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: