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| ====== Lecture 33: Uniqueness of solutions I ====== | ====== Lecture 33: Uniqueness of solutions I ====== | ||
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| + | < | ||
| + | <iframe width=" | ||
| + | </ | ||
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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]): | ||
| + | $$ | ||
| + | \begin{aligned} | ||
| + | u_t &= \kappa u_{xx}, \\ | ||
| + | u(x, 0) &= 0, | ||
| + | \end{aligned} | ||
| + | $$ | ||
| + | with either Dirichlet conditions [D] | ||
| + | $$ | ||
| + | \begin{aligned} | ||
| + | u &= T_0 \quad (x = 0), \\ | ||
| + | u(L, t) &= T_1\quad (x = L), | ||
| + | \end{aligned} | ||
| + | $$ | ||
| + | or Neumann conditions [N] | ||
| + | $$ | ||
| + | \begin{aligned} | ||
| + | u_x &= G_0 \quad (x = 0), \\ | ||
| + | u_x &= G_1\quad (x = L), | ||
| + | \end{aligned} | ||
| + | $$ | ||
| + | or mixed conditions [M] | ||
| + | $$ | ||
| + | \begin{gathered} | ||
| + | A_0 u + B_0 u_x = H_0 \quad (x = 0), \\ | ||
| + | A_1 u + B_1 u_x = H_1 \quad (x = L). | ||
| + | \end{gathered} | ||
| + | $$ | ||
| ===== Section 19.1: Uniqueness for zero Dirichlet heat equation | ===== Section 19.1: Uniqueness for zero Dirichlet heat equation | ||
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| E'(t) = \kappa w w_x \biggr\rvert_0^L - \kappa \int_0^L w_x^2 \, \mathrm{d}x = - \kappa \int_0^L w_x^2 \, \mathrm{d}x | E'(t) = \kappa w w_x \biggr\rvert_0^L - \kappa \int_0^L w_x^2 \, \mathrm{d}x = - \kappa \int_0^L w_x^2 \, \mathrm{d}x | ||
| $$ | $$ | ||
| - | where the second equality occurs because $w = 0atx = 0andx = L$. So in the end, we have the fact that | + | where the second equality occurs because $w(0, t) = 0 = w(L, t)$ [*]. So in the end, we have the fact that |
| $$ | $$ | ||
| E'(t) = - \kappa \int_0^L w_x^2 \, \mathrm{d}x \leq 0, | E'(t) = - \kappa \int_0^L w_x^2 \, \mathrm{d}x \leq 0, | ||
| $$ | $$ | ||
| - | 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. | + | 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. |
| + | - E′(t)≤0 for all time, | ||
| + | - E(0)=0, | ||
| + | - E(t)≥0 for all time, | ||
| + | |||
| + | and you would conclude that it has to remain at its initial value, and therefore | ||
| $$ | $$ | ||
| E(t) \equiv 0 | E(t) \equiv 0 | ||
| $$ | $$ | ||
| - | for all time. | + | 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 |
| + | $$ | ||
| + | w^2(x, t) = 0 | ||
| + | $$ | ||
| + | for all x∈[0,L] and for all t≥0. So w=u−v≡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 | ||
| + | |||
| + | The proof of uniqueness for the Neumann condition [N] is done identically to the above, except in the step marked with [*] we use the fact that wx(0,t)=0=wx(L,t). But you would arrive at the same conclusions as above. | ||