Obtain the temperature distribution u(x,t) in a literally 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 find the temperature distribution, u(x,t), in the insulated rod, we can use the heat equation. The heat equation describes how heat (or temperature) is transferred through a medium over time.

The heat equation is given by:

∂u/∂t = α * ∂²u/∂x²

where u(x,t) is the temperature distribution, α is the thermal diffusivity of the material, and ∂ represents partial differentiation with respect to the variable.

In this case, the rod is insulated, which means that there is no heat exchange with the surroundings. Therefore, the boundary conditions are:

u(0, t) = 0 (left end of the rod)
u(L, t) = 0 (right end of the rod)

The initial temperature distribution in the rod is given by:

u(x, 0) = 6sin(πx/L)

To solve the heat equation, we need to use separation of variables. We can assume that the solution can be written as a product of two functions:

u(x,t) = X(x) * T(t)

Substituting this into the heat equation gives:

X(x) * dT/dt = α * d²X/dx² * T(t)

Dividing both sides of the equation by X(x) * T(t) gives:

1/T(t) * dT/dt = α/X(x) * d²X/dx²

Since the left side depends only on time t and the right side depends only on position x, both sides must be equal to a constant, which we will call -λ²:

1/T(t) * dT/dt = -λ²
α/X(x) * d²X/dx² = -λ²

Now we have two separate ordinary differential equations: one for T(t) and one for X(x).

For T(t), the equation becomes:

1/T(t) * dT/dt = -λ²

This is a simple first-order ordinary differential equation with the general solution:

T(t) = Ce^(-λ²t)

For X(x), the equation becomes:

α/X(x) * d²X/dx² = -λ²

This is a second-order ordinary differential equation with the general solution:

X(x) = A*sin(λx) + B*cos(λx)

Applying the boundary conditions, u(0,t) = 0 and u(L,t) = 0, we have:

u(0,t) = X(0) * T(t) = 0
u(L,t) = X(L) * T(t) = 0

From these conditions, we can conclude that sin(0) = 0 and sin(λL) = 0. The only possible values for λ to satisfy these conditions are λ = nπ/L, where n is an integer. Therefore, the general solution for X(x) becomes:

X(x) = A*sin(nπx/L)

Finally, combining the solutions for T(t) and X(x), the temperature distribution u(x,t) in the insulated rod is given by:

u(x,t) = Σ(A*sin(nπx/L) * Ce^(-n²π²αt/L²))

where the summation is over all possible values of n.

To find the specific temperature distribution u(x,t) given the initial condition u(x,0) = 6sin(πx/L), we need to find the coefficients A and C. This can be done by applying the initial condition to the general solution and using Fourier series or other techniques to solve for A.

Note: The above explanation gives a general framework for solving the problem, but the specific solution will depend on the values of L and α, and the time t at which you want to find the temperature distribution.