Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: GENERAL AND PARTICULAR SOLUTIONS◮An infrared laser delivers a pulse of (heat) energy E to a point P on a large insulatedsheet of thickness b, thermal conductivity k, specific heat s and density ρ. The sheet isinitially at a uniform temperature. If u(r, t) is the excess temperature a time t later, at apoint that is a distance r (≫ b) from P , then show that a suitable expression for u isu(r, t) = α t exp (− r22βt), (20.37)where α and β are constants. (Note that we use r instead of ρ to denote the radial coordinatein plane polars so as to avoid confusion with the density.)Further, (i) show that β =2k/(sρ); (ii) demonstrate that the excess heat energy in thesheet is independent of t, and hence evaluate α; and (iii) prove that the total heat flow pastany circle of radius r is E.The equation to be solved is the heat diffusion equationk∇ 2 u(r,t)=sρ ∂u(r,t) .∂tSince we only require the solution for r ≫ b we can treat the problem as two-dimensionalwith obvious circular symmetry. Thus only the r-derivative term in the expression for ∇ 2 uis non-zero, giving(k ∂r ∂u )= sρ ∂ur ∂r ∂r ∂t , (20.38)where now u(r,t)=u(r, t).(i) Substituting the given expression (20.37) into (20.38) we obtain( ))2kα r2βt 2 2βt − 1 exp(− r2= sρα ( ) r22βt t 2 2βt − 1 exp(− r22βtfrom which we find that (20.37) is a solution, provided β =2k/(sρ).(ii) The excess heat in the system at any time t is∫ ∞∫ ∞( )rbρs u(r, t)2πr dr =2πbρsα00 t exp − r2dr2βt=2πbρsαβ.The excess heat is therefore independent of t and so must be equal to the total heatinput E, implying thatEα =2πbρsβ = E4πbk .(iii) The total heat flow past a circle of radius r is−2πrbk∫ ∞0∂u(r, t)dt = −2πrbk∂r[= E exp∫ ∞0(− r2E4πbkt),( ))−rexp(− r2dtβt 2βt∞= E for all r.2βt)]0As we would expect, all the heat energy E deposited by the laser will eventually flow pasta circle of any given radius r. ◭698