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I know that the "hand-wavy" definition of the $\delta (x)$ function is $$ \delta(x) = \begin{cases} \infty &\quad\ x=0 \\ 0 &\quad\text{otherwise} \end{cases} $$

and the more rigorous definition is that it's the limit of a sequence of functions $f_n$ for which $f_n(x) \rightarrow 0$ for all $x \neq 0$, and $f_n \rightarrow \infty$ for $x=0$, and (edited to add) $\int f_n = 1$ for all $n$. From this perspective, I see why the integral should be one, because the integral of all of the $f_n$ is equal to $1$.

Now, suppose I want to construct a function $f(x,y)$ in the plane for which

$$ \nabla ^2f(x,y) = \begin{cases} a &\quad\ (x,y) \in D \\ 0 &\quad\text{otherwise} \end{cases} $$ where $D$ is some simply connected region.

I can definitely solve $\nabla ^2f(x,y) = \delta(\|(x,y) - (x_0,y_0)\|)$ for any point $(x_0,y_0)$. This is just done by using the fundamental solution $$f(x,y) = \frac{-1}{2\pi} \ln\left( \|(x,y)-(x_0,y_0)\|\right)$$

My question is whether I can do the following:

Because I want the Laplacian of $f$ to be as described above, can I write

$$ f(x,y) = a \int_D \frac{-1}{2\pi} \ln\left( \|(x,y)-(x_0,y_0)\|\right) \,dA \quad ?$$

where $dA$ refers to integration with respect to $(x_0,y_0)$ over the area of $D$.

My confusion is coming from the fact that: The Laplacian of $f$ will be the Laplacian of a sum of (infinitely) many $\delta$ functions, so intuition tells me it should be infinite; on the other hand, integrating a $\delta$ function gives $1$, so the factor of $a$ in front of the integral should give the desired result, no?

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  • $\begingroup$ Your "more rigorous" definition is still pretty hand-wavy. You may say that $\delta$ is the limit of a certain kind of sequence of functions, but if you multiply all of those functions by $3$ then you get another sequence of the same kind, but it doesn't converge to $\delta$, but rather to $3\delta$, and that's different since $$ \int_{-\infty}^\infty g(x) \Big( 3\delta(x)\Big)\, dx = 3g(0) \ne g(0). $$ $\endgroup$ – Michael Hardy May 2 '17 at 14:40
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    $\begingroup$ @MichaelHardy He did mention the integral of each $f_n$ is $1$. $\endgroup$ – Paichu May 2 '17 at 14:41
  • $\begingroup$ @Paichu : ok. But still there's the issue of what happens if you use one sequence of functions meeting the desiderata and I use another such sequence. $\endgroup$ – Michael Hardy May 2 '17 at 15:29
  • $\begingroup$ @MichaelHardy I agree that the rigorous definition is still not very rigorous. $\endgroup$ – Paichu May 2 '17 at 15:31
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You should solve Laplace's equation using a Fourier transform. You can write the actual Delta function as

$$\delta(x)=\int_{-\infty}^{\infty}\!e^{-ikx}\,dk$$

Further, when you Fourier transform the Laplace equation you get

$$\mathcal{F}[\nabla^2f(x,y)]=-k^2\hat{f}(x,y)$$

If you solve the problem in $k$-space for $\hat{f}(x,y)$ and then perform a reverse Fourier transform, you will have your solution.

To be clear: Laplace's equation is NOT solved by integrating a sum of delta functions.

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