I want to implement an algorithm to solve a heat equation, i.e. \begin{align*} \partial_t u - \Delta u = f \text{ in } \Omega\times(0,T)\\ \partial_nu = 0 \text{ in } \partial\Omega \times (0,T)\\ u(0) = u_0 \end{align*} I'm looking for some reasonable functions $f$ and $u_0$ to model a very hot area (e.g. a box in the middle of $\Omega = [0,1]^3$), which cools down slowly since the rest of the room has room temperature. Can you give me a proper example for this problem and explain me the physical meaning of $f$ and $u_0$?


1 Answer 1


The problem you proposes must be treated with care.

The function $f$ must have zero mean in the domain $\Omega$,i.e. $$\int_\Omega{f\, dV}=0$$ Otherwise the problem is not well posed.

Physically the terms of your equation mean:

$\partial_tu$: the energy change of your system.

$-\Delta u$:the diffusion of the temperature through the domain.

$f$: Heat sources/sinks in the domain.

For the example you proposes: the function $f=0$, since there are not sources/sinks, and the initial condition, given by $u_0$ can be any distribution you want. It can be discontinuous. The only restriction is that $u_0\in L^2(\Omega)$ i.e. it must be square integrable.

Finally the BC you want is not what you have written but: $$\partial_n u=-h(u-u_\infty)$$

Where $h$ may be a function of $u$.

This BC takes into account the fact that in equilibrium there is no heat loss through the boundaries to the outside (the room). Otherwise, if the flux is zero, the temperature reaches a value (homogeneous) being its energy a constant through time and it will never be at room temperature.

The BC has also a physical meaning: the heat loss is proportional to the temperature differente between the hot object and its surroundings, i.e. the hotter the object is the greater the amount of heat that is loss. For example ice is melt quicker when is outside the fridge than outside because the $u-u_\infty$ in the first case is greater than the second given $u$ as the ice temperature.

  • 3
    $\begingroup$ You only need the integral of $f$ to equal the integral of $\partial nu$ over the boundary in the steady-state case. Of course, the solution will increase or decrease without bound in the limit as $T \to \infty$ if the two are not equal, but the problem is still well-posed on a finite time interval. $\endgroup$ Commented Mar 3, 2018 at 20:19
  • $\begingroup$ In this case, the OP ask for an equilibrium. Since the system reaches equilibrium when $T\to \infty$ it is ill-posed. $\endgroup$
    – HBR
    Commented Mar 4, 2018 at 11:18
  • $\begingroup$ The existence of a steady state only require that $\|f(\cdot, t)\|_{L_2}\rightarrow 0$ as $t\rightarrow \infty$. $\endgroup$ Commented Mar 5, 2018 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.