Math Assignment 14

Transcription

1 Math Assignment 14 Dylan Zwick Spring 2014 Section 9.5-1, 3, 5, 7, 9 Section 9.6-1, 3, 5, 7, 14 Section 9.7-1, 2, 3, 4 1

2 Section Heat Conduction and Separation of Variables Solve the boundary value problem: u t = 3u xx, 0 < x < π, t > 0; u(0, t) = u(π, t) = 0, u(x, 0) = 4 sin(2x). Solution - The five problems for this section are all very similar. So, I m going to solve this one in detail, and then reference it for the other problems. We assume that our solution is separable, and so can be written in the form u(x, t) = X(x)T(t). Taking partial derivatives of u(x, t) and plugging them into our partial differential equation we get X(x)T (t) = 3X (x)t(t). We can rewrite this as T (t) 3T(t) = X (x) X(x) = λ, where λ is a constant. We know both quotients must be constants because one is a function of t only, and the other is a function of x only. We can rewrite the second equality as 2

3 X (x) + λx(x) = 0. The boundary condition u(0, t) = 0 gives us u(0, t) = X(0)T(t) = 0. As u(x, t) is not the 0 function we cannot have T(t) = 0, and so we must have X(0) = 0. Similarly, given u(π, t) = 0 we get X(π) = 0. So, our differential equation for X(x) must satisfy X(0) = X(π) = 0. If λ < 0 we would have the solution X(x) = Ae λx + Be λx. Plugging in X(0) = 0 gives us X(0) = A + B = 0 A = B. This means our function X(x) has the form X(x) = A(e λx e λx ). If X(x) is not the trivial solution X(x) = 0 then we must have A 0. If X(π) = A(e λπ e λπ ) = 0 then as A 0 we would have e λπ = e λπ. 3

4 As λ < 0, this cannot be. Therefore, there are no nontrivial solutions for λ < 0. If λ = 0 then our solution is X(x) = Ax + B. If X(0) = 0 we must have B = 0. If B = 0 and X(π) = 0 we must have A = 0, and so there are no nontrivial solutions for λ = 0. If λ > 0 then our solution has the form ( ) ( ) X(x) = A sin λx + B cos λx. If X(0) = 0 we must have B = 0, and if X(π) = 0 we get ( ) X(π) = A sin λπ. We re looking for nontrivial solutions, so we want A 0. If this is the case we must have λ = n, where n is a positive integer. So, the eigenvalues for this endpoint value problem are λ n = n 2, with eigenfunctions X n (x) = sin (nx). The differential equation for T(t) is T (t) + 3λT(t) = 0. 4

5 The solution to this differential equation is T(t) = Ce 3λt, where C is some constant. If we have λ = n 2 then T n (t) = Ce 3n2t. So, the solutions to our partial differential equation are of the form u n (x, t) = A n sin (nx)e 3n2t. Our final solution will be a linear combination of these solutions u(x, t) = A n sin (nx)e 3n2t. n=1 We want to find coefficients A n such that u(x, 0) = A n sin (nx) = 4 sin(2x). n=1 From this we can see immediately that A 2 = 4, while all other A n terms are 0, and our solution is u(x, t) = 4 sin(2x)e 12t. 5

6 Solve the boundary value problem: u t = 2u xx, 0 < x < 1, t > 0; u(0, t) = u(1, t) = 0, u(x, 0) = 5 sin (πx) 1 sin (3πx). 5 Solution - Using the same argument as in Problem 9.5.1, mutatis mutandis, we get that our solution will be of the form u(x, t) = A n sin (nπx)e 2n2 π 2t. n=1 We want the coefficients to be such that u(x, 0) = A n sin (nπx) = 5 sin (πx) 1 sin (3πx). 5 n=1 From this we can see immediately that A 1 = 5, A 3 = 1, and all 5 other A n terms are 0. So, our solution is u(x, t) = 5 sin (πx)e 2π2t 1 5 sin (3πx)e 18π2t. 6

7 Solve the boundary value problem: u t = 2u xx, 0 < x < 3, t > 0; ( ) ( ) 2 4 u x (0, t) = u x (3, t) = 0, u(x, 0) = 4 cos 3 πx 2 cos 3 πx. Solution - The solution to our boundary value problem in this case will be the cosine terms instead of the sine terms, but otherwise things work out, making the appropriate adjustments, the same as they do in Problem The solution to our partial differential equation will be u(x, t) = ( nπx A n cos )e 2n2 π n=0 We want to find coefficients such that u(x, 0) = ( nπx ) A n cos 3 n=0 ( ) ( ) 2 4 = 4 cos 3 πx 2 cos 3 πx. From this we can immediately see that A 2 = 4, A 4 = 2, and all other A n terms are 0. So, our solution is ( ) 2πx u(x, t) = 4 cos e 8π2 t 9 2 cos 3 ( 4πx 3 ) e 32π2 t 9. 7

8 Solve the boundary value problem: 3u t = u xx, 0 < x < 2, t > 0; u x (0, t) = u x (2, t) = 0, u(x, 0) = cos 2 (2πx). Solution - This problem is solved the same way, mutatis mutandis, as Problem Our solution will be u(x, t) = ( nπx A n cos )e n2 π 2 12 t. 2 n=0 To find the A n coefficients we note u(x, 0) = ( nπx ) A n cos = cos 2 (2πx) = 2 n=0 1 + cos (4πx). 2 So, we have A 0 = 1 2, A 8 = 1 2, and all other A n terms are 0. Our solution is u(x, t) = cos (4πx)e 16π2 3 t. 8

9 Solve the boundary value problem: 10u t = u xx, 0 < x < 5, t > 0; u(0, t) = u(5, t) = 0, u(x, 0) = 25. Solution - To find the solution we, again, follow the same approach as in Problem 9.5.1, mutatis mutandis, and we get u(x, t) = A n sin n=1 ( nπx )e n2 π t. 5 We want to find values for the coefficients such that u(x, 0) = A n sin n=1 ( nπx ) = Or, more specifically, u(x, 0) = 25 on the interval from 0 to 5. If we take the odd extension of u(x, 0) and then solve for the Fourier coefficients we get A n = ( nπx ) 25 sin dx = 50 ( nπx ) 5 5 nπ cos 5 = 50 (1 cos (nπ)) = nπ { 100 nπ n odd 0 n even 0 So, our solution is u(x, t) = 100 π n odd 1 ( nπx ) n sin e n2 π 2 t

10 Section Vibrating Strings and the One-Dimensional Wave Equation Solve the boundary value problem: y tt = 4y xx, 0 < x < π, t > 0; y(0, t) = y(π, t) = 0, y(x, 0) = 1 10 sin (2x), y t(x, 0) = 0. 10

11 More room for Problem

12 Solve the boundary value problem: 4y tt = y xx, 0 < x < π, t > 0; y(0, t) = y(π, t) = 0, y(x, 0) = y t (x, 0) = 1 sin (x)

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14 Solve the boundary value problem: y tt = 25y xx, 0 < x < 3, t > 0; y(0, t) = y(3, t) = 0, y(x, 0) = 1 4 sin (πx), y t(x, 0) = 10 sin (2πx). 14

15 More room for Problem

16 Solve the boundary value problem: y tt = 100y xx, 0 < x < 1, t > 0; y(0, t) = y(1, t) = 0, y(x, 0) = 0, y t (x, 0) = x. 16

17 More room for Problem

18 Given the differentiable odd period 2L function F(x), show that the function y(x, t) = 1 [F (x + at) + F (x at)] 2 satisfies the conditions y(0, t) = y(l, t) = 0, y(x, 0) = F(x), y t (x, 0) = 0. 18

19 Section Steady-State Temperature and Laplace s Equation Solve the Dirichlet problem for the rectangle 0 < x < a, 0 < y < b consisting of Laplace s equation u xx + u yy = 0 and the boundary value conditions: u(x, 0) = u(x, b) = u(0, y) = 0, u(a, y) = g(y). 19

20 More room for Problem

21 Solve the Dirichlet problem for the rectangle 0 < x < a, 0 < y < b consisting of Laplace s equation u xx + u yy = 0 and the boundary value conditions: u(x, 0) = u(x, b) = u(a, y) = 0, u(0, y) = g(y). 21

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23 Solve the Dirichlet problem for the rectangle 0 < x < a, 0 < y < b consisting of Laplace s equation u xx + u yy = 0 and the boundary value conditions: u(x, 0) = u(0, y) = u(a, y) = 0, u(x, b) = f(x). 23

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25 Consider the boundary value problem u xx + u yy = 0; u x (0, y) = u x (a, y) = u(x, 0) = 0, u(x, b) = f(x) corresponding to the rectangular plate 0 < x < a, 0 < y < b with the edges x = 0 and x = a insulated. Derive the solution u(x, y) = a 0y 2b + ( nπx a n (cos a a n = 2 a a 0 n=1 ( nπx f(x) cos a )) ( sinh ( nπy a sinh ( nπb a )) ) dx (n = 0, 1, 2,...). ), where (Suggestion: Show first that λ 0 is an eigenvalue with X 0 (x) 1 and Y 0 (y) = y.) 25

26 More room for Problem