X 1 would be a better choice of general solution to x

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x1would be a better choice of general solution tox2n′′xn22x2nx0for then it would yieldu2x,yn1Enxsinhn21xFnxsinhn1x1nyand nowu21/2,yn112Fnsinhnnyg1yyielding12sinhnFn01g1ynydy01n2ydy201g1ynydyorFn4sinhn01g1ynydywithout the need to solve a system of two equations, andu21,yn1Ensinhnnyg2yyieldingsinhnEn01g2ynydy01n2ydy201g2ynydyorEn2sinhn01g2ynydyalso without the need to solve a system of two equations. Thus we would haveu2x,yn1Enxsinhn21xFnxsinhn1x1nywithEn2sinhn01g2ynydyandFn4sinhn01g1ynydy.14
It is very much like using sinhnyand sinhn1yis the solution foru1x,y.———————————————————————————————————————Solution to Problem#3Forux,y,z,t, we have2ux,y,z,tx22ux,y,z,ty22ux,y,z,tz21v22ux,y,z,tt2in the boxed regionRx,y,z|0xa,0yb,0zc,given thatux,y,z,tsatisfies the sixboundary conditionsof the form:u0,y,z,t0,ua,y,z,t0andux,0,z,t0,ux,b,z,t0andux,y,0,t0,ux,y,c,t0and the twoinitial conditions:ux,y,z,0Fx,y,z,ux,y,z,ttt0Gx,y,z.Now we letux,y,zxyzt, and get′′xyztx′′yztxy′′zt1v2xyz′′tor′′xx′′yy′′zz′′tv2t.This says that′′xxc1,′′yyc2,′′zzc3,′′tc2tc4withc1c2c3c4. Using the boundary condition, these become′′xc1x0with00anda0and′′yc2y0with00andb0and′′zc3z0with00andc0.These lead toc1ka2andkxsinkxaandc2mb2andmysinmyband15
c3nc2andnzsinnzcandc4ka2mb2nc2kmn2so that′′tc4v2t0orkmn′′tvkmn2kmnt0.Then we havekmntGkmncosvkmntHkmnsinvkmntandwkmnx,y,z,tGkmncosvkmntHkmnsinvkmntsinkxasinmybsinnzcandwx,y,z,tk1m1n1GkmncosvkmntHkmnsinvkmntsinkxasinmybsinnzc.The first initial condition yieldsFx,y,zk1m1n1Gkmnsinkxasinmybsinnzc1so thatGkmn8abc0c0b0aFx,y,zsinkxasinmybsinnzcdxdydz8abc0c0b0asinkxasinmybsinnzcdxdydzwhich reduces toGkmn8abc0asinkxadx0bsinm

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Term
Fall
Professor
RIMMER
Tags
Boundary value problem, Partial differential equation, M Carchidi, u x, z

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