Math 293 solution set 12

Math 293 solution set 12 - Math 293 Solutions to Problem...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 293 Solutions to Problem Set 12. § 10.5 #4 . In general : The equation u t = ku xx with boundary conditions u x (0 , t ) = 0 = u x ( L, t ) and initial condition u ( x, 0) = f ( x ) has solution u ( x, t ) = a 0 2 + X n =0 a n e - kn 2 π 2 t/L 2 cos( nπx/L ) where the a n ’s are the coefficients in the Fourier sine series for f ( x ), hence a n = 2 L Z L 0 f ( x ) cos( nπx/L ) dx In the present problem we have k = 2, L = 1, and f ( x ) = x - x 2 . To compute a n one can use integration by parts, but it is much quicker to use tabular integration, explained in Thomas & Finney on page 566. Whichever method you use, you should get a n = 2 ( - cos - 1 n 2 π 2 ) . Since cos = ( - 1) n , we see that a n is 0 for n odd and - 4 n 2 π 2 for n even. This calculation doesn’t work for n = 0 since it would involve dividing by 0, so a separate calculation gives a 0 = 2 R 1 0 x - x 2 dx = 1 3 . Now we plug these values into the general formula for u ( x, t ) above, replacing n by 2 n since we only want the even values of n , and we get u ( x, t ) = 1 6 - 4 π 2 X n =1 1 (2 n ) 2 e - 2(2 n ) 2 π 2 t cos 2 nπx #7 . In general : The equation u t = ku xx + P ( x ) with boundary conditions u (0 , t ) = A , u ( L, t ) = B , and initial condition u ( x, 0) = f ( x ), has solution u ( x, t ) = v ( x ) + w ( x, t ) where: (1) v satisfies 0 = kv
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern