AEM_3e_Chapter_13

AEM_3e_Chapter_13 - 13 13 Boundary-Value Problems in...

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13 13 Boundary-Value Problems in Rectangular Coordinates EXERCISES 13.1 Separable Partial Differential Equations 3. Substituting u ( x, y )= X ( x ) Y ( y ) into the partial diferential equation yields X 0 Y + XY 0 = . Separating variables and using the separation constant λ we obtain X 0 X = Y Y 0 Y = λ. Then X 0 + λX = 0 and Y 0 (1 + λ ) Y =0 so that X = c 1 e λx and Y = c 2 e (1+ λ ) y . A particular product solution oF the partial diferential equation is u = = c 3 e y + λ ( y x ) . 6. Substituting u ( x, y X ( x ) Y ( y ) into the partial diferential equation yields yX 0 Y + xXY 0 = 0. Separating variables and using the separation constant λ we obtain X 0 xX = Y 0 yY = λ. When λ 6 X 0 + λxX = 0 and Y 0 λyY so that X = c 1 e λx 2 / 2 and Y = c 2 e λy 2 / 2 . A particular product solution oF the partial diferential equation is u = = c 3 e λ ( x 2 y 2 ) / 2 . When λ = 0 the diferential equations become X 0 = 0 and Y 0 = 0, so in this case X = c 4 , Y = c 5 , and u = = c 6 . 9. Substituting u ( x, t X ( x ) T ( t ) into the partial diferential equation yields kX 0 T XT = 0 . Separating variables and using the separation constant λ we obtain 0 X X = T 0 T = λ. Then X 0 + λ 1 k X = 0 and T 0 + λT . 220
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13.1 Separable Partial Diferential Equations The second diferential equation implies T ( t )= c 1 e λt . For the ±rst diferential equation we consider three cases: I. I² ( λ 1) /k = 0 then λ =1, X 0 = 0, and X ( x c 2 x + c 3 ,so u = XT = e t ( A 1 x + A 2 ) . II. I² ( λ 1) /k = α 2 < 0, then λ =1 2 , X 0 α 2 X = 0, and X ( x c 4 cosh αx + c 5 sinh αx , so u = =( A 3 cosh αx + A 4 sinh αx ) e (1 2 ) t . III. I² ( λ 1) /k = α 2 > 0, then λ =1+ λα 2 , X 0 + α 2 X = 0, and X ( x c 6 cos αx + c 7 sin αx u = A 5 cos αx + A 6 sin αx ) e (1+ λα 2 ) t . 12. Substituting u ( x, t X ( x ) T ( t ) into the partial diferential equation yields a 2 X 0 T = 0 +2 kXT 0 . Separating variables and using the separation constant λ we obtain X 0 X = T 0 kT 0 a 2 T = λ. Then X 0 + λX = 0 and T 0 0 + a 2 λT =0 . We consider three cases: I. λ = 0 then X 0 = 0 and X ( x c 1 x + c 2 . Also, T 0 0 = 0 and T ( t c 3 + c 4 e 2 kt u = c 1 x + c 2 )( c 3 + c 4 e 2 kt ) . II. λ = α 2 < 0, then X 0 α 2 X = 0, and X ( x c 5 cosh αx + c 6 sinh αx . The auxiliary equation T 0 0 α 2 a 2 T =0i s m 2 km α 2 a 2 = 0. Solving ²or m we obtain m = k ± k 2 + α 2 a 2 , so T ( t c 7 e ( k + k 2 + α 2 a 2 ) t + c 8 e ( k k 2 + α 2 a 2 ) t . Then u = c 5 cosh αx + c 6 sinh αx ) ³ c 7 e ( k + k 2 + α 2 a 2 ) t + c 8 e ( k k 2 + α 2 a 2 ) t ´ . λ = α 2 > 0, then X 0 + α 2 X = 0, and X ( x c 9 cos αx + c 10 sin αx . The auxiliary equation T 0 0 + α 2 a 2 T =0is m 2 + α 2 a 2 = 0. Solving ²or m we obtain m = k ± k 2 α 2 a 2 .W e consider three possibilities ²or the discriminant k 2 α 2 a 2 : (i) k 2 α 2 a 2 = 0 then T ( t c 11 e kt + c 12 te kt and u = c 9 cos αx + c 10 sin αx )( c 11 e kt + c 12 te kt ) .
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This note was uploaded on 01/03/2011 for the course BIS 511 at Yale.

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AEM_3e_Chapter_13 - 13 13 Boundary-Value Problems in...

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