Construction of Σ 0

By the standard Stable Manifold Theorems, there is a strongest unstable manifold Ω1 tangent to e1

at (K, 0, 0, 0), and a parametric representation for the one-dimension strongest unstable manifold Ω1

in a small neighborhood of (K, 0, 0, 0) is given by

F1(ε1) = (K, 0, 0, 0) + ε1e1+ O(|ε¹|²).

There is also a 2-dimensional strongly unstable manifold Ω2 tangent to the linear subspace spanned by e₁ and e₂ at (K, 0, 0, 0), and a parametric representation for the 2-dimension strongly unstable manifold Ω₂ in a small neighborhood of (K, 0, 0, 0) is given by

F₂(ε₁, ε₂) = (K, 0, 0, 0) + ε₁e₁+ ε₂e₂+ O(|ε¹|²+|ε²|²).

Finally, the 3-dimension unstable manifold Ω3at (K, 0, 0, 0) has a parametric representation in a small neighborhood of (K, 0, 0, 0) given by

F3(ε1, ε2, ε3) = (K, 0, 0, 0) + ε1e1+ ε2e2+ ε3e3+ O(|ε¹|²+|ε²|²+|ε³|²).

Throughout the rest of this article, y(s; y0) stands for the solution of (5.31) with initial value y0 = (u₀, v₀, w₀, z₀); u(s; y₀) stands for the u-coordinate of of y(s; y₀), and similarly for the other three coordinate y.

Lemma 5.5 Consider a solution y(s, y₀) with y₀ ∈ Ω¹ and u₀< K. Then there is a finite s₀ such that u(s0; y0) < u_∗ and v(s, y0) < 0 for all s < s0. That is, the solution enters the region R.

Proof. Since e1 ∈ H which is an invariant manifold, obviously Ω¹ is contained in H. Thus, to consider the dynamics on Ω1 we may let w = z = 0 in (5.31). Fix a y0 ∈ Ω¹ closed to (K, 0, 0, 0).

The parametrization F1of Ω1 implies that there exists m > n > 0 such that y0lies between the two curves v = m(h(u)− h(K)) and v = n(h(u) − h(K)). We claim that that y(s, y⁰) is between the two curves v = m(h(u)− h(K)) and v = n(h(u) − h(K)) until u = u^∗ provided m large enough and n small enough. We prove the claim by contradiction. If there is s₁> 0 such that v = m[h(u)− h(K)]

and [v− m(h(u) − h(K))]⁰ < 0 at s = s₁ (where u < u^∗< K), then we will have

0 > v⁰− mh⁰(u)v|^s=s1

= cv− h(u)p(u) − mh⁰(u)v|^s=s1

= −h⁰(u)[h(u)− h(K)]m²+ c[h(u)− h(K)]m − p(u)h(u)|^s=s1

> 0,

for m large enough. This is a contradiction. Therefore, the trajectory of a solution starting on Ω₁ is never below curve v = m[h(u)− h(K)] until u = u^∗.

v = m[h(u) − h(K)]

v = n[h(u) − h(K)]

u v

u = u

_∗

K

Figure 8: Phase plane diagram for Lemma 5.5.

Similarly, if there is s₂ > 0 such that v = n[h(u)− h(K)] and [v − n(h(u) − h(K))]⁰ > 0 at s = s₂ (where u < u^∗< K). then we will have

0 < v⁰− nh⁰(u)v|^s=s2

= cv− h(u)p(u) − nh⁰(u)v|^s=s2

= −h⁰(u)[h(u)− h(K)]n²+ c[h(u)− h(K)]n|^s=s2

< 0

for n small enough. This is a contradiction. Therefore, the trajectory of a solution starting on Ω1

never above the line v = n[h(u)− h(K)] until u = u^∗.

With y(s; y0) bounded by the curves: v = m[h(u)− h(K)], v = n[h(u) − h(K)], it follows that v(s; y0) < 0 and u(s, y0) decreases until u(s, y0) < u^∗.

Lemma 5.6 A solution y(s, y0) with y0 ∈ Ω¹, u0 > K and v0 > 0 will remain in that region {u > K, v > 0} for all s.

Proof. Let y₀∈ Ω¹be closed to (K, 0, 0, 0). With u₀> K and v₀> 0, we have y₀∈ S, which implies that v⁰(s; y₀) > 0 for all s > 0. This prove the assertion.

Lemma 5.7 A solution y(s, y0) with 0 < u0< K, w0> 0, and z0> ^c₂w0 will remain in the region {w > 0, z > 2^cw} whenever 0 < u(s; y⁰) < K.

Proof. Suppose that y(s, y0) leaves the region{w > 0, z > 2^cw} at some s > 0 when 0 < u(s) < K.

Let

s1= inf{s > 0 : 0 < u(s) < K, z(s) = c 2w(s)}.

Then for s∈ [0, s¹), we have

w(0) > 0 and w⁰(s) = z(s) > c 2w(s),

which implies w(s₁) > 0. It must be that z⁰(s₁)− (c/2)w⁰(s₁)≤ 0, which yields to (by (A2) and (A3)) 0 ≥ cz(s¹) + `(w(s1))q(u(s1))−c

2z(s1)

≥ c²

4 w(s1) + `(w(s1))q(K))

≥ c²

4 + `⁰(0)q(K)
w(s1)

This contradicts to the assumption c > c^∗ and the assertion follows.

On Ω2, we parametrize a small circle with radius ε and center at (K, 0, 0, 0) by

G(θ) = (K, 0, 0, 0) + ε cos(θ + ψ0)e1+ ε sin(θ + ψ0)e2+ O(|ε|²), (5.37) where the constant phase ψ0 satisfies that G(0) lies on Ω1 in the region u < K, and the θ ∈ [0, 2π).

Denote by

A :={θ : ∃s⁰> 0 such that u(s0; G(θ)) < u_∗ and v(s; G(θ)) < 0 on 0 < s≤ s⁰}.

Then θ = 0∈ A is nonempty by Lemma 5.5. Let

θ₁:= sup{θ ∈ A : [0, θ) ⊂ A}, and y¹:= G(θ₁).

Remark 5.8 1. ψ0 is close to zero provide with ≈ 0.

2. According to Lemma 5.5, there exist a s₀> 0 for G(0) (on Ω₁ with u < K) such that u(s0; G(0)) < u∗ and v(s; G(0)) < 0 for 0 < s≤ s⁰.

The continuous dependence of solution on initial condition implies that A contains all θ1> 0.

3. By definition of A, v₁ (the v component of y₁) must be negative, which implies θ₁+ ψ₀≤ π, i.e.

θ₁. π. It follows that w1> 0.

Proof. Fix a θ∈ (0, θ¹), then there exists s0 such that

u(s0; G(θ)) = u∗and v(s; G(θ)) < 0, ∀s ≤ s⁰. If g(w)− p(u)|^s=s0 < 0, we have

dv⁰(s₀) = cv + h(u)(g(w)− p(u))|^s=s0 < 0,

then v(s⁺₀) < 0 and u(s⁺₀) < u_∗. That is, the trajectory immediately enters the region R.

If g(w)− p(u)|s^s0 > 0, then u(s0) = u_∗ and w(s0) > w_∗. The fact that v(s0) < 0 implies u⁺ < u_∗. That is, the trajectory enters the region P .

If g(w)−p(u)|^s=s0= 0, we have u(s(θ)) = u∗w(s0) = w∗. The fact that v(s0) < 0 implies u(s⁺₀) < u∗. Since θ + ψ0< θ1+ ψ0≤ π, it follows that w(0; G(θ)) > 0 and

z(0; G(θ)) = λ₂w(0; G(θ)) > c

2w(0; G(θ)).

Lemma 5.7 implies that w(s₀) > 0 and z(s₀) > ^c₂w(s₀) > 0. Thus w(s⁺₀) > w_∗. That is, the trajectory enters the region P .

The next lemma shows that there is a “last” trajectory on Ω2 which has u(s) decreasing to the value u = u_∗.

Lemma 5.10 There exists an s0> 0 such that

u(s0; y1) = u∗, and v(s0; y1) = 0.

Moreover

g(w)− p(u)|^s0 > 0, and w(s₀; y₁) > w_∗.

Proof. Recall that u_∗ < u(0; y₁) < K (for ε & 0), v(0; y1) < 0 (by the definition of A) and w(0; y₁) > 0 (by Remark 5.8.3). First we prove that

u(s; y1)≤ u^∗ or v(s; y1)≥ 0 for somes > 0.

If not, we have

u(s; y1) > u_∗, and v(s; y1) < 0 for all s > 0,

which implies u^∗ < u(s; y1) < K and v(∞; y¹) = 0. However, since w⁰ > c/2w and w(0; y1) > 0, it follow from Lemma 5.7 that w(∞; y¹) = ∞. Therefore, dv⁰ = cv + g(w)− p(u) > 0 for large s > 0.

This contradicts to v(∞, y¹) = 0.

Now either u(s; y1) reaches u∗ before v(s; y1) reach 0, or v(s; y1) reach u∗ before u(s; y1) reach 0. It can be shown that v(s; y1) reach u∗before u(s; y1) reach 0, i.e. there is a s0> 0 such that

v(s0; y1) = 0 and u(s0; y1)≥ u^∗, v(s0; y1) < 0 on [0, s0) If not this case, we have

u(s₀; y₁) = u_∗ and u(s₀; y₁) < u_∗, v(s₀; y₁) < 0 on [0, s₀).

It follows from Implicit Function Theorem that there exist s0(θ) for θ≈ θ¹ such that u(s₀(θ); G(θ)) = u^∗

and from continuous dependence of solution on θ that

v(s₀(θ); G(θ)) < 0 on [0, s₀(θ)).

That is, there is a θ > θ1 in A. This contradicts to the definition of θ1.

Since v(s0; y1) < 0, it must be that v⁰(s0; y1)≥ 0. At s = s⁰, we have 0≤ dv⁰ = cv + h(u)[g(w)− p(u)] = h(u)[g(w) − p(u)],

which implies [g(w)− p(u)]|^s0 ≥ 0. Should the equality hold, it follow from Lemma 5.7 that at s = s⁰ dv⁰⁰= h(u)g⁰(w)z > h(u)g⁰(w)c

2w > 0.

The fact that v(s0) = v⁰(s0) = 0 and v⁰⁰(s) > 0 implies that v(s) > 0 for all s near s0 which is a contradiction. Therefore g(w)− p(u)|^s0 > 0 and v⁰(s0; y1) > 0.

Now we prove that u(s0; y1) = u_∗. Since v(s0; y1) = 0 and v⁰(s0; y1) > 0, Implicit Function Theorem implies that there exists s0(θ) for θ≈ θ¹such that v(s0(θ); G(θ)) = 0. If u(s0; y1) > u∗, the continuous dependence of solution on initial value implies that that for θ≈ θ¹,

v(s; g(θ)) < 0 for s∈ [0, s⁰(θ)), v⁰(s₀(θ); G(θ)) > 0, and u(s₀(θ), G(θ)) > u_∗, which implies θ /∈ A for θ ≈ θ¹, a contradiction. The proof is complete.

u w u = u

_∗

K w

_∗

(u(s

0

, y

1

), w(s

0

, y

1

))

Figure 9: Projection to the uw-plane of the trajectory y(s; y1) in Lemma 5.10.

Lemma 5.11 There is a θ2∈ (θ¹, π) such that the v coordinate of y2:= G(θ2) has v coordinate equal to zero.

Proof. By (5.37), the v coordinate of G(θ) is given by

v = −ε(λ¹cos(θ + ψ₀) + λ₂sin(θ + ψ₀))

= −ε q

λ²₁+ λ²₂sin(θ + ψ0+ ψ1)

where sin ψ1 = λ1/pλ²₁+ λ²₂, ψ1 ∈ (0, π/2). Obviously v = 0 at some θ² ∈ (0, π). Recall that v coordinate of G(θ1) is negative. It follows that θ2> θ1.

On Ω₃, we consider a small sphere centered at (K, 0, 0, 0) with radius ε spherically parametrized by

Q(θ, φ) = (K, 0, 0, 0) + ε cos(θ + ψ₀) sin(φ + φ₀)e₁+ ε sin(θ + ψ₀) sin(φ + φ₀)e₂

+ε cos(φ + φ0)e3+ O(|ε|²), (5.38)

where for θ ∈ [0, 2π), φ ∈ [0, π]; the constant phase ψ⁰ is the one for (5.37) and φ₀ is selected to let Q(θ, π/2) lies in Ω₂. It follows that Q(θ, π/2) = G(θ). From Lemma 5.11 we know that the sphere intersects the hyperplane v = 0 at least one point l(θ₂, π/2). The next lemma shows that this intersection is actually a smooth closed curve. Without misunderstanding, we will drop the constant phase ψ0and φ0 for the parametrization of Q(θ, φ) in the sequel.

Lemma 5.12 The sphere which is the image of Q(θ, φ) intersects the hyperplane v = 0 in a smooth closed curve.

Proof. The intersection of the sphere and the hyperplane v = 0 gives the equation

M (θ, φ) := λ1cos θ sin φ + λ2sin θ sin φ + λ3cos φ + O(ε) = 0. (5.39) We show by Implicit Function Theorem that this equation solve φ as a C¹function of θ for θ∈ [0, 2π].

Indeed, since the v coordinate of G(θ₂) is zero, we have M (θ₂, π/2) = 0, i.e. (θ₂, π/2) solves (5.39).

At the point (θ, φ) satisfying (5.39), ∂M/∂φ is calculated as

∂M

∂φ = λ₁cos θ cos φ + λ₂sin θ cos φ− λ³sin φ + O(ε)

= −λ³csc φ + O(ε),

which is not zero for ε small enough. It follows from Implicit Function Theorem that we can solve φ = φ(θ), a C¹ function for θ∈ [0, 2π]. The proof is complete.

Lemma 5.13 The sphere which is the image of Q(θ, φ) intersects the hyperplane z = 0 in a smooth closed curve.

Proof. The intersection of the sphere and the hyperplane z = 0 gives the equation

N (θ, φ) := λ2ψ(λ2) sin θ sin φ + λ3ψ(λ3) cos φ + O(ε) = 0 (5.40) We show by Implicit Function Theorem that this equation solves φ as a C¹function of θ for θ∈ [0, 2π].

Indeed, since the z coordinate of G(0) is zero, we have N (0, π/2) = 0, i.e. (0, π/2) solves (5.40). At the point (θ, φ) satisfying (5.40), ∂N/∂φ is calculated as

∂N

∂φ = λ₂ψ(λ₂) sin θ cos φ− λ³ψ(λ₃) sin φ + O(ε)

= −λ³ψ(λ3) csc φ + O(ε),

which is not zero for ε small enough. It follows from Implicit Function Theorem that we can solve φ = φ(θ), a C¹ function for θ∈ [0, 2π]. The proof is complete.

Lemma 5.14 There is a point y₃on the sphere such that the v and z coordinates of y₃ are both zero.

Moreover, y3= Q(θ3, π/2) for some θ3∈ (0, π)

Proof. Equation (5.40) gives sin θ = O(ε) as φ = π/2. This implies that when φ = π/2, we have θ ≈ 0, π. By Equation (5.38), the v coordinate is positive for (θ ≈ 0, φ = π/2), and negative for (θ≈ π, φ = π/2). Now on the close curve defined by the intersection of the sphere and {z = 0}, the v coordinate takes both positive and negative values. Therefore the v coordinate must takes zero at some point of the curve. This prove the assertion.

We are read to give the definition of Σ. Figure will help readers to understand the definition.

Denote by y0 := G(θ0) the intersection of the sphere with Ω1 in the region 0 < u < K. According to Lemma According to previous lemmas. The yd0yi, i = 1, 2 is the portion of intersection which the sphere intersects with Ω2 between y0 and yi. The arc [y2y3 is the portion of which the sphere intersects with{v = 0} between y²and y₃. The arc [y₃y₀is the portion of which the sphere intersects with{z = 0} between y³ and y₀. Let B be a small ball around y₀in the space spanned by e_i’s. Then B intersects the arcs [y₀y₂ and [y₃y₀ at y₄ and y₅, resp. Then the arc [y₄y₅ is the portion of which the sphere intersects B. Define Σ to be the subset of the sphere bounded the arcs \yiyi+1, i = 1, 2, 3, 4 and [y5y1

Remark 5.15 1. According the definition of yi’s, it is easy to verify that the point in the interior of Σ have components satisfying

0 < u < K, v < 0, w > 0, and z > 0. (5.41)

2. For y∈ Σ, the component of y · s satisfies (5.41) for s ≤ 0.

y

0

e

1

e

₂

y

₄

y

5

y

1

y

2

y

₃

Σ e

3

Figure 10: Construction of the Σ.

44

在文檔中 反應擴散方程之進行波解的存在性 (頁 45-54)