,y,z,t) dV $$.The total flux for the cube will be a double integral over the closed surface of the cube volume.

$$ \mbox{Net Flux} = \int \int_{S_B} -k \nabla u_i \cdot n dA + \int_{B} f(x,y,z,t) dV$$

There are three results for the integration: 1) The net flux is positive, which implies more flux is leaving the cube than entering it. An example would be the radiation from a radioactive source enclosed in a block of cement. 2) The net flux is zero, which implies the in flux equals the out flux. Even if there are sources in the cube there are also sinks that absorb the addition flux. An example would be putting water in a barrel while it flows out through the bottom at the same rate as the entry rate. 3) The net flux is negative, which implies more flux is entering the cube than leaving it. An example would be a material that absorbs heat applied to it's surface, such as melting ice.

Another way of looking at the heat energy stored in a piece of material: A mass will hold heat. The amount of heat in an incremental mass is estimated by the expression:

$$ \displaystyle \Delta H = c u \Delta m = c u \rho \Delta V $$ c - specific heat capacity is C per unit mass so j/(gm deg) The total heat, $$ \displaystyle H $$, in the total mass will be $$\displaystyle H = \int_B c u \rho dV $$

The change in heat per unit time is $$ \frac {dH} {dt} = \frac {d} {dt} \int_B c u \rho dV = \int_B c \frac {du} {dt}  \rho dV$$. The change in the heat is the net flux $$ \mbox{Net Flux} = \int \int_{S_p} -k \nabla u \cdot n dA$$ so by equivalence:

$$ \int_B c \frac {du} {dt} \rho dV = \int \int_{S_p} -k \nabla u \cdot n dA  + \int_B  f dV$$.

The divergence theorem is applied to the first term on the right so that all three terms are volume integrals.

$$ \int_B c \frac {du} {dt} \rho dV = \int_{B} \nabla \cdot (k \nabla u )dV + \int_B  f dV$$.

Dividing through by $$ c \rho $$ and labeling the factor $$ \alpha^2 = \frac {k} {c \rho } $$ the integral equation becomes.

$$ \int_B \frac {du} {dt}   dV = \int_{B} \alpha^2  \nabla \cdot  \nabla u dV + \int_B  \frac {f}{\rho c} dV$$.

Pull all the integrals together on the right: $$ \int_B \frac {du} {dt} dV = \int_{B} \alpha^2  \nabla^2 u  +  \frac {f}{\rho c} dV $$.

Without proving it we will state $$ \frac {du} {dt} = \alpha^2 \nabla^2 u  +  \frac {f}{\rho c} $$. This is the heat equation without convection (strictly conduction).

$$ \alpha^2 $$ is the material's diffusivity. And $$ F = \frac {f} {\rho c} $$ is the heat source/sink density in the material.

The temperature $$ u(x,y,z,t) $$ is a scalar field associated with the points in the volume. As with ODE's we consider the initial state of the system (initial value)as part of the solution process. For this problem the initial value is associated with every point in the volume at $$ t=0 $$.

The initial value is $$ u(x,y,z,t=0)=g(x,y,z) $$ where $$ g $$ is a given function.

In addition to the initial conditions there will be equations that define the flow of heat across the boundary of the surface. Three things can happen at a boundary:

1) Flux can flow in a positive direction from the body out across the boundary at a point. If $$ \nabla u \cdot \hat{n} > 0 $$ at a point on the surface then the flow is from the interior to the exterior of the volume, across the surface. This condition occurs if the interior is at a higher temperature than the exterior. 2) Flux can flow in across the boundary at a point on the surface, If $$ \nabla u \cdot \hat{n} < 0 $$ at a point on the surface then the flow is from the exterior to the interior of the volume, across the surface. This occurs if the exterior temperature is higher than the interior.

3) Flux does not flow cross the boundary.If $$ \nabla u \cdot \hat{n} = 0 $$. Interior temperature at the boundary equals the exterior temperature at the boundary.

These cases occur for fixed boundary conditions: $$\displaystyle u_{\delta \Omega}(t)= T $$, the temperature at the boundary $$\displaystyle u_{\delta \Omega}(t) $$ is held at a constant exterior temperature, $$\displaystyle T $$.