This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Partial Differential Equations If the subject of ordinary differential equations is large, this is enormous. I am going to examine only one corner of it, and will develop only one tool to handle it: Separation of Variables. Another major tool is the method of characteristics and I’ll not go beyond mentioning the word. When I develop a technique to handle the heat equation or the potential equation, don’t think that it stops there. The same set of tools will work on the Schroedinger equation in quantum mechanics and on the wave equation in its many incarnations. 10.1 The Heat Equation The flow of heat in one dimension is described by the heat conduction equation P = κA ∂T ∂x (10 . 1) where P is the power in the form of heat energy flowing toward positive x through a wall and A is the area of the wall. κ is the wall’s thermal conductivity. Put this equation into words and it says that if a thin slab of material has a temperature on one side different from that on the other, then heat energy will flow through the slab. If the temperature difference is big or the wall is thin ( ∂T/∂x is big) then there’s a big flow. The minus sign says that the energy flows from hot toward cold. When more heat comes into a region than leaves it, the temperature there will rise. This is described by the specific heat, c . dQ = mcdT, or dQ dt = mc dT dt (10 . 2) Again in words, the temperature rise in a chunk of material is proportional to the amount of heat added to it and inversely proportional to its mass. P ( x,t ) x A P ( x + Δ x,t ) x + Δ x For a slab of area A , thickness Δ x , and mass density ρ , let the coordinates of the two sides be x and x + Δ x . m = ρA Δ x, and dQ dt = P ( x,t ) P ( x + Δ x,t ) The net power into this volume is the power in from one side minus the power out from the other. Put these three equations together. dQ dt = mc dT dt = ρA Δ xc dT dt = κA ∂T ( x,t ) ∂x + κA ∂T ( x + Δ x,t ) ∂x If you let Δ x → here, all you get is 0 = 0 , not very helpful. Instead divide by Δ x first and then take the limit. ∂T ∂t = + κA ρcA ∂T ( x + Δ x,t ) ∂x ∂T ( x,t ) ∂x 1 Δ x James Nearing, University of Miami 1 10—Partial Differential Equations 2 and in the limit this is ∂T ∂t = κ cρ ∂ 2 T ∂x 2 (10 . 3) I was a little cavalier with the notation in that I didn’t specify the argument of T on the left side. You could say that it was ( x + Δ x/ 2 ,t ) , but in the limit everything is evaluated at ( x,t ) anyway. I also assumed that κ , the thermal conductivity, is independent of x . If not, then it stays inside the derivative, ∂T ∂t = 1 cρ ∂ ∂x κ ∂T ∂x (10 . 4) In Three Dimensions In three dimensions, this becomes ∂T ∂t = κ cρ ∇ 2 T (10 . 5) Roughly speaking, the temperature in a box can change because of heat flow in any of three directions....
View
Full
Document
This note was uploaded on 01/30/2012 for the course PHYS 315 taught by Professor Nearing during the Fall '08 term at University of Miami.
 Fall '08
 Nearing

Click to edit the document details