16.ut=kuxxux(0, t) = 0ux(L, t) = 0u(x,0) =f(x)u(x, t) =∞n=0uncosn π xLe−k(n πL)2tu(x,0) =∞n=0uncosn π xL≡f(x)unare the coeﬃcients of expandingf(x) in terms of Fourier cosine series.u0=1LL0f(x)dxun=2LL0f(x) cosnπLxdxb.The equilibrium is whenuxx= 0 subject to the same boundary conditions. The solutionis then obtained by integration with respect toxux=KandK= 0 because of the boundary conditions.Now integrate again to getu=C. This means that the temperature is constant. There is no other condition to fix this
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