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).