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If Ω is bounded 3-dimensional volume use the Maximum Principle to show the solution to the following Poisson Problem is unique −DΔu=f(x,y,z);(x,y,z)∈Ω u(x,y,z)=g(x,y,z);(x,y,z)∈∂Ω

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Final Answer:

The solution to the Poisson Problem −DΔu=f(x,y,z);(x,y,z)∈Ω, u(x,y,z)=g(x,y,z);(x,y,z)∈∂Ω, is unique in the bounded 3-dimensional volume Ω by applying the Maximum Principle.

Step-by-step explanation:

To establish the uniqueness of the solution, we employ the Maximum Principle. Suppose there exist two solutions, u₁ and u₂, both satisfying the given Poisson Problem. Define their difference as v = u₁ - u₂. Since both u₁ and u₂ satisfy the same boundary conditions, v vanishes on ∂Ω. Now, consider the interior of Ω.

The Maximum Principle states that if v attains a maximum or minimum in the interior of Ω, then it must be constant throughout Ω. Assume, for contradiction, that v attains a non-zero maximum at some point (x₀, y₀, z₀) in Ω.

This leads to a contradiction because Δv = Δ(u₁ - u₂) = Δu₁ - Δu₂ = f - f = 0 in Ω. Since v is non-constant and Δv = 0, the Maximum Principle implies that v must attain its maximum at the boundary ∂Ω. However, on ∂Ω, v = 0. Thus, the assumption that v has a non-zero maximum in the interior is false.

By a similar argument, we can show that v cannot have a non-zero minimum in the interior. Therefore, v must be identically zero throughout Ω, establishing the uniqueness of the solution.

User Giuseppe Urso
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