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Problem of the Week #20 - October 15th, 2012

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Chris L T521

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Jan 26, 2012
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I realized that I had posted solutions last night to the POTWs, but forgot to create the new ones last night...I guess that not sleeping well the night before travelling all day can make you do these kinds of things. Anyways, here's this week's problem.

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Problem: Consider the upper half plane with its standard hyperbolic metric $\frac{1}{y^2}(dx^2+dy^2)$. For $k$ a fixed real number, compute the Laplacian of the function $y^k$ relative to this metric.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
I realized that I had posted solutions last night to the POTWs, but forgot to create the new ones last night...I guess that not sleeping well the night before travelling all day can make you do these kinds of things. Anyways, here's this week's problem.

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Problem: Consider the upper half plane with its standard hyperbolic metric $\frac{1}{y^2}(dx^2+dy^2)$. For $k$ a fixed real number, compute the Laplacian of the function $y^k$ relative to this metric.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
No one answered this week's question. Here's my solution below.

Recall that for any metric $g_{ij}$, the Laplacian of a scalar function $f$ is given by the formula
\[\Delta_g f = \frac{1}{\sqrt{\det g_{ij}}}\partial_i\left(\sqrt{\det g_{ij}}g^{ij}\partial_jf\right)\]
Since we're working with the standard hyperbolic metric $ds^2=E\,dx^2+2F\,dx\,dy + G\,dy^2=\frac{1}{y^2}\,dx^2+\frac{1}{y^2}\,dy^2$, we have
\[g_{ij}=\begin{bmatrix}E & F\\ F & G\end{bmatrix}=\begin{bmatrix}1/y^2 & 0 \\ 0 & 1/y^2\end{bmatrix}\implies g^{ij} = g_{ij}^{-1} = \begin{bmatrix}y^2 & 0\\ 0 & y^2\end{bmatrix}\]
and thus $\det g_{ij}=\dfrac{1}{y^4}\implies \dfrac{1}{\sqrt{\det g_{ij}}}=y^2$. We now have that
\[\begin{aligned}\Delta_g (y^k) &= y^2 \partial_y\left(\frac{1}{y^2}\cdot y^2\cdot ky^{k-1}\right)\\ &= y^2k(k-1)y^{k-2}\\ &= k(k-1)y^k.\end{aligned}\]
Therefore, $\Delta_g y^k = k(k-1)y^k$.
 
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