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In a mathematical physical problem, I came across the following partial differential equation involving a delta Dirac function: $$ a \, \frac{\partial^2 w}{\partial x^2} + b \, \frac{\partial^2 w}{\partial y^2} + \delta^2(x,y) = 0 \, , $$ subject to the boundary conditions $w(x = \pm 1, y) = w(x, y = \pm 1) = 0$. Here $a, b \in \mathbb{R}_+$ and $\delta^2(x,y) = \delta(x)\delta(y)$ is the two-dimensional delta Dirac function.

While solutions for ODEs with delta Dirac functions can readily be obtained using the standard approach, I am not aware of any resolution recipe for PDEs with delta Dirac functions.

Any help or hint is highly desirable and appreciated.

Thank you

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    $\begingroup$ The Dirac terms indicate that the source of the signal is at the origin in your problem. Maybe switch to polar coordinates, then $\delta^2(x,y) \to \delta(r)$? I know it makes the boundary conditions harder, but might be an ok trade off. $\endgroup$ – Mnifldz Jan 16 '19 at 13:45
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    $\begingroup$ Try separation of variables (say $w=X(x)Y(y)$). Apply the boundary conditions including (1) continuity of $X$ at $x=0$ and of $Y$ at $y=0$ and (2) the appropriate discontinuity of $X'$ at $x=0$ and $Y'$ at $y=0$. $\endgroup$ – Mark Viola Jan 16 '19 at 13:49
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    $\begingroup$ Another way of dealing with the delta-function here is using that the Greens-function for the 2D Laplacian is $\frac{\log(\sqrt{x^2+y^2})}{2\pi}$. Thus if we change variables to $X = x/\sqrt{a}$ and $Y = y/\sqrt{b}$ then we can write $w = -\frac{\log(X^2+Y^2)}{4\pi \sqrt{ab}} + w_0$ where $\frac{\partial^2 w_0}{\partial X^2} + \frac{\partial^2 w_0}{\partial Y^2} = 0$. The BC for $w_0$ follows from the BC for $w$. This procedure reduces to solving the Laplace equation on a square with BC $w_0(X=\pm \sqrt{a},Y) = f(Y)$ and $w_0(X,Y=\pm \sqrt{b}) = g(X)$. $\endgroup$ – Winther Jan 17 '19 at 17:09
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Use an ansatz of the form

$$ w(x,y) = \sum_{n,m=1}^\infty c_{n,m} \sin \left(n\pi \frac{1+x}{2}\right)\sin\left(m\pi \frac{1+y}{2}\right) $$

Decomposing the delta function into its Fourier series gives

$$ \delta(x,y) = \sum_{n,m=1}^\infty \sin \left(\frac{n\pi}{2}\right)\sin\left( \frac{m\pi}{2}\right)\sin \left(n\pi \frac{1+x}{2}\right)\sin\left(m\pi \frac{1+y}{2}\right) $$

Plugging the above expressions into the equation gives

$$ -\left[a\left(\frac{n\pi}{2}\right)^2 + b\left(\frac{m\pi}{2}\right)^2\right]c_{n,m} = -\sin \left(\frac{n\pi}{2}\right)\sin\left( \frac{m\pi}{2}\right) $$

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  • $\begingroup$ Thanks for the answer. The series are convergent as required. $\endgroup$ – Daddi-Moussa-Ider Jan 16 '19 at 14:26
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  1. The method of images with Dirichlet boundary conditions for a square region $[-1,1]^2$ implies that the 2D Dirac delta distribution $$\delta(x)\delta(y)$$ is part of an alternating Dirac comb/Shah function $$A(x)A(y),$$ where $$\begin{align}A(x)&~=~\sum_{n\in\mathbb{Z}}(-1)^n\delta(x\!-\!2n)~=~III_4(x)-III_4(x+2) ~=~\frac{1}{4}\sum_{n\in\mathbb{Z}}e^{i\pi n x/2} -\frac{1}{4}\sum_{n\in\mathbb{Z}}e^{i\pi n(x+2)/2}\cr &~=~\frac{1}{2}\sum_{n\in\mathbb{Z}}\frac{1-(-1)^n}{2}e^{i\pi n x/2} ~\stackrel{n=2k-1}=~\frac{1}{2}\sum_{k\in\mathbb{Z}}e^{i\pi(k-1/2) x} ~=~\sum_{k\in\mathbb{N}}\cos(\pi(k\!-\!1/2) x)\cr &~=~\sum_{k\in\mathbb{N}-\frac{1}{2}}\cos(\pi k x).\end{align}$$

  2. Therefore the solution to OP's BVP becomes $$ w(x,y)~=~\frac{1}{\pi^2}\sum_{n,m\in\mathbb{N}-\frac{1}{2}}\frac{\cos(\pi n x)\cos(\pi m y)}{a n^2+b m^2}.$$ We leave it to the reader to analyze convergence properties of the double sum.

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  • $\begingroup$ Thanks for the answer. Indeed, i think that the series given in the answer by @Dylan can further be improved by noting that the terms with even $m$ and $n$ vanish. This accelerate the convergence of the series. Still, the solution is singular at $x=y=0$ since the series is divergent but this is not an issue in the physical problem i am trying to solve. See a related question: math.stackexchange.com/questions/3074662/… $\endgroup$ – Daddi-Moussa-Ider Jan 17 '19 at 11:00
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  1. Idea: If we rescale the coordinates $$(x,y)~=~(\sqrt{a}X,\sqrt{b}Y),$$ the new problem is a 2D electrostatic problem $$ \left(\frac{\partial^2 }{\partial X^2}+\frac{\partial^2 }{\partial Y^2}\right)w~=~-\frac{1}{\sqrt{ab}}\delta(X)\delta(Y), $$ $$ w(X = \pm 1/\sqrt{a}, Y) ~=~0~=~w(X, Y = \pm1/\sqrt{b}) ,$$ for a rectangular with Dirichlet boundary conditions. The potential $w$ is expected to diverge logarithmically at the location of the point charge.

  2. The solution can then be formally obtained via the method of images $$w~=~-\frac{1}{4\pi\sqrt{ab}} \sum_{n,m\in\mathbb{Z}}(-1)^{n+m}\ln\left\{(X\!-\!\frac{2n}{\sqrt{a}})^2+(Y\!-\!\frac{2m}{\sqrt{b}})^2\right\}$$ $$~=~-\frac{1}{4\pi\sqrt{ab}} \sum_{n,m\in\mathbb{Z}}(-1)^{n+m}\ln\left\{\frac{(x\!-\!2n)^2}{a}+\frac{(y\!-\!2m)^2}{b}\right\}.$$ Hm. The double sum is divergent as the general term doesn't go to zero as $|n|,|m|\to\infty$, cf. below comment by user Winther. We speculate that it may be possible to group alternating terms together to achieve a conditionally convergent series.

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    $\begingroup$ It's not much to analyze: the series clearly diverges as the general term doesn't go to zero as $n,m\to\infty$. $\endgroup$ – Winther Jan 16 '19 at 14:07
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    $\begingroup$ @MathStudent Feel free to explain. There might be ways of regularizing it (or a way of performing the summation as to make it convergent), however as a normal double sum then it's clearly divergent. $\endgroup$ – Winther Jan 16 '19 at 14:42

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