The Laplacian in Polar Coordinates R. C. Trinity University Partial Differential Equations March 17, 15
To solve boundary value problems on circular regions, it is convenient to switch from rectangular (x,y) to polar (r,θ) spatial coordinates: x = r cosθ, r x θ y y = r sinθ, x +y = r. This requires us to express the rectangular Laplacian u = u xx +u yy in terms of derivatives with respect to r and θ.
The chain rule For any function f(r,θ), we have the familiar tree diagram and chain rule formulae: f x = f r r x + f θ θ x f f y = f r r y + f θ θ y r θ or x y x y f x = f r r x +f θ θ x f y = f r r y +f θ θ y
First take f = u to obtain u x = u r r x +u θ θ x u xx = u r r xx +(u r ) x r x +u θ θ xx +(u θ ) x θ x. Applying the chain rule with f = u r and then with f = u θ yields u xx = u r r xx +(u rr r x +u rθ θ x )r x +u θ θ xx +(u θr r x +u θθ θ x )θ x = u r r xx +u rr rx +u rθr x θ x +u θ θ xx +u θθ θx. An entirely similar computation using y instead of x also gives u yy = u r r yy +u rr ry +u rθr y θ y +u θ θ yy +u θθ θy. If we add these expressions and collect like terms we get ( u =u r (r xx +r yy )+u rr r x +ry) +urθ (r x θ x +r y θ y ) ( +u θ (θ xx +θ yy ) +u θθ θ x +θy).
Differentiate x +y = r with respect to x and then y: x = rr x r x = x r y = rr y r y = y r Now differentiate tanθ = y x r xx = r xr x r = r x r 3 r yy = r yr y r = r y r 3 with respect to x and then y: = y r 3, = x r 3. sec θθ x = y x θ x = y cos θ x = y r θ xx = y r 3 r x = xy r 4, sec θθ y = 1 x θ y = cos θ x = x r θ yy = x r 3 r y = xy r 4.
Together these yield r xx +r yy = y +x r 3 = 1 r, r x +ry = x +y r = 1. θ xx +θ yy = xy r 4 + xy r 4 =, θ x +θ y = y +x r 4 = 1 r, r x θ x +r y θ y = xy r 3 + yx r 3 =, and we finally obtain ( u =u r (r xx +r yy )+u rr r x +ry) +urθ (r x θ x +r y θ y ) ( +u θ (θ xx +θ yy ) +u θθ θ x +θy ) = 1 r u r +u rr + 1 r u θθ = u rr + 1 r u r + 1 r u θθ.
Example Use polar coordinates to show that the function u(x,y) = is harmonic. We need to show that u =. In polar coordinates we have so that and thus u(r,θ) = r sinθ r = sinθ r u r = sinθ r, u rr = sinθ r 3, u θθ = sinθ, r u = u rr + 1 r u r + 1 r u θθ = sinθ r 3 sinθ r 3 sinθ r 3 =. y x +y
The Dirichlet problem on a disk Goal: Solve the Dirichlet problem on a disk of radius a, centered at the origin. In polar coordinates this has the form u = u rr + 1 r u r + 1 r u θθ =, r < a, u = f y x u(a,θ) = f(θ), θ π. a Remarks: We will require that f is π-periodic. Likewise, we require that u(r,θ) is π-periodic in θ.
Separation of variables If we assume that u(r,θ) = R(r)Θ(θ) and plug into u =, we get R Θ+ 1 r R Θ+ 1 r RΘ = r R R +rr R + Θ Θ = r R R +rr R = Θ Θ = λ. This yields the pair of separated ODEs r R +rr λr = and Θ +λθ =. We also have the boundary conditions Θ is π-periodic and R(+) is finite.
Solving for Θ The solutions of Θ +λθ = are periodic only if λ = µ Θ = acos(µθ)+bsin(µθ). In order for the period to be π we also need 1 = cos(µ) = cos(πµ) πµ = πn µ = n N. Hence λ = n and Θ = Θ n = a n cos(nθ)+b n sin(nθ), n N. It follows that R satisfies r R +rr n R =, which is called an Euler equation.
Interlude Euler equations An Euler equation is a second order ODE of the form x y +αxy +βy =. Its solutions are determined by the roots of its indicial equation ρ +(α 1)ρ+β =. Case 1: If the roots are ρ 1 ρ, then the general solution is y = c 1 x ρ 1 +c x ρ. Case : If there is only one root ρ 1, then the general solution is y = c 1 x ρ 1 +c x ρ 1 lnx.
Solving for R The indicial equation of r R +rr n R = is ρ +(1 1)ρ n = ρ n = ρ = ±n. This means that R = c 1 r n +c r n (n ), R = c 1 +c lnr (n = ). These will be finite at r = only if c =. Setting c 1 = a n gives ( r ) n, R = R n = n N. a
Separated solutions and superposition We therefore obtain the separated solutions ( r ) n(an u n (r,θ) = R n (r)θ n (θ) = cos(nθ)+b n sin(nθ)), n N. a Noting that ( r ) (a u (r,θ) = cos+b sin) = a, a superposition gives the general solution u(r,θ) = a + ( r ) n (an cos(nθ)+b n sin(nθ)). a n=1
Boundary values and conclusion Imposing our Dirichlet boundary conditions gives f(θ) = u(a,θ) = a + (a n cos(nθ)+b n sin(nθ)), n=1 which is just the ordinary π-periodic Fourier series for f! Theorem The solution of the Dirichlet problem on the disk of radius a centered at the origin, with boundary condition u(a,θ) = f(θ) is u(r,θ) = a + n=1( r a) n(an cos(nθ)+b n sin(nθ)), where a n = 1 π π a = 1 π π f(θ)dθ, f(θ)cos(nθ)dθ, b n = 1 π π f(θ)sin(nθ)dθ.
Example Find the solution to the Dirichlet problem on a disk of radius 3 with boundary values given by 3 π (π +θ) if π θ <, f(θ) = 3 π (π θ) if θ < π, π if θ < 3π. We have a = 3. The graph of f is
Changing to polar coordinates The Dirichlet problem on a disk According to exercise.3.8 (with p = π, c = 3 and d = π/): 15 1 X 1 cos(nπ/) f (θ) = + cos(nθ). π n n=1 Hence, the solution to the Dirichlet problem is u(r, θ) = 15 1 X r n 1 cos(nπ/) + cos(nθ). π 3 n n=1 5 15 1 5 3 3 1 1 1 3 1 Examples
Example Solve the Dirichlet problem on a disk of radius with boundary values given by f(θ) = cos θ. Express your answer in cartesian coordinates. We have a = and f(θ) = cos θ = 1+cos(θ) = 1 + 1 cos(θ), which is a finite π-periodic Fourier series (i.e. a = 1/, a = 1/, and all other coefficients are zero). Hence u(r,θ) = 1 ( r ) + 1 cos(θ) = 1 + r cos(θ). 8
Changing to polar coordinates The Dirichlet problem on a disk Since cos(θ) = cos θ sin θ, we find that r cos(θ) = r cos θ r sin θ = x y and hence u= 1 r cos(θ) 1 x y + = +. 8 8 1.8.6.4. 1 1 1 1 Examples
Example Solve the Dirichlet problem on a disk of radius 1 if the boundary value is 5 in the first quadrant, and zero elsewhere. We are given a = 1, f(θ) = 5 for θ π/ and f(θ) = otherwise. The Fourier coefficients of f are so that a = 1 π a n = 1 π b n = 1 π u(r,θ) = 5 +5 π π/ π/ π/ n=1 5dθ = 5, 5cos(nθ)dθ = 5sin(nπ/), nπ 5sin(nθ)dθ = 5(1 cos(nπ/)), nπ r n ( sin(nπ/) n cos(nθ)+ (1 cos(nπ/)) n ) sin(nθ).
Remarks: One can frequently use identities like (valid for r < 1) n=1 n=1 r n cos(nθ) n r n sin(nθ) n = 1 ln( 1 r cosθ +r ), ( ) r sinθ = arctan, 1 r cosθ to convert series solutions in polar coordinates to cartesian expressions. Using the second identity, one can show that the solution in the preceding example is u(x,y) = 5 + 5 ( ( ) ( )) y x arctan +arctan. π 1 x 1 y