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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

You should solve Laplace's equation using a Fourier transform. You can write the actual Delta function as


Further, when you Fourier transform the Laplace equation you get


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.