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4.10 Dirichlet problem in the circle and the Poisson kernel

Note: 2 lectures, Β§9.7 in [EP], Β§10.8 in [BD]

4.10.1 Laplace in polar coordinates

A more natural setting for the Laplace equation Ξ”u = 0 is the circle rather than the square. On the other hand, what makes the problem somewhat more difficult is that we need polar coordinates.


Recall that the polar coordinates for the (x,y) -plane are (r,πœƒ) :

x = rco sπœƒ, y = rsinπœƒ,

where r β‰₯ 0 and βˆ’ Ο€ < πœƒ ≀ Ο€ . So (x,y) is distance r from the origin at angle πœƒ .

Now that we know our coordinates, let us give the problem we wish to solve. We have a circular region of radius 1, and we are interested in the Dirichlet problem for the Laplace equation for this region. Let u(r,πœƒ) denote the temperature at the point (r,πœƒ) in polar coordinates. We have the problem: PIC

Ξ”u = 0, for r < 1, u(1,πœƒ) = g(πœƒ), for βˆ’ Ο€ < πœƒ ≀ Ο€.

The first issue we face is that we do not know what the Laplacian is in polar coordinates. Normally we would find uxx and uyy in terms of the derivatives in r and πœƒ . We would need to solve for r and πœƒ in terms of x and y . While this is certainly possible, it happens to be more convenient to work in reverse. Let us instead compute derivatives in r and πœƒ in terms of derivatives in x and y and then solve. The computations are easier this way. First

xr = cosπœƒ, xπœƒ = βˆ’ rsinπœƒ, yr = sinπœƒ, yπœƒ = r cosπœƒ.

Next by chain rule we obtain

ur = uxxr + uyyr = cos(πœƒ)ux + sin(πœƒ)uy, 2 2 urr = cos(πœƒ)(uxxxr + uxyyr) + sin (πœƒ)(uyxxr + uyyyr) = cos (πœƒ)uxx + 2 cos(πœƒ)sin (πœƒ)uxy + sin (πœƒ
Similarly for the πœƒ derivative. Note that we have to use product rule for the second derivative.
uπœƒ = uxxπœƒ + uyyπœƒ = βˆ’rsin(πœƒ)ux + rco s(πœƒ)uy, uπœƒπœƒ = βˆ’r cos(πœƒ)ux βˆ’ rsin(πœƒ)(uxxxπœƒ + uxyyπœƒ) βˆ’ rsin(πœƒ)uy + rcos(πœƒ)(uyxxπœƒ + uyyyπœƒ) = βˆ’r cos(πœƒ)ux βˆ’ rsin(πœƒ)uy + r2sin2(πœƒ)uxx βˆ’ r22sin(πœƒ)cos(πœƒ)uxy + r2cos2(πœƒ)uyy.
Let us now try to solve for uxx + uyy . We start with 1- r2uπœƒπœƒ to get rid of those pesky  2 r . If we add urr and use the fact that cos2(πœƒ) + sin 2(πœƒ) = 1 , we get
-1 1- 1- r2 uπœƒπœƒ + urr = uxx + uyy βˆ’ r cos(πœƒ)ux βˆ’ r sin(πœƒ)uy.

We’re not quite there yet, but all we are lacking is 1u r r . Adding it we obtain the Laplacian in polar coordinates:

|----------------------------------| | 1- 1- | | r2uπœƒπœƒ + rur + urr = uxx + uyy = Ξ”u. ------------------------------------

Notice that the Laplacian in polar coordinates no longer has constant coefficients.

4.10.2 Series solution

Let us separate variables as usual. That is let us try u(r,πœƒ) = R (r)Θ (πœƒ) . Then

0 = Ξ”u = 1-RΘ β€²β€² + 1Rβ€²Ξ˜ + R β€²β€²Ξ˜. r2 r

Let us put R on one side and Θ on the other and conclude that both sides must be constant.

 1 (1 ) --R Ξ˜β€²β€² = βˆ’ --Rβ€² + R β€²β€² Θ. r2 r Ξ˜β€²β€² rRβ€² + r2R β€²β€² ---= βˆ’ ---------- = βˆ’Ξ». Θ R
We get two equations:
Θ β€²β€² + λΘ = 0, 2 β€²β€² β€² r R + rR βˆ’ Ξ»R = 0.
Let us first focus on Θ . We know that u(r,πœƒ) ought to be 2Ο€ -periodic in πœƒ , that is, u (r,πœƒ) = u(r,πœƒ + 2Ο€) . Therefore, the solution to  β€²β€² Θ + Ξ» Θ = 0 must be 2Ο€ -periodic. We conclude that  2 Ξ» = n for a nonnegative integer n = 0,1,2,3,... . The equation becomes Ξ˜β€²β€² + n2Θ = 0 . When n = 0 the equation is just Θ β€²β€² = 0 , so we have the general solution A πœƒ + B . As Θ is periodic, A = 0 . For convenience let us write this solution as
 a Θ0 = -0- 2

for some constant a0 . For positive n , the solution to Θ β€²β€² + n2Θ = 0 is

Θn = ancos(nπœƒ) + bn sin(nπœƒ),

for some constants an and bn .

Next, we consider the equation for R ,

2 β€²β€² β€² 2 rR + rR βˆ’ n R = 0.

This equation appeared in exercises beforeβ€”we solved it in Exercise 2.1.6 and Exercise 2.1.7. The idea is to try a solution rs and if that does not work out try a solution of the form rslnr . When n = 0 we obtain

 0 0 R0 = Ar + Br lnr = A + B ln r,

and if n > 0 , we get

 n βˆ’n Rn = Ar + Br .

The function u(r,πœƒ) must be finite at the origin, that is, when r = 0 . Therefore, B = 0 in both cases. Let us set A = 1 in both cases as well, the constants in Θn will pick up the slack so we do not lose anything. Therefore let

R 0 = 1, and Rn = rn.

Hence our building block solutions are

 a u 0(r,πœƒ) = -0, un(r,πœƒ) = anrnco s(nπœƒ) + bnrnsin(nπœƒ). 2
Putting everything together our solution is:
|--------------------------------------------| | βˆ‘βˆž | | u(r,πœƒ) = a-0+ anrncos(nπœƒ) + bnrnsin(nπœƒ).| | 2 n=1 | ---------------------------------------------

We look at the boundary condition in (4.25),

 ∞ a0- βˆ‘ g (πœƒ) = u(1,πœƒ) = 2 + ancos(nπœƒ) + bn sin(nπœƒ). n=1

Therefore, the solution to (4.25) is to expand g(πœƒ ) , which is a 2Ο€ -periodic function, as a Fourier series, and then the nth coordinate is multiplied by rn . In other words, to compute an and bn from the formula we can, as usual, compute

 ∫ ∫ 1- Ο€ 1- Ο€ an = Ο€ g(πœƒ)cos(nπœƒ) dπœƒ, and bn = Ο€ g(πœƒ)sin(nπœƒ) dπœƒ. βˆ’Ο€ βˆ’Ο€

Example 4.10.1: Suppose we wish to solve

Ξ”u = 0, 0 ≀ r < 1, βˆ’ Ο€ < πœƒ ≀ Ο€, u(1,πœƒ) = cos(1 0πœƒ), βˆ’ Ο€ < πœƒ ≀ Ο€.

The solution is

u(r,πœƒ) = r10cos(10πœƒ).

See the plot in Figure 4.23. The thing to notice in this example is that the effect of a high frequency is mostly felt at the boundary. In the middle of the disc, the solution is very close to zero. That is because r10 rather small when r is close to 0.


Figure 4.23: The solution of the Dirichlet problem in the disc with cos(10πœƒ) as boundary data.

Example 4.10.2: Let us solve a more difficult problem. Suppose we have a long rod with circular cross section of radius 1 and we wish to solve the steady state heat problem. If the rod is long enough we simply need to solve the Laplace equation in two dimensions. Let us put the center of the rod at the origin and we have exactly the region we are currently studyingβ€”a circle of radius 1. For the boundary conditions, suppose in Cartesian coordinates x and y , the temperature is fixed at 0 when y < 0 and at 2y when y > 0 .

We set the problem up. As y = rsin(πœƒ) , then on the circle of radius 1 we have 2y = 2sin(πœƒ) . So

Ξ”u = 0, 0 ≀ r < 1, βˆ’ Ο€ < πœƒ ≀ Ο€, (| u(1,πœƒ) = |{|2sin(πœƒ) if 0 ≀ πœƒ ≀ Ο€, |(0 if βˆ’Ο€ < πœƒ < 0.

We must now compute the Fourier series for the boundary condition. By now the reader has plentiful experience in computing Fourier series and so we simply state that

 2 βˆ‘βˆž βˆ’4 u(1,πœƒ) = --+ sin (πœƒ) + -----2----co s(2nπœƒ). Ο€ n=1Ο€(4n βˆ’ 1)

Exercise 4.10.1: Compute the series for u(1,πœƒ) and verify that it really is what we have just claimed. Hint: Be careful, make sure not to divide by zero.

We now simply write the solution (see Figure 4.24) by multiplying by  n r in the right places.

 ∞ 2- βˆ‘ --βˆ’4r2n--- u(r,πœƒ ) = Ο€ + rsin(πœƒ) + Ο€(4n2 βˆ’ 1) cos(2nπœƒ). n=1


Figure 4.24: The solution of the Dirichlet problem with boundary data 0 for y < 0 and 2y for y > 0 .

4.10.3 Poisson kernel

There is another way to solve the Dirichlet problem with the help of an integral kernel. That is, we will find a function P(r,πœƒ,Ξ±) called the Poisson kernel7 such that

 ∫ 1 Ο€ u(r,πœƒ) = 2Ο€- P (r,πœƒ,Ξ± )g(Ξ± ) dΞ±. βˆ’Ο€

While the integral will generally not be solvable analytically, it can be evaluated numerically. In fact, unless the boundary data is given as a Fourier series already, it may be much easier to numerically evaluate this formula as there is only one integral to evaluate.

The formula also has theoretical applications. For instance, as P(r,πœƒ,Ξ±) will have infinitely many derivatives, then via differentiating under the integral we find that the solution u(r,πœƒ) has infinitely many derivatives, at least when inside the circle, r < 1 . By infinitely many derivatives what you should think of is that u(r,πœƒ) has β€œno corners” and all of its partial derivatives of all orders exist and also have β€œno corners.”

We will compute the formula for P (r,πœƒ,Ξ±) from the series solution, and this idea can be applied anytime you have a convenient series solution where the coefficients are obtained via integration. Hence you can apply this reasoning to obtain such integral kernels for other equations, such as the heat equation. The computation is long and tedious, but not overly difficult. Since the ideas are often applied in similar contexts, it is good to understand how this computation works.

What we do is start with the series solution and replace the coefficients with the integrals that compute them. Then we try to write everything as a single integral. We must use a different dummy variable for the integration and hence we use Ξ± instead of πœƒ .

 ∞ a0- βˆ‘ n n u(r,πœƒ) = 2 + anr cos(nπœƒ) + bnr sin(nπœƒ) ∫ n=Ο€1 1-- = 2Ο€ βˆ’Ο€ g (Ξ± ) dΞ± βˆ‘βˆž ( ∫ Ο€ ) ( ∫ Ο€ ) + 1- g(Ξ±)cos(nΞ± ) dΞ± rncos(nπœƒ) + 1- g(Ξ±)sin(nΞ±) d Ξ± rnsin(nπœƒ) n=1 Ο€ βˆ’Ο€ Ο€ βˆ’Ο€ ∫ Ο€( ∞ ) 1-- |||| βˆ‘ n n |||| = 2Ο€ βˆ’Ο€|(g(Ξ± ) + 2 g(Ξ±)cos(nΞ±)r co s(nπœƒ) + g(Ξ±)sin(n Ξ±)r sin (nπœƒ )|) dΞ± ∫ ( n=1 ) 1-- Ο€||| βˆ‘βˆž n( )||| = 2Ο€ ||(1 + 2 r co s(nΞ± )cos(nπœƒ) + sin(nΞ±)sin(nπœƒ)||) g(Ξ±) dΞ± βˆ’Ο€ n=1

OK, so we have what we wanted, the expression in the parentheses is the Poisson kernel, P(r,πœƒ,Ξ±) . However, we can do a lot better. It is still given as a series, and we would really like to have a nice simple expression for it. We must work a little harder. The trick is to rewrite everything in terms of complex exponentials. Let us work just on the kernel.

 βˆ‘βˆž P (r,πœƒ,Ξ±) = 1 + 2 rn(cos(nΞ±) cos(nπœƒ ) + sin(nΞ± )sin(nπœƒ)) n=1 ∞ βˆ‘ n ( ) = 1 + 2 r cos n(πœƒ βˆ’ Ξ±) n=1 βˆžβˆ‘ n( in(πœƒβˆ’Ξ±) βˆ’in(πœƒβˆ’Ξ±)) = 1 + r e + e n=1 βˆžβˆ‘ ( i(πœƒβˆ’Ξ±))n βˆ‘βˆž ( βˆ’i(πœƒβˆ’Ξ±))n = 1 + re + re . n=1 n=1

In the above expression we recognize the geometric series. That is, recall from calculus that as long as |z| < 1 , then

βˆ‘βˆž zn = --z--. 1 βˆ’ z n=1

Note that n starts at 1 and that is why we have the z in the numerator. It is the standard geometric series multiplied by z . Let us continue with the computation.

 βˆ‘βˆž βˆ‘βˆž P (r,πœƒ,Ξ± ) = 1 + (rei(πœƒβˆ’Ξ±))n + (reβˆ’i(πœƒβˆ’Ξ±))n n=1 n=1 i(πœƒβˆ’Ξ±) βˆ’i(πœƒβˆ’Ξ±) = 1 + --re------ + --re-------- 1 βˆ’ rei(πœƒβˆ’Ξ±) 1 βˆ’ reβˆ’i(πœƒβˆ’Ξ±) (1 βˆ’ rei(πœƒβˆ’Ξ±))(1 βˆ’ reβˆ’i(πœƒβˆ’Ξ±)) + (1 βˆ’ reβˆ’i(πœƒβˆ’Ξ±))rei(πœƒβˆ’Ξ±) + (1 βˆ’ rei(πœƒβˆ’Ξ±))reβˆ’i(πœƒβˆ’ = --------------------(------i(πœƒβˆ’Ξ±))(------βˆ’i(πœƒβˆ’Ξ±))--------------------- 1 βˆ’ re 1 βˆ’ re ---------1 βˆ’-r2--------- = 1 βˆ’ rei(πœƒβˆ’Ξ±) βˆ’ reβˆ’i(πœƒβˆ’Ξ±) + r2 2 = -------1 βˆ’-r--------. 1 βˆ’ 2r cos(πœƒ βˆ’ Ξ±) + r2

Now that’s a formula we can live with. The solution to the Dirichlet problem using the Poisson kernel is

|------------∫--Ο€------------2-----------------| | u(r,πœƒ) =-1- --------1 βˆ’-r-------g (Ξ± ) dΞ±. | | 2 Ο€ βˆ’Ο€1 βˆ’ 2rco s(πœƒ βˆ’ Ξ±) + r2 | -----------------------------------------------

Sometimes the formula for the Poisson kernel is given together with the constant 1- 2Ο€ , in which case we should of course not leave it in front of the integral. Also, often the limits of the integral are given as 0 to 2Ο€ ; everything inside is 2Ο€ -periodic in Ξ± , so this does not change the integral.


Let us not leave the Poisson kernel without explaining its geometric meaning. Let s be the distance from (r,πœƒ) to (1,Ξ± ) . You may recall from calculus that this distance s in polar coordinates is given precisely by the square root of 1 βˆ’ 2r cos(πœƒ βˆ’ Ξ±) + r2 . That is, the Poisson kernel is really the formula

1 βˆ’-r2 s2 .

One final note we make about the formula is that it is really a weighted average of the boundary values. First let us look at what happens at the origin, that is when r = 0 .

 ∫ 1 Ο€ 1 βˆ’ 02 u(0,0) = --- ---------------------2g(Ξ±) d Ξ± 2Ο€ βˆ«βˆ’Ο€Ο€ 1 βˆ’ 2(0 )cos(πœƒ βˆ’ Ξ±) + 0 = -1- g(Ξ±) dΞ±. 2Ο€ βˆ’Ο€

So u(0,0) is precisely the average value of g(πœƒ) and therefore the average value of u on the boundary. This is a general feature of harmonic functions, the value at some point p is equal to the average of the values on a circle centered at p .

What the formula says is that the value of the solution at any point in the circle is a weighted average of the boundary data g(πœƒ) . The kernel is bigger when (1,Ξ±) is closer to (r,πœƒ) . Therefore when computing u(r,πœƒ) we give more weight to the values g(Ξ± ) when (1,Ξ±) is closer to (r,πœƒ) and less weight to the values g(Ξ±) when (1,Ξ± ) far from (r,πœƒ) .

4.10.4 Exercises

Exercise 4.10.2: Using series solve Ξ”u = 0 , u(1, πœƒ) = |πœƒ | , for βˆ’ Ο€ < πœƒ ≀ Ο€ .

Exercise 4.10.3: Using series solve Ξ”u = 0 , u(1,πœƒ) = g (πœƒ) for the following data. Hint: trig identities.

a) g(πœƒ) = 1βˆ•2 + 3sin(πœƒ) + cos(3πœƒ) b) g(πœƒ) = 3cos(3πœƒ) + 3 sin(3πœƒ) + sin(9πœƒ)
c) g(πœƒ) = 2 cos(πœƒ + 1) d) g(πœƒ) = sin2(πœƒ)

Exercise 4.10.4: Using the Poisson kernel, give the solution to Ξ”u = 0 , where u(1,πœƒ) is zero for πœƒ outside the interval [βˆ’Ο€βˆ•4,Ο€βˆ•4] and u(1,πœƒ) is 1 for πœƒ on the interval [βˆ’ Ο€βˆ•4,Ο€βˆ•4] .

Exercise 4.10.5: a) Draw a graph for the Poisson kernel as a function of Ξ± when r = 1βˆ•2 and πœƒ = 0 . b) Describe what happens to the graph when you make r bigger (as it approaches 1). c) Knowing that the solution u (r,πœƒ) is the weighted average of g(πœƒ) with Poisson kernel as the weight, explain what your answer to part b means.

Exercise 4.10.6: Take the function g(πœƒ) to be the function xy = cosπœƒ sin πœƒ on the boundary. Use the series solution to find a solution to the Dirichlet problem Ξ”u = 0 , u(1,πœƒ) = g (πœƒ) . Now convert the solution to Cartesian coordinates x and y . Is this solution surprising? Hint: use your trig identities.

Exercise 4.10.7: Carry out the computation we needed in the separation of variables and solve r2Rβ€²β€² + rRβ€² βˆ’ n 2R = 0 , for n = 0,1,2,3, ... .

Exercise 4.10.8 (challenging): Derive the series solution to the Dirichlet problem if the region is a circle of radius ρ rather than 1. That is, solve Ξ”u = 0 , u(ρ,πœƒ) = g (πœƒ) .

Exercise 4.10.9 (challenging): a) Find the solution for Ξ”u = 0 , u(1,πœƒ) = x2y3 + 5x2 . Write the answer in Cartesian coordinates.
b) Now solve Ξ”u = 0 ,  k β„“ u(1, πœƒ) = x y . Write the solution in Cartesian coordinates.
c) Suppose you have a polynomial  βˆ‘m βˆ‘n j k P(x,y) = j=0 k=0cj,kx y , solve Ξ”u = 0 , u(1,πœƒ) = P(x,y) (that is, write down the formula for the answer). Write the answer in Cartesian coordinates.
Notice the answer is again a polynomial in x and y . See also Exercise 4.10.6.

Exercise 4.10.101: Using series solve Ξ”u = 0 ,  βˆ‘βˆž u(1,πœƒ) = 1 + 1n2 sin(n πœƒ) n=1 .

Exercise 4.10.102: Using the series solution find the solution to Ξ”u = 0 , u(1,πœƒ) = 1 βˆ’ co s(πœƒ) . Express the solution in Cartesian coordinates (that is, using x and y ).

Exercise 4.10.103: a) Try and guess a solution to Ξ”u = βˆ’1 , u(1,πœƒ) = 0 . Hint: try a solution that only depends on r . Also first, don’t worry about the boundary condition. b) Now solve Ξ”u = βˆ’ 1 , u(1,πœƒ) = sin(2πœƒ) using superposition.

Exercise 4.10.104 (challenging): Derive the Poisson kernel solution if the region is a circle of radius ρ rather than 1. That is, solve Ξ”u = 0 , u(ρ,πœƒ) = g(πœƒ) .

7Named for the French mathematician SimΓ©on Denis Poisson (1781–1840).