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Find the spherically symmetric solution to $$\nabla^2u=1$$ in the region $r=|\mathbf{r}|\le a$ for $a>0$ that satisfies the following boundary condition at $r=a$:

$\frac{\partial u}{\partial n}=0$

The solution I have looked at states to begin with $\frac{\partial u}{\partial r}=0$ and I can go from here, my question is not actually the above question (but rather I'm using that to illustrate my actualy question) which is:

Why does $\frac{\partial u}{\partial n}=0 \implies \frac{\partial u}{\partial r}=0$? I thought maybe the chain rule, i.e. $\frac{\partial r}{\partial n}\frac{\partial u}{\partial r}=\frac{\partial u}{\partial n}=0$ but I have no idea what $\frac{\partial r}{\partial n}$ actually is and so I don't know that it's not equal to $0$, in fact I don't have a great grasp on what $\frac{\partial u}{\partial n}$ actually $is$, so i would also greatly appreciate if someone could just explain to me what these actually represent- at the moment they're very much just notation for me.

Sorry one more thing: we also arrive at $\frac{\partial u}{\partial r}=\frac13a$ at $r=a$ which I'm fine with, but than it says $>0$ so contradiction. Where has this come from?

Thanks in advance and sorry there are so many parts to my question, they just all are related to the above question and hence why I did not ask each separately.

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  • $\begingroup$ The normal to the surface of spherically symmetric body is always in the direction of the radius i.e. $\mathbf{n}$ points in the direction of $\mathbf{r}$. $\endgroup$ – Chinny84 Apr 17 '17 at 20:28
  • $\begingroup$ Ah, yeah actually I think I knew that after all, thanks. I must have had a mental block $\endgroup$ – Aka_aka_aka_ak Apr 17 '17 at 20:32
  • $\begingroup$ Happens to the best of us. $\endgroup$ – Chinny84 Apr 17 '17 at 20:34
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$\frac{\partial u}{\partial n}$ denotes $(\nabla u)n$. So in your case of acircular domain, it is equal to $\frac{\partial u}{\partial r}$ at the boundary. I guess you'll now understand the whole solution, or you need to give more details about what our problem is with $a/3$ being $ > 0$.

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  • $\begingroup$ Thanks, I think we get the general solution $u=\frac16a^2+A$ for some constant $A$... oh I'm an idiot, god it's late I must be getting tired. I get it all now. Thanks $\endgroup$ – Aka_aka_aka_ak Apr 17 '17 at 20:35
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In general it is not true that $\frac{\partial u }{\partial n}=0 \Rightarrow \frac{\partial u }{\partial r}=0$. But in your case the domain is a sphere centered at the origin so the normal $\vec n$ to the boundary surface is parallel to the radius $r$ from the origin to the surface.

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