METO 621Prototype Problem I: Differential Equation ApproachPrototype Problem ISlide 4Slide 5Slide 6Slide 7Two stream solution for uniform illumination (Problem 1)Prototype problem 2Slide 10Slide 11Two-stream solution for an imbedded sourceSource function, S, divided by BPrototype problem 3Slide 15Slide 16Slide 17Eddington ApproximationSlide 19Slide 20Slide 21METO 621Lesson 14Prototype Problem I: Differential Equation Approach• In this problem we will ignore the thermal emission term•First add and subtract the two two-stream equations)()())(1()(−+−+−+−+−=++−=−IIdIIdIIadIIdτμτμPrototype Problem I•Differentiating the second equation with respect to  and substituting for d(I+-I-)/d from the first equation we get)()1()(222−+−++−=+IIadIIdμτ• Similarly, differentiate the first equation, and substitute for d(I+-I-)/d)()1()(222−+−+−−=−IIadIIdμτPrototype Problem I•We have the same differential equation to solve for both quantities. Calling the unknown Y, we obtain a simple second order diffusion equation.μτ/)a-1( where222≡ΓΓ= YdYd• for which the general solution is a sum of positive and negative exponentialsττ ''''Γ−Γ+= eBeAY• A’ and B’ are arbitrary constants to be determinedPrototype Problem I•Since the sum and differences of the two intensities are both expressed as sums of exponentials, each intensity component must also be expressed in the same way. ττττττΓ−Γ−Γ−Γ++=+= DeCeIBeAeI )( ,)(• where A, B, C.and D are additional arbitrary constants.• We now introduce boundary conditions at the top and bottom of the medium. For prototype problem 1 these are0*)( constant )0( ===+−ττ IIPrototype Problem I•The solution of this two stream problem has the simplest analytic form.•The equation shows four constants of integration, but in fact only two of these are independent.][)(][)(are solutions the1122)*(2)*()*()*(ττττττττρτρτρμμμ−Γ−∞−Γ−−Γ−−Γ∞+∞−=−==Γ+Γ−=Γ+−==eeIIeeIIaaDBACDDPrototype Problem I*2* whereττρΓ−∞Γ−≡ eeDTwo stream solution for uniform illumination (Problem 1)Parameters I=1.0, τ*=1.0, a=0.4μ=0.5 , p=1Downward flux = solid lineUpward flux = dotted lineMean intensity = dashed lineParameters as above except a=1.0Prototype problem 2•Consider the only source of radiation is thermal emission within the slab. The two stream equations are 0*)()0( conditionsboundary with the)1()(2)(2)()()1()(2)(2)()(==−−−−=−−−−−=+−−+−−−−++++ττττττμτττττμIIBaIaIaIddIBaIaIaIddI• Note the extra inhomogeneous term on the RHSPrototype problem 2•Solving these simultaneous equation starts by seeking the homogeneous solution, and then a particular equation that satisfies the whole equation – using the boundary conditions. We getττττρτρτΓ−Γ∞−Γ−∞Γ++=+= DeAeIDeAeI )( ,)(• The particular solution is obtained by guessing that I+=B and I-=B are solutionsPrototype problem 2•This gives for the intensities{ }{ }BeeeeBIBeeeeBI+−+−=+−+−=ΓΓ−∞−Γ−Γ−∞−−Γ−Γ−∞ΓΓ−∞+][)(][)()*()*(2)*()*(2ττττττττττττρρτρρτDDTwo-stream solution for an imbedded sourceParameters B=100, τ*=1.0, a=0.4, μ=0.5 , p=1Downward flux = solid lineUpward flux = dotted lineMean intensity = dashed lineParameters as above except a=1.0Source function, S, divided by BPrototype problem 3•Assume an isotropically scattering homogeneous atmosphere with a black lower boundary. The appropriate two-stream equations are00//422422μτμτπτμπτμ−−+−−−−+++−−−=−−−−=eFaIaIaIddIeFaIaIaIddISddddSddddPrototype problem 3•As before we differentiate and substitute into the equations and get two simultaneous equations0/2224))(1()(μτπτμ−−+−+−+−=+eFaIIadIIdSdddd0/2224))(1()(μτπτμ−−+−+−−−=−eFaIIadIIdSddddPrototype problem 3Prototype problem 3•Using the same solution method as for problem 2 we consider the homogeneous solutionττττρρΓ−Γ∞−Γ−∞Γ++=+= DeAeIDeAeIdd ,• We now guess that a particular solution will be proportional to exp(-We get00//Z Z μτττμτττρρ−−Γ−Γ∞−−+Γ−∞Γ+++=++=eDeAeIeDeAeIdd• Z+ and Z- can be determined by substitutionEddington Approximation•Two stream approximations are used primarily to compute fluxes and mean intensities in plane geometry. These quantities depend only on the azimuthally averaged radiation field. We are interested in solutions valid for anisotropic scattering),()',(),'('2),(),(*11uSuIuupduauIdudIudddτττττ−−=∫−Eddington Approximation•Another approach is to approximate the angular dependence of the intensity by a polynomial in u. •We choose I(,u)=uI1()•This approach is referred to as the Eddington approximation. Upon substitution we get0/010111010),(4 )'(),'('2)(μτμπτ−−−−+−+=+∫eupaFIuIuupduauIIduIIduSEddington Approximation•Remember that the phase function can be expanded in terms of Legendre polynomialsglluPuPuupduuPuPluuplllllllll=====−=∫∑−∞=χχχχχ 1for ,1 for )( )'(),'('21bygiven are moments thewhere)'()()12(),'(11Eddington Approximation•If we only retain these first two terms then the equation becomes∫∫−−=〉〈〉〈+=+1122120111021