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The Laplacian in polar and spherical coordinates.
Polar coordinates. The Laplacian is defined with respect the canonical base of RN^. Let us consider, for instance, the following problem
−4u = 0 , in B¯r(0) ,
where Br(0) := {x ∈ R^2 : |x| < r¯} is the ball of radius ¯r centered at the origin. Since the set Br(0) has a spherical symmetry, it is more convenient to describe it with spherical coordinates (radius r and angle θ). That is, we describe points (x, y) ∈ R^2 as { x = r cos θ , y = r sin θ.
DISEGNO!!! Let us consider the change of variable P : (0, ∞) × [0, 2 π) → R^2 \ {(0, 0)} given by
P (r, θ) := (r cos θ, r sin θ).
Notice that, for technical reasons of the change of variable, we are not considering the origin in the target space. This will not affect our computations. Now, define the function v(r, θ) := u (P (r, θ)) = u (r cos θ, r sin θ). We want to understand what equation v has to solve in the rectangle [0, r¯) × [0, 2 π) (that is, the set that describes the ball B¯r(0)), in order for the function u to solve the Laplace equation. In other terms, we want to understand how to write the Laplacian in polar coordinates. Namely, we would like to write
−4u = −∂ xx^2 u − ∂^2 yyu ,
as something involving only derivatives with respect to r and θ. For, let’s reason as follows: we first write the first derivatives ∂xu and ∂yu in terms of ∂rv and ∂θv. By applying the chain rule, we have that { ∂rv = ∂xu ∂rx + ∂yu ∂ry , ∂θv = ∂xu ∂θx + ∂yu ∂θy.
That is (^) { ∂rv = ∂xu cos θ + ∂yu sin θ , ∂θv = −r∂xu sin θ + r∂yu cos θ. That is, in a matrix form, ( ∂rv ∂θv
cos θ sin θ −r sin θ r cos θ
∂xu ∂yu
= DP (r, θ)
∂xu ∂yu
where DP (r, θ) denotes the differential of the map P at the point (r, θ). Since the above equation is true for every u (and the correspondent v), we can simply write it as an equality between differential operators as ( ∂r ∂θ
cos θ sin θ −r sin θ r cos θ
∂x ∂y
= DP (r, θ)
∂x ∂y
What we want is ∂x and ∂yu in terms of ∂rv and ∂θv. so, we have to invert the above equality, that is (^) ( ∂x ∂y
= (DP (r, θ))−^1
∂r ∂θ
where (DP (r, θ))−^1 is the inverse of the matrix DP (r, θ), that is
(DP (r, θ))−^1 =
cos θ − sin r θ sin θ cos r θ
Thus, we get (^) ( ∂x ∂y
cos θ − sin r θ sin θ cos r θ
∂r ∂θ
that is (^)
∂x = cos θ∂r − sinr^ θ∂θ ,
∂y = sin θ∂r + cosr θ∂θ.
We now want to compute ∂^2 xx. We have that
∂ xx^2 = ∂x∂x =
cos θ∂r − sin^ θ r
∂θ
cos θ∂r − sin^ θ r
∂θ
= cos θ∂r (cos θ∂r) + cos θ∂r
− sin^ θ r
∂θ
− sin^ θ r
∂θ (cos θ∂r) − sin^ θ r
∂θ
− sin^ θ r
∂θ
= cos θ(∂r cos θ)∂r + cos^2 θ∂r∂r + cos θ∂r
sin θ r
∂θ −
cos θ sin θ r ∂r∂θ
− sin^ θ r
∂θ (cos θ) ∂r − sin^ θ^ cos^ θ r
∂θ∂r − sin^ θ r
∂θ
− sin^ θ r
∂θ + sin
(^2) θ r^2
∂θ∂θ
= 0 + cos^2 θ∂ rr^2 +
r^2 cos^ θ^ sin^ θ∂θ^ −^
cos θ sin θ r ∂
(^2) rθ
(^2) θ r ∂r^ −^
cos θ sin θ r ∂ rθ^2 + sin^ θ^ cos^ θ r^2 ∂θ^ +
sin^2 θ r^2 ∂ θθ^2.
With similar computations, we get
∂^2 yy = ∂y∂y =
sin θ∂r +
cos θ r ∂θ
sin θ∂r +
cos θ r ∂θ
= sin^2 θ∂ rr^2 − 1 r^2
cos θ sin θ∂θ + cos^ θ^ sin^ θ r
∂^2 rθ
cos^2 θ r ∂r^ +
cos θ sin θ r ∂ rθ^2 −^ sin^ θ^ cos^ θ r^2 ∂θ^ +
cos^2 θ r^2 ∂
(^2) θθ.
Thus, we can write
4 = ∂ rr^2 +
r ∂r^ +
r^2 ∂θθ^.
The above is the expression of the Laplacian in polar coordinates. Notice that it is made by a radial component
∂ rr^2 +
r ∂r^ , and by an angular one ∂θθ. In our example, this means that, u solves the Laplace equation in the ball Br(0) if and only if v solves the equation
∂ rr^2 v +^1 r
∂rv +^1 r^2
∂θθv = 0 ,
in the rectangle [0, r) × [0, 2 π). Even if at a first glance this does not seem like a good simplification of the problem we will see that it is possible to solve the equation for v.
Spherical coordinates. We would like to perform the same computation in dimension N = 3 with the spherical
Again, also here we notice that the radial and the angular part are separated. This last one, is called the Laplace-Beltrami operator, and functions w defined on the sphere (the boundary of the ball!) for which
∂ ϕϕ^2 w +
cos ϕ sin ϕ ∂ϕw^ +^
sin^2 ϕ
∂ θθ^2 w = 0 ,
are called spherical harmonics.
