Chem Differential Eq HW Solutions Fall 2011 122

Chem Differential Eq HW Solutions Fall 2011 122 - 122...

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122 Chapter 7 The Fourier Transform and its Applications Solutions to Exercises 7.5 1. To solve the Dirichlet problem in the upper half-plane with the given boundary function, we use formula (5). The solution is given by u ( x, y )= y π ± -∞ f ( s ) ( x - s ) 2 + y 2 ds = 50 y π ± 1 - 1 ds ( x - s ) 2 + y 2 = 50 π ² tan - 1 ³ 1+ x y ´ + tan - 1 ³ 1 - x y ´µ , where we have used Example 1 to evaluate the deFnite integral. 5. Appealing to (4) in Section 7.5, with y = y 1 ,y 2 1 + y 2 ,weFnd F ( P y 1 )( w e - y 1 | w | , F ( P y 2 )( w e - y 2 | w | , F ( P y 1 + y 2 )( w e - ( y 1 + y 2 ) | w | . Hence F ( P y 1 )( w ) ·F ( P y 2 )( w e - y 1 | w | e - y 2 | w | = e - ( y 1 + y 2 ) | w | = F ( P y 1 + y 2 )( w ) . But F ( P y 1 )( w ) ( P y 2 )( w F ( P y 1 * P y 2 )( w ) , Hence F ( P y 1 + y 2 )( w F ( P y 1 * P y 2 )( w ); and so P y 1 + y 2 = P y 1 * P y 2 . 9. Modify the solution of Example 1(a) to obtain that, in the present case, the solution is u ( x, y T 0 π tan - 1 ³ a + x y ´ + tan - 1 ³ a - x y ´· . To Fnd the isotherms, we must determine the points ( x, y ) such that u ( x, y T . As in the solution of Example 1(b), these points satisfy x 2 + ³ y - a cot( πT T 0 ) ´ 2 = ³ a csc( T 0 ) ´ 2 . Hence the points belong to the arc in the upper half-plane of the circle with center (0 ,a cot( πT T 0 )) and radius a csc( πT T 0 ). The isotherm corresponding to T = T 0 2 is the arc of the circle x 2 + ¸ y - a cot( π 2 ) ¹ 2 = ¸ a csc( π 2 ) ¹ 2 , or x 2 + y 2 = a 2 . Thus the isotherm in this case is the upper semi-circle of radius
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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