Chapter_5_Solutions

# Chapter_5_Solutions - CHAPTER 5 PDEs in The Life Sciences 1...

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Unformatted text preview: CHAPTER 5 PDEs in The Life Sciences 1. Age-Structured Models Exercise 1: Write 1 = integraldisplay 8 3 4 e − ( r +0 . 03) a da =- 4 r + 0 . 03 parenleftBig e − 8( r +0 . 03)- e − 3( r +0 . 03) parenrightBig , and use a software package or calculator to solve for r . Exercise 2: First note that u ( a,t ) = 0 for a > t + δ, since f ( a ) = 0 for a > δ. For (a) observe that the renewal equation (5.9) is B ( t ) = integraldisplay t βB ( t- a ) e − γa da + integraldisplay ∞ βf ( a- t ) e − γt da. The first integral becomes, upon changing variables to s = t- a , integraldisplay t βB ( t- a ) e − γa da = integraldisplay t βB ( s ) e − γ ( t − s ) ds. The second integral is integraldisplay ∞ βf ( a- t ) e − γt da = integraldisplay t + δ t βu e − γt da = βu δe − γt . For (b), differentiate (using Leibniz rule) B ( t ) = integraldisplay t βB ( s ) e − γ ( t − s ) ds + βu δe − γt to get B ′ ( t ) = integraldisplay t βB ( s ) e − γ ( t − s ) ds (- γ ) + βB ( t )- γβu δe − γt = ( β- γ ) B ( t ) . For (c) note that the last equation is the differential equation for growth- decay and has solution B ( t ) = B (0) e ( β − γ ) t . Therefore the solution from (5.7)–(5.8) is given by u ( a,t ) = , a > t + δ u e − γt , t < a < t + δ B (0) e ( β − γ ) t e − βa , < a < t. 1 2 5. PDES IN THE LIFE SCIENCES Finally, for part (d) we have, using part (c), N ( t ) = integraldisplay t u ( a,t ) da + integraldisplay t + δ t u ( a,t ) da = integraldisplay t B (0) e ( β − γ ) t e − βa da + integraldisplay t + δ t u e − γt da = B (0) β e ( β − γ ) t (1- e − βt ) + δu e − γt ....
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## This note was uploaded on 03/08/2012 for the course MATH 1010 taught by Professor Mihai during the Fall '08 term at UOIT.

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Chapter_5_Solutions - CHAPTER 5 PDEs in The Life Sciences 1...

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