Example 1

Solve the following set of 3 simultaneous equations by the Neumann series.

⎛ ⎜ ⎜ ⎜ ⎝ 1.1, 0, −0.1 0.2, 0.9, 0 −0.1, 0.2, 1 ⎞ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎝ x y z ⎞ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎝ 2 −3 1 ⎞ ⎟ ⎟ ⎟ ⎠ .

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Solution The decomposition is

M = L − I = ⎛ ⎜ ⎜ ⎜ ⎝ 1.1, 0, −0.1 0.2, 0.9, 0 −0.1, 0.2, 1 ⎞ ⎟ ⎟ ⎟ ⎠ − ⎛ ⎜ ⎜ ⎜ ⎝ 1, 0, 0 0, 1, 0 0 0, 1 ⎞ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎝ 0.1, 0, −0.1 0.2, −0.1, 0 −0.1, 0.2, 0 ⎞ ⎟ ⎟ ⎟ ⎠ . (2)

The norm of M is less than 1 (verify!). The power of the matrix, M, can be easily computed as

M2 = ⎛ ⎜ ⎜ ⎜ ⎝ 0.02, −0.02, −0.01 0, 0.01, −0.02 0.03, −0.02, 0.01 ⎞ ⎟ ⎟ ⎟ ⎠ ,   M3 = ⎛ ⎜ ⎜ ⎜ ⎝ −0.001, 0, −0.002 0.004, −0.005, 0 −0.002, 0.004, −0.003 ⎞ ⎟ ⎟ ⎟ ⎠ , M4 = ⎛ ⎜ ⎜ ⎜ ⎝ 0.0001, −0.0004, 0.0001 −0.0006, 0.0005, −0.0004 0.0009, −0.001, 0.0002 ⎞ ⎟ ⎟ ⎟ ⎠ (3)

Taking the first four terms of the Neumann series yields

u  ∼  (I − M + M2 − M3 + M4 ) c = c − Mc + M2c − M3c + M4 c = ⎛ ⎜ ⎜ ⎜ ⎝ 1.9955 −3.7761 1.954 ⎞ ⎟ ⎟ ⎟ ⎠ . (4)

The exact solution is

u = L−1c = ⎛ ⎜ ⎜ ⎜ ⎝ 1.999591 −3.77687 1.95496 ⎞ ⎟ ⎟ ⎟ ⎠ .

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Example 2 (Fredholm's integral equation of the second kind)

The following integral equation,

u(x) = f(x) + λ ⌠ ⌡ b a  K(x, y) u(y) dy,     |λ| < 1,

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can be written as

u = f + λM u,

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where M is an integral operator defined for a function, ϕ(x), as

M ϕ(x) ≡ ⌠ ⌡ b a  K(x, y) ϕ(y) d y.

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The Neumann series solution is

u(x) = (1 − λM)−1 f = (1 + λM + (λM)2 + (λM)3 + …) f = f + λM f + λ2 M2 f + λ3 M3 f + … = f(x) + λ ⌠ ⌡ b a  K(x, y) f(y) dy + λ2 ⌠ ⌡ b a  K(x, y) ⎛ ⎝ ⌠ ⌡ b a  K(y, y′) f(y′) dy′ ⎞ ⎠ dy + … (9)

Example 3

u(t) = f(t) + λ ⌠ ⌡ 1 0  u(τ) dτ,     |λ| < 1.

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The Neumann series solution is

u = (1 − λM)−1 f = (1 + λM + λ2 M2 + λ3 M3 + …) f = f + λ ⌠ ⌡ 1 0  f(τ) dτ+ λ2 ⌠ ⌡ 1 0  ⎛ ⎝ ⌠ ⌡ 1 0  f(τ′) dτ′ ⎞ ⎠ dτ+ λ3 ⌠ ⌡ 1 0  ⎛ ⎝ ⌠ ⌡ 1 0  ⎛ ⎝ ⌠ ⌡ 1 0  f(τ") dτ" ⎞ ⎠ d τ′ ⎞ ⎠ dτ+ … (11)

Let

c ≡ ⌠ ⌡ 1 0  f(τ) dτ,

u = f + λ ⌠ ⌡ 1 0  f(τ) dτ+ λ2 ⌠ ⌡ 1 0  ⎛ ⎝ ⌠ ⌡ 1 0  f(τ′) dτ′ ⎞ ⎠ dτ+λ3 ⌠ ⌡ 1 0  ⎛ ⎝ ⌠ ⌡ 1 0  ⎛ ⎝ ⌠ ⌡ 1 0  f(τ") dτ" ⎞ ⎠ d τ′ ⎞ ⎠ dτ+ … = f + c λ(1 + λ+ λ2 + λ3 + …) = f + c λ 1−λ . (12)

Example 4

ϵy" + y = sin x,     |ϵ| < 1.

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can be written as

L y = c,     L = 1 + M,     M ≡ ϵ d2 dx2 .

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The Neumann series solution (particular solution) is

y = (1 + M)−1 sin x = (1 − M + M2 − M3 + M4 − …) sin x = sin x − ϵ(sin x)" + ϵ2 (sin x)"" − … = (1 + ϵ+ ϵ2 + ϵ3 + …) sin x = 1 1 − ϵ sin x. (15)

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On 18 Jan 2021, 22:24.