by Alain Busser Siméon Denis Poisson (1781-1860) was lecturer in "mechanics" (today we would say "theoretical physics") at the french Ecole Polytechnique. He was also a member of the french Academy of Sciences, which allowed him to try to steal the works claimed by other people in mathematics... Two of his victims were Joseph Fourier, who happened to be Napoleon's friend (alas for Poisson), and Evariste Galois who died too soon to claim whatever Poisson could have stolen to him... Poisson is not only famous for his statistical law, but also because he seems to have been the first (for once!) to solve a boundary value problem by a convolution of the boundary conditions with a so-called "kernel", which is now called the "Poisson kernel" and has found in the 1960s an interesting application as a model for the plucked string! This kernel is
This is a function of two variables, expressed in polar coordinates r and theta, defined where the series converges, which means the disk of radius 1, and which looks like this:
>plot2d("(1-abs(x+%i*y)*cos(arg(x+%i*y)))/(1-2*x+x^2+y^2)*(x^2+y^2<1)", ... > contour=true,hue=true,title="The famous Poisson kernel"); ... >insimg;
The equipotential lines constitute a bundle of circles which have a common point at (1,0); then the line fields are the circles which are orthogonal to these circles: They make up another bundle of circles passing through (1,0) but whoses axis is vertical. The Laplace problem consists in finding a potential (a function which laplacian is null in the unit disk) which takes known values at the boundary which here is the circle. Poisson's method is simple to state: First, compute the Fourier coefficients of the function which describes the boundary potential, then multiply them by factors of the form r^n*cos(n*theta), and sum the products up to infinity. Whenever the series converges, its sum is the solution of the boundary value problem. It remains only to represent it (especially the equipotentials). In this worksheet, we choose electrical units so that the electric field is equal, and not only proportionnal, to the opposite of the gradient of the potential. For example, if only one electrical charge is present at (1,0), so that the repartition of the charges is a Dirac delta function at (1,0), the solution of the BVP is the Poisson kernel itself, so that the equipotential were drawn just above. As an other example, let's consider the case where an unit electric charge is uniformly reparted over the right half of the unit circle, and its opposite uniformly reparted over the left half. Now the repartition of the electric charges is a square signal, which Fourier series is
if n is odd, 0 otherwise (known result). Then the potential of this circular capacity is the
>function pot(x,y) ...
provsum=0; for n=0 to 100 provsum=provsum+(-abs(x+%i*y))^n*cos((2*n+1)*arg(x+%i*y))/(2*n+1); end return 4/%pi*provsum endfunction
>plot2d("pot(x,y)*(x^2+y^2<1)",contour=true,hue=1,polar=1); ... >insimg;
As one can see, the potential is maximal near the center of the capacity. It is also constant, as it should, on the right half of the capacity.
>plot3d("pot(x,y)",hue=1,user=1,polar=1,angle=220°); insimg;
As an other example, we can look at a condenser with 3 plates, which means that the angular dependency is three times the preceding one. Then the theta above is to be replaced by 3*theta, which yields
>function pot3(x,y) ...
provsum=0; for n=0 to 100 provsum=provsum+(-abs(x+%i*y))^n*cos(3*(2*n+1)*arg(x+%i*y))/(2*n+1); end return 4/%pi*provsum endfunction
>plot2d("pot3(x,y)*(x^2+y^2<1)",hue=true,contour=true,polar=true); ... >insimg;
If now the electric charge is proportionnal to the square of the angular position, the Fourier series is still well known, but converges more rapidly:
Because the convergence is more rapid, we do not need to sum up to 100; n=20 should be enough:
>function pot2(x,y) ...
provsum=%pi/4; for n=1 to 20 provsum=provsum+(-abs(x+%i*y))^n*cos(n*arg(x+%i*y))/n^2; end return 4/%pi*provsum; endfunction
>plot2d("pot2(x,y)*(x^2+y^2<1)",hue=true,polar=true,contour=true); ... >insimg;
The field lines are almost concentric circles.
>plot3d("pot2(x,y)",hue=1,polar=1); insimg;
For our last example, we will look at a condenser charged so that the electric charge is equal to the cosine of the angle. In this case, there is only one non zero term in the Fourier series:
>function pot1(x,y) ...
return abs(x+%i*y)*cos(arg(x+%i*y)); endfunction
>plot2d("pot1(x,y)*(x^2+y^2<1)",hue=true,contour=true,polar=true); ... >insimg;
We just found the recipe to generate a uniform field! Finally, to show how the Poisson kernel can generate a plucked string sound, we remark that, for any value of r,
is 2*%pi-periodic in theta, which in music means "definite pitch", whereas the wave shape depends from r. If now we make r vary slowly, when r decreases say from 0.9 to 0, the sound gets poorer and less loud at the same time, which is caracteristic of the plucked vibrating strings:
>plot2d("(1-(0.9-0.9*x/256)^2*cos(x/2))"| ... > "/(1+(0.9-0.9*x/256)^4-2*(0.9-0.9*x/256)^2*cos(x/2))", ... > xmin=0,xmax=256); ... >insimg;
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