Resonant boundary-value problem and the Fredholm alternative
Consider the boundary-value problem y''+y=f(x), 0<x<π, y(0)=y(π)=0. A mechanical model is driven at frequency 1, exactly the first natural frequency of the homogeneous problem. Three forcing cases are proposed: f1(x)=sin 2x, f2(x)=sin x, and f3(x)=sin x+sin 2x. The engineer must decide, before computing, which cases are solvable, then solve the solvable ones and exhibit the non-uniqueness or non-existence clearly. The homogeneous solutions are sin x and cos x; the boundary conditions kill cos x but not sin x, so the adjoint solvability condition is expected to involve sin x. A numerical colleague suggests simply adding a small damping term εy' and taking ε→0; the engineer must explain whether the limit recovers the undamped conclusion.
1. State the Fredholm solvability condition for this problem and use it to classify f1, f2, and f3.
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Write L[y]=y''+y. The homogeneous solutions are sin x and cos x; the boundary conditions eliminate cos x but leave sin x, so λ=0 is an eigenvalue and φ=sin x is the adjoint homogeneous solution. The Fredholm alternative says L[y]=f with y(0)=y(π)=0 is solvable iff ∫_0^π f(x) sin x dx=0. For f1=sin2x, ∫_0^π sin2x sinx dx=2∫_0^π sin²x cosx dx=(2/3)sin³x|_0^π=0, so f1 is solvable. For f2=sinx, ∫_0^π sin²x dx=π/2≠0, so f2 is not solvable. For f3=f2+f1, the integral is π/2≠0, so f3 is not solvable.
2. Solve the solvable case f1 explicitly and describe the solution set.
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For f1=sin2x, try y_p=C sin2x. Then y_p''=-4C sin2x, so y_p''+y_p=-3C sin2x=sin2x, giving C=-1/3. The general solution is y=A sinx+B cosx-(1/3)sin2x. The condition y(0)=0 gives B=0. The condition y(π)=0 gives A sinπ-(1/3)sin2π=0, which imposes no condition on A. Hence y=A sinx-(1/3)sin2x for arbitrary A: there are infinitely many solutions. The free term A sinx is exactly the homogeneous resonance mode.
3. Explain what happens when a damping term εy' is added and ε→0, for both f1 and f2.
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With damping, the equation is y''+εy'+y=f. For f2=sinx, the forcing is at resonance and the steady-state particular amplitude scales like 1/ε, so no finite limit exists as ε→0+. For f1=sin2x, the forcing is off resonance, and the damped particular solution tends to the undamped one as ε→0+. Thus the damping limit recovers the Fredholm conclusion: a finite limit exists exactly when the forcing is orthogonal to sinx. At ε=0 the undamped problem has either infinitely many solutions or none; the damped solutions blow up in the non-orthogonal case.