Quotient transformation invariant continuation method

In this subsection, we will introduce the quotient transformation invariant continuation method [7]. And we replace standard continuation method with the quotient transformation invariant continuation method. In the process of computing, some problems will occur. We need to overcome these problems.

First, we will face the problem about the predicting directions. Before discussing the problem, we consider the solution curve of standard continu-ation method. From equcontinu-ation (4.4), we know that there are four variables and three equations. Consequently, we can get the solution which is a two-dimensional manifold. We denote the solution set of equation (4.4) as M.

Hence, we suppose that the solution will be a pipeline.

M

Figure 4: standard continuation method

From Figure 4, we can see that why the standard continuation method will fail. We will find that the predicting directions will move along the manifold. As a result, we can not get meaningful predictions of standard continuation method. Hence, we want to use quotient transformation in-variant continuation method which can improve the disadvantage. For the purpose of letting the solution of equation (4.4) can be a straight line, we add a plane to equation (4.4). Thus, there are four variables and four equations.

M

C

'( ) u

prediction

P

( ) u

θ

θ

Figure 5: quotient transformation invariant continuation method

From Figure 5, we can see that the direction of predictions will be on the plane. And the quotient solution curve C is on the plane P . We will explain the details of the process of transformation.

From equation (4.5), we can rewrite into the forms as follows.

G(y(s)) = 0,

where y(s) = (u(s), γ(s)). And we differentiate G(y(s)) = 0 with respect to s. Thus, we can get

JG(y(s)) ˙y(s) = 0, (4.8)

where J_G(y(s)) is the Jacobian matrix of G. And we denote a tangent vector of C at y(s) as ˙y(s) = ( ˙u(s)^>, ˙γ(s))^>. We let C be the quotient solution curve.

We show that the conception of quotient solution. Let u(θ) = u · e^iθ,

= (u₁+ iu₂) · e^iθ,

= u₁cos θ + iu₁sin θ + iu₂cos θ − u₂sin θ,

= (u₁cos θ − u₂sin θ) + i(u₁sin θ + u₂cos θ).

We define

u(θ) =

· u1cos θ − u2sin θ u₁sin θ + u₂cos θ

¸ .

Let equation (2.3) be multiplied in both right and left sides by e^iθ. We can obtain the equation as follows.

[−∆u + iγ∂_θ1u + V u + α₁|u|²u] · e^iθ = λ₁u · e^iθ.

We know that u will satisfy equation (2.3) if u is a solution. And we suppose that u · e^iθ will be a solution, that is,

[−∆ + iγ∂_θ1 + V + α₁|u · e^iθ|²]u · e^iθ = λ₁u · e^iθ. (4.9) We put |u · e^iθ|² = |u|²|e^iθ|² = |u|² into equation (4.9). Then, we can get

[−∆ + iγ∂_θ1 + V + α₁|u|²]u · e^iθ = λ₁u · e^iθ,

⇒[−∆ + iγ∂_θ1 + V + α₁|u|²]u = λ₁u.

And we consider the normalized constraint.

Z

|u · e^iθ|²dx = Z

|u|²dx = 1.

Hence, we can verify the assumption, that is, if u is a solution, then ue^iθ is also a solution. We can see that the solution M contains transformation invariant solutions u(θ).

We denote the quotient solution as C. Now, we want to get the tangent vector of u(θ). Hence, we will differentiate u(θ) with respect to θ. We can get

u⁰(θ) =

· −u₁sin θ − u₂cos θ u₁cos θ − u₂sin θ

¸

. (4.10)

We choose θ = 0 and put it into equation (4.10). Then we can get u⁰(0) =

∂u

∂θ(0).

∂u

∂θ(0) =

· −u₂ u₁

¸

= (−u^>₂, u^>₁)^>.

Consequently, we can obtain the normal vector of the plane which we add, that is, the tangent vector of u(θ). We denote (^∂u_∂θ(0)^>, 0)^> as the tangent vector.

Then, we begin the further exploration of the quotient transformation invariant continuation method. First, we discuss the part of prediction. We know that the prediction vector ˙y(s) = ( ˙u(s)^>, ˙γ(s))^>. And ( ˙u(s)^>)^> is perpendicular to the tangent vector (^∂u_∂θ(0)^>, 0)^>. Next, we consider the part of correction. We can see that the vector of correction will be on the plane which we add. And the normal vector of the plane is (^∂u_∂θ(0)^>, 0, 0)^>.

• predictions

By using fixed point iteration method, we can get the initial solution (u₀, γ₀)^> ∈ R^{(2N +2)×1}. We let y₀ = (u₀, γ₀)^>. We discuss how we find the vectors of predictions. In the part of predictions, we know that the vector of predictions will be on the plane P . The plane has the normal vector (^∂u_∂θ(0)^>, 0, 0)^>. The vector of predictions ˙y(s) = ( ˙u(s)^>, ˙γ(s))^>

should satisfy the equation (4.8). And ( ˙u(s)^>)^> is perpendicular to the tangent vector (^∂u_∂θ(0)^>, 0)^>. Therefore, we add a plane as follows.

a^>_θ ˙u = 0, (4.11)

where

a_θ = (∂u

∂θ(0)^>, 0)^>

= (((−u^>₂, u^>₁)^>)^>, 0)^>

= (−u^>₂, u^>₁, 0)^>.

Originally, we solve the system of (4.7). We let y(s) = (u(s), γ(s)) and ˙y(s) = ( ˙u(s)^>, ˙γ(s))^>. Hence, we can rewrite it into the form as

follows. £

G_u G_γ ¤ ˙y(s) = 0.

Since we add a plane which is satisfied equation (4.11), we have to solve the system as follows.

· Gu Gγ

a^>_θ 0

¸

˙y =

· 0 0

¸ .

By computing above, we can get the vectors of predictions. Conse-quently, we will get the new solution (u⁽¹⁾₁ , γ₁⁽¹⁾)^> = (u₀, γ₀)^>+ ¯s · ˙y, where k ˙yk₂ = 1.

In the process of predictions, we let y_i = (u_i, γ_i)^> be the approximate solution every time. And we have to solve the system as follows.

· G_u G_γ a^>_θ 0

¸

˙y_i =

· 0 0

¸

. (4.12)

Then, we denote the vector of prediction as ˙y_i = ( ˙u^>_i , ˙γ_i)^>. As a result, we will get the new solution (u⁽¹⁾_i , γ_i⁽¹⁾)^>= (u_i−1, γ_i−1)^>+ ¯s_i· ˙y_i, where i = 1, 2, · · · , ¯s_i is the length of step every time and the two norm of ˙y_iis equivalent to one. After computing the vector of predictions, we hope that the new solution will converge to the solution by using Newton’s method. Hence, we introduce the steps of corrections.

• corrections

We continue discussing the correction part of quotient transforma-tion invariant continuatransforma-tion method. Now, we have the new solutransforma-tion (u⁽¹⁾_i , γ_i⁽¹⁾)^> which is on the plane P_i. Here, P_i represents the plane which has the normal vector (^∂u_∂θ(0)^>, 0)^>. And we know that the vec-tor of corrections will also be on the plane P_i. The solutions which will proceed to converge are also on the plane. We want to get the vectors of correction. As a result, we have to consider the system as follows.





G : G(u, γ) = 0, P : a^>_θ ˙u = 0,

N_i : ˙u^>_i−1u + ˙γ_i−1γ = ˙u^>_i−1u⁽¹⁾_i + ˙γ_i−1γ_i⁽¹⁾

It means that the solution curve is related to the system above. We know that the point which we want to find has to satisfy the system,

that is, (u_i, γ_i)^>. We consider the system as follows.

H =





G : G(u_i, γ_i) = 0, P : a^>_θ ˙u_i−1= 0,

N_i : ˙u^>_i−1u_i+ ˙γ_i−1γ_i = ˙u^>_i−1u⁽¹⁾_i + ˙γ_i−1γ_i⁽¹⁾

For the purpose of getting the vectors of corrections, we still use New-ton’s method. Therefore, we consider the Jacobian matrix of H, that is,

J_H =



 G_u G_γ a^>_θ 0

˙u^>_i−1 ˙γ_i−1



 ∈ R(2N +3)×(2N +2). (4.13)

Then, we use Newton’s method to solve it iteratively. And we put it into the formula as follows.

(u⁽ⁿ⁺¹⁾_i , γ_i⁽ⁿ⁺¹⁾) = (u⁽ⁿ⁾_i , γ_i⁽ⁿ⁾) − H(u⁽ⁿ⁾_i , γ_i⁽ⁿ⁾) · J_H⁻¹(u⁽ⁿ⁾_i , γ_i⁽ⁿ⁾), where n = 1, 2, · · · . Hence, we can get the solutions which will converge gradually. We stop the steps of correction until the solutions converge.

• improvement in computing

Now, we have the system H and J_H. In the process of computing, we observe the last row of G_u. We will find that the norm of the last row is much smaller than other rows. It will result in error in calculation.

The last row will be ignored. To review the system H which we want to solve, we observe G₃, that is, the normalized constraint. We let G3 be multiplied in both sides by accuracy. And we let accuracy = max(k(u, γ)k, 1). We also let the planes P and N_i be multiplied in both sides by accuracy. Now, we have new normalized constraint, in other words, we rewrite G3 to accuracy · (C^>U⁰ − 1). And we rewrite the planes into the forms as following.

P : accuracy · (a^>_θ ˙u_i−1) = 0,

N_i : accuracy · ( ˙u^>_i−1u_i+ ˙γ_i−1γ_i) = accuracy · ( ˙u^>_i−1u⁽¹⁾_i + ˙γ_i−1γ_i⁽¹⁾).

By using the skill above, we can obtain H and J_H accurately. And we have to rewrite equation (4.5) as follows.

G_u =





A + α¯ ₁(3u^°₁² + u^°₂² ) 2α₁(u₁◦ u₂) − γS −u₁ 2α1(u1 ◦ u2) + γS A + α¯ 1(3u^°₂² + u^°₁² ) −u2

accuracy · (2C^>◦ u^>₁) accuracy · (2C^>◦ u^>₂) 0



 ∈ R(2N +1)×(2N +1).

In the part of prediction, system (4.12) has to rewrite into the form as

Furthermore, we consider the part of prediction. We rewrite equation (4.13) into the form as follows.

J_H =

In this section, we will introduce the algorithm of continuation method. We can see that the algorithm consists of predictions and corrections. We show the Algorithm as follows.

Step1. Computing initial value Compute initial value: (u₀, γ₀).

Set the length of step.

Set count = the number of possible solutions.

Step2. Predictions for i = 0, 1, · · · ,count

Compute Jacob_F =

· G_u(u_i, γ_i) G_γ(u_i, γ_i) accuracy · a^>_θ 0

¸ .

Find the desired singular values and singular vectors of Jacob_F.

Let the smallest singular vector be the vector of predictions: P red uγ.

end

Step4. Corrections (Newton’s method) Compute Del uγ = −J_H⁻¹· H(u⁽¹⁾_i+1, γ_i+1⁽¹⁾)

(u⁽²⁾_i+1, γ_i+1⁽²⁾) = (u⁽¹⁾_i+1, γ_i+1⁽¹⁾) + Del uγ. ComputeJ_H, N, G, H

If(kH(u⁽¹⁾_i+1, γ_i+1⁽¹⁾)k/kH(u⁽²⁾_i+1, γ_i+1⁽²⁾)k < 10 and kH(u⁽²⁾_i+1, γ_i+1⁽²⁾)k > 10⁻⁸·accuracy

在文檔中 National University of Kaohsiung Repository System:Item 310360000Q/10839 (頁 24-29)