Homework Eight.

82

$$u_t = \alpha^2 u_{xx},$$

$$u(0,t)=T_1, u(l,t) = T_2, u(x,0) =f(x),$$

$\alpha^2($silver$) = 1.71$ cm$^2$/sec,

$T_1 = 30 C, T_2 = 50C,$

$$f(x) = A x^2 + B x + C,$$

Where constants $A$, $B$ and $C$ are determined as following:

$A= -20$ degrees Centigrade over meter $^2$,

$B= 50$ degrees Centigrade over meter,

$C = 30$ degrees Centigrade.

83

$$u(x,t) = 2 e^{-\frac{\pi^2 t}{16}}\sin \frac{\pi x}{2} - e^{-\frac{\pi^2 t}{4}}\sin \pi x + 4 e^{-\pi^2 t}\sin 2\pi x.$$

84

$$X''(x)-\lambda [X'(x)+X(x)] =0,$$

$$\dot T(t)+\lambda T(t)=0.$$

85

$$x X(x) -\lambda X(x) = 0,$$

$$\dot T(t)+\lambda T(t)=0.$$

86

Solution can not be presented as a product of function of $x$ only and a function of $t$ only.

87

$$u(x,t)=\sum\limits_{n=1}^\infty c_n e^{-\frac{n^2\pi^2\alpha^2}{l^2}t}\sin\frac{n \pi x}{l},$$

$$c_n = \frac{100}{n\pi}\Big(\cos\frac{n\pi}{4}-\cos\frac{3 n\pi}{4} \Big).$$

88

$$u(x,t) = \sum\limits_{n=1}^\infty c_n e^{-\frac{n^2 \pi^2 \alpha^2 t}{l}}\sin\frac{n\pi x}{l},$$

$$c_n = \frac{400}{(2m+1)\pi}, n =2 m+1,$$

$$u(x,t)=\frac{400}{\pi}\sum\limits_{m=0}^{\infty} e^{-\frac{(2 m+1)^2\pi^2\alpha^2 t}{400}} \sin\frac{(2m+1)\pi x}{20},$$

(a) Silver $\alpha^2 = 1.71 {\rm cm}^2/{\rm sec},$ $u(10,30)=35.9$ Centigrade

(b) Aluminum $\alpha^2 = 0.86 {\rm cm}^2/{\rm sec},$ $u(10,30)=67.2$ Centigrade

(c) Cast iron $\alpha^2 = 0.12 {\rm cm}^2/{\rm sec},$ $u(10,30)=97.4$ Centigrade

$$u(x,t)\simeq \frac{400}{\pi}e^{-\frac{\pi^2\alpha^2 t}{400}}\sin\frac{x\pi}{20},$$

$$t=\frac{400}{\alpha^2\pi^2}\log\frac{400}{25\pi},$$

(a) $t=38.6$ sec

(b) $t =76.7$ sec

(c) $t=550$ sec