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The PDE and its BCs (the physical problem is given below the page break):

$$u_{xx}+u_{yy}=0\tag{1}$$ $$u(x,0)=0\tag{2}$$ $$u(x,H)=T_0\tag{3}$$ $$u_x(0,y)=-hu(0,y)\tag{4}$$ $$u_x(W,y)=-hu(W,y)\tag{5}$$ So we're looking for a function $u(x,y)$. Simple separation of variables does not work here because of the inhomogeous BCs.

Several resources I consulted for similar problems suggest to set:

$$u(x,y)=v(x,y)+u_0(x,y)$$

where $u_0(x,y)$ is a particular solution, which satisfies the BCs (but not necessarily the PDE itself).

And $v(x,y)$ a 'remainder' function which satisfies the homogeneous PDE.

The examples I've seen were for 1D, time-dependent problems with very simple BCs, allowing the construction of a $v_p$ quite easily. But here the BCs are mixed and I can't seem to find a $v_p$ that satisfies all BS?

Any help is much appreciated.


A rectangular, thin, flat plate is perfectly insulated on top and bottom sides. The plate is held in a bath at $0$ temperature, except for one edge which is rigorously kept at $T_0$. The left and right edges lose heat through convection only.

enter image description here

What is the steady state temperature distribution $u(x,y)$ of the plate?

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So first you should correct the BCs: from the physical specification of the problem you have

$$u_y(x,0)=0 \\ u(x,H)=T_0 \\ u_x(0,y)=hu(0,y) \\ u_x(W,y)=-hu(W,y)$$

(assuming $h>0$). This change in sign is because of the change in the direction of the normal vector out of the domain. The first equation is because they told you the bottom side is insulated.

That said, notice that for a $C^1$ function to satisfy these BCs (putting the PDE aside), $u_x$ is zero on the top, which means it goes to zero as you move to the top along the left edge. But in this case $u$ must also go to zero as you move to the top along the left edge. But then $u$ is discontinuous at the corner. That means this problem has no strong solution at all.

In view of that, you can consider a regularized problem with the "strong thermal contact" boundary condition on the top:

$$u_y(x,H)=-M(u(x,H)-T_0)$$

where $M$ is a large positive parameter. The $M \to \infty$ limit of this problem gives the weak solution to your problem.

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  • $\begingroup$ Thanks for your correction and help. But how do we find $M$? $\endgroup$
    – Gert
    Commented Nov 19, 2019 at 16:05
  • $\begingroup$ @Gert To actually solve your problem you don't, you solve the BVP with $M$ as a parameter and send $M \to \infty$. $\endgroup$
    – Ian
    Commented Nov 19, 2019 at 16:59
  • $\begingroup$ @Gert Actually I may be confused about what "top" and "bottom" mean, is it top wrt $y$ or wrt $z$? But if that is the case then it is not clear what one of the horizontal BCs is... $\endgroup$
    – Ian
    Commented Nov 19, 2019 at 17:12
  • $\begingroup$ WRT $z$. Like a cheese sandwich where the bread is the insulation and the cheese the $xy$ plate. $\endgroup$
    – Gert
    Commented Nov 19, 2019 at 17:34
  • $\begingroup$ @Gert Then I actually am not certain what the intended BCs are, maybe three sides are in thermal contact with the bath and a fourth is held at $T_0$? This still has the same discontinuity though. $\endgroup$
    – Ian
    Commented Nov 19, 2019 at 17:37

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