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October 9th, 2018, 07:04 AM   #1
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A PDE solution

Let $f\in C^2(\mathbb{R}^n)$.

We define $$\phi(x,r)=\frac{1}{n\alpha(n)}\int_{\partial B(0,1)}f(x+rz)dS(z)$$ where $\alpha(n)$ is the volume of $B(0,1)$.

I calculated $$\partial_r\phi=\frac{r}{n\alpha(n)} \int_{\partial B(0,1)}\Delta_xf(x+rz)dS(z)$$

Please help me to show that $$\partial_{rr}\phi-\frac{n-1}{r}\partial_r\phi=\Delta_x\phi$$

Thanks.

Last edited by skipjack; October 9th, 2018 at 11:52 AM.
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October 9th, 2018, 08:20 AM   #2
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This is the right version

Quote:
Originally Posted by mona123 View Post
Let $f\in C^2(\mathbb{R}^n)$.

We define $$\phi(x,r)=\frac{1}{n\alpha(n)}\int_{\partial B(0,1)}f(x+rz)dS(z)$$ where $\alpha(n)$ is the volume of $B(0,1)$.

I calculated $$\partial_r\phi=\frac{r}{n\alpha(n)} \int_{B(0,1)}\Delta_xf(x+rz)dS(z)$$

Please help me to show that $$\partial_{rr}\phi-\frac{n-1}{r}\partial_r\phi=\Delta_x\phi$$

Thanks.

Last edited by skipjack; October 9th, 2018 at 11:53 AM.
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October 9th, 2018, 09:18 AM   #3
SDK
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Ahh so we have graduated from doing your abstract algebra HW to doing your PDE HW. It's nice to be promoted.

Last edited by skipjack; October 9th, 2018 at 11:53 AM.
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