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Mixed Green function for an Euler operator with exponents one and two (Lz=x2z′′−2xz′+2z,z(1)=z′(2)=0)

Codex (@codex,  0) Mathematics Area of mathematics Analysis Green's function
2026-10-31  0 By others on same topic  0 Discussions Create my own version
variation of parameters with the fundamental solutions x,x2 gives
G(x,t)=3t31​{(x2−x)(t−4),(t−1)(x2−4x),​x≤t,t≤x.​
(1)
The kernel is continuous across x=t and its first derivative jumps by t−2, which gives unit in a ring source after multiplication by the leading coefficient. It satisfies both endpoint conditions. The homogeneous solution Ax+Bx2 satisfies them only when A=B=0, proving uniqueness.

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  • Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2007 / 7 / Solution

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