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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 $$ \int_{-\infty}^\infty g(x) \Big( 3\delta(x)\Big)\, dx = 3g(0) \ne g(0). $$ $\endgroup$ Commented May 2, 2017 at 14:40
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    $\begingroup$ @MichaelHardy He did mention the integral of each $f_n$ is $1$. $\endgroup$
    – Paichu
    Commented May 2, 2017 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. $\endgroup$ Commented May 2, 2017 at 15:29
  • $\begingroup$ @MichaelHardy I agree that the rigorous definition is still not very rigorous. $\endgroup$
    – Paichu
    Commented May 2, 2017 at 15:31

2 Answers 2

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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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Yes, this is a solution to your equation. This is an instance of a more general method: convolving with the fundamental solution (see e.g. the Wikipedia article about it).

Notes about notation: single $x$ denotes a point $(x_1,x_2) \in \mathbb{R}^2$, $dx$ denotes integration over a domain in $\mathbb{R}^2$, $\Delta$ is the Laplace operator (so that it's not confused with the hessian),

How it's applied. Let $g$ be the fundamental solution, meaning that $\Delta g = \delta$ (in our case, $g(x) = \frac{1}{2\pi} \ln|x|$). and $h$ is an integrable function, then $$ f(x) := \int_{\mathbb{R}^2} g(x-y)h(y) \, dy \qquad \text{(often denoted as $(g * h)(x)$)} $$ solves $\Delta f = h$ in a weak sense. If $h$ has better regularity, then $f$ is also a classical solution (a $C^2$ function with $\Delta f = h$ pointwise). If you take $h$ to be $a \cdot \mathbf{1}_{D}$ (the function equal $a$ in $D$ and $0$ elsewhere), then $f$ is exactly the function you described: $f(x) = \int_D g(x-y) \, dy$. On the other hand, $\Delta f = \mathbf{1}_{D}$ is the equation you wanted to solve

How it works. Since this topic (solving the Poisson equation by convolving with the fundamental solution) is covered in many places, let me just direct you to L. Evans's Partial Differential Equations. One word of caution: in our example the function $h$ is not continuous and so $f$ cannot possibly have $C^2$ regularity. For that reason, one should aim for the weak sense of $\Delta f = h$, which is even easier to prove than the classical case, if only one already knows what weak means. Again, Evans gives a nice account of weak solutions.

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