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**Unformatted text preview: **Theorem 12.1. Suppose u ( t, x, y ) is a solution to the forced heat equation u t = γ Δ u + F ( t, x, y ) , for ( x, y ) ∈ Ω , < t < c, where Ω is a bounded domain, and γ > . Suppose F ( t, x, y ) ≤ for all ( x, y ) ∈ Ω and ≤ t ≤ c . Then the global maximum of u on the set { ( t, x, y ) | ( x, y ) ∈ Ω , ≤ t ≤ c } , occurs either when t = 0 , or at a boundary point ( x, y ) ∈ ∂ Ω . Proof : First, let us prove the result under the assumption that F ( t, x, y ) < 0 ev- erywhere. At a local interior maximum, u t = 0, and, since its Hessian matrix must be negative semi-definite, u xx , u yy ≤ 0, which would imply that u t − γ Δ u ≥ 0. If the maxi- mum were to occur when t = c , then u t ≥ 0 there, and also u xx , u yy ≤ 0, leading again to a contradiction. To generalize to the case when F ( t, x, y ) ≤ 0, which includes the heat equation when F ( t, x, y ) ≡ 0, set v ( t, x, y ) = u ( t, x, y ) + ε ( x 2 + y 2 ) , where ε > . Then, ∂v ∂t = γ Δ v − 4 γ ε + F ( t, x, y ) = γ Δ v + tildewide F ( t, x, y ) , where tildewide F ( t, x, y ) = F ( t, x, y ) − 4 γ ε < . Thus, by the previous paragraph, the maximum of v occurs either when t = 0 or at a boundary point ( x, y ) ∈ ∂ Ω. Now we let ε → 0 and conclude the same for u . More precisely, let u ( t, x, y ) ≤ M on t = 0 or ( x, y ) ∈ ∂ Ω. Then v ( t, x, y ) ≤ M + C ε where C = max braceleftbig x 2 + y 2 vextendsingle vextendsingle ( x, y ) ∈ ∂ Ω bracerightbig is finite since Ω is a bounded domain. Thus, u ( t, x, y ) ≤ v ( t, x, y ) ≤ M + C ε....

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