Boundary-value problems
Objective. Use eigenfunctions and Green functions.
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- 2. Worked example
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Learn: the key idea
Boundary-value problems ask for a solution of a PDE that also satisfies conditions imposed on the boundary of the domain, and eigenfunction methods expand the solution in a basis that automatically respects those conditions. For −u'' = λu on (0,L) with u(0) = u(L) = 0, the eigenfunctions are sin(nπx/L) with eigenvalues λₙ = (nπ/L)², and a general forcing problem's solution is expanded in this basis. The Green's function G(x,ξ) encodes the response at x to a point source at ξ, so that u(x) = ∫G(x,ξ)f(ξ)dξ solves −u'' = f with the given boundary conditions. Example: for −u'' = 1 on (0,1) with u(0) = u(1) = 0, the solution u(x) = x(1−x)/2 can be recovered from either method.
Worked example
Solve the boundary-value problem −u''(x) = 1 on (0,1) with u(0) = u(1) = 0.
- 1Integrate once: −u'(x) = x + A, so u'(x) = −x − A.
- 2Integrate again: u(x) = −x²/2 − Ax + B.
- 3Apply u(0) = 0: B = 0. Apply u(1) = 0: −1/2 − A = 0, so A = −1/2.
- 4Thus u(x) = −x²/2 + x/2 = x(1−x)/2, which satisfies both boundary conditions.
Answer: u(x) = x(1 − x)/2
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Start practisingCommon mistake. A frequent error is using eigenfunctions that do not satisfy the problem's own boundary conditions, such as using cos(nπx/L) terms when the condition is u = 0 at both ends; the basis must be matched to the specific boundary data.
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