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Uniqueness of solutions

We seek to show that if ϕ satisfies [link] with fixed boundary conditions, then ϕ is unique. To do this we will need the following two lemmas:

Lemma 1. Let a function ϕ = ϕ ( θ , t ) be defined on the rectangle [ 0 , 2 π ] × [ 0 , 1 ] , with boundary condition ϕ ( θ , 0 ) = 0 . Then, letting ϕ denote the gradient of ϕ ,

0 2 π 0 1 ϕ 2 d t d θ 1 2 0 2 π 0 1 | ϕ | 2 d t d θ

with equality only when ϕ is identically zero.

Proof: Motivated by the Fundamental Theorem of Calculus, we express ϕ 2 = ( 0 t ϕ t d s ) 2 . Then by the Cauchy - Schwarz Inequality:

0 2 π 0 1 ϕ 2 d t d θ = 0 2 π 0 1 0 t ϕ t d s 2 d t d θ
0 2 π 0 1 0 t ϕ t 2 d s 1 2 0 t 1 d s 1 2 2 d t d θ

Evaluating 0 t 1 d s = t and noting that ϕ t 2 | ϕ | 2 produces:

0 2 π 0 1 t 0 t ϕ t 2 d s d t d θ 0 2 π 0 1 t 0 t | ϕ | 2 d s d t d θ 0 2 π 0 1 t 0 1 | ϕ | 2 d s d t d θ

Since ϕ is a function of θ and s , we may regard it as a constant with respect to t , and so performing the t integration yields the desired result.

However, 0 2 π 0 1 t 0 t | ϕ | 2 d s d t d θ 0 2 π 0 1 t 0 1 | ϕ | 2 d s d t d θ is an equality only when | ϕ | 2 is identically zero. If this is the case, then ϕ is constant, and the boundary conditions imply that ϕ is identically zero.

Corollary 1. With the additional assumption that ϕ ( θ , 1 ) = 0 , this result can be improved to

0 2 π 0 1 ϕ 2 d t d θ 1 8 0 2 π 0 1 | ϕ | 2 d t d θ :

Proof: We write ϕ ( θ , t ) as ϕ ( θ , t ) = 0 t ϕ t d s = - t 1 ϕ t d s .

0 1 ϕ 2 d t = 0 1 / 2 ϕ 2 d t + 1 / 2 1 ϕ 2 d t = 0 1 / 2 0 t ϕ t d s 2 + 1 / 2 1 t 1 ϕ t d s 2 0 1 / 2 0 t ϕ t 2 d s 0 t 1 d s d t + 1 / 2 1 t 1 ϕ t 2 d s t 1 1 d s d t = 0 1 / 2 t 0 t ϕ t 2 d s d t + 1 / 2 1 ( 1 - t ) t 1 ϕ t 2 d s d t 0 1 / 2 t 0 t | D ϕ | 2 d s d t + 1 / 2 1 ( 1 - t ) t 1 | D ϕ | 2 d s d t 0 1 / 2 t 0 1 / 2 | D ϕ | 2 d s d t + 1 / 2 1 ( 1 - t ) 1 / 2 1 | D ϕ | 2 d s d t = 1 8 0 1 / 2 | D ϕ | 2 d s + 1 8 1 / 2 1 | D ϕ | 2 d s = 1 8 0 1 | D ϕ | 2 d s

Again, equality holds only when ϕ is identically zero.

Corollary 2. The inequality

0 2 π 0 h ϕ 2 d t d θ 0 2 π 0 h | ϕ | 2 d t d θ

holds for 0 < h 8 .

Proof: This follows from the calculations above.

Lemma 2. Suppose x 1 and x 2 are real numbers. Then

| sin ( x 1 ) - sin ( x 2 ) | | x 1 - x 2 | .

Proof: Follows from the Fundamental Theorem of Calculus.

We are now ready to address the uniqueness of solutions to [link] .

Theorem 1. Suppose a function ϕ ( θ , t ) : [ 0 , 2 π ] × [ 0 , 1 ] R is periodic in θ and satisfies

Δ ϕ + sin ( 2 ϕ ) 2 = 0

with fixed boundary conditions. Then ϕ is unique.

Proof: Suppose there are two functions ϕ 1 , ϕ 2 : [ 0 , 2 π ] × [ 0 , 1 ] R , periodic in θ , which both satisfy [link] . Because the boundary conditions are fixed, we can suppose that

ϕ 1 ( θ , 0 ) = ϕ 2 ( θ , 0 ) = f ( θ ) ϕ 1 ( θ , 1 ) = ϕ 2 ( θ , 1 ) = g ( θ )

where f and g are real valued functions. Then:

0 = Δ ϕ 1 - ϕ 2 + sin ( 2 ϕ 1 ) 2 - sin ( 2 ϕ 2 ) 2 .

Multiplying [link] by ( ϕ 1 - ϕ 2 ) sets up a situation in which we may use integration by parts:

0 2 π 0 1 ( ϕ 1 - ϕ 2 ) Δ ( ϕ 1 - ϕ 2 ) + ( ϕ 1 - ϕ 2 ) 2 ( sin ( 2 ϕ 1 ) - sin ( 2 ϕ 2 ) ) d t d θ = 0

Note that ϕ 1 - ϕ 2 vanishes on the boundary of the square since the boundary conditions are equal and both are periodic in θ . Thus, letting u = ϕ 1 - ϕ 2 and d v = Δ ϕ , the integration by parts of [link] becomes clear:

( ϕ 1 - ϕ 2 ) ( ϕ 1 - ϕ 2 ) | boundary - 0 2 π 0 1 ( ϕ 1 - ϕ 2 ) 2 d t d θ + 0 2 π 0 1 ϕ 1 - ϕ 2 2 ( sin ( 2 ϕ 1 ) - sin ( 2 ϕ 2 ) ) d t d θ = 0

Applying Lemma 1 to the first integration term, and Lemma 2 to the second integration term, we see that

0 - 2 0 2 π 0 1 ( ϕ 1 - ϕ 2 ) 2 + 0 2 π 0 1 ( ϕ 1 - ϕ 2 ) 2 = - 0 2 π 0 1 ( ϕ 1 - ϕ 2 ) 2 0 .

But since the integrand is non-negative, it must be that ϕ 1 - ϕ 2 = 0 and thus ϕ 1 = ϕ 2 , implying the uniqueness of a solution ϕ as desired.

Theorem 2. Let S be a cylinder of unit radius and height h < 8 . It is clear that ϕ ( θ , t ) = 0 satisfies equation [link] with boundary conditions ϕ ( θ , 0 ) = ϕ ( θ , h ) = 0 . This function is a stable local minimum of the energy equation [link] .

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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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