em2sp13e1s

# D dx 1 x 2 y ? 1 x 2 y 0 y 0 y 1 0 thus p x 1 x 2 q x

• Notes
• 3

This preview shows page 2 - 3 out of 3 pages.

d dx ( (1 + x 2 ) y ) + λ (1 + x 2 ) y = 0 , y (0) = y (1) = 0 . Thus: p ( x ) = 1 + x 2 , q ( x ) = 0, σ ( x ) = 1 + x 2 . (b) (5 points) Explain why there canNOT be any negative eigenvalues. Solution. The Rayleigh quotient works out to λ = 1 0 (1 + x 2 ) y ( x ) 2 dx 1 0 (1 + x 2 ) y ( x ) 2 dx , which can clearly not be negative. (c) (5 points) Can λ = 0 be an eigenvalue? JUSTIFY! Solution. The answer is YES. y = 1 is a non-zero solution of the problem for λ = 0. 3. Solve by separation of variables the following problems. You do not have to go through a detailed analysis of any Sturm-Liouville problem that you already solved several times before, assuming you remember what you did. Or have it on your cheat sheet. But explain from where it comes! (For example, Write, “As seen in class, the solution of this problem is given by. . . ”) (a) (20 points) u t = u xx , 0 < x < π, t > 0 , u (0 , t ) = 0 , u ( π, t ) = 0 , t > 0 u ( x, 0) = 3 sin 5 x 4 sin 7 x, 0 < x < π. Solution. The solution of the PDE solving the boundary conditions obtained by separation of variables is u ( x, t ) = n =1 c n e n 2 t sin nx. The c n ’s must satisfy 3 sin 5 x 4 sin 7 x = n =1 c n sin nx. Thus c 5 = 3 , c 7 = 4, c n = 0 if n ̸ = 5 , 7. The solution is u ( x, t ) = 3 e 25 t sin 5 x 4 e 49 t sin 7 x. (b) (20 points) u t = 4 u xx , 0 < x < 1 , t > 0 , u x (0 , t ) = 0 , u (1 , t ) = 0 , t > 0 u ( x, 0) = 5 cos 9 πx 2 , 0 < x < 1 . Solution. The solution of the PDE solving the boundary conditions obtained by separation of vari- ables is u ( x, t ) = n =1 c n e (2 n 1) 2 π 2 t/ 4 cos (2 n

Subscribe to view the full document.

• Spring '13
• Schonbek
• Sin, Boundary value problem, Partial differential equation, Sturm–Liouville theory

{[ snackBarMessage ]}

###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern