Answer to Question #91730 in Classical Mechanics for ss

Question

Solve the heat conduction equation:
for the following boundary and initial conditions:

u(0,t) = u(5,t) = 0,
u(x,0) =2sin(πx)− 4sin(2πx)

Answer 1

$$\frac{\partial u}{\partial t} = -\kappa \Delta u$$ $$u(0, t) = u(5, t) = 0$$ $$u(x, 0) = 2\sin(\pi x) -4\sin(2\pi x)$$

Let's find the solution in such a way:

$$u = T(t) X(x)$$ $$\frac{T'}{T}(t) = -\kappa \frac{X''}{X}(x) = -\kappa\lambda$$

where lambda is a constant.

$$X'' + \lambda X = 0$$ $$X(0) = X(5) = 0$$

hence

$$X = \sin \frac{n\pi}{5}x$$

while n is integer and

$$\lambda = \bigg(\frac{n\pi}{5}\bigg)^2$$

Thus

$$T' = - \kappa\lambda T$$ $$T = C e^{-\kappa\lambda t}$$

The heat conduction equation is linear, therefore a sum of solutions is also a solution.

$$u(x, 0) = 2\sin(\pi x) -4\sin(2\pi x)$$ $$\lambda_1 = \bigg(\frac{5\pi}{5}\bigg)^2, \qquad \lambda_2 = \bigg( \frac{5\cdot 2\pi}{5}\bigg)^2$$ $$u(x, t) = 2e^{-\kappa \pi^2 t} \sin(\pi x) -4e^{-4\kappa\pi^2 t}\sin(2\pi x)$$