Laplace’s equation in the Polar Coordinate System
As I mentioned in my lecture, if you want to solve a partial differential equation (PDE) on the domain whose shape is a 2D disk, it is much more convenient
to represent the solution in terms of the polar coordinate system than in terms of
the usual Cartesian coordinate system. For example, the behavior of the drum
surface when you hit it by a stick would be best described by the solution of the
wave equation in the polar coordinate system. In this note, I would like to derive
Laplace’s equation in the polar coordinate system in details.
Recall that Laplace’s equation in R2 in terms of the usual (i.e., Cartesian)
(x, y) coordinate system is:
∂2 u ∂2 u
+
= u xx + u y y = 0.
∂ x2 ∂ y 2
(1)
The Cartesian coordinates can be represented by the polar coordinates as follows:
(
x = r cos θ;
(2)
y = r sin θ.
Let us first compute the partial derivatives of x, y w.r.t. r, θ :

∂x



= cos θ,
∂r
∂y



= sin θ,
∂r
∂x
= −r sin θ;
∂θ
∂y
= r cos θ.
∂θ
(3)
∂u
first. We will use the Chain Rule since (x, y) are
To do so, let’s compute
∂r
functions of (r, θ) as shown in (2).
∂u
∂r
Now, let’s compute
∂u ∂x ∂u ∂y
+
∂x ∂r ∂y ∂r
∂u
∂u
=
cos θ +
sin θ
∂x
∂y
∂u
∂u
= cos θ
+ sin θ .
∂x
∂y
=
using (3)
(4)
∂2 u
∂u
∂u
.
Noticing
that
both
and
are functions of (x, y)
∂r 2
∂x
∂y
1
and using (3), we have
∂2 u
∂r 2
∂ ∂u
∂ ∂u
+ sin θ
∂r ∂x
∂r ∂y
µ
¶
µ
¶
∂ ∂u ∂x
∂ ∂u ∂y
∂ ∂u ∂x
∂ ∂u ∂y
= cos θ
+
+ sin θ
+
∂x ∂x ∂r ∂y ∂x ∂r
∂x ∂y ∂r ∂y ∂y ∂r
= cos θ
2
∂2 u
∂2 u
2 ∂ u
= cos θ 2 + 2 cos θ sin θ
+ sin θ 2 .
∂x
∂x∂y
∂y
2
Similarly, let’s compute
∂u
∂θ
∂2 u
∂ θ2
(5)
∂u
∂2 u
and 2 .
∂θ
∂θ
∂u ∂x ∂u ∂y
+
∂x ∂θ ∂y ∂θ
∂u
∂u
(−r sin θ) +
(r cos θ)
=
∂x
∂y
∂u
∂u
= −r sin θ
+ r cos θ .
∂x
∂y
=
∂ ∂u
∂u
∂ ∂u
∂u
− r sin θ
− r sin θ
+ r cos θ
∂x
∂θ ∂x
∂y
∂θ ∂y
µ
µ
¶
¶
∂ ∂u ∂x
∂ ∂u ∂y
∂ ∂u ∂x
∂ ∂u ∂y
∂u
∂u
− r sin θ
+
+ r cos θ
+
= −r cos θ
− r sin θ
∂x
∂x ∂x ∂θ ∂y ∂x ∂θ
∂y
∂x ∂y ∂θ ∂y ∂y ∂θ
µ 2
¶
∂u
∂ u
∂2 u
= −r cos θ
− r sin θ
(−r sin θ) +
r cos θ
∂x
∂ x2
∂x∂y
µ 2
¶
∂u
∂ u
∂2 u
− r sin θ
+ r cos θ
(−r sin θ) +
r cos θ
∂y
∂x∂y
∂ y2
µ
¶
µ
2
2 ¶
∂u
∂u
∂2 u
2
2 ∂ u
2 ∂ u
+ r sin θ 2 − 2 cos θ sin θ
= −r cos θ
+ sin θ
+ cos θ 2
∂x
∂y
∂x
∂x∂y
∂y
= −r cos θ
Dividing both sides by r 2 and using (4), we have
2
2
1 ∂2 u
1 ∂u
∂2 u
2 ∂ u
2 ∂ u
=
−
−
2
cos
θ
sin
θ
+
sin
θ
+
cos
θ
r 2 ∂ θ2
r ∂r
∂ x2
∂x∂y
∂ y2
(6)
Finally, adding (5) and (6), using the obvious relation cos2 θ + sin2 θ = 1, we have
∂2 u 1 ∂2 u
1 ∂u ∂2 u ∂2 u
+
=
−
+
+
,
∂ r 2 r 2 ∂ θ2
r ∂r ∂ x 2 ∂ y 2
2
which can be cleaned up as:
∂2 u ∂2 u ∂2 u 1 ∂u 1 ∂2 u
+
=
+
+
.
∂ x 2 ∂ y 2 ∂ r 2 r ∂r r 2 ∂ θ 2
Hence, Laplace’s equation (1) becomes:
1
1
u xx + u y y = u r r + u r + 2 u θθ = 0.
r
r
Once we derive Laplace’s equation in the polar coordinate system, it is easy to
represent the heat and wave equations in the polar coordinate system. For the heat
equation, the solution u(x, y, t ) = u(r, θ, t ) satisfies
¶
µ
1
1
u t = k(u xx + u y y ) = k u r r + u r + 2 u θθ ,
r
r
k > 0: diffusivity,
whereas for the wave equation, we have
µ
¶
1
1
u t t = c (u xx + u y y ) = c u r r + u r + 2 u θθ
r
r
2
2
3
c > 0: wave velocity.