Paper 3, Section II, C

Partial Differential Equations | Part II, 2013

Define the parabolic boundary parΩT\partial_{p a r} \Omega_{T} of the domain ΩT=[0,1]×(0,T]\Omega_{T}=[0,1] \times(0, T] for T>0T>0.

Let u=u(x,t)u=u(x, t) be a smooth real-valued function on ΩT\Omega_{T} which satisfies the inequality

utauxx+bux+cu0u_{t}-a u_{x x}+b u_{x}+c u \leqslant 0

Assume that the coefficients a,ba, b and cc are smooth functions and that there exist positive constants m,Mm, M such that maMm \leqslant a \leqslant M everywhere, and c0c \geqslant 0. Prove that

max(x,t)ΩˉTu(x,t)max(x,t)par ΩTu+(x,t).\max _{(x, t) \in \bar{\Omega}_{T}} u(x, t) \leqslant \max _{(x, t) \in \partial_{\text {par }} \Omega_{T}} u^{+}(x, t) .

[Here u+=max{u,0}u^{+}=\max \{u, 0\} is the positive part of the function uu.]

Consider a smooth real-valued function ϕ\phi on ΩT\Omega_{T} such that

ϕtϕxx(1ϕ2)ϕ=0,ϕ(x,0)=f(x)\phi_{t}-\phi_{x x}-\left(1-\phi^{2}\right) \phi=0, \quad \phi(x, 0)=f(x)

everywhere, and ϕ(0,t)=1=ϕ(1,t)\phi(0, t)=1=\phi(1, t) for all t0t \geqslant 0. Deduce from ()(*) that if f(x)1f(x) \leqslant 1 for all x[0,1]x \in[0,1] then ϕ(x,t)1\phi(x, t) \leqslant 1 for all (x,t)ΩT(x, t) \in \Omega_{T}. [Hint: Consider u=ϕ21u=\phi^{2}-1 and compute utuxx]\left.u_{t}-u_{x x} \cdot\right]

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