Question

The ends 𝑥 = 0 and 𝑥 = 10 of a thin aluminum bar (𝛼 2 = 0.86) are kept at 0 𝑜𝐶, while the surface of the bar is insulated. Find an expression for the temperature 𝑢(𝑥,𝑡) in the bar if initially

𝑢(𝑥, 0) = { 10𝑥 0 < 𝑥 < 5 10(10 − 10𝑥) 5 ≤ 𝑥 < 10 

Answer 1

Solution;

x=0 ; x=10

The heat equation is;

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

The general solution of the heat equation is in the form;

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

The boundary conditions are;

$(i)u(0,t)=0$ ; $t\geq0$

$(ii)u(10,t)=0$ ; $t>0$

$(iii)u(x,0)=f(x)=10x$ ; $0<x<5$ ;$10(10-10x);5\leq x<10$

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

$0=Ae^{-0.86\lambda^2t}$

Hence;

$A=0$

Equation (2) reduces to ;

$u(x,t)=Bsin\lambda xe^{-0.86\lambda^2t}...(3)$

Apply condition (ii) in (3);

$0=Bsin10\lambda e^{-0.86\lambda^2t}$

$i.e$ $10\lambda=nπ$ ;$\lambda=\frac{nπ}{10}$ (n is an integer)

Substituting in (3);

$u(x,t)=Bsin\frac{nπx}{10}e^{-0.86\frac{n^2π^2}{100}t}$

The most general solution is;

$u(x,t)=\displaystyle\sum_{n=1}^{\infin}B_nsin\frac{nπx}{10}e^{\frac{-0.86n^2π^2}{100}t}$ .......(4)

Apply condition (iii) in (4);

$u(x,0)=\displaystyle\sum_{n=1}^{\infin}B_nsin\frac{nπx}{10}=f(x)$

The LHS series is the half range Fourier sine series of the RHS functions.

Therefore;

$B_n=\frac{2}{10}[\int_0^5f(x)sin\frac{nπx}{10}dx+\int_5^{10}f(x)sin\frac{nπx}{10}dx]$

$B_n=\frac{2}{10}[\int_0^510xsin\frac{nπx}{10}dx+\int_5^{10}10(10-10x)sin\frac{nπx}{10}dx]$

Integrate and simplify to obtain;

$B_n=\frac{(90x-100)cos\frac{nπx}{10}}{πn}-\frac{900sin\frac{nπx}{10}}{n^2π^2}$

Replace B_n in (3) ;

$u(x,t)=(\frac{(90x-100)cos\frac{nπx}{10}}{πn}-\frac{900sin\frac{nπx}{10}}{n^2π^2})e^{\frac{-0.86n^2π^2t}{100}}$