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Mathematics
Course
MATH 251
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Unknown
Semester
Spring

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Second Order Linear Partial Differential Equations Part III One▯dimensional Heat Conduction Equation revisited; temperature distribution of a bar with insulated ends; nonhomogeneous boundary conditions; temperature distribution of a bar with ends kept at arbitrary temperatures; steady▯state solution Previously, we have learned that the general solution of a partial differential equation is dependent of boundary conditions. Thesame equation will have different general solutions under different sets of boundary conditions. We shall witness this fact, by examining additional examples of heat conduction problems with new sets of boundary conditions. Keep in mind that, throughout this section, we will be solving the same partial differential equation, the homogeneous one▯dimensional heat conduction equation: 2 α u =xx t where u(x,t) is the temperature distribution function of a thin bar, which has 2 length L, and the positive constant α is the thermo diffusivity constant of the bar. The equation will now be paired up with new sets of boundary conditions. © 2008 Zachary S Tseng E▯3 ▯ 1 Bar with both ends insulated Now let us consider the situation where, instead of them being kept at constant 0 degree temperature, the two ends of the bar are also sealed with perfect insulation so that no heat could escape to the outside environment (recall that the side of the bar is always perfectly insulated in the one▯ dimensional assumption), or vice versa. The new boundary conditions are u (0,t) = 0 and u (L,t) = 0, reflecting the fact that there will be no heat x x transferring, spatially, across the points x = 0 and x = L. The heat conduction problem becomes the initial▯boundary value problem below. 2 (Heat conduction eq.) α u xxu t , 0 < x < L, t > 0, (Boundary conditions) ux(0,t) = 0, and ux(L,t) = 0, (Initial condition) u(x,0) = f(x). The first step is the separation of variables. The equation is the same as before. Therefore, it will separate into the exact same two ordinary differential equations as in the first heat conduction problem seen earlier. The new boundary conditions separate into ux(0,t) = 0 → X′(0)T(t) = 0 → X′(0) = 0 or T(t) = 0 ux(L,t) = 0 → X′(L)T(t) = 0 → X′(L) = 0 or T(t) = 0 As before, we cannot choose T(t) = 0. Else we could only get the trivial solution u(x,t) = 0, rather than the general solution. Hence, the new boundary conditions should be X′(0) = 0 and X′(L) = 0. Again, we end up with a system of two simultaneous ordinary differential equations. Plus a set of two boundary conditions that goes with the spatial independent variable x: X″ + λX = 0, X′(0) = 0 and X′(L) = 0 , 2 T′ + α λT = 0 . © 2008 Zachary S Tseng E▯3 ▯ 2 The second step is to solve the eigenvalue problem X″ + λX = 0 , X′(0) = 0 and X′(L) = 0 . The result is summarized below. Case 1: If λ < 0: No such λ exists. Case 2: If λ = 0: Zero is an eigenvalue. Its eigenfunction is the constant function X 0= 1 (or any other nonzero constant). Case 3: If λ > 0: The positive eigenvalues λ are n π 2 λ = 2 , n = 1, 2, 3, … L The corresponding eigenfunctions that satisfy the said boundary conditions are n π x X n =cos , n = 1, 2, 3, … L The third step is to substitute the positive eigenvalues found above into the equation of t and solve: n π 2 T + α 2 T =0 . L 2 Notice that this is exactly the same equation as in the first (both ends kept at 0 degree) heat conduction problem, due to the fact that both problems have the same set of eigenvalues (but with different eigenfunctions). © 2008 Zachary S Tseng E▯3 ▯ 3 As a result, the solutions of the second equation are just the ones we have gotten the last time 2 2 2 2 −α n π t /L Tn(t) =C e n , n = 1, 2, 3, … There is this extra eigenvalue of λ = 0 that also needs to be accounted for. It has as an eigenfunction the consta0t X = 1. Put λ = 0 into the second equation and we get T′ = 0, which has only constant solut0ons T0(t) = C . Thus, we get the (arbitrary) constant func0ion u) =0X (0)T (t)0= C as a solution. Therefore, the solutions of the one▯dimensional heat conduction equation, with the boundary conditioxs u (0,t) = 0xand u (L,t) = 0, are in the form u 0(x,t) = C 0 −α n π t/ L nπ x u n t, ) = X tnT t n( ) = C n cos , L n = 1, 2, 3, … The general solution is their linear combination. Hence, for a bar with both ends insulated, the heat conduction problem has general solution: ∞ −α n π t / L n π u ( , ) = C 0 + ∑ C n cos n = L . © 2008 Zachary S Tseng E▯3 ▯ 4 Now set t = 0 and equate it with the initial condition u(x,0) = f(x): ∞ n π u ( ,0) = C0 + ∑ C ncos = f (x). n 1 L We see that the requirement is that the initial temperature distribution f(x) must be a Fourier cosine series. That is, it needs to be an even periodic function of period 2L. If f(x) is not already an even periodic function, then we will need to expand it into one and use the resulting even periodic extension of f) in its place in the above equation. Once this is done, the coefficients C’s in the particular solution are just the corresponding Fourier cosine coefficients of the initial condition f(x). (Except for the constant term, where the relatio0 C 0 a /2 holds, instead.) The explicit formula fnr C is, therefore, L 2 n x C n a =n L ∫ f (x)cos L dx , n = 1, 2, 3, … 0 C 0= a0/2 © 2008 Zachary S Tseng E▯3 ▯ 5 Example: Solve the heat conduction problem 3 uxx u t , 0 < x < 8, t > 0, u x0,t) = 0, and ux(8,t) = 0, u(x,0) = 9 − 3cos(πx/4) − 6cos(2πx). 2 First note that = 3 and L = 8, and the fact that the boundary conditions indicating this is a bar with both ends perfectly insulated. Substitute them into the formula we have just derived to obtain the general solution for this problem: ∞ 2 2 −3n π t /64 nπ x u(x,t) C= 0 + ∑ C n cos . n=1 8 Check the initial condition f(x), and we see that it is already in the require form of a Fourier cosine series of period 16. Therefore, there is no need to find its even periodic extension. Instead, we just need to extract the correct Fourier cosine coefficients from f(x): C0= a 02 = 9, C2= a 2 −3, C = a = −6, 16 16 Cn= a n 0, for all other n, n ≠ 0, 2, or 16. Note that C0is actually 0 /2, due to the way we write the constant term of the Fourier series. But that shouldn’t present any more difficulty. Since when you see a Fourier series, its constant term is already expressed in the form 0 /2. Therefore, you could just copy it down directly to be the C0term without thinking. Finally, the particular solution is 2 2 2 2 u(x,t) =9−3e −3(2 )π t /6cos( π x ) −6e −3(16 )π t /6cos(2π x) 4 © 2008 Zachary S Tseng E▯3 ▯ 6 Bar with two ends kept at arbitrary temperatures: An example of nonhomogeneous boundary conditions In both of the heat conduction initial▯boundary value problems we have seen, the boundary conditions are homogeneous − they are all zeros. Now let us look at an example of heat conduction problem with simple nonhomogeneous boundary conditions. The general set▯up is the same as the first example (where the both ends of the bar were kept at constant 0 degree, but were not insulated), except now the ends are kept at arbitrary (but constant) temperatures of T degrees at the left end, and T degrees at 1 2 the right end. The initial condition, as usual, is arbitrary. The heat conduction problem is therefore given by the initial▯boundary value problem: 2 α u =xx t , 0 < x < L, t > 0, u(0,t) = T 1 and u(L,t) = T 2 u(x,0) = f(x). The boundary conditions is now nonhomogeneous (unless T and T are both 1 2 0, then the problem becomes identical to the earlier example), because at least one of the boundary values are nonzero. The nonhomogeneous boundary conditions are rather easy to work with, more so than we might have reasonably expected. First, let us be introduced to the concept of the steady▯state solution. It is the part of the solution u(x,t) that is independent of the time variable t. Therefore, it is a function of the spatial variable alone. We can thusly rewrite the solution u(x,t) as a sum of 2 parts, a time▯independent part and a time▯dependent part: u(x, t) = v(x) + w(x, t) . Where v(x) is the steady▯state solution, which is independent of t, and w(x,t) is called the transient solution, which does vary with t. © 2008 Zachary S Tseng E▯3 ▯ 7 The Steady-State Solution The steady▯state solution, v(x), of a heat conduction problem is the part of the temperature distribution function that is independent of time t. It represents the equilibrium temperature distribution. To find it, we note the fact that it is a function of x alone, yet it has to satisfy the heat conduction equation. Since vxx = v″ andv t= 0, substituting them into the heat conduction equation we get 2 α v xx= 0. Divide both sides by α and integrate twice with respect to x, we find that v(x) must be in the form of a degree 1 polynomial: v(x) = Ax + B. Then, rewrite the boundary conditions in terms of v: u(0,t) = v(0) =1T , and u(L,t) = v(L) = 2 . Apply those 2 conditions to find that: v(0) = T 1 A(0) + B = B → B = T 1 v(L) = T 2 AL + B = AL + T 1 → A = (T 2 T )/1 Therefore, T2−T 1 v(x) = x + T 1. L Thing to remember: The steady▯state solution is a time▯independent function. It is obtained by setting the partial derivative(s) with respect to t in the heat equation (or, later on, the wave equation) to constant zero, and then solving the equation for a function that depends only on the spatial variable x. © 2008 Zachary S Tseng E▯3 ▯ 8 Comment: Another way to understand the behavior of v(x) is to think from the perspective of separation of variables. You could think of the steady▯ state solution as, during the separation of variables, the solution you would have obtained if T(t) = 1, the constant function 1. Therefore, the solution is independent of time, or time▯invariant. Hence, u(x, t) = X(x)T(t) = X(x) = v(x). We can, in addition, readily see the substitutions required for rewriting the boundary conditions prior to solving for the steady▯state solution: u(0, t) = X(0) = v(0) = T ,1and u(L,t) = X(L) = v(L) = T . 2 That is, just rename the function u as v, ignore the time variable t, and put whatever x▯coordinate specified directly into v(x). © 2008 Zachary S Tseng E▯3 ▯ 9 The solution of bar with two ends kept at arbitrary temperatures Once the steady▯state solution has been found, we can set it aside for the time being and proceed to find the transient part of solution, w(x,). First we will need to rewrite the given initial▯boundary value problem slightly. Keep in mind that the initial and boundary conditions as originally given were meant for the temperature distribution function u(x,t) = v(x) + w(x,t). Since we have already found v(x), we shall now subtract out the contribution of v(x) from the initial and boundary values. The results will be the conditions that the transient solution w(x,) alone must satisfy. Change in the boundary conditions: u(0, t) = 1 = v(0) + w(0,t) → w(0,t) = T 1 v(0) = 0 u(L,t) = T 2 v(L) + w(L,t) → w(L,t) = T 2 v(L) = 0 Note: Recall that u(0,t) = v(0) = T1, and u(L,t) = v(L) = T2. Change in the initial condition: u(x,0) = f(x) = v(x) + w(x, 0) → w(x, 0) = f(x) − v(x) Consequently, the transient solution is a function of both x and t that must satisfy the n
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