{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Worksheet 12 Solutions

Worksheet 12 Solutions - MA 2930 Worksheet 12 1 Lets solve...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MA 2930, April 20, 2011 Worksheet 12 1. Let’s solve a heat equation problem similar to the one on the prelim in a step-by-step fashion: u t = u xx - 2 u x (a) Separate the time and space variables and obtain two ODEs. How are the solutions of the two ODEs connected? (b) Solve the ODEs. (c) Write down a fundamental solution of the heat equation. (d) Apply the spatial boundary conditions: u x (0 , t ) = 0 = u (5 , t ) . (To which of the two ODEs do these conditions apply?) Write down the non- trivial solutions of the ODEs first, then the list of fundamental solutions of the heat equation, and finally its general solution. In constructing the general solution from the list of fundamental solutions what is the principle you are using? What kind of differential equations does that principle apply to? (e) Now apply the temporal initial condition: u ( x, 0) = 7 e - x cos( πx/ 10) . What is the resulting solution’s behavior as t → ∞ , or in other words, what would be the final temperature distribution along the bar? (a) To separate the variables let u ( x, t ) = X ( x ) T ( t ). Then, u t = u xx - 2 u x XT 0 = X 00 T - 2 X 0 T XT 0 = T ( X 00 - 2 X 0 ) T 0 T = X 00 - 2 X 0 X Since the two sides are functions of different variables and yet they always manage to remain equal, they must equal a constant, say - λ . Equating each side to - λ gives us two ODEs: T 0 + λT = 0 and X 00 - 2 X 0 + λX = 0
Image of page 1

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

View Full Document Right Arrow Icon
Note that the two ODEs - and their solutions - are linked through the com- mon parameter λ . For each value of λ we get a solution of the x -ODE and of the t -ODE. Only these solutions can be multiplied together to get a solution u ( x, t ); x - and t - solutions for different values of λ do not go together. (b) The solutions of the t -ODE are clearly T λ ( t ) = ce - λt where c is any constant. The solutions of the x -ODE are constructed from its characteristic roots, which are the roots of r 2 - 2 r + λ = 0. So, r = - 1 ± 1 - λ . The (real) solutions are: X λ ( x ) = c 1 e ( - 1+ 1 - λ ) x + c 2 e ( - 1 - 1 - λ ) x , if λ < 1 c 1 e - x + c 2 xe - x , if λ = 1 e - x ( c 1 cos λ - 1 x + c 2 sin λ - 1 x ) , if λ > 1 Note that solutions depend upon the parameter λ . (c) A fundamental solution of the heat equation is u λ ( x, t ) = e - λt [ c 1 e ( - 1+ 1 - λ ) x + c 2 e ( - 1 - 1 - λ ) x ] , if λ < 1 e - λt [ c 1 e - x + c 2 xe - x ] , if λ = 1 e - λt [ e - x ( c 1 cos λ - 1 x + c 2 sin λ - 1 x )] , if λ > 1 (d) We can apply these boundary conditions to u or X ; it’s equivalent because, e.g., u x (0 , t ) = X 0 (0) T ( t ) we want T λ ( t ) to be non-trivial. Applying them to X we get 0 = X 0 λ (0) = c 1 ( - 1 + 1 - λ ) + c 2 ( - 1 - 1 - λ ) , if λ < 1 - c 1 + c 2 , if λ = 1 c 1 + c 2 λ - 1 , if λ > 1 and 0 = X (5) = c 1 e 5( - 1+ 1 - λ ) + c 2 e 5( - 1 - 1 - λ ) , if λ < 1 c 1 e - 5 + c 2 5 e - 5 , if λ = 1 e - 5 ( c 1 cos 5 λ - 1 + c 2 sin 5 λ - 1) , if λ > 1 Solving these two equations for c 1 and c 2 simultaneously for all λ ’s we get X ( x ) = 0 , if λ < 1 0 , if λ = 1 0 , if λ > 1 and λ - 1 6 = 10 , n = 1 , 3 , 5 , . . .
Image of page 2
Image of page 3
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