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Unformatted text preview: Page 1 of 6 Y O U R N A M E Introduction to Partial Differential Equations AM 3815 - Test 3 - Solution Solve u xx + u yy = 0 for 0 < x < π , 0 < y < 1 subject to u ( x , 0) = cos x u ( x , 1) = 1 2 cos 2 x + 1 2 , u x (0, y ) = 0 u x ( π , y ) = 0. If applicable, you may use results from BVPs studied in class or on an assignment without showing how those results were obtained. Solution: We posit a separable solution, u ( x , y ) = X ( x ) Y ( y ). Substituting this into the given PDE we get- X 00 X = Y 00 Y = λ . The BVP problem corresponding to the X equation has Neumann BCs, and so λ = n 2 , n = 0, 1, 2, ... with eigenfunctions X = 1 2 , X n = cos nx since x ∈ (0, π ). For n = 0 the solution to the Y equation is Y ( y ) = A + B y . For n ≥ 1 the solution to the Y equation is Y n ( y ) = A n cosh ny + B n sinh ny , thus the separated solution is u ( x , y ) = 1 2 A + 1 2 B y + ∞ X n =1 A n cosh ny + B n sinh ny cos nx Page 2 of 6 Y O U R N A M E The coefficients A n must satisfy...
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- Spring '11
- Fourier Series, Cos, Partial differential equation, Bn sinh ny