The Heat Equation for Two-Dimensional Objects

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

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

Solution The Dirichlet problem can be solved by separation of variables. 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 =0 𝑇 𝑥,𝑦 =𝑓 𝑥 𝑔 𝑦 𝑇 𝑥,0 =100 𝑓 ′′ 𝑥 𝑔 𝑦 +𝑓 𝑥 𝑔 ′′ 𝑦 =0 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑓 ′′ 𝑥 𝑓 𝑥 =− 𝑔 ′′ 𝑦 𝑔 𝑦 −𝜆= 𝑇 𝑥,4 =0 𝑓 ′′ 𝑥 +𝜆𝑓 𝑥 =0 𝑔 ′′ 𝑦 −𝜆𝑔 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 𝜆= 𝜆 𝑛 = 𝑛 2 𝜋 2 25 𝑓 𝑥 = 𝑓 𝑛 𝑥 = sin 𝑛𝜋 5 𝑥

Solution Let’s discuss the situation we’re in. We guessed 𝑇 𝑥,𝑦 =𝑓 𝑥 𝑔 𝑦 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 =0 In order to satisfy the the left and right boundary conditions 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑇 𝑥,0 =100 We were forced to take 𝜆= 𝜆 𝑛 = 𝑛 2 𝜋 2 25 , and 𝑓 𝑥 = sin 𝑛𝜋 5 𝑥 . We had no choice in the matter. 𝑇 0,𝑦 =𝑇 5,𝑦 =0 Now we’re faced with an upsetting situation: 𝑇 𝑥,4 =0 𝑔 0 = 100 𝑓 𝑥 , 𝑔 4 =0 𝑔 ′′ 𝑦 −𝜆𝑔 𝑦 =0 We cannot possibly satisfy 𝑔 0 = 100 𝑓 𝑥 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 𝑥,0 =100. 𝑔 0 =1, 𝑔 4 =0

Solution 𝜆 𝑛 = 𝑛 2 𝜋 2 25 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 =0 𝑔 ′′ 𝑦 −𝜆𝑔 𝑦 =0 𝑔 0 =1, 𝑔 4 =0 𝑇 𝑥,0 =100 𝑟 2 − 𝑛 2 𝜋 2 25 =0 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑇 𝑥,4 =0 𝑟=± 𝑛𝜋 5 𝑔 𝑛 𝑦 = 𝐶 1 𝑒 𝑛𝜋 5 𝑦 + 𝐶 2 𝑒 − 𝑛𝜋 5 𝑦 𝑇 𝑛 𝑥,𝑦 = 𝑓 𝑛 𝑥 𝑔 𝑛 𝑦 𝐶 1 + 𝐶 2 =1 𝑇 𝑛 𝑥,𝑦 = sin 𝑛𝜋 5 𝑥 𝛼 𝑛 𝑒 𝑛𝜋 5 𝑦 + 𝛽 𝑛 𝑒 − 𝑛𝜋 5 𝑦 𝐶 1 𝑒 4𝑛𝜋 5 + 𝐶 2 𝑒 − 4𝑛𝜋 5 =0 Where 𝛼 𝑛 = 1 1− 𝑒 8𝑛𝜋 5 , 𝛽 𝑛 = 𝑒 8𝑛𝜋 5 𝑒 8𝑛𝜋 5 −1 𝑇 𝑥,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin 𝑛𝜋 5 𝑥 𝛼 𝑛 𝑒 𝑛𝜋 5 𝑦 + 𝛽 𝑛 𝑒 − 𝑛𝜋 5 𝑦 𝑔 𝑛 𝑦 = 1 1− 𝑒 8𝑛𝜋 5 𝑒 𝑛𝜋 5 𝑦 + 𝑒 8𝑛𝜋 5 𝑒 8𝑛𝜋 5 −1 𝑒 − 𝑛𝜋 5 𝑦

Solution 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 =0 𝑇 𝑥,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin 𝑛𝜋 5 𝑥 𝛼 𝑛 𝑒 𝑛𝜋 5 𝑦 + 𝛽 𝑛 𝑒 − 𝑛𝜋 5 𝑦 𝑇 𝑥,0 =100 Where 𝛼 𝑛 = 1 1− 𝑒 8𝑛𝜋 5 , 𝛽 𝑛 = 𝑒 8𝑛𝜋 5 𝑒 8𝑛𝜋 5 −1 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑇 𝑥,4 =0 Let’s satisfy the boundary condition 𝑇 𝑥,0 =100 𝑇 𝑥,0 = 𝑛=1 ∞ 𝑏 𝑛 sin 𝑛𝜋 5 𝑥 𝛼 𝑛 + 𝛽 𝑛 =100 Remember when we had that boundary condition 𝑔 0 = 100 𝑓 𝑥 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 𝑛𝜋 5 𝑥 =100, 0≤𝑥≤5 (Odd 10-periodic function)

Solution 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 =0 𝑇 𝑥,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin 𝑛𝜋 5 𝑥 𝛼 𝑛 𝑒 𝑛𝜋 5 𝑦 + 𝛽 𝑛 𝑒 − 𝑛𝜋 5 𝑦 𝑇 𝑥,0 =100 Where 𝛼 𝑛 = 1 1− 𝑒 8𝑛𝜋 5 , 𝛽 𝑛 = 𝑒 8𝑛𝜋 5 𝑒 8𝑛𝜋 5 −1 𝑇 0,𝑦 =𝑇 5,𝑦 =0 𝑇 𝑥,4 =0 It’s interesting to note that The center 2.5 , 2 of the slab is 32.6 𝑜 C

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!

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

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:

Example A 5m×4m steel slab is initially at 500 𝑜 C throughout when it is subjected to the following boundary conditions: 500 𝑜 C Will the center ever cool to 160 𝑜 C? If so, when? Use 𝛼=9.88× 10 −5 . Solution Temperature depends on 𝑥,𝑦,𝑡: 𝑇 𝑥,𝑦,𝑡 . Let’s write 𝑇 𝑥,𝑦,𝑡 = 𝑇 0 𝑥,𝑦 + 𝑇 𝑥,𝑦,𝑡 where 𝑇 0 𝑥,𝑦 is the steady-state solution. From previous work, we find that the steady state solution 𝑇 0 𝑥,𝑦 is: 𝑇 0 𝑥,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin 𝑛𝜋 5 𝑥 𝛼 𝑛 𝑒 𝑛𝜋 5 𝑦 + 𝛽 𝑛 𝑒 − 𝑛𝜋 5 𝑦 Where 𝛼 𝑛 = 1 1− 𝑒 8𝑛𝜋 5 , 𝛽 𝑛 = 𝑒 8𝑛𝜋 5 𝑒 8𝑛𝜋 5 −1 𝑏 𝑛 = 2 10 0 10 𝑓 𝑥 sin 2𝑛𝜋 10 𝑥 𝑓 𝑥 = 500 0≤𝑥≤5 −500 5≤𝑥≤10

Solution Let’s write 𝑇 𝑥,𝑦,𝑡 = 𝑇 0 𝑥,𝑦 + 𝑇 𝑥,𝑦,𝑡 where 𝑇 0 𝑥,𝑦 is the steady-state solution. 𝑇 0 𝑥,𝑦 = 𝑛=1 ∞ 𝑏 𝑛 sin 𝑛𝜋 5 𝑥 𝛼 𝑛 𝑒 𝑛𝜋 5 𝑦 + 𝛽 𝑛 𝑒 − 𝑛𝜋 5 𝑦 Substitute 𝑇 𝑥,𝑦,𝑡 = 𝑇 0 𝑥,𝑦 + 𝑇 𝑥,𝑦,𝑡 into the 2D-heat equation 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 𝑡 Where 𝛼 𝑛 = 1 1− 𝑒 8𝑛𝜋 5 , 𝛽 𝑛 = 𝑒 8𝑛𝜋 5 𝑒 8𝑛𝜋 5 −1 𝑇 0 𝑥𝑥 + 𝑇 0 𝑦𝑦 + 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 0 𝑡 + 1 𝛼 𝑇 𝑡 𝑏 𝑛 = 2 10 0 10 𝑓 𝑥 sin 2𝑛𝜋 10 𝑥 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 𝑡 𝑓 𝑥 = 500 0≤𝑥≤5 −500 5≤𝑥≤10 Since 𝑇 0 satisfies the inhomogeneous boundary conditions, 𝑇 has to satisfy homogeneous boundary conditions. 𝑇 𝑥,𝑦,0 =500 since the slab starts at 500 𝑜 C 𝑇 𝑥,𝑦,0 + 𝑇 0 𝑥,𝑦 =500 𝑇 𝑥,𝑦,0 =500− 𝑇 0 𝑥,𝑦

Solution 𝑇 𝑥,𝑦,𝑡 =𝑓 𝑥 𝑔 𝑦 ℎ(𝑡) 𝑇 𝑥𝑥 + 𝑇 𝑦𝑦 = 1 𝛼 𝑇 𝑡 𝑓 ′′ 𝑥 𝑔 𝑦 ℎ 𝑡 +𝑓 𝑥 𝑔 ′′ 𝑦 ℎ 𝑡 = 1 𝛼 𝑓 𝑥 𝑔 𝑦 ℎ ′ 𝑡 𝑇 𝑥,𝑦,0 =500− 𝑇 0 𝑥,𝑦 𝑓 ′′ 𝑥 𝑓 𝑥 + 𝑔 ′′ 𝑦 𝑔 𝑦 = 1 𝛼 ℎ ′ 𝑡 ℎ 𝑡 =−𝜆 0 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑖𝑒𝑠 𝑓 ′′ 𝑥 𝑓 𝑥 + 𝑔 ′′ 𝑦 𝑔 𝑦 =−𝜆 𝑥=0,𝑥=5,𝑦=0,𝑦=4 1 𝛼 ℎ ′ 𝑡 ℎ 𝑡 =−𝜆 𝑓 ′′ 𝑥 𝑓 𝑥 =−𝜇 𝜆 𝑛,𝑚 = 𝜇 𝑛 + 𝜈 𝑚 = 4 2 𝑛 2 𝜋 2 + 5 2 𝑚 2 𝜋 2 4 2 5 2 𝑔 ′′ 𝑦 𝑔 𝑦 =−𝜈 ℎ 𝑛,𝑚 ′ 𝑡 +𝛼 𝜆 𝑛,𝑚 ℎ 𝑛,𝑚 𝑡 =0 ℎ 𝑛,𝑚 𝑡 = 𝑒 −𝛼 𝜆 𝑛,𝑚 𝑡 𝜇+𝜈=𝜆 𝑇 𝑥,𝑦,𝑡 = 𝑛=1 ∞ 𝑚=1 ∞ 𝑏 𝑛,𝑚 𝑓 𝑛 𝑥 𝑔 𝑚 𝑦 ℎ 𝑛,𝑚 𝑡 𝑓 ′′ 𝑥 +𝜇𝑓 𝑥 =0, homog. at 0 and 5 𝑔 ′′ 𝑦 +𝜈𝑔 𝑦 =0, homog at 0 and 4 𝜇= 𝜇 𝑛 = 𝑛 2 𝜋 2 5 2 𝑓 𝑛 𝑥 = sin 𝑛𝜋 5 𝑥 𝜈= 𝜈 𝑚 = 𝑚 2 𝜋 2 4 2 𝑔 𝑚 𝑦 = sin 𝑛𝜋 4 𝑦

Solution 𝑇 𝑥,𝑦,0 =500− 𝑇 0 𝑥,𝑦 𝑇 𝑥,𝑦,𝑡 = 𝑛=1 ∞ 𝑚=1 ∞ 𝑏 𝑛,𝑚 𝑓 𝑛 𝑥 𝑔 𝑚 𝑦 ℎ 𝑛,𝑚 𝑡 𝑛=1 ∞ 𝑚=1 ∞ 𝑏 𝑛,𝑚 sin 𝑛𝜋𝑥 5 sin 𝑚𝜋𝑦 4 =500− 𝑇 0 𝑥,𝑦 “Fix 𝑦” 𝑛=1 ∞ 𝑚=1 ∞ 𝑏 𝑛,𝑚 sin 𝑚𝜋 𝑦 0 4 sin 𝑛𝜋𝑥 5 =500− 𝑇 0 𝑥, 𝑦 0