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Problem Of The Week # 356 - Jan 21, 2020

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Euge

MHB Global Moderator
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Jun 20, 2014
1,890
Here is this week's POTW:

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Let $\phi_1$ and $\phi_2$ be harmonic functions on a bounded domain $\Omega \subset \mathbb R^3$ such that \[\phi_1 \frac{\partial \phi_1}{\partial n} + \phi_2 \frac{\partial \phi_2}{\partial n} = \phi_2 \frac{\partial \phi_1}{\partial n} + \phi_1 \frac{\partial \phi_2}{\partial n}\quad \text{on}\quad \partial \Omega\]
Prove that $\phi_1 = \phi_2$ everywhere in $\Omega$. [The operator $\frac{\partial}{\partial n}$ denotes the normal derivative on $\partial \Omega$.]

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
No one answered this week's problem correctly, but that's ok since it was tacitly assumed that $\phi_1$ and $\phi_2$ agree at some point in $\Omega$. (Smile) You can read my solution below.

If $\phi := \phi_1 - \phi_2$, then $\phi \frac{\partial \phi}{\partial n} = 0$ on $\partial \Omega$ and $\nabla^2 \phi = 0$ in $\Omega$. By Green's formula, $$\int_\Omega \lvert \nabla \phi\rvert^2\, dx = \oint_{\partial \Omega} \phi \frac{\partial \phi}{\partial n}\, dS - \int_{\Omega} \phi \nabla^2 \phi\, dx = 0 - 0 = 0$$ Hence $\lvert \nabla \phi\rvert^2 = 0$. Since $\Omega$ is connected $\phi$ is constant. As $\phi_1$ and $\phi_2$ agree at some point in $\Omega$, $\phi$ must be zero at that point, making $\phi$ identically zero. Hence $\phi_1$ and $\phi_2$ are identical.
 
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