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Example of the Method of Reduction

Example: The differential equation for a viscous fluid, contained between two infinitely

long, vertical, concentric cylinders, where the inner cylinder is rotating at a rate, ω, and

the outer cylinder is fixed is given by:

r d²y/dr² + dy/dr - y/r = 0 (1)

Let the inner cylinder have radius, a, and the outer cylinder have radius, b. The boundary

conditions are:

y(a) = a ω and y(b) = 0 (2) and (3)

Strategy: To obtain one solution of this differential equation try a power form of

solution such as y = rⁿ . The simplest one, if it works, is for n = 1. i.e. y₁(r) = C₁ r.

So y₁^'' = 0, y₁^' = C₁ and - y₁ /r = - C₁ . Thus C₁ - C₁ = 0 and

Therefore C₁ r is one solution to the d.e. (1).

For the second solution using the method of reduction assume

y₂(r) = r w(r) and put into d.e. (1) where w(r) is an unknown function

y₂'(r) = w + r w ' and y₂''(r) = 2 w + r w ''

So r d²y₂/dr² + dy₂/dr - y₂/r = 2 r w' + r² w'' + w + rw - w = 0

r² w'' + 3r w' = 0 or r w'' + 3 w' = 0

Now let w' = v so r v' + 3 v = 0 or v' + (3/r) v = 0

which can be integrated using an integration factor, IF: If = e ^{∫(3/r) dr} = e ^{3 ln r} = r³

And r³ [ v' + (3/r) v ] = 0 or r³ v' + (3r²) v ] = 0

Finally d/dr [r³ v] = 0 r³ v = C₁ , v = C₁ / r³ = w'

Integrate to obtain: w(r) = C₂/r² + C₃

Now y₂(r) = r w(r) = C₂/r + C₃ r

If C₃ = 0 and C₂ = 1 y₂(r) = 1/r and if C₃ = 1 and C₂ = 0 y₂(r) = r