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vpde_lecture32 [2020/04/16 12:34] (current) trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] |
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| ====== Lecture 33: Uniqueness of solutions I ====== | ====== Lecture 33: Uniqueness of solutions I ====== | ||
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| In this lecture, we'll prove uniqueness for two variants of the heat equation. The different variants you should consider as as follows (all posed on x∈[0,L]): | In this lecture, we'll prove uniqueness for two variants of the heat equation. The different variants you should consider as as follows (all posed on x∈[0,L]): | ||
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| so the energy is always decreasing. But note that E′(0)=0 since w(x,0)=0. Finally, note that E(t) is always ≥0 by its form (the integral of a squared quantity). So the energy is always decreasing, begins from zero, and can never be negative. We have the three statements: | so the energy is always decreasing. But note that E′(0)=0 since w(x,0)=0. Finally, note that E(t) is always ≥0 by its form (the integral of a squared quantity). So the energy is always decreasing, begins from zero, and can never be negative. We have the three statements: | ||
| - | - E′(t)≤0 for all time | + | - E′(t)≤0 for all time, |
| - | - E(0)=0 | + | - E(0)=0, |
| - | - E(t)≥0 for all time | + | - E(t)≥0 for all time, |
| - | Therefore | + | and you would conclude that it has to remain at its initial value, and therefore |
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| E(t) \equiv 0 | E(t) \equiv 0 | ||
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| - | for all time. Looking at the form of the integrand, you would conclude that | + | for all time. Looking at the form of the integrand, you would conclude that the only way this occurs is if the integrand is itself zero, or |
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| w^2(x, t) = 0 | w^2(x, t) = 0 | ||
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| - | for all x∈[0,L] and for all t≥0. So u(x,t)≡v(x,t) and the solutions must be the same. | + | for all x∈[0,L] and for all t≥0. So $w = u - v \equiv 0$ and thus u(x, t) \equiv v(x, t)$ and the solutions must be the same. |
| ===== Section 19.2: Uniqueness for other BCs of the heat equation | ===== Section 19.2: Uniqueness for other BCs of the heat equation | ||