### Author Topic: Web bonus problem : Week 4 (#4)  (Read 2284 times)

#### Victor Ivrii

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##### Web bonus problem : Week 4 (#4)
« on: October 04, 2015, 05:52:01 AM »

#### Zaihao Zhou

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##### Re: Web bonus problem : Week 4 (#4)
« Reply #1 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
$$-\frac{\beta_1}{\beta_0} = \frac{b}{c^2} \ \ \ \rightarrow \ \ \ b = - \frac{\beta_1}{\beta_0}c^2$$
$$-\frac{\beta_1}{\beta_0} = -\frac{a}{c^2} \ \ \ \rightarrow \ \ \ a = \frac{\beta_1}{\beta_0}c^2$$