Answer to Question #91732 in Differential Equations for ROHIT SHARMA

Question

Solve the heat conduction equation:
for the following boundary and initial conditions:
8(d^2u/dt^2)=(du/dt), for 0<X< 5 and t> 0,
u(0,t) = u(5,t) = 0,
u(x,0) =2sin(πx)− 4sin(2πx)

Answer 1

$$\frac{\partial u}{\partial T} = 8 \frac{\partial^2 u}{\partial x^2}$$ $$u(0, t) = u(5, t) = 0 \qquad u(x,0) = 2 \sin \pi x - 4 \sin 2\pi x$$

Let's find the solution in the form

$$u(x,t) = X(x) T(t)$$ $$X T' = 8 X'' T$$ $$\frac{T'}{T}(t) = 8\frac{X''}{X}(x) = -8\lambda = \mathrm{const}$$

where lambda is an unknown constant.

From the boundary conditions

$$X'' + \lambda X = 0 \qquad X(0) = X(5) = 0$$ $$X(x) = C \sin \frac{n \pi x}{5} \qquad \lambda = \bigg(\frac{\pi n}{5}\bigg)^2$$

where n is integer.

Hence

$$T' = -8\lambda T \rightarrow T(t) = C \exp(-8\lambda t)$$

The heat equation is linear, so the solution can be written in the form

$$u(x,t) = \sum_{n=1}^{\infty} C_n X_n (x) T_n(t)$$

where C_n are constants determined from the initial conditions and X_n, T_n are the functions found above with the integer parameter n (for n = 0 X₀(x)= 0).

From the initial conditions

$$u(x,t) = 2 \exp \bigg(-8\frac{\pi^2 5^2}{5^2}t\bigg)\sin \pi x - 4 \bigg(-8\frac{\pi^2 10^2}{5^2}t\bigg) \sin 2\pi x$$ $$u(x,t) = 2 e^{-8\pi^2 t}\sin \pi x - 4 e^{-32\pi^2 t} \sin 2\pi x$$