Solve the PDE partial differential upartial differential t

Solve the P.D.E. partial differential u/partial differential t = k partial differential^2 u/partial differential x^2 - alpha u, -L lessthanorequalto x lessthanorequalto L, t > 0, (with k > 0) subject to the following conditions u(-L, t) = u(L, t) and partial differential u/partial differential x (-L, t) = partial differential u/partial differential x (L, t), u(x, 0) = cos^2 (pi x/L) + sin^2 (2 pi x/L), -L lessthanorequalto x lessthanorequalto L. This problem corresponds to the temperature distribution in a thin circular ring with heat loss through the lateral sides (alpha > 0). How long does it take for the average temperature on the rod to become half of the initial average temperature?

We consider a thin wire (with lateral sides insulated) of length 2L which is bent into the shape of a circle. The model problem is in this case u t = k 2u x2 , L < x < L (1) with the boundary conditions u(L, t) = u(L, t), t > 0 (2) u x(L, t) = u x(L, t), t > 0 (3) and the initial condition u(x, 0) = f(x), L x L (4) We apply the method of separation of variables and seek for product solutions

u(x, t) = (x)G(t) (5) Using similar arguments as in the previous case, we obtain G(t) = cekt

and the boundary value problem for d 2 dx2 = (7) (L) = (L) (8) d dx (L) = d dx (L) (9) The boundary conditions (8) and (9) are referred to as periodic boundary conditions.

up vote1down vote

let us(x)limtu(x,t)us(x)limtu(x,t) ( steady state is time independent)

clearly ust=k2usx2=0ust=k2usx2=0

so the only solutions take the form us(x)=ax+bus(x)=ax+b which can only satisfy the boundary conditions if a=0a=0

us(x)=Const.us(x)=Const.

The specific value you have for the constant makes perfect sense for f(x)=Const.f(x)=Const.

You may be able to justify it in general by appealing to conservation of molecular kinetic energy.

Otherwise you might need to revisit your boundary conditions, if they are truly periodic they should read something like

u(x,t)=u(x+L,t)