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The Heat Equation for Two-Dimensional Objects

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1 The Heat Equation for Two-Dimensional Objects

2 Example A thin iron square 5 m by 4 m is kept at 100 π‘œ C at the bottom 5 m side, and kept at 0 π‘œ C at the other three sides. Find the steady-state solution. Solution Temperature now depends on three variables, π‘₯,𝑦,𝑑: so we write 𝑇=𝑇 π‘₯,𝑦,𝑑 . It satisfies the heat equation: Two-dimensional heat equation 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 𝑑 Definition A steady-state solution is a solution to a partial differential equation with the property that 𝑇 𝑑 =0 --hence it depends only on π‘₯ and 𝑦. If this is the case, then we have The Dirichlet Problem 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 =0 An interesting consequence: the steady-state solution does not depend on the material constant 𝛼.

3 Example A thin iron square 5 m by 4 m is kept at 100 π‘œ C at the bottom 5 m side, and kept at 0 π‘œ C at the other three sides. Find the steady-state solution. Solution 𝑇=𝑇 π‘₯,𝑦 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 =0 Bottom: 𝑇 π‘₯,0 =100 Left & Right: 𝑇 0,𝑦 =𝑇 5,𝑦 =0 Top: 𝑇 π‘₯,4 =0

4 Solution The Dirichlet problem can be solved by separation of variables. 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 =0 𝑇 π‘₯,𝑦 =𝑓 π‘₯ 𝑔 𝑦 𝑇 π‘₯,0 =100 𝑓 β€²β€² π‘₯ 𝑔 𝑦 +𝑓 π‘₯ 𝑔 β€²β€² 𝑦 =0 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑓 β€²β€² π‘₯ 𝑓 π‘₯ =βˆ’ 𝑔 β€²β€² 𝑦 𝑔 𝑦 βˆ’πœ†= 𝑇 π‘₯,4 =0 𝑓 β€²β€² π‘₯ +πœ†π‘“ π‘₯ =0 𝑔 β€²β€² 𝑦 βˆ’πœ†π‘” 𝑦 =0 𝑇 0,𝑦 =0 ⟹ 𝑓 0 𝑔 𝑦 =0 ⟹ 𝑓 0 =0 𝑇 5,𝑦 =0 ⟹ 𝑓 5 𝑔 𝑦 =0 ⟹ 𝑓 5 =0 πœ†= πœ† 𝑛 = 𝑛 2 πœ‹ 2 25 𝑓 π‘₯ = 𝑓 𝑛 π‘₯ = sin π‘›πœ‹ 5 π‘₯

5 Solution Let’s discuss the situation we’re in. We guessed 𝑇 π‘₯,𝑦 =𝑓 π‘₯ 𝑔 𝑦 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 =0 In order to satisfy the the left and right boundary conditions 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑇 π‘₯,0 =100 We were forced to take πœ†= πœ† 𝑛 = 𝑛 2 πœ‹ , and 𝑓 π‘₯ = sin π‘›πœ‹ 5 π‘₯ . We had no choice in the matter. 𝑇 0,𝑦 =𝑇 5,𝑦 =0 Now we’re faced with an upsetting situation: 𝑇 π‘₯,4 =0 𝑔 0 = 100 𝑓 π‘₯ , 𝑔 4 =0 𝑔 β€²β€² 𝑦 βˆ’πœ†π‘” 𝑦 =0 We cannot possibly satisfy 𝑔 0 = 100 𝑓 π‘₯ because a constant cannot be equal to a non-constant. So we will ignore that boundary condition, and deal with it later with superposition. For reasons that will be clear later, we will instead demand that 𝑔 0 =1 and 𝑔 4 =0. Thus, we will have satisfied all our boundary conditions except 𝑇 π‘₯,0 =100. 𝑔 0 =1, 𝑔 4 =0

6 Solution πœ† 𝑛 = 𝑛 2 πœ‹ 2 25 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 =0 𝑔 β€²β€² 𝑦 βˆ’πœ†π‘” 𝑦 =0 𝑔 0 =1, 𝑔 4 =0 𝑇 π‘₯,0 =100 π‘Ÿ 2 βˆ’ 𝑛 2 πœ‹ =0 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑇 π‘₯,4 =0 π‘Ÿ=Β± π‘›πœ‹ 5 𝑔 𝑛 𝑦 = 𝐢 1 𝑒 π‘›πœ‹ 5 𝑦 + 𝐢 2 𝑒 βˆ’ π‘›πœ‹ 5 𝑦 𝑇 𝑛 π‘₯,𝑦 = 𝑓 𝑛 π‘₯ 𝑔 𝑛 𝑦 𝐢 1 + 𝐢 2 =1 𝑇 𝑛 π‘₯,𝑦 = sin π‘›πœ‹ 5 π‘₯ 𝛼 𝑛 𝑒 π‘›πœ‹ 5 𝑦 + 𝛽 𝑛 𝑒 βˆ’ π‘›πœ‹ 5 𝑦 𝐢 1 𝑒 4π‘›πœ‹ 5 + 𝐢 2 𝑒 βˆ’ 4π‘›πœ‹ 5 =0 Where 𝛼 𝑛 = 1 1βˆ’ 𝑒 8π‘›πœ‹ 5 , 𝛽 𝑛 = 𝑒 8π‘›πœ‹ 5 𝑒 8π‘›πœ‹ 5 βˆ’1 𝑇 π‘₯,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin π‘›πœ‹ 5 π‘₯ 𝛼 𝑛 𝑒 π‘›πœ‹ 5 𝑦 + 𝛽 𝑛 𝑒 βˆ’ π‘›πœ‹ 5 𝑦 𝑔 𝑛 𝑦 = 1 1βˆ’ 𝑒 8π‘›πœ‹ 5 𝑒 π‘›πœ‹ 5 𝑦 + 𝑒 8π‘›πœ‹ 5 𝑒 8π‘›πœ‹ 5 βˆ’1 𝑒 βˆ’ π‘›πœ‹ 5 𝑦

7 Solution 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 =0 𝑇 π‘₯,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin π‘›πœ‹ 5 π‘₯ 𝛼 𝑛 𝑒 π‘›πœ‹ 5 𝑦 + 𝛽 𝑛 𝑒 βˆ’ π‘›πœ‹ 5 𝑦 𝑇 π‘₯,0 =100 Where 𝛼 𝑛 = 1 1βˆ’ 𝑒 8π‘›πœ‹ 5 , 𝛽 𝑛 = 𝑒 8π‘›πœ‹ 5 𝑒 8π‘›πœ‹ 5 βˆ’1 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑇 π‘₯,4 =0 Let’s satisfy the boundary condition 𝑇 π‘₯,0 =100 𝑇 π‘₯,0 = 𝑛=1 ∞ 𝑏 𝑛 sin π‘›πœ‹ 5 π‘₯ 𝛼 𝑛 + 𝛽 𝑛 =100 Remember when we had that boundary condition 𝑔 0 = 100 𝑓 π‘₯ that we couldn’t solve? But, miraculously, 𝛼 𝑛 + 𝛽 𝑛 =1. Why? So we arbitrarily made it 𝑔 0 =1 instead? Now we are reaping the benefits of that decision. 𝑛=1 ∞ 𝑏 𝑛 sin π‘›πœ‹ 5 π‘₯ =100, 0≀π‘₯≀5 (Odd 10-periodic function)

8 Solution 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 =0 𝑇 π‘₯,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin π‘›πœ‹ 5 π‘₯ 𝛼 𝑛 𝑒 π‘›πœ‹ 5 𝑦 + 𝛽 𝑛 𝑒 βˆ’ π‘›πœ‹ 5 𝑦 𝑇 π‘₯,0 =100 Where 𝛼 𝑛 = 1 1βˆ’ 𝑒 8π‘›πœ‹ 5 , 𝛽 𝑛 = 𝑒 8π‘›πœ‹ 5 𝑒 8π‘›πœ‹ 5 βˆ’1 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑇 π‘₯,4 =0 It’s interesting to note that The center 2.5 , 2 of the slab is π‘œ C

9 Discussion (Dealing with other boundary conditions)
The technique we just used works for the following situation: f(x) You might ask, what about other boundary conditions? No problem!

10 General Boundary Conditions
𝑓 2 π‘₯ 𝑔 1 𝑦 𝑔 2 𝑦 𝑓 1 π‘₯ Just solve four easier problems: 𝑓 2 π‘₯ 𝑔 1 𝑦 𝑔 2 𝑦 𝑓 1 π‘₯ And add your four solutions together to get the solution to the original problem

11 Discussion We now have a satisfactory technique for finding a steady-state solution for any boundary conditions. But we’re not just interested in steady state solutions. Consider, for example, the following problem:

12 Example A 5mΓ—4m steel slab is initially at 500 π‘œ C throughout when it is subjected to the following boundary conditions: 500 π‘œ C Will the center ever cool to 160 π‘œ C? If so, when? Use 𝛼=9.88Γ— 10 βˆ’5 . Solution Temperature depends on π‘₯,𝑦,𝑑: 𝑇 π‘₯,𝑦,𝑑 . Let’s write 𝑇 π‘₯,𝑦,𝑑 = 𝑇 0 π‘₯,𝑦 + 𝑇 π‘₯,𝑦,𝑑 where 𝑇 0 π‘₯,𝑦 is the steady-state solution. From previous work, we find that the steady state solution 𝑇 0 π‘₯,𝑦 is: 𝑇 0 π‘₯,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin π‘›πœ‹ 5 π‘₯ 𝛼 𝑛 𝑒 π‘›πœ‹ 5 𝑦 + 𝛽 𝑛 𝑒 βˆ’ π‘›πœ‹ 5 𝑦 Where 𝛼 𝑛 = 1 1βˆ’ 𝑒 8π‘›πœ‹ 5 , 𝛽 𝑛 = 𝑒 8π‘›πœ‹ 5 𝑒 8π‘›πœ‹ 5 βˆ’1 𝑏 𝑛 = 𝑓 π‘₯ sin 2π‘›πœ‹ 10 π‘₯ 𝑓 π‘₯ = ≀π‘₯≀5 βˆ’500 5≀π‘₯≀10

13 Solution Let’s write 𝑇 π‘₯,𝑦,𝑑 = 𝑇 0 π‘₯,𝑦 + 𝑇 π‘₯,𝑦,𝑑 where 𝑇 0 π‘₯,𝑦 is the steady-state solution. 𝑇 0 π‘₯,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin π‘›πœ‹ 5 π‘₯ 𝛼 𝑛 𝑒 π‘›πœ‹ 5 𝑦 + 𝛽 𝑛 𝑒 βˆ’ π‘›πœ‹ 5 𝑦 Substitute 𝑇 π‘₯,𝑦,𝑑 = 𝑇 0 π‘₯,𝑦 + 𝑇 π‘₯,𝑦,𝑑 into the 2D-heat equation 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 𝑑 Where 𝛼 𝑛 = 1 1βˆ’ 𝑒 8π‘›πœ‹ 5 , 𝛽 𝑛 = 𝑒 8π‘›πœ‹ 5 𝑒 8π‘›πœ‹ 5 βˆ’1 𝑇 0 π‘₯π‘₯ + 𝑇 0 𝑦𝑦 + 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 0 𝑑 + 1 𝛼 𝑇 𝑑 𝑏 𝑛 = 𝑓 π‘₯ sin 2π‘›πœ‹ 10 π‘₯ 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 𝑑 𝑓 π‘₯ = ≀π‘₯≀5 βˆ’500 5≀π‘₯≀10 Since 𝑇 0 satisfies the inhomogeneous boundary conditions, 𝑇 has to satisfy homogeneous boundary conditions. 𝑇 π‘₯,𝑦,0 =500 since the slab starts at 500 π‘œ C 𝑇 π‘₯,𝑦,0 + 𝑇 0 π‘₯,𝑦 =500 𝑇 π‘₯,𝑦,0 =500βˆ’ 𝑇 0 π‘₯,𝑦

14 Solution 𝑇 π‘₯,𝑦,𝑑 =𝑓 π‘₯ 𝑔 𝑦 β„Ž(𝑑) 𝑇 π‘₯π‘₯ + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 𝑑 𝑓 β€²β€² π‘₯ 𝑔 𝑦 β„Ž 𝑑 +𝑓 π‘₯ 𝑔 β€²β€² 𝑦 β„Ž 𝑑 = 1 𝛼 𝑓 π‘₯ 𝑔 𝑦 β„Ž β€² 𝑑 𝑇 π‘₯,𝑦,0 =500βˆ’ 𝑇 0 π‘₯,𝑦 𝑓 β€²β€² π‘₯ 𝑓 π‘₯ + 𝑔 β€²β€² 𝑦 𝑔 𝑦 = 1 𝛼 β„Ž β€² 𝑑 β„Ž 𝑑 =βˆ’πœ† 0 π‘œπ‘› π‘‘β„Žπ‘’ π‘π‘œπ‘’π‘›π‘‘π‘Žπ‘Ÿπ‘–π‘’π‘  𝑓 β€²β€² π‘₯ 𝑓 π‘₯ + 𝑔 β€²β€² 𝑦 𝑔 𝑦 =βˆ’πœ† π‘₯=0,π‘₯=5,𝑦=0,𝑦=4 1 𝛼 β„Ž β€² 𝑑 β„Ž 𝑑 =βˆ’πœ† 𝑓 β€²β€² π‘₯ 𝑓 π‘₯ =βˆ’πœ‡ πœ† 𝑛,π‘š = πœ‡ 𝑛 + 𝜈 π‘š = 𝑛 2 πœ‹ π‘š 2 πœ‹ 𝑔 β€²β€² 𝑦 𝑔 𝑦 =βˆ’πœˆ β„Ž 𝑛,π‘š β€² 𝑑 +𝛼 πœ† 𝑛,π‘š β„Ž 𝑛,π‘š 𝑑 =0 β„Ž 𝑛,π‘š 𝑑 = 𝑒 βˆ’π›Ό πœ† 𝑛,π‘š 𝑑 πœ‡+𝜈=πœ† 𝑇 π‘₯,𝑦,𝑑 = 𝑛=1 ∞ π‘š=1 ∞ 𝑏 𝑛,π‘š 𝑓 𝑛 π‘₯ 𝑔 π‘š 𝑦 β„Ž 𝑛,π‘š 𝑑 𝑓 β€²β€² π‘₯ +πœ‡π‘“ π‘₯ =0, homog. at 0 and 5 𝑔 β€²β€² 𝑦 +πœˆπ‘” 𝑦 =0, homog at 0 and 4 πœ‡= πœ‡ 𝑛 = 𝑛 2 πœ‹ 𝑓 𝑛 π‘₯ = sin π‘›πœ‹ 5 π‘₯ 𝜈= 𝜈 π‘š = π‘š 2 πœ‹ 𝑔 π‘š 𝑦 = sin π‘›πœ‹ 4 𝑦

15 Solution 𝑇 π‘₯,𝑦,0 =500βˆ’ 𝑇 0 π‘₯,𝑦 𝑇 π‘₯,𝑦,𝑑 = 𝑛=1 ∞ π‘š=1 ∞ 𝑏 𝑛,π‘š 𝑓 𝑛 π‘₯ 𝑔 π‘š 𝑦 β„Ž 𝑛,π‘š 𝑑 𝑛=1 ∞ π‘š=1 ∞ 𝑏 𝑛,π‘š sin π‘›πœ‹π‘₯ 5 sin π‘šπœ‹π‘¦ =500βˆ’ 𝑇 0 π‘₯,𝑦 β€œFix 𝑦” 𝑛=1 ∞ π‘š=1 ∞ 𝑏 𝑛,π‘š sin π‘šπœ‹ 𝑦 sin π‘›πœ‹π‘₯ 5 =500βˆ’ 𝑇 0 π‘₯, 𝑦 0


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