MATH 285 E1 F1 GRADED HOMEWORK SET 6 DUE FRIDAY NOVEMBER 21 IN LECTURE This time the homework has just one part Please staple your homework together and put your name and section on it Thank you 1 15 points Consider the eigenvalue problem 00 0 y 2y y 0 y 0 0 y 1 0 Find the eigenvalues and eigenfunctions for this problem That is find the values of for which the problem has a nontrivial solution and find those nontrivial solutions Hint The smallest eigenvalue is 2 1 with associated eigenfunction e x sin x In your answer you should verify this Considering first the differential equation the corresponding characteristic equation is r2 2r 0 with solutions 2 4 4 1 1 2 Thus the nature of the solutions depends on whether 1 0 1 0 or 1 0 First consider the case 1 0 Then the function y has the form y Ae 1 1 x Be 1 1 x r We ask what restriction the endpoint conditions place on A and B y 0 0 0 A B y 1 0 0 Ae 1 1 If we multiply the latter equation by 2 1 0 Ae Be 1 e1 1 1 we obtain B In conjunction with 0 A B we obtain A Ae2 1 This can only happen if A 0 But then B 0 as well Thus the only solution is y 0 the trivial uninteresting solution We conclude that is not an eigenvalue 1 2 GRADED HOMEWORK 6 Now we consider the case 1 0 which is to say 1 The characteristic roots r 1 1 are both equal to 1 so we have a repeated root Then the function y has the form y Ae x Bxe x Considering the endpoint conditions y 0 0 0 A y 1 0 0 Ae 1 Be 1 Thus we find directly A 0 and hence 0 Be 1 so B 0 as well Thus the only solution is y 0 and we conclude that is not an eigenvalue Lastly we consider the case 1 0 Since now 1 is a positive number the characteristic roots may be written r 1 1 1 i 1 Then the function y has the form y Ae x cos 1x Be x sin 1x Considering the endpoint conditions y 0 0 0 A y 1 0 0 Ae x cos 1 Be 1 sin 1 The first equation gives directly that A 0 so the second one simplifies to 0 Be 1 sin 1 The number e 1 is not zero so this equation implies that either B 0 or sin 1 0 Now we have two subcases either sin 1 0 or not If not then B 0 so the entire solution y 0 and we conclude that is not an eigenvalue But if the number does satisfy the condition sin 1 0 then B can take any value and we have an nontrivial interesting x solution namely y Be sin 1x so is an eigenvalue Thus the condition for to be an eigenvalue is sin 1 0 This is equivalent to 1 n n 1 2 3 n 2 1 n 1 2 3 The eigenfunction associated to the eigenvalue n 2 1 is y e x sin n x GRADED HOMEWORK 6 3 2 10 points Find the solution of the heat problem on the interval 0 x 5 u 2 2 xu2 t u 0 t 0 x u 5 t 0 x u x 0 cos2 10 x This is the heat equation problem with insulated ends which was discussed in lecture see lecture 33 It corresponds to the case k 2 and L 5 We quote from that lecture the eigenfunctions are the constant function 1 and the cosine functions cos n x 5 The separable 2t n x 2 n 5 solutions are un x t e cos 5 The general solutions is u x t c0 X 2 cn e 2 n 5 t cos n 1 n x 5 with initial value u x 0 c0 X cn cos n 1 n x 5 We need this to match f x cos2 10 x so we need to find the Fourier cosine series with period 10 L 5 of f x That can be done by using the double angle formula cos 2 2 cos2 1 or cos2 1 cos 2 2 100 x 1 1 f x cos2 10 x 1 cos 20 x 2 cos 2 2 5 Thus in order to match the initial data we must take c0 1 2 c100 1 2 and all other coefficients zero c0 1 2 c1 0 c2 0 c99 0 c100 1 2 c101 0 Putting it all together the solution is 1 1 100 x 2 u x t e 2 100 5 t cos 2 2 5 or simplified 1 1 2 u x t e 800 t cos 20 x 2 2 3 15 points Find the solution of the heat problem on the interval 0 x 1 u 2 5 xu2 t u 0 t 0 u 1 t 1 u x 0 x2 You should use the following strategy 4 GRADED HOMEWORK 6 a Find a steady state solution u0 of all parts of the problem except the initial condition That is find a function u0 x t that is constant in time u t 0 that satisfies the heat equation and that satisfies the conditions u0 0 t 0 and u0 1 t 1 2 u0 0 Assuming u t 0 and the heat equation we obtain x2 0 so u0 x t must be independent of t and linear in x u0 x t Ax B In order to satisfy the boundary conditions we must have 0 u0 0 t B and 1 u0 1 t A B hence B 0 and A 1 Thus the steady state solution is u0 x t x b Posit w u u0 and show that w must solve a slighty different heat problem w 2 5 xw2 t w 0 t 0 w 1 t 0 w x 0 x2 u0 x 0 w is called the transient term We are assuming that u x t solves the problem and we want to determine the relevant condtions on w x t First of all w satisfies the heat equation w u u0 2u 2 u0 2 2w u u0 5 2 5 2 5 2 u u0 5 2 t t t t x x x x Next w satisfies w 0 t 0 w 0 t u 0 t u0 0 t 0 0 0 Next w satisfies w 1 t 0 w 1 t u 1 t u0 1 t 1 1 0 Lastly w satisfies w x 0 x2 u0 x 0 w x 0 u x 0 u0 x 0 x2 u0 x 0 c Using the methods described in the lectures find the solution w x t and hence u x t Using the steady state solution u0 x 0 x we need to solve the heat problem w …
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