1 Laplace’s Equation on Circular Domains
Laplace’s equation on rotationally symmetric domains can be solved using a change of variables to
polar coordinates. The two dimensional Laplace operator in its Cartesian and polar forms are
1 1
∆u(x, y) = uxx + uyy and ∆u(r, θ) = urr + ur + 2 uθθ .
r r
We are interested in finding bounded solutions to Laplace’s equation, so we often have that implicit
assumption. The “radial” problem will be an Euler ODE which has the following solution.
Euler Equations: An ODE of the form
ax2 y 00 + bxy 0 + cy = 0
are called Euler ODEs. The ODE is solved by finding the roots r1 and r2 of the characteristic
polynomial
C(x) = ax(x − 1) + bx + c = 0
and the general form of the solution is given by
r1 r2
C1 x + C2 x
r1 , r2 ∈ R, r1 6= r2
y(x) = C1 xr + C2 log(x)xr r1 = r2 = r ∈ R
α α
C1 x cos(β log x) + C2 x sin(β log x) r1 = α + iβ, r2 = α − iβ, β 6= 0
1.1 Example Problems
Problem 1.1. (?) (Interior of a Disk) Solve
(
∆u = 0 for r < a, − π ≤ θ ≤ π,
u|r=a = 1 + 2 sin(θ),
Solution 1.1. After converting to polar coordinates, our PDE can be written as the following problem
on the circle
1 1
urr + r ur + r2 uθθ = 0 0 < r < a, −π ≤ θ ≤ π
u(r, −π) = u(r, π) 0<r<a
uθ (r, −π) = uθ (r, π) 0<r<a
u(a, θ) = 1 + 2 sin(θ) −π ≤ θ ≤ π
r→0 u(r, θ) < ∞ −π ≤ θ ≤ π
lim
The condition that limr→0 u(r, θ) < ∞ is an implicit assumption of this problem.
Step 1 — Separation of Variables: The PDE has periodic homogeneous angular boundary conditions,
so we look for a solution of the form u(r, θ) = R(r)Θ(θ). For such a solution, the PDE implies
1 1 r2 R00 + rR0 Θ00
∆u = R00 Θ + R0 Θ + 2 RΘ00 = 0 =⇒ − = = −λ.
r r R Θ
This results in the ODEs
Since the solution should be regular at 0 (limr→0 R(r) < ∞), we need Cn = 0 for all n ≥ 0, so our
solution is of the form
R0 (r) = D0 , Rn (r) = Dn rn , n = 1, 2, . . .
for some arbitrary coefficients D0 and Dn .
Step 4 — General Solution: Using the principle of superposition, and summing all the eigenfunc-
tions gives us the general solution
∞
X
u(r, θ) = A0 + rn (An cos(nθ) + Bn sin(nθ))
n=1
where A0 , An and Bn are yet to be determined coefficients.
Step 5 — Particular Solution: To find constants A0 , An and Bn we need to use the boundary condition.
Using the boundary condition we get
∞
X
1 + 2 sin(θ) = u(a, θ) = A0 + an (An cos(nθ) + Bn sin(nθ)) .
n=1
Instead of solving for the Fourier series like usual, we can just equate coefficients to see that
2
A0 = 1, a1 A1 = 2 =⇒ A1 = ,
a
and the rest of the coefficients are 0.
Step 6 — Final Answer: To summarize, the solution to the PDE is given by
2
u(r, θ) = 1 + r sin(θ).
a
Page 2 of 7
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller 9kfhgia89h1. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $7.99. You're not tied to anything after your purchase.
