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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 differential equation yields X Y + XY = XY . Separating variables and using the separation constant λ we obtain X X = Y Y Y = λ. Then X + λX = 0 and Y (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 differential equation is u = XY = c 3 e y + λ ( y x ) . 6. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields yX Y + xXY = 0. Separating variables and using the separation constant λ we obtain X xX = Y yY = λ. When λ = 0 X + λxX = 0 and Y λyY = 0 so that X = c 1 e λx 2 / 2 and Y = c 2 e λy 2 / 2 . A particular product solution of the partial differential equation is u = XY = c 3 e λ ( x 2 y 2 ) / 2 . When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c 4 , Y = c 5 , and u = XY = c 6 . 9. Substituting u ( x, t ) = X ( x ) T ( t ) into the partial differential equation yields kX T XT = XT . Separating variables and using the separation constant λ we obtain kX X X = T T = λ. Then X + λ 1 k X = 0 and T + λT = 0 . 220

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13.1 Separable Partial Differential Equations The second differential equation implies T ( t ) = c 1 e λt . For the first differential equation we consider three cases: I. If ( λ 1) /k = 0 then λ = 1, X = 0, and X ( x ) = c 2 x + c 3 , so u = XT = e t ( A 1 x + A 2 ) . II. If ( λ 1) /k = α 2 < 0, then λ = 1 2 , X α 2 X = 0, and X ( x ) = c 4 cosh αx + c 5 sinh αx , so u = XT = ( A 3 cosh αx + A 4 sinh αx ) e (1 2 ) t . III. If ( λ 1) /k = α 2 > 0, then λ = 1 + λα 2 , X + α 2 X = 0, and X ( x ) = c 6 cos αx + c 7 sin αx , so u = XT = ( 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 differential equation yields a 2 X T = XT +2 kXT . Separating variables and using the separation constant λ we obtain X X = T + 2 kT a 2 T = λ. Then X + λX = 0 and T + 2 kT + a 2 λT = 0 . We consider three cases: I. If λ = 0 then X = 0 and X ( x ) = c 1 x + c 2 . Also, T + 2 kT = 0 and T ( t ) = c 3 + c 4 e 2 kt , so u = XT = ( c 1 x + c 2 )( c 3 + c 4 e 2 kt ) . II. If λ = α 2 < 0, then X α 2 X = 0, and X ( x ) = c 5 cosh αx + c 6 sinh αx . The auxiliary equation of T + 2 kT α 2 a 2 T = 0 is m 2 + 2 km α 2 a 2 = 0. Solving for 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 = XT = ( c 5 cosh αx + c 6 sinh αx ) c 7 e ( k + k 2 + α 2 a 2 ) t + c 8 e ( k k 2 + α 2 a 2 ) t . III. If λ = α 2 > 0, then X + α 2 X = 0, and X ( x ) = c 9 cos αx + c 10 sin αx . The auxiliary equation of T +2 kT + α 2 a 2 T = 0 is m 2 +2 km + α 2 a 2 = 0. Solving for m we obtain m = k ± k 2 α 2 a 2 . We consider three possibilities for the discriminant k 2 α 2 a 2 : (i) If k 2 α 2 a 2 = 0 then T ( t ) = c 11 e kt + c 12 te kt and u = XT = ( c 9 cos αx + c 10 sin αx )( c 11 e kt + c 12 te kt ) .
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AEM_3e_Chapter_13 - 13 13 Boundary-Value Problems in...

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