Discretized Classical Model of the New Equation

H(p;x) = [1／(2M)]Σ_k[p^k - QA^k(x)]² + Qφ(x).
M = mεα², Q = m.
A^k(x) = α(x^k+1 - x^k),
φ(x) = -[1／(2ε)]Σ_k(x^k+1 - x^k)².

dx^j/dt = ∂H/∂p^j = (p^j - QA^j)／M,
dp^j/dt = -∂H/∂x^j = (Q/M)Σ_k(p^k - QA^k)∂A^k/∂x^j - Q∂φ/∂x^j.

p^j = Mdx^j/dt + QA^j,
M(d/dt)²x^j + QdA^j/dt
= (Q/M)Σ_k(p^k - QA^k)∂A^k/∂x^j - Q∂φ/∂x^j
= (Q/M)Σ_k(Mdx^k/dt)∂A^k/∂x^j - Q∂φ/∂x^j,
dA^j/dt = Σ_k(∂A^j/∂x^k)(dx^k/dt),
∴ M(d/dt)²x^j = (Q/M)Σ(Mdx^k/dt)∂A^k/∂x^j - QΣ(∂A^j/∂x^k)(dx^k/dt) - Q∂φ/∂x^j,
∴ M(d/dt)²x^j = QΣ_k(dx^k/dt)[∂A^k/∂x^j -∂A^j/∂x^k] - Q∂φ/∂x^j.

M(d/dt)²x^j = Qα(dx^j-1/dt - dx^j+1/dt) + Q(2/ε){x^j - (1/2)[x^j-1 + x^j+1]}.
This equation is different from the following equation which represents an automatic collision avoidance system.
M(d/dt)²x^j = Qα[(dx^j-1/dt - dx^j/dt) + (dx^j+1/dt - dx^j/dt)] - Q(2/ε){x^j - (1/2)[x^j-1 + x^j+1]}.

Multiplying both sides of the equation by dx^j/dt and summing over j, we get the following equation representing the energy conservation law.
(d/dt)Σ_a≦j≦b(M/2)(dx^j/dt)² = Qα[(dx^a-1/dt)(dx^a/dt) - (dx^b/dt)(dx^b+1/dt)] + Q(1/ε)(d/dt)Σ_a≦j≦b-1[(x^j)² - x^jx^j+1] - (Q/ε)(x^a-1dx^a/dt + x^b+1dx^b/dt) + (Q/ε)(d/dt)(x^b)².
∴ (d/dt)[Σ_a≦j≦b(M/2)(dx^j/dt)² - Σ_a≦j≦b-1(Q/ε)x^j(x^j - x^j+1)]
= Q(dx^a/dt)(αdx^a-1/dt - x^a-1/ε) - Q(dx^b/dt)(αdx^b+1/dt + x^b+1/ε) + (Q/ε)(d/dt)(x^b)².
∴ (d/dt)[Σ_a≦j≦b(M/2)(dx^j/dt)² - Σ_a≦j≦b-1(Q/ε)x^j(x^j - x^j+1) - (Q/ε)(x^b)²]
= Q(dx^a/dt)(αdx^a-1/dt - x^a-1/ε) - Q(dx^b/dt)(αdx^b+1/dt - x^b+1/ε) - 2(Q/ε)(dx^b/dt)x^b+1.

I want to let the equation reduce to the form meaning that energy of a≦j≦b varies by energy flow at j=a and j=b.
However the term - 2(Q/ε)(dx^b/dt)x^b+1 prevents it.
It is of course because of the interaction between j=a-1 and j=a and the interaction between j=b and j=b+1.
An equation of such a form may be made in the continuous case.

To know the character of this problem, I omit the scalar potential term and try solving the following equation.
M(d/dt)²x^j = Qα(dx^j-1/dt - dx^j+1/dt).

M(d/dt)²x^j = QαΣ_kA_j,kdx^k/dt,
A_j,k = δ_j-1,k - δ_j+1,k

dx/dt = a[exp(QαtA/M)]b
where a is a real number, and b is an infinite column vector.

(A²)_j,k
= Σ_sA_j,sA_s,k
= Σ_s(δ_j-1,s - δ_j+1,s)(δ_s-1,k - δ_s+1,k)
= δ_j-2,k - 2δ_j,k + δ_j+2,k

(A³)_j,k
= Σ_s(δ_j-2,s - 2δ_j,s + δ_j+2,s)(δ_s-1,k - δ_s+1,k)
= δ_j-3,k - 3δ_j-1,k + 3δ_j+1,k - δ_j+3,k

(A⁴)_j,k
= Σ_s(δ_j-3,s - 3δ_j-1,s + 3δ_j+1,s - δ_j+3,s)(δ_s-1,k - δ_s+1,k)
= δ_j-4,k - 4δ_j-2,k + 6δ_j,k - 4δ_j+2,k + δ_j+4,k

(A⁵)_j,k
= Σ_s(δ_j-4,s - 4δ_j-2,s + 6δ_j,s - 4δ_j+2,s + δ_j+4,s)(δ_s-1,k - δ_s+1,k)
= δ_j-5,k - 5δ_j-3,k + 10δ_j-1,k - 10δ_j+1,k + 5δ_j+3,k - δ_j+5,k

(Aⁿ)_j,k
= Σ_0≦s≦n/2(-1)^sc₁(s,n)δ_j-n+2s,k + Σ_0≦s＜n/2(-1)^n+sc₂(s,n)δ_j+n-2s,k,
c_r(s,n+1) = c_r(s,n) + c_r(s-1,n),
c_r(0,n) = 1,

If a = 1, b_j = δ_j,0 then:
dx^j/dt = Σ_n=0^∞(1/n!)(Qαt/M)ⁿ[Σ_0≦s≦n/2(-1)^sc₁(s,n)δ_j-n+2s,0 - Σ_0≦s＜n/2(-1)^sc₂(s,n)δ_j+n-2s,0].

From the point of view of energy conservation, velocity does not increase or decrease as a whole.
This is expected to be represented well by
dx/dt = a exp(it×somematrix) b,
but the above solution does not include the imaginary unit.
Perhaps the matrix A has a property like a pure imaginary number.

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Last edited at 2014/05/15/16:53JST