### Instruction:
Solve the heat equation ∂u/∂t = α^2∂^2u/∂x^2 with boundary conditions u(0,t) = 0 and u(L,t) = 0.

### Response:

To solve the heat equation with given boundary conditions, we will use the method of separation of variables. We assume that the solution u(x,t) can be written as a product of two functions, one depending only on x and the other depending only on t:

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

Now let's plug this into the heat equation:

∂u/∂t = α^2 ∂²u/∂x²
X(x) dT/dt = α^2 T(t) d²X/dx²

Now divide both sides by α^2 XT:

(1/α^2T) dT/dt = (1/X) d²X/dx²

Notice that the left side depends only on t and the right side depends only on x. Since these are independent variables, both sides must be equal to a constant. Let's call this constant -λ:

(1/α^2T) dT/dt = -λ
(1/X) d²X/dx² = -λ

Now we have two ordinary differential equations to solve:

d²X/dx² + λX = 0
dT/dt + α^2λT = 0

Let's first solve the second equation for T(t):

dT/dt + α^2λT = 0

This is a first-order linear ODE, which has the general solution:

T(t) = Ce^(-α^2λt), where C is an arbitrary constant.

Now let's solve the first equation for X(x):

d²X/dx² + λX = 0

This is a second-order linear homogeneous ODE with constant coefficients. The form of the solution depends on the sign of λ:

1. If λ > 0, then X(x) = A cos(√λx) + B sin(√λx)
2. If λ = 0, then X(x) = Ax + B
3. If λ < 0, then X(x) = A e^(√(-λ)x) + B e^(-√(-λ)x)

Now we apply the boundary conditions u(0,t) = 0 and u(L,t) = 0:

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

From the first boundary condition, we can see that only cases 1 and 3 satisfy this condition. However, case 3 leads to exponential growth or decay in x, which is not physically reasonable for a heat conduction problem with fixed endpoints at zero temperature. Therefore, we consider only case 1.

So, X(x) = B sin(√λx), and since X(0) = 0, we don't need to worry about the cosine term.

Now let's apply the second boundary condition, X(L) = 0:

B sin(√λL) = 0

Since B cannot be zero (otherwise the solution would be trivial), we must have sin(√λL) = 0. This occurs when √λL = nπ, where n is an integer. Thus, λ_n = (nπ/L)², and the corresponding eigenfunctions are:

X_n(x) = sin(nπx/L)

The general solution of the heat equation is given by a linear combination of these eigenfunctions multiplied by their time-dependent parts:

u(x,t) = Σ C_n sin(nπx/L)e^(-α^2(nπ/L)²t)

To find the coefficients C_n, we need to know the initial temperature distribution u(x,0). Let's call this function f(x):

f(x) = Σ C_n sin(nπx/L)

To find the coefficients C_n, we can use Fourier sine series:

C_n = (2/L) ∫[0,L] f(x)sin(nπx/L) dx

Finally, the solution of the heat equation with given boundary conditions is:

u(x,t) = Σ (2/L) (∫[0,L] f(x)sin(nπx/L) dx) sin(nπx/L)e^(-α^2(nπ/L)²t)