Someone tries to build the Green function for on with the insulated (Neumann) conditions

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Partial differential equations and boundary-value problems · Partial differential equations and boundary-value problems

Conditions of Validity, Ch Green functions: inverting the operator

Problem

Someone tries to build the Green function for −u′′=f(x)-u'' = f(x) on [0,L][0, L] with the insulated (Neumann) conditions u′(0)=u′(L)=0u'(0) = u'(L) = 0, for a rod of length L=0.60 mL = 0.60\,\mathrm{m} carrying the uniform source f(x)=4.0 K m−2f(x) = 4.0\,\mathrm{K\,m^{-2}} (the source divided by the conductivity, so that p=1p = 1). Their working: "Take ya(x)=1y_a(x) = 1, which satisfies ya′(0)=0y_a'(0) = 0, and yb(x)=1y_b(x) = 1, which satisfies yb′(L)=0y_b'(L) = 0. The matching formula gives G(x,ξ)=−ya(x<)yb(x>)/WG(x,\xi) = -y_a(x_<)y_b(x_>)/W with W=yayb′−ya′ybW = y_ay_b' - y_a'y_b. Substituting ya=yb=1y_a = y_b = 1 gives W=0W = 0; as with the taut string, where the Wronskian W=−LW = -L is just a constant setting the scale, replace it by the constant W=−1/LW = -1/L, which has the right dimensions. Then G(x,ξ)=LG(x,\xi) = L for all x,ξx, \xi, andu(x)=∫0LG(x,ξ)f(ξ) dξ=L∫0Lf(ξ) dξ=1.44 Ku(x) = \int_0^L G(x,\xi)f(\xi)\,\mathrm{d}\xi = L\int_0^L f(\xi)\,\mathrm{d}\xi = 1.44\,\mathrm{K}solves the problem: a constant temperature of 1.44 K1.44\,\mathrm{K} everywhere along the rod." ((a)) Identify the specific step where this argument fails, and explain why W=0W = 0 cannot be patched by a rescaling convention here. ((b)) Derive, directly from the differential equation and the boundary conditions, the condition that ff must satisfy for a steady solution to exist at all, and check whether the stated f(x)=4.0 K m−2f(x) = 4.0\,\mathrm{K\,m^{-2}} meets it. ((c)) State what object should be used instead of an ordinary Green function when this condition fails, and why no rescaling of WW can substitute for it.

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