The equation under consideration is

(d²φ/dx²) = −k²(x)φ

If k is constant then the solution is φ(x)=A·sin(kx)+B·cos(kx).

The single second order equation may be replaced by a pair of first order equations by defining

ψ = (dφ/dx)

Thus

(dφ/dx) = ψ
(dψ/dx) = −k²(x)φ

In matrix form these are (dV/dx)=KV(x) where

		| φ |
V	=	|    |
		| ψ |

 

		|  0	1 |
K	=	|	|
		| −k²	0 |

If K is constant the solution to the system of the two equations can be represented as

V(x) = exp(Kx)V(0)

More generally the solution is

V(x)=exp(L(x))V(0)
where
L(x) = ∫₀^xK(ζ)dζ

The matrix function L(x) takes the form

		|     0	x |
L(x)	=	|	|
		| −l(x)	0 |

where l(x) is equal to ∫₀^xk²(ζ)dζ.

Proof that the above is indeed the solution
to the System of First Order Differential Equations

Note that the exponential of an n×n matrix M is defined as

exp(M) = I/0! + M/1! + M²/2! + …

where I is the n×n identity matrix.

Consider now

(dV(x)/dx) = lim_Δx→0[V(x+Δx)−V(x)]/Δx

Note that

V(x+Δx) = exp(L(x) + ΔL)V(0) = exp(ΔL + L(x))V(0) = exp(ΔL)exp(L(x))V(0) = exp(ΔL)V(x)

Furthermore

exp(ΔL) = I + ΔL + higher order terms in ΔL

Thus

V(x+Δx) − V(x) = (I+ΔL)V(x) − I·V(x) + higher order terms in ΔL
hence
V(x+Δx) − V(x) = ΔL·V(x) + higher order terms in ΔL

Ignoring the higher order terms ΔL takes the form

		|                    0	Δx |
ΔL	=	|	|
		| −∫xx+Δxk²(ζ)dζ	0 |

Therefore

		|       0	1 |
ΔL/Δx	=	|	|
		| −k²(x)	0 |

where x is a value between x and x+Δx. As Δx→0 the higher order terms divided by Δx go to zero and

		|       0	1 |
dL/dx	=	|	|
		| −k²(x)	0 |

The RHS of this is the matrix K(x). Thus

(dV/dx) = (dL/dx)V(x) = K(x)V(x)

which is what was to be proven.

(To be continued.)