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How to solve a second order partial differential equation involving a delta Dirac function?
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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
calculus pde dirac-delta greens-function elliptic-equations
asked Jan 15 at 9:02
Math StudentMath Student
130521
$endgroup$
add a comment |
$begingroup$
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
calculus pde dirac-delta greens-function elliptic-equations
asked Jan 15 at 9:02
Math StudentMath Student
130521
$endgroup$
1
$begingroup$
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 at 13:45
1
$begingroup$
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 at 13:49
1
$begingroup$
Another way of dealing with the delta-function here is using that the Greens-function for the 2D Laplacian is $frac{log(sqrt{x^2+y^2})}{2pi}$. Thus if we change variables to $X = x/sqrt{a}$ and $Y = y/sqrt{b}$ then we can write $w = -frac{log(X^2+Y^2)}{4pi sqrt{ab}} + w_0$ 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 at 17:09
add a comment |
$begingroup$
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
calculus pde dirac-delta greens-function elliptic-equations
asked Jan 15 at 9:02
Math StudentMath Student
130521
$endgroup$
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
calculus pde dirac-delta greens-function elliptic-equations
calculus pde dirac-delta greens-function elliptic-equations
asked Jan 15 at 9:02
Math StudentMath Student
130521
asked Jan 15 at 9:02
Math StudentMath Student
130521
edited Jan 16 at 13:38
Math Student
asked Jan 15 at 9:02
Math StudentMath Student
130521
asked Jan 15 at 9:02
Math StudentMath Student
130521
asked Jan 15 at 9:02
Math StudentMath Student
130521
130521
1
$begingroup$
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 at 13:45
1
$begingroup$
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 at 13:49
1
$begingroup$
Another way of dealing with the delta-function here is using that the Greens-function for the 2D Laplacian is $frac{log(sqrt{x^2+y^2})}{2pi}$. Thus if we change variables to $X = x/sqrt{a}$ and $Y = y/sqrt{b}$ then we can write $w = -frac{log(X^2+Y^2)}{4pi sqrt{ab}} + w_0$ 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 at 17:09
add a comment |
1
$begingroup$
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 at 13:45
1
$begingroup$
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 at 13:49
1
$begingroup$
Another way of dealing with the delta-function here is using that the Greens-function for the 2D Laplacian is $frac{log(sqrt{x^2+y^2})}{2pi}$. Thus if we change variables to $X = x/sqrt{a}$ and $Y = y/sqrt{b}$ then we can write $w = -frac{log(X^2+Y^2)}{4pi sqrt{ab}} + w_0$ 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 at 17:09
1
1
$begingroup$
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 at 13:45
$begingroup$
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 at 13:45
1
1
$begingroup$
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 at 13:49
$begingroup$
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 at 13:49
1
1
$begingroup$
Another way of dealing with the delta-function here is using that the Greens-function for the 2D Laplacian is $frac{log(sqrt{x^2+y^2})}{2pi}$. Thus if we change variables to $X = x/sqrt{a}$ and $Y = y/sqrt{b}$ then we can write $w = -frac{log(X^2+Y^2)}{4pi sqrt{ab}} + w_0$ 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 at 17:09
$begingroup$
Another way of dealing with the delta-function here is using that the Greens-function for the 2D Laplacian is $frac{log(sqrt{x^2+y^2})}{2pi}$. Thus if we change variables to $X = x/sqrt{a}$ and $Y = y/sqrt{b}$ then we can write $w = -frac{log(X^2+Y^2)}{4pi sqrt{ab}} + w_0$ 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 at 17:09
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Use an ansatz of the form
$$ w(x,y) = sum_{n,m=1}^infty c_{n,m} sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Decomposing the delta function into its Fourier series gives
$$ delta(x,y) = sum_{n,m=1}^infty sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right)sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Plugging the above expressions into the equation gives
$$ -left[aleft(frac{npi}{2}right)^2 + bleft(frac{mpi}{2}right)^2right]c_{n,m} = -sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right) $$
answered Jan 16 at 13:57
DylanDylan
12.9k31026
$endgroup$
$begingroup$
Thanks for the answer. The series are convergent as required.
$endgroup$
– Math Student
Jan 16 at 14:26
add a comment |
$begingroup$
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
$$begin{align}A(x)&~=~sum_{ninmathbb{Z}}(-1)^ndelta(x!-!2n)~=~III_4(x)-III_4(x+2)
~=~frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n x/2}
-frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n(x+2)/2}cr
&~=~frac{1}{2}sum_{ninmathbb{Z}}frac{1-(-1)^n}{2}e^{ipi n x/2}
~stackrel{n=2k-1}=~frac{1}{2}sum_{kinmathbb{Z}}e^{ipi(k-1/2) x}
~=~sum_{kinmathbb{N}}cos(pi(k!-!1/2) x)cr
&~=~sum_{kinmathbb{N}-frac{1}{2}}cos(pi k x).end{align}$$Therefore the solution to OP's BVP becomes
$$ w(x,y)~=~frac{1}{pi^2}sum_{n,minmathbb{N}-frac{1}{2}}frac{cos(pi n x)cos(pi m y)}{a n^2+b m^2}.$$
We leave it to the reader to analyze convergence properties of the double sum.
answered Jan 17 at 10:27
QmechanicQmechanic
5,07211856
$endgroup$
$begingroup$
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/…
$endgroup$
– Math Student
Jan 17 at 11:00
add a comment |
$begingroup$
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.The solution can then be formally obtained via the method of images
$$w~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{(X!-!frac{2n}{sqrt{a}})^2+(Y!-!frac{2m}{sqrt{b}})^2right}$$
$$~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{frac{(x!-!2n)^2}{a}+frac{(y!-!2m)^2}{b}right}.$$
Hm. The double sum is divergent as the general term doesn't go to zero as $|n|,|m|toinfty$, cf. below comment by user Winther. We speculate that it may be possible to group alternating terms together to achieve a conditionally convergent series.
answered Jan 16 at 14:02
QmechanicQmechanic
5,07211856
$endgroup$
1
$begingroup$
It's not much to analyze: the series clearly diverges as the general term doesn't go to zero as $n,mtoinfty$.
$endgroup$
– Winther
Jan 16 at 14:07
1
$begingroup$
@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 at 14:42
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use an ansatz of the form
$$ w(x,y) = sum_{n,m=1}^infty c_{n,m} sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Decomposing the delta function into its Fourier series gives
$$ delta(x,y) = sum_{n,m=1}^infty sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right)sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Plugging the above expressions into the equation gives
$$ -left[aleft(frac{npi}{2}right)^2 + bleft(frac{mpi}{2}right)^2right]c_{n,m} = -sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right) $$
answered Jan 16 at 13:57
DylanDylan
12.9k31026
$endgroup$
$begingroup$
Thanks for the answer. The series are convergent as required.
$endgroup$
– Math Student
Jan 16 at 14:26
add a comment |
$begingroup$
Use an ansatz of the form
$$ w(x,y) = sum_{n,m=1}^infty c_{n,m} sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Decomposing the delta function into its Fourier series gives
$$ delta(x,y) = sum_{n,m=1}^infty sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right)sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Plugging the above expressions into the equation gives
$$ -left[aleft(frac{npi}{2}right)^2 + bleft(frac{mpi}{2}right)^2right]c_{n,m} = -sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right) $$
answered Jan 16 at 13:57
DylanDylan
12.9k31026
$endgroup$
$begingroup$
Thanks for the answer. The series are convergent as required.
$endgroup$
– Math Student
Jan 16 at 14:26
add a comment |
$begingroup$
Use an ansatz of the form
$$ w(x,y) = sum_{n,m=1}^infty c_{n,m} sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Decomposing the delta function into its Fourier series gives
$$ delta(x,y) = sum_{n,m=1}^infty sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right)sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Plugging the above expressions into the equation gives
$$ -left[aleft(frac{npi}{2}right)^2 + bleft(frac{mpi}{2}right)^2right]c_{n,m} = -sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right) $$
answered Jan 16 at 13:57
DylanDylan
12.9k31026
$endgroup$
Use an ansatz of the form
$$ w(x,y) = sum_{n,m=1}^infty c_{n,m} sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Decomposing the delta function into its Fourier series gives
$$ delta(x,y) = sum_{n,m=1}^infty sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right)sin left(npi frac{1+x}{2}right)sinleft(mpi frac{1+y}{2}right) $$
Plugging the above expressions into the equation gives
$$ -left[aleft(frac{npi}{2}right)^2 + bleft(frac{mpi}{2}right)^2right]c_{n,m} = -sin left(frac{npi}{2}right)sinleft( frac{mpi}{2}right) $$
answered Jan 16 at 13:57
DylanDylan
12.9k31026
edited Jan 16 at 14:06
answered Jan 16 at 13:57
DylanDylan
12.9k31026
answered Jan 16 at 13:57
DylanDylan
12.9k31026
answered Jan 16 at 13:57
DylanDylan
12.9k31026
12.9k31026
$begingroup$
Thanks for the answer. The series are convergent as required.
$endgroup$
– Math Student
Jan 16 at 14:26
add a comment |
$begingroup$
Thanks for the answer. The series are convergent as required.
$endgroup$
– Math Student
Jan 16 at 14:26
$begingroup$
Thanks for the answer. The series are convergent as required.
$endgroup$
– Math Student
Jan 16 at 14:26
$begingroup$
Thanks for the answer. The series are convergent as required.
$endgroup$
– Math Student
Jan 16 at 14:26
add a comment |
$begingroup$
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
$$begin{align}A(x)&~=~sum_{ninmathbb{Z}}(-1)^ndelta(x!-!2n)~=~III_4(x)-III_4(x+2)
~=~frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n x/2}
-frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n(x+2)/2}cr
&~=~frac{1}{2}sum_{ninmathbb{Z}}frac{1-(-1)^n}{2}e^{ipi n x/2}
~stackrel{n=2k-1}=~frac{1}{2}sum_{kinmathbb{Z}}e^{ipi(k-1/2) x}
~=~sum_{kinmathbb{N}}cos(pi(k!-!1/2) x)cr
&~=~sum_{kinmathbb{N}-frac{1}{2}}cos(pi k x).end{align}$$Therefore the solution to OP's BVP becomes
$$ w(x,y)~=~frac{1}{pi^2}sum_{n,minmathbb{N}-frac{1}{2}}frac{cos(pi n x)cos(pi m y)}{a n^2+b m^2}.$$
We leave it to the reader to analyze convergence properties of the double sum.
answered Jan 17 at 10:27
QmechanicQmechanic
5,07211856
$endgroup$
$begingroup$
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/…
$endgroup$
– Math Student
Jan 17 at 11:00
add a comment |
$begingroup$
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
$$begin{align}A(x)&~=~sum_{ninmathbb{Z}}(-1)^ndelta(x!-!2n)~=~III_4(x)-III_4(x+2)
~=~frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n x/2}
-frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n(x+2)/2}cr
&~=~frac{1}{2}sum_{ninmathbb{Z}}frac{1-(-1)^n}{2}e^{ipi n x/2}
~stackrel{n=2k-1}=~frac{1}{2}sum_{kinmathbb{Z}}e^{ipi(k-1/2) x}
~=~sum_{kinmathbb{N}}cos(pi(k!-!1/2) x)cr
&~=~sum_{kinmathbb{N}-frac{1}{2}}cos(pi k x).end{align}$$Therefore the solution to OP's BVP becomes
$$ w(x,y)~=~frac{1}{pi^2}sum_{n,minmathbb{N}-frac{1}{2}}frac{cos(pi n x)cos(pi m y)}{a n^2+b m^2}.$$
We leave it to the reader to analyze convergence properties of the double sum.
answered Jan 17 at 10:27
QmechanicQmechanic
5,07211856
$endgroup$
$begingroup$
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/…
$endgroup$
– Math Student
Jan 17 at 11:00
add a comment |
$begingroup$
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
$$begin{align}A(x)&~=~sum_{ninmathbb{Z}}(-1)^ndelta(x!-!2n)~=~III_4(x)-III_4(x+2)
~=~frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n x/2}
-frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n(x+2)/2}cr
&~=~frac{1}{2}sum_{ninmathbb{Z}}frac{1-(-1)^n}{2}e^{ipi n x/2}
~stackrel{n=2k-1}=~frac{1}{2}sum_{kinmathbb{Z}}e^{ipi(k-1/2) x}
~=~sum_{kinmathbb{N}}cos(pi(k!-!1/2) x)cr
&~=~sum_{kinmathbb{N}-frac{1}{2}}cos(pi k x).end{align}$$Therefore the solution to OP's BVP becomes
$$ w(x,y)~=~frac{1}{pi^2}sum_{n,minmathbb{N}-frac{1}{2}}frac{cos(pi n x)cos(pi m y)}{a n^2+b m^2}.$$
We leave it to the reader to analyze convergence properties of the double sum.
answered Jan 17 at 10:27
QmechanicQmechanic
5,07211856
$endgroup$
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
$$begin{align}A(x)&~=~sum_{ninmathbb{Z}}(-1)^ndelta(x!-!2n)~=~III_4(x)-III_4(x+2)
~=~frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n x/2}
-frac{1}{4}sum_{ninmathbb{Z}}e^{ipi n(x+2)/2}cr
&~=~frac{1}{2}sum_{ninmathbb{Z}}frac{1-(-1)^n}{2}e^{ipi n x/2}
~stackrel{n=2k-1}=~frac{1}{2}sum_{kinmathbb{Z}}e^{ipi(k-1/2) x}
~=~sum_{kinmathbb{N}}cos(pi(k!-!1/2) x)cr
&~=~sum_{kinmathbb{N}-frac{1}{2}}cos(pi k x).end{align}$$Therefore the solution to OP's BVP becomes
$$ w(x,y)~=~frac{1}{pi^2}sum_{n,minmathbb{N}-frac{1}{2}}frac{cos(pi n x)cos(pi m y)}{a n^2+b m^2}.$$
We leave it to the reader to analyze convergence properties of the double sum.
answered Jan 17 at 10:27
QmechanicQmechanic
5,07211856
edited Jan 17 at 17:27
answered Jan 17 at 10:27
QmechanicQmechanic
5,07211856
answered Jan 17 at 10:27
QmechanicQmechanic
5,07211856
answered Jan 17 at 10:27
QmechanicQmechanic
5,07211856
5,07211856
$begingroup$
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/…
$endgroup$
– Math Student
Jan 17 at 11:00
add a comment |
$begingroup$
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/…
$endgroup$
– Math Student
Jan 17 at 11:00
$begingroup$
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/…
$endgroup$
– Math Student
Jan 17 at 11:00
$begingroup$
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/…
$endgroup$
– Math Student
Jan 17 at 11:00
add a comment |
$begingroup$
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.The solution can then be formally obtained via the method of images
$$w~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{(X!-!frac{2n}{sqrt{a}})^2+(Y!-!frac{2m}{sqrt{b}})^2right}$$
$$~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{frac{(x!-!2n)^2}{a}+frac{(y!-!2m)^2}{b}right}.$$
Hm. The double sum is divergent as the general term doesn't go to zero as $|n|,|m|toinfty$, cf. below comment by user Winther. We speculate that it may be possible to group alternating terms together to achieve a conditionally convergent series.
answered Jan 16 at 14:02
QmechanicQmechanic
5,07211856
$endgroup$
1
$begingroup$
It's not much to analyze: the series clearly diverges as the general term doesn't go to zero as $n,mtoinfty$.
$endgroup$
– Winther
Jan 16 at 14:07
1
$begingroup$
@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 at 14:42
add a comment |
$begingroup$
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.The solution can then be formally obtained via the method of images
$$w~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{(X!-!frac{2n}{sqrt{a}})^2+(Y!-!frac{2m}{sqrt{b}})^2right}$$
$$~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{frac{(x!-!2n)^2}{a}+frac{(y!-!2m)^2}{b}right}.$$
Hm. The double sum is divergent as the general term doesn't go to zero as $|n|,|m|toinfty$, cf. below comment by user Winther. We speculate that it may be possible to group alternating terms together to achieve a conditionally convergent series.
answered Jan 16 at 14:02
QmechanicQmechanic
5,07211856
$endgroup$
1
$begingroup$
It's not much to analyze: the series clearly diverges as the general term doesn't go to zero as $n,mtoinfty$.
$endgroup$
– Winther
Jan 16 at 14:07
1
$begingroup$
@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 at 14:42
add a comment |
$begingroup$
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.The solution can then be formally obtained via the method of images
$$w~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{(X!-!frac{2n}{sqrt{a}})^2+(Y!-!frac{2m}{sqrt{b}})^2right}$$
$$~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{frac{(x!-!2n)^2}{a}+frac{(y!-!2m)^2}{b}right}.$$
Hm. The double sum is divergent as the general term doesn't go to zero as $|n|,|m|toinfty$, cf. below comment by user Winther. We speculate that it may be possible to group alternating terms together to achieve a conditionally convergent series.
answered Jan 16 at 14:02
QmechanicQmechanic
5,07211856
$endgroup$
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.The solution can then be formally obtained via the method of images
$$w~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{(X!-!frac{2n}{sqrt{a}})^2+(Y!-!frac{2m}{sqrt{b}})^2right}$$
$$~=~-frac{1}{4pisqrt{ab}} sum_{n,minmathbb{Z}}(-1)^{n+m}lnleft{frac{(x!-!2n)^2}{a}+frac{(y!-!2m)^2}{b}right}.$$
Hm. The double sum is divergent as the general term doesn't go to zero as $|n|,|m|toinfty$, cf. below comment by user Winther. We speculate that it may be possible to group alternating terms together to achieve a conditionally convergent series.
answered Jan 16 at 14:02
QmechanicQmechanic
5,07211856
edited Jan 17 at 10:27
answered Jan 16 at 14:02
QmechanicQmechanic
5,07211856
answered Jan 16 at 14:02
QmechanicQmechanic
5,07211856
answered Jan 16 at 14:02
QmechanicQmechanic
5,07211856
5,07211856
1
$begingroup$
It's not much to analyze: the series clearly diverges as the general term doesn't go to zero as $n,mtoinfty$.
$endgroup$
– Winther
Jan 16 at 14:07
1
$begingroup$
@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 at 14:42
add a comment |
1
$begingroup$
It's not much to analyze: the series clearly diverges as the general term doesn't go to zero as $n,mtoinfty$.
$endgroup$
– Winther
Jan 16 at 14:07
1
$begingroup$
@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 at 14:42
1
1
$begingroup$
It's not much to analyze: the series clearly diverges as the general term doesn't go to zero as $n,mtoinfty$.
$endgroup$
– Winther
Jan 16 at 14:07
$begingroup$
It's not much to analyze: the series clearly diverges as the general term doesn't go to zero as $n,mtoinfty$.
$endgroup$
– Winther
Jan 16 at 14:07
1
1
$begingroup$
@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 at 14:42
$begingroup$
@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 at 14:42
add a comment |
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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 at 13:45
1
$begingroup$
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 at 13:49
1
$begingroup$
Another way of dealing with the delta-function here is using that the Greens-function for the 2D Laplacian is $frac{log(sqrt{x^2+y^2})}{2pi}$. Thus if we change variables to $X = x/sqrt{a}$ and $Y = y/sqrt{b}$ then we can write $w = -frac{log(X^2+Y^2)}{4pi sqrt{ab}} + w_0$ 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 at 17:09