Obtain the temperature distribution u(x,t) in a laterally insulated rod of length L if both ends on the road are kept at 0°c and the initial temperature distribution in the bar is:
u(x,0) = 6sin[πx/L]
To obtain the temperature distribution u(x,t) in a laterally insulated rod of length L with both ends kept at 0°C, we can use the heat equation and apply appropriate boundary conditions and initial conditions.
The heat equation is given by:
∂u/∂t = α * (∂^2u/∂x^2)
where α is the thermal diffusivity, and u(x, t) represents the temperature distribution at position x and time t.
Boundary conditions:
- Both ends of the rod are kept at 0°C, which means the temperature at the ends remains constant throughout. This can be written as u(0, t) = u(L, t) = 0.
Initial condition:
- The initial temperature distribution in the bar is given as u(x, 0) = 6sin(πx/L).
To solve this problem, we need to apply the method of separation of variables. We assume that the solution can be written as a product of two functions:
u(x, t) = X(x) * T(t)
We substitute this into the heat equation:
X(x) * dT/dt = α * d^2X/dx^2 * T(t)
Then, we divide through by X(x) * T(t) and rearrange terms to separate the variables:
1/(α * T(t)) * dT/dt = 1/(X(x)) * d^2X/dx^2 = -λ^2
where λ is a separation constant.
Rearranging the first equation, we get:
1/(α * T(t)) * dT/dt = -λ^2
This can be solved by separation of variables:
1/(α * T(t)) * dT = -λ^2 * dt
Integrating both sides, we have:
(1/(α * T(t))) * dT = -λ^2 * dt
Integrating again, we get:
ln(T(t))/(α * λ^2) = -t + C_1
where C_1 is the constant of integration.
Solving for T(t), we have:
T(t) = C_2 * exp(-(α * λ^2 * t))
where C_2 = exp(C_1)
Next, we solve the second ODE:
1/(X(x)) * d^2X/dx^2 = -λ^2
This is a second-order linear ordinary differential equation (ODE) with constant coefficients. The general solution is given by:
X(x) = A * cos(λx) + B * sin(λx)
Applying the boundary conditions u(0, t) = u(L, t) = 0, we have:
X(0) = A * cos(0) + B * sin(0) = A = 0
X(L) = A * cos(λL) + B * sin(λL) = B * sin(λL) = 0
Since sin(λL) = 0, this implies that λL = nπ, where n is an integer.
Therefore, λ = nπ/L
Substituting this back into the X(x) equation, we have:
X(x) = B * sin(nπx/L)
Now, combining the solutions for T(t) and X(x), we have:
u(x, t) = Σ [B * sin(nπx/L) * exp(-(α * (nπ/L)^2 * t))]
where Σ denotes a summation over all values of n.
To find the specific values of B for each n, we can use the initial condition u(x, 0) = 6sin(πx/L).
Substituting t = 0 into the above equation, we have:
u(x, 0) = Σ [B * sin(nπx/L) * exp(0)] = Σ [B * sin(nπx/L)] = 6sin(πx/L)
Comparing the functions on both sides, we need to match coefficients of sin(nπx/L) term by term.
Therefore, B = 6 for n = 1, and B = 0 for all other values of n.
Finally, we have:
u(x, t) = 6 * sin(πx/L) * exp(-(α * (π/L)^2 * t))
This is the temperature distribution u(x, t) in the laterally insulated rod of length L with both ends kept at 0°C and the initial temperature distribution given as u(x, 0) = 6sin(πx/L).