# Problem Of The Week # 332 - Nov 27, 2018

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Show that, for every compactly supported, smooth, real valued function $f : \Bbb R^3 \to \Bbb R$,

$$\iiint_{\Bbb R^3} \nabla^2\left(\frac{1}{\| \mathbf{x} - \mathbf{y}\|}\right) f(\mathbf{x})\, d\mathbf{x} = -4\pi f(\mathbf{y})$$

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#### Euge

##### MHB Global Moderator
Staff member
This week’s problem was correctly solved by Ackbach . You can read his solution below.

The Green's function for the Laplacian operator $\nabla^2$ is
$$G(\mathbf{x},\mathbf{y})=-\frac{1}{4\pi\|\mathbf{x}-\mathbf{y}\|},$$
or
$$-4\pi G(\mathbf{x},\mathbf{y})=\frac{1}{\|\mathbf{x}-\mathbf{y}\|}.$$
By the properties of the Green's function, we have that
$$\iiint_{\mathbb{R}^3}\nabla^2\left(\frac{1}{\|\mathbf{x}-\mathbf{y}\|}\right)f(\mathbf{x})\,d\mathbf{x}= \iiint_{\mathbb{R}^3}\nabla^2\left(-4\pi G(\mathbf{x},\mathbf{y})\right)f(\mathbf{x})\,d\mathbf{x}= -4\pi \iiint_{\mathbb{R}^3}\nabla^2G(\mathbf{x},\mathbf{y})f(\mathbf{x})\,d\mathbf{x}= -4\pi f(\mathbf{y}),$$
as needed.

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