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Introductory Physics Homework Help
Solve initial-value problem for heat equation and find relaxation time
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[QUOTE="mmo115, post: 2005942, member: 159954"] [h2]Homework Statement [/h2] Solve the initial-value problem for the heat equation u[SUB]t[/SUB] = K[tex]\nabla[/tex][SUP]2[/SUP]u in the column 0< x < L[SUB]1[/SUB], 0< y < L[SUB]2[/SUB] with the boundary conditions u(0,y;t)=0, u[SUB]x[/SUB](L[SUB]1[/SUB],y,t)=0, u(x,0;t)=0, u[SUB]y[/SUB](x,L[SUB]2[/SUB];t)=0 and the initial condition u(x,y;0)=1. Find the relaxation time. Can anyone please explain how to get this solution.. I really don't understand how to arrive at the solution. I'm hoping that if i can learn specifically how to do this i can apply it to other similar problems. I have searched around the internet for some information, but it seems like most of it is using actual situations as opposed to theoretical. i mean the book doesn't even explain what relaxation time is or how to derive it.. =/ [h2]Homework Equations[/h2] We are told that : where m,n=1 to infinity u(x,y;t)= [tex]\Sigma[/tex] B[SUB]mn[/SUB]sin (m*pi*x / L[SUB]1[/SUB]) sin (n*pi*y)/L[SUB]2[/SUB]) * e^-lambda[SUB]mn[/SUB]Kt We can use this to solve initial-value problems for the heat equation.[h2]The Attempt at a Solution[/h2] I really don't get how to solve for the B[SUB]mn[/SUB]... the book really doesn't give a good explanation. The solution is u(x,y;t) = 4/pi[SUP]2[/SUP] [tex]\Sigma[/tex][SUB]m,n=1[/SUB] [ sin[(m-(1/2))([tex]\pi[/tex]x/L[SUB]1[/SUB])] / (m-1/2) ] * [ sin[n-1/2)(pi y/L[SUB]2[/SUB])\ / (n-1/2) ] * [ e^-lambda[SUB]mn[/SUB]Kt ] lambda[SUB]mn[/SUB]= (m-1/2)[SUP]2[/SUP] (pi/L[SUB]1[/SUB])[SUP]2[/SUP] + (n-1/2)[SUP]2[/SUP](pi/L[SUB]1[/SUB])[SUP]2[/SUP] relaxation time = (4/pi[SUP]2[/SUP] K)[L1[SUP]2[/SUP]L2[SUP]2[/SUP]/L1[SUP]2[/SUP]+L2[SUP]2[/SUP])] [/QUOTE]
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Solve initial-value problem for heat equation and find relaxation time
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