The following equation is considered: $$ \frac{\partial u}{\partial t} - a\Delta u + \mathbf v \cdot \nabla u = f. $$ I have difficulties in formulating boundary conditions for this equation. If $\mathbf v$ is absent, the boundary conditions are $$ a\frac{\partial u}{\partial n} + \beta(u - u_b) = 0 $$ where $u_b$ is prescribed temperature field on the boundary. But if $\mathbf v$ is present? If we suggest the following boundary condition: $$ a\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n) + \beta(u - u_b) = 0 $$ then we will have problems with analysing the equation when $(\mathbf v \cdot \mathbf n) > 0$ (where the fluid outflows). How to set correct boundary conditions?
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$\begingroup$ I'm no expert here, but I seem to remember just the opposite: if $\mathbf{n}$ is the exterior normal then you just impose zero Neumann BC where $\mathbf{v}\cdot \mathbf{n}\geq 0$, whereas you have to prescribe $a\frac{\partial u}{\partial n}+\beta(u-u_b)=F$ for given $F$ where $\mathbf{v}\cdot \mathbf{n}<0$. This makes physical sense to me: in the outflow region the boundary conditions are just self-adjusting to what comes "from the inside", but in the inflow region you have to know what comes from the outside (the PDE cannot know this information). But I have no reference, sorry... $\endgroup$– leo monsaingeonCommented Jun 6, 2014 at 9:19
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$\begingroup$ Just found a reference in my archives: look at this paper and in particular their equation (1.5). I guess the references [5,10] therein should help $\endgroup$– leo monsaingeonCommented Jun 6, 2014 at 9:36
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1 Answer
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your second boundary condition is missing a factor $u-u_b$:
$$a\frac{\partial u}{\partial n} - \gamma(u-u_b)(\mathbf v \cdot \mathbf n) + \beta(u - u_b) = 0$$
the coefficient $\beta$ gives the strength of the heat transfer at the boundary; the coefficients $a$ and $\gamma$ are the same as in the diffusion-convection equation,
$$\frac{\partial u}{\partial t} - a\Delta u + \gamma \mathbf v \cdot \nabla u = f.$$
see, for example, The convective-diffusion equation and its use in building physics (2000).
answered Jun 6, 2014 at 10:39
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$\begingroup$ Thank you! This was a typo. I meant the condition $a\dfrac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)u + \beta(u - u_b) = 0$. The heat flux is $\mathbf q = -a\nabla u + \mathbf v u$. Why $(\mathbf v \cdot \mathbf n)(u-u_b)$ instead of $(\mathbf v \cdot \mathbf n)u$? $\endgroup$ Commented Jun 6, 2014 at 11:26
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1$\begingroup$ if the temperature $u$ equals the temperature $u_b$ of the surroundings, there can be no outgoing heat flow, or you'd be violating the second law of thermodynamics... $\endgroup$ Commented Jun 6, 2014 at 11:42
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$\begingroup$ In this case we should impose the condition $\beta \geq (\mathbf v \cdot \mathbf n)$. Why is it so (physically)? $\endgroup$ Commented Jun 7, 2014 at 6:02
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$\begingroup$ In the case $\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)u + \beta(u - u_b)$ there can be outgoing heat flow due to convective transfer ($\mathbf v$), can't it? $\endgroup$ Commented Jun 7, 2014 at 6:06
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1$\begingroup$ just take $u=u_b$, so no temperature difference between inside and outside at the boundary; you would hope that that then $du/dn=0$ irrespective of $v$, wouldn't you? $\endgroup$ Commented Jun 7, 2014 at 15:53