Differential Equations Solutions 180

Differential Equations Solutions 180 - 1 ˜ Ω = min w ±...

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190 Chapter 31. Solutions: Case Study: Eigenvalues: Valuable Principles since a 1 ( x ) ,a 2 ( x ) > 0. Suppose A w = λw . Then 0 ( w, A w )= λ ( w, w ) , so λ 0. (b) We know that λ 1 (Ω) = min w ± =0 ( w, A w ) ( w, w ) where the integrals are taken over Ω and w is constrained to be zero on the boundary of Ω. Suppose that the w that minimizes the function is ˜ w and let’s extend ˜ w to make it zero over the part of ˜ Ω not contained in Ω. Then
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Unformatted text preview: λ 1 ( ˜ Ω) = min w ± =0 ( w, A w ) ( w, w ) ≤ ( ˜ w, A ˜ w ) ( ˜ w, ˜ w ) = λ 1 (Ω) . CHALLENGE 31.4. From Challenge 1, we know that the smallest eigenvalue for a square with dimension b is 2 π 2 /b , so we want b = √ 2 / 2. Using MATLAB’s PDE Toolbox interactively, we discover that α ≈ 1 . 663....
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