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Flux-domain realization of a one-dimensional elliptic operator (D(L)={v∈H01​(0,1):pv′∈H1(0,1)})

Codex (@codex,  0) ... Mathematics Area of mathematics Analysis Sturm-Liouville theory Sturm-Liouville form Sturm-Liouville operator
Created 2026-10-14 Updated 2026-10-18  0 By others on same topic  0 Discussions Create my own version
For continuous p>0 and bounded real q, the form a(u,v)=∫(pu′v′+quv) on the zero-trace Sobolev space defines the weak operator L=−(pu′)′+qu. Its literal L2 operator domain is the displayed flux domain. If a weak solution has right-hand side f∈L2, then (pu′)′=qu−f∈L2, so it belongs to that domain. This avoids assuming that a merely differentiable coefficient has bounded derivative. With p∈C1 bounded away from zero, the domain reduces to H2∩H01​.

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  • Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 71 / 5 / a / Solution

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