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

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  • $\begingroup$ Do you really mean Laplace transform and not Laplacian? the term in (2) is the laplacian of ku(x,t) $\endgroup$
    – BCLC
    Commented Nov 18, 2015 at 2:22
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    $\begingroup$ @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 $k\nabla ^2 u(x,t)$ $\endgroup$
    – sam_rox
    Commented Nov 18, 2015 at 2:29
  • $\begingroup$ Big difference. I guess you really do mean laplace transform. See my answer $\endgroup$
    – BCLC
    Commented Nov 18, 2015 at 3:04

1 Answer 1

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  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 (enter image description here),

$$\nabla[k \nabla u(\underline x,t)]$$

$$= \nabla k \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$

$$= (0) \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$

$$= (0) \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$

$$= k \nabla \nabla u(\underline x,t)]$$

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.

enter image description 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 $Q_r(x) = 0$


enter image description here


enter image description here

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  • $\begingroup$ 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 $k\nabla ^2 U(x,s)$ $\endgroup$
    – sam_rox
    Commented Nov 18, 2015 at 3:25
  • $\begingroup$ @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) :| $\endgroup$
    – BCLC
    Commented Nov 18, 2015 at 3:54
  • $\begingroup$ U(x,s) is the Laplace transform of u(x,t) $\endgroup$
    – sam_rox
    Commented Nov 18, 2015 at 4:09

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