# Laplace transform to bio heat equation

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

$$\rho c \frac{\partial u(\mathbf{x},t)}{\partial t} = k \nabla^2 u(\mathbf{x},t) +\omega_b \rho_b c_b[ u_a - u(\mathbf{x},t) ] + Q_m + Q_r(\mathbf{x},t).$$

1. Here does $\nabla[k \nabla u(x,t)]$ become $k \nabla^2 u(\mathbf{x},t)$ due to this constant $k$?

2. When taking the Lapalace transform what is the Laplace transform of the term $k \nabla^2 u(\mathbf{x},t)$?

3. 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
$$pc \frac{\partial u(x)}{\partial t} = \frac{\partial^2 u(x)}{\partial^2 x} + \omega_b p_b c_b[ u_a - u(x) ] +Q(m) + Q(x)?$$

• Do you really mean Laplace transform and not Laplacian? the term in (2) is the laplacian of ku(x,t) – BCLC Nov 18 '15 at 2:22
• @BCLC what is the difference of laplace transform and laplacian? All I want is to take the Laplace transform of the second equation in the post. When doing that I don't know how to handle the term $k\nabla ^2 u(x,t)$ – sam_rox Nov 18 '15 at 2:29
• Big difference. I guess you really do mean laplace transform. See my answer – BCLC Nov 18 '15 at 3:04

1. I think yes if $$\nabla[k \nabla u(\underline x,t)] = \nabla \cdot [k \nabla u(\underline x,t)]:$$

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:

$$\nabla \times \nabla u(\underline x,t) = 0$$

1. Try (10) and (11) here.

1. Steady state temperature means $t = 0$?

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 $Q_r(x) = 0$

• I was referring the article you had mentioned for the question 2. In that I still don't understand how $k\nabla ^2 u(x,t)$ becomes $k\nabla ^2 U(x,s)$ – sam_rox Nov 18 '15 at 3:25
• @sam_rox How do you know that that is the Laplace transform? What is U? How about you show your work so far? It would've helped if you provided the context in the first place (i.e. the article) :| – BCLC Nov 18 '15 at 3:54
• U(x,s) is the Laplace transform of u(x,t) – sam_rox Nov 18 '15 at 4:09