A function f : U → R is said to be twice continuously differentiable if all of the second partial derivatives of f exist and continuous on U

1. Laplace Operator and the Laplace Equation

Let U be an open subset of Rⁿ. A function f : U → R is said to be twice continuously differentiable if all of the second partial derivatives of f exist and continuous on U. The set of all real valued twice continuously differentiable functions on U is denoted by C²(U ).

Let f ∈ C²(U ). The Laplacian of f is defined to be

∆f =

n

X

i=1

fxixi.

Definition 1.1. A function f ∈ C²(U ) is said to be harmonic if ∆f = 0.

Let g : (0, ∞) → R be a C²-function. We define a function f : R²\ {0} → R by f (x, y) = g(r), r =p

x²+ y². Let us compute ∆f. By chain rule,

f_x= g⁰(r)x

r, f_y = g⁰(r)y r and

fxx= g⁰⁰(r)

x r

2

+ g⁰(r)r²− x² r³ , fyy= g⁰⁰(r)

y r

2

+ g⁰(r)r²− y² r³ . We obtain that

∆f = f_xx+ f_yy= g⁰⁰(r)x²+ y²

r² + g⁰(r)2r²− (x²+ y²) r³

= g⁰⁰(r) +1 rg⁰(r).

Hence ∆f = 0 if and only if

g⁰⁰(r) + 1

rg⁰(r) = 0, r > 0.

Let h(r) = g⁰(r) for r > 0 We have rh⁰(r) + h(r) = 0, for r > 0 i.e. (rh(t))⁰ = 0 for r > 0.

By mean value theorem, rh⁰(r) = c1 for r > 0. Therefore g⁰(r) = h(r) = c1

r for r > 0.

By integrating g⁰(r), we have

g(r) = Z r

r0

c1

tdt = c1ln r + c2

where c₂ = −c₁ln r₀ and r₀ > 0. This shows that f (x, y) = c1

2 log(x²+ y²) + c2. Let us try to solve the Laplace equation

∆f = 0 on B(0, 1).

1

2

Proposition 1.1. Let f ∈ C²(B(0, 1)) ∩ C(D(0, 1)). We set

u(r, θ) = f (r cos θ, r sin θ), 0 < r < 1, θ ∈ R.

Then u is a periodic function in θ with period 2π.

∆f = urr+1

rur+ 1 r²u_θθ.

Assume that u(r, θ) = R(r)Θ(θ) is a nonzero solution to ∆f = 0. Then R⁰⁰(r)Θ(θ) +R⁰(r)

r Θ(θ) +R(r)

r² Θ⁰⁰(θ) = 0.

Dividing the above equation by R(r)Θ(θ) and multiplying the above by r², we obtain r²R⁰⁰(r) + rR⁰(r)

R(r) = −Θ⁰⁰(θ) Θ(θ) .

The left hand side of the equation is a function of r which is independent of θ while the right hand side of the equation a function of θ which is independent of r. Thus both of the right hand side and the left hand side of the equation must be a constant. Set the constant to be λ We obtain two differential equation

r²R⁰⁰(r) + rR⁰(r) −λR(r) = 0,

Θ⁰⁰(θ) + λΘ(θ) = 0.

When λ = 0, Θ(θ) = aθ + b for some a, b. Since Θ is a period function, a = 0. Therefore r²R⁰⁰(r) + rR⁰(r) = 0. We have seen that R(r) = c₁log r + c₂. Since we assume that f is continuously differentiable at 0, it is continuous at 0. This implies that c1 = 0.

When λ 6= 0, for each n ≥ 1, the function c_n(θ) = cos nθ and s_n(θ) = sin nθ are both periodic of period 2π and satisfy Θ⁰⁰+ n²Θ = 0. In this case, we obtain that for each n ≥ 1,

r²R⁰⁰(r) + rR⁰(r) − n²R(r) = 0.

To solve this equation, we define a new function φ(t) = R(e^t), i.e. we make a change of variable r = e^t. Then we obtain that

φ⁰⁰(t) − n²φ(t) = 0.

By theory of O.D.E., we find that φ(t) is of the form φ(t) = C₁e^nt+ C₂e^−nt. This implies that

R(r) = C₁rⁿ+ C₂r⁻ⁿ.

Again, by the continuity of f, we find C2 = 0. Set un(r, θ) = rⁿcos nθ and vn(r, θ) = rⁿsin nθ. Then {1, u_n(r, θ), v_n(r, θ)} are all solutions to the Laplace equation. Fourier’s idea: combine all these solutions into one. More precisely, we set

(1.1) u(r, θ) = a0

2 +

∞

X

n=1

rⁿ(ancos nθ + bnsin nθ).

How do we determine a0, an, bn? Observe that (1.2) u(1, θ) = f (cos θ, sin θ) = a₀

2 +

∞

X

n=1

(a_ncos nθ + b_nsin nθ).

3

Let us define g(θ) = f (cos θ, sin θ) for θ ∈ R. It follows from the definition that g is a periodic function of period 2π, in fact,

g(θ + 2π) = f (cos(θ + 2π), sin(θ + 2π)) = f (cos θ, sin θ) = g(θ)

for any θ ∈ R. We can also use the notation f |S¹ to denote g. We call g the restriction of f to S¹ = ∂B(0, 1). Hence if we know g, the restriction of f to the boundary of S¹, and express g as an infinite series of the form (1.2), then we can find (solve for) f. This leads to the following Dirichlet problem

(1.3)  ∆f = 0 on B(0, 1)

f |_S¹ = g Hence if g has the following series representation

(1.4) g(θ) = a₀

2 +

∞

X

n=1

a_ncos nθ + b_nsin nθ, then f is given by (1.1).

Example 1.1. Solve (1.3) with the boundary condition g(θ) = 1 − cos θ + sin θ + 3 sin 3θ.

The harmonic function is given by

f (r cos θ, sin θ) = 1 − r cos θ + r sin θ + 3r³sin 3θ.

We can also express f in terms of x, y as follows. By De Moivre’s law, we know (x + iy)ⁿ= rⁿ(cos nθ + i sin nθ).

We set C_n(x, y) = Re(x + iy)ⁿ and S_n(x, y) = Im(x + iy)ⁿ. The above function can be rewritten as

f (x, y) = 1 − C₁(x, y) + S₁(x, y) + 3S₃(x, y).

Since C1(x, y) = x and S1(x, y) = y and S3(x, y) = 3xy²− y³, f (x, y) = 1 − x + y + 9x²y − 3y³.

In general, if the boundary condition is given by (1.4), then the solution to (1.3) is given by

(1.5) f (x, y) = a0

2 +

∞

X

n=1

(anCn(x, y) + bnSn(x, y)).

Later, we will show that any continuous periodic function of period 2π always possess such a series representation as (1.4) and discuss the convergence of (1.5). We need the notion of uniform convergence.

Example 1.2. Solve (1.3) with the boundary condition

g(θ) = 1 + 2 cos θ − 3 sin θ + 4 sin²θ + cos²θ + 5 sin θ cos θ + cos³θ + 6 sin³θ.