# Toronto Math Forum

## APM346-2015F => APM346--Lectures => Web Bonus = Oct => Topic started by: Victor Ivrii on October 04, 2015, 05:52:01 AM

Title: Web bonus problem : Week 4 (#4)
Post by: Victor Ivrii on October 04, 2015, 05:52:01 AM
Problem 3 http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.7.P.html#problem-2.7.P.3 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.7.P.html#problem-2.7.P.3)
Title: Re: Web bonus problem : Week 4 (#4)
Post by: Zaihao Zhou on October 21, 2015, 10:02:55 PM
a)
$$\frac{\partial E(t)}{\partial t} = \int_0^\infty (u_tu_{tt} +c^2u_xu_{xt}) dx + au(0)u_t(0)$$
$$\frac{\partial E(t)}{\partial t} = \int_0^\infty (c^2u_tu_{xx} +c^2u_xu_{xt}) dx + au(0)u_t(0)$$
$$\frac{\partial E(t)}{\partial t} = c^2\int_0^\infty (u_tu_x)_x dx + au(0)u_t(0)$$
$$\frac{\partial E(t)}{\partial t} = u_t(au - c^2u_x)|_{x=0}$$
We need
$$\frac{\alpha_1}{\alpha_0} = \frac{a}{-c^2} \ \ \ \rightarrow \ \ \ a = - \frac{\alpha_1}{\alpha_0}c^2$$

b)
$$\frac{\partial E(t)}{\partial t} = \int_0^l (u_tu_{tt} +c^2u_xu_{xt}) dx + auu_t|_{x=0} + buu_t|_{x=l}$$
$$\frac{\partial E(t)}{\partial t} =u_t( c^2u_x +bu)|_{x=l} + u_t(au - c^2u_x)|_{x=0}$$
Thus we need
$$-\frac{\beta_1}{\beta_0} = \frac{b}{c^2} \ \ \ \rightarrow \ \ \ b = - \frac{\beta_1}{\beta_0}c^2$$
$$-\frac{\beta_1}{\beta_0} = -\frac{a}{c^2} \ \ \ \rightarrow \ \ \ a = \frac{\beta_1}{\beta_0}c^2$$