Answer to Question #261404 in Calculus for mikes

Question

The ends and sides of a thin copper bar (𝛼 2 = 1.14) of length 2 are insulated so that no heat can pass through them. Find the temperature 𝑢(𝑥,𝑡) in the bar if initially 𝑢(𝑥, 0) = { 60𝑥 0 < 𝑥 < 1 60(2 − 𝑥) 1 ≤ 𝑥 < 2

Answer 1

The heat equation;

$\frac{\delta u}{\delta t}=\alpha^2\frac{\delta^2u}{\delta x^2}......(1)$

The generalized solution of the equation is;

$u(x,t)=(Acos\lambda x+Bsin\lambda x)e^{-\delta ^2\lambda^2t}.......(2)$

The boundary conditions are;

$(i)u(0,t)=0$

$(ii)u(2,t)=0$

$u(x,0)=60x;0<x<1$

$u(x,0)=60(2-x);1\leq x<2$

Applying condition (i) in (2),we have;

$u(0,t)=Ae^{-\alpha^2\lambda^2t}$

Hence ;

A=0

Equation (2) becomes;

$u(x,t)=Bsin\lambda x)e^{-\alpha^2\lambda^2t}$ ......(2a)

Apply condition (ii) into (2a),we have;

$0=Bsin(2\lambda)e^{-\alpha^2\lambda^2t}$

Take ;

$2\lambda=nπ$ ; $\lambda=\frac{nπ}{x}$

By substitution;

$u(x,t)=B_nsin(\frac{nπ}{2}x)e^{\frac{-\alpha^2n^2π^2}{4}t}$

The most general solution of the above equation is a series sum (n is an integer)

$u(x,t)=\displaystyle\sum_{n=1}^{\infin}B_nsin\frac{nπx}{2}e^{\frac{-1.14n^2π^2t}{4}}...(3)$

Using half ranger sin Fourier series;

$B_n=\frac{2}{l}[\int f(x)sin\frac{nπx}{2}dx]$

$B_n=\int_0^160xsin\frac{nπx}{2}dx+\int_1^260(2-x)sin\frac{nπx}{2}dx$

Integrate by parts and simplify;

$B_n=\frac{2}{nπ}$

Substitute $B_n$ into (3);

$u(x,t)=\displaystyle\sum_{n=1}^{\infin}\frac{2}{nπ}sin\frac{nπx}{2}e^{\frac{-1.14n^2π^2t}{4}}$