What's $\nabla \cdot [k \cdot \nabla T]$ formulation in Laplace's eq. really?

Question

What's $\nabla \cdot [k \cdot \nabla T]$ formulation in Laplace's eq. really?

(https://en.wikiversity.org/wiki/Introduction_to_finite_elements/Weak_form_of_heat_equation)

I read that this is like

$$\left(\frac{d}{dx_1}, \frac{d}{dx_2}, ...\right) \cdot \bigg[ k \cdot \bigg(\frac{dT}{dx_1}, \frac{dT}{dx_2},... \bigg) \bigg]$$

The inner dot product can be expanded as:

$$k\frac{dT}{dx_1}+k\frac{dT}{dx_2}+...$$

Then one takes the outer dot product to get:

$$\frac{d}{dx_1}k\frac{dT}{dx_1}+\frac{d}{dx_2}k\frac{dT}{dx_2}+...$$

Since $k$ is constant and $\frac{d}{dx_i}$ is the conventional derivative. Then since derivative is linear operator, then one can move the constant to the front?

And then then this formulation is actually equivalent to:

$$k\nabla^2T$$

Am I right?

But what's the reason for using the formulation $\nabla \cdot [k \cdot \nabla T]$?

Answer 1

In the more general problem, $k$ can be a function of space, at which point $\nabla\cdot[k\nabla T] \ne k\nabla^2T$

It's also possible that $k$ is meant to be a tensor quantity (and thus should be written $\mathbf{k}$), so $\nabla\cdot[\mathbf{k}\cdot\nabla T]$ is really the only way of writing it. In particular, your statement that the inner dot product can be expanded as $$ k\frac{dT}{dx_1}+k\frac{dT}{dx_2}+... $$ is incorrect. For starters, it is clear this is wrong because this is a scalar quantity, not a vector one, and the divergence operator takes a vector argument. The correct expansion, taking into account $\mathbf{k}$'s tensor nature, is $$ \left(k_{11}\frac{\partial T}{\partial x_1} + k_{12}\frac{\partial T}{\partial x_2} + ...\right)\hat{\mathbf{x}}_1 + \left(k_{21}\frac{\partial T}{\partial x_1} + k_{22}\frac{\partial T}{\partial x_2} + ...\right)\hat{\mathbf{x}}_2 + ... $$ If $\mathbf{k}$ is constant, then the divergence can be expanded as $$ k_{11}\frac{\partial^2 T}{\partial x_1^2} + k_{12}\frac{\partial ^2T}{\partial x_1\partial x_2} + ... + k_{21}\frac{\partial T}{\partial x_2 \partial x_1} + k_{22}\frac{\partial T}{\partial x_2^2} + ... + ... $$ which is not equal to $k\nabla^2T$ for any scalar $k$.

Answer 2

Eyeballfrog has a good answer. K (or $\bf \kappa$) is likely a tensor of conduction, especially as it is bold and other non-scalar quantities also seem to be. The tensor contains information like in which orientation is there how much conductance at this point and (complementarily) which orientation do we have good heat insulation.

---

I would also like to add that $\nabla^2$ is not very good notation. We need to know if gradient is applied as inner or outer product. If it is outer+inner (expansion+contraction) then it is Laplacian, but it can also be outer+outer and then it would be more like a Hessian.