Question

solve the heat conduction equation: 8*∂^2u(x,t) /∂x^2 = ∂u(x,t)/∂t; for0<x<5 and t>0 the following boundary and initial conditions: u(0,t) = u(5,t) = 0, u(x,0) = 2sin(πx) − 4sin(2πx)

Answer 1

$$\partial u/\partial t=8\ (\partial^2\ u)/(\partial x^2\ )\ \ \ \ \ \ (0<x<5,\ t>0)$$

$$u(0,t)=u(5,t)=0\ \ \ \ (\ t\geq0),\ u(x,0)=2\sin \pi x-4\sin 2\pi x\ \ \ \ \ \ (0<x<5).$$

Using separation of variables, with $u(x,t)=X(x)T(t)$ :

$$X(x)\ T\dot (t)=kT(t) X^{\prime\prime} (x)\Rightarrow(T \dot(t))/kT(t)\ =(X^{\prime\prime} (x))/X(x)\ =-\lambda$$

$$u(0,t)=u(5,t)=0\ \ \ \ (\ t\geq0)\ \ \ \ \ \ \ \ \ \ \ X(0)=X(5)=0$$

We got an eigenvalue problem for X (x):

$$X^{\prime\prime} (x)+\lambda X(x)=0, X(0)=X(5)=0 \Rightarrow X_n (x)=\sin{\pi nx/5} ,\ \ \lambda_n=(n^2 \pi^2)/25,\ n=1,2,\dots$$

On substituting eigenvalue to differential equation, we obtain

$$(\dot T_n(t))/(8T_n(t))=-\lambda_n=\frac{n^2\pi^2}{25} \Rightarrow \dot T(t)+8 \frac{n^2\pi^2}{25} T_n (t)=0$$ $$T_n(t) = a_n \exp{\left(-8 \frac{n^2\pi^2}{25}t\right)}$$

General Solution:

$$u\left(x,t\right)=\sum_{n=0}^{\infty\ \ }{T_n\left(t\right)X_n\left(x\right)}=\sum_{n=0}^{\infty\ \ }{a_n\exp{\left(-8 \frac{n^2\pi^2}{25}t\right)}\sin{\left(\pi nx/5\right)}}$$

Applying initial condition:

$$u\left(x,0\right)=\sum_{n=0}^{\infty\ \ }{a_n\sin{\left(\pi nx/5\right)}} = 2\sin{(\pi x)} - 4 \sin{(2\pi x)}$$

As X_N(x) is orthonormal basis, $a_5 = 2, a_{10} = -4,\, and \, a_n=0$ all other possible values of n.

Finally:

$$u\left(x,t\right)=2\exp{(-8\pi^2 t)} \sin{\left(\pi x\right)} - 4\exp{(-32\pi^2 t)} \sin{\left(2 \pi x\right)}$$