# Diffusion equation (derivation)

1. Dec 21, 2005

### Benny

Hi, I'm not understanding an example in my book. Can someone please shed some light on it?

> Derive the equation satisfied by the temperature $$u\left( {\mathop r\limits^ \to ,t} \right)$$ at time t for a material of uniform conductivity k, specific heat capacity s and density $$\rho$$. Express the equation in Catersian coordinates.

Let us consider an arbitrary volume V lying within the solid and bounded by a surface S (this may coincide with the surface of the solid if so desired). At any point in the solid the rate of heat flow per unit area in any given direction $$\mathop {\mathop r\limits^ \to }\limits^\^$$ is proportional to minus the component of the temperature gradient in that direction and so is given by $$\left( { - k\nabla u} \right) \bullet \mathop {\mathop r\limits^ \to }\limits^\^$$.

Since r has 3 components x,y,z then I understand that it is directed out of the volume. But what about u? Is it a function of 2 or 3 variables? u = u(r,t) = u(x,y,z,t) ? Certainly, if u is a function of 3 variables then it's gradient is directed out of the volume and then the above would make sense. Otherwise I don't really get what's going on here.

The total flux of heat out of the volume V per unit time is given by

$$- \frac{{dQ}}{{dt}} = \int\limits_{}^{} {\int\limits_S^{} {\left( { - k\nabla u} \right) \bullet ndS} }$$

$$= \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_V^{} {\nabla \bullet \left( { - k\nabla u} \right)dV} } }$$

Ok this is just the divergence theorem...

where Q is the total heat energy in V at time t and n is the outward-pointing unit normal to S.

We can also express Q as a volume integral over V,

$$Q = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_V^{} {s\rho udV} } }$$

and its rate of change is given by:

$$\frac{{dQ}}{{dt}} = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_V^{} {s\rho \frac{{\partial u}}{{\partial t}}} dV} }$$

I don't understand this bit. I know that it's got something to do with differentiating inside the integral. However, I can't think of a suitable formula to apply to this situation to carry out the differentiation. The only relevant things which come to mind apply to single integrals - not sure if there is an equivalent for a triple integral.

Comparing the two expression for dQ/dt and remembering that the volume V is arbitrary, we obtain the three-dimensional diffusion equation

$$\kappa \nabla ^2 u = \frac{{\partial u}}{{\partial t}}$$

Where kappa k is a suitable constant.

I'm not sure what's been done in this step either. It looks like they've just taken the two expressions for dQ/dt, integrated with respect to t but that doesn't really seem to make sense. I also have no idea as to how an arbitrary volume allows the last step to be performed. Wouldn't the expressions for the volume integrals differ depending on limits of integration?

Any explanations are appreciated, thanks.

Edit: The formatting is a little hard to read. -_-

Last edited: Dec 21, 2005
2. Dec 22, 2005

### Tom Mattson

Staff Emeritus

You yourself said that $u=u(\vec{r},t)$. Since the argument $\vec{r}$ is a vector, then yes $u=u(x,y,z,t)$.

Yes, they transported the differentiation operator inside the triple integral. It only acts on $u$ because that's the only factor in the integrand that depends on time. And the total derivative operator changed to a partial derivative operator because the derivative operator inside the integrand acts on a function of several variables, whereas outside it acts on a function of only one variable, namely $t$.

Yes, they compared the two expressions for $\frac{dQ}{dt}$, and then set the integrals equal. Since the integrals are equal for arbitrary volumes it means that the integrands must be equal, and so they simply set them equal to obtain the PDE.

If you want your LaTeX to be in a sentence, then replace the tex tags with itex (inline tex) tags.

3. Dec 22, 2005