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Solving Differential Equations in Terms of Bessel Functions

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48 CHAPTER 3. SOLVING IN TERMS OF BESSEL FUNCTIONS<br />

(a) If p is a zero <strong>of</strong> f with multiplicity m ∈ N, then p is a regular s<strong>in</strong>gularity<br />

<strong>of</strong> M and ∆(M, p) = 2mν.<br />

(b) If p is a pole <strong>of</strong> f with multiplicity m ∈ N such that<br />

f =<br />

then p is an irregular s<strong>in</strong>gularity <strong>of</strong> M and<br />

∆(M, p) = 2<br />

∞<br />

∑ fit<br />

i=−m<br />

i p, (3.3)<br />

−1<br />

∑<br />

i=−m<br />

i fit i p. (3.4)<br />

Pro<strong>of</strong>. Let t be the local parameter tp.<br />

(a) Let p be a zero <strong>of</strong> f with multiplicity m > 0, then f has the representation<br />

f = t m ∑ ∞ i=0 fit i with fi ∈ k and f0 = 0. Furthermore, let y ∈ V (LB) be a local<br />

solution at x = 0 <strong>of</strong> the form<br />

∞<br />

ν<br />

y = x ∑<br />

i=0<br />

If we now replace x by f , we get<br />

aix i , ai ∈ k,a0 = 0.<br />

∞<br />

ν<br />

z = f ∑ ai f<br />

i=0<br />

i<br />

which is a local solution <strong>of</strong> M at p. Hence, we can rewrite z as a series <strong>in</strong> t, i.e.<br />

∞<br />

e<br />

z = t ∑<br />

i=0<br />

(3.5)<br />

bit i , bi ∈ k,b0 = 0 (3.6)<br />

and e is the exponent <strong>of</strong> z. Now f i = tmi ¯f where the constant coefficient <strong>of</strong> ¯f ∈<br />

k[[t]] is non-zero and compar<strong>in</strong>g the representations (3.5) and (3.6) <strong>of</strong> z yields<br />

e = mν.<br />

Similarly, for the second <strong>in</strong>dependent local solution <strong>of</strong> LB at x = 0, which has<br />

exponent −ν, we obta<strong>in</strong> the generalized exponent e = −mν. Hence, the s<strong>in</strong>gularity<br />

p is regular and ∆(M, p) = 2mν.<br />

If ν ∈ Z the second <strong>in</strong>dependent solution conta<strong>in</strong>s a logarithm ln(x). However,<br />

we can still do the same computations. The solution z would then <strong>in</strong>volve ln(t)<br />

and the result for the exponent is still true.<br />

(b) A similar approach works <strong>in</strong> second case. Let p be a pole <strong>of</strong> f with multiplicity<br />

m ∈ N. Then representation (3.3) can also be written as f = t−m ∑ ∞ i=0 fi−mti with fi ∈ k, f−m = 0.

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