Worksheet 12 - MA 2930 Worksheet 12 1 Lets solve a heat equation problem similar to the one on the prelim in a step-by-step fashion ut = uxx 2ux(a

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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
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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
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This note was uploaded on 06/10/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).

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Worksheet 12 - MA 2930 Worksheet 12 1 Lets solve a heat equation problem similar to the one on the prelim in a step-by-step fashion ut = uxx 2ux(a

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