Mathematics · Partial Differential Equations
Let u = u(x, y) be the solution of the boundary value problem: ∂2u∂x2 + ∂2u∂y2 = 0, (x, y) ∈ (0, 1) × (0, 1) u(x, 0) = eπx, u(x, 1) = -eπx, x ∈ [0, 1] u(0, y) = cos(πy) + sin(πy), y ∈ [0, 1] u(1, y) = eπ(cos(πy) + sin(πy)), y ∈ [0, 1] Then there exists a point (x0, y0) ∈ (0, 1) × (0, 1) such that:
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 1u(x0, y0) = -eπ
- 2u(x0, y0) = -1
- 3u(x0, y0) = eπ
- 4u(x0, y0) = √2 eπ✓
Correct answer
u(x0, y0) = √2 eπ
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