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Lecture Notes in Differential Equations - Bruce E. Shapiro

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416 APPENDIX C. SUMMARY OF METHODS<br />

To solve<br />

a(t)y ′′ + b(t)y ′ + c(t)y = g(t)<br />

about a po<strong>in</strong>t t 0 where a(t) = 0 but the limits b(t)/a(t) and c(t)/a(t) exist<br />

as t → 0 (a regular s<strong>in</strong>gularity), solve the <strong>in</strong>dicial equation<br />

r(r − 1) + rp 0 + q 0 = 0<br />

for r where p 0 = lim t→0 b(t 0 )/a(t 0 ) and and q 0 = lim t→0 c(t 0 )/a(t 0 ). Then<br />

one solution to the homogeneous equation is<br />

y(t) = (t − t 0 ) r<br />

∞<br />

∑<br />

k=0<br />

c k (t − t 0 ) k<br />

for some unknown coefficients c k .Determ<strong>in</strong>e the coefficients by l<strong>in</strong>ear <strong>in</strong>dependance<br />

of the powers of t. The second solution is found by reduction of<br />

order and the particular solution by variation of parameters.<br />

Method of Frobenius<br />

To solve<br />

(t − t 0 ) 2 y ′′ + (t − t 0 )p(t)y ′ + q(t)y = 0<br />

where p and q are analytic at t 0 , let p 0 = p(0) and q 0 = q(0) and f<strong>in</strong>d the<br />

roots α 1 ≥ α 2 of<br />

α 2 + (p 0 − 1)α + q 0 = 0<br />

Def<strong>in</strong>e ∆ = α 1 − α 2 . Then for some unknowns c k , a first solution is<br />

y 1 (t) = (t − t 0 ) α1<br />

∞ ∑<br />

k=0<br />

c k (t − t 0 ) k<br />

If ∆ ∈ R is not an <strong>in</strong>teger or the roots are complex,<br />

y 2 (t) = (t − t 0 ) α2<br />

∞ ∑<br />

k=0<br />

a k (t − t 0 ) k<br />

If α 1 = α 2 = α, then y 2 = ay 1 (t) ln |t − t 0 | + (t − t 0 ) α<br />

If ∆ ∈ Z, then y 2 = ay 1 (t) ln |t − t 0 | + (t − t 0 ) α2<br />

∞<br />

∑<br />

k=0<br />

∞<br />

∑<br />

k=0<br />

a k (t − t 0 ) k<br />

a k (t − t 0 ) k

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