This is the bio heat equation and I have several questions when trying to work with it.
$$ \rho c \frac{\partial u(x,t)}{\partial t} = \nabla[k \nabla u(x,t)] + \omega_b \rho_b c_b [u_a - u(x,t)] + Q_m + Q_r(x,t) $$
where the first expression on the right hand side describes the conduction of heat induced by temperature gradient. Then by letting $k$ to be a constant through out the process this equation has been written as
$$ \rho c \frac{\partial u(\mathbf{x},t)}{\partial t} = k \nabla^2 u(\mathbf{x},t) +\omega_b \rho_b c_b[ u_a - u(\mathbf{x},t) ] + Q_m + Q_r(\mathbf{x},t). $$
Here does $\nabla[k \nabla u(x,t)]$ become $k \nabla^2 u(\mathbf{x},t)$ due to this constant $k$?
When taking the Lapalace transform what is the Laplace transform of the term $k \nabla^2 u(\mathbf{x},t)$?
The bio heat equation at the top of the post is for a 2-D case. That is $\mathbf{x} = (x_1,x_2)$. But if this equation is written for a 1-D case at steady state temperature does the bio heat equation become
$$ pc \frac{\partial u(x)}{\partial t} = \frac{\partial^2 u(x)}{\partial^2 x} + \omega_b p_b c_b[ u_a - u(x) ] +Q(m) + Q(x)? $$
By one of the product rules (
),
$$\nabla[k \nabla u(\underline x,t)]$$
$$= \nabla k \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$
$$= (0) \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$
$$= (0) \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$
$$= k \nabla \nabla u(\underline x,t)]$$
But what does $$\nabla \nabla u(\underline x,t)$$
even mean?
By definition, $\nabla^2 u(\underline x,t) = \nabla \cdot \nabla u(\underline x,t)$.
Also for 'well-behaved' u(\underline x,t), we have:
$$\nabla \times \nabla u(\underline x,t) = 0$$
If so, I think so except I think the last terms should be: $Q_m + Q_r(x)$ based on (1) and (4) here. In the link, it seems that $Q_r(x) = 0$