The given partial differential equation (PDE) is the heat equation:

$\frac{\partial u}{\partial t} = \alpha^2 \frac{\partial^2 u}{\partial x^2}$

with the following boundary and initial conditions:

1. Boundary condition (i): $u$ is not infinite as $t \to \infty$.
2. Boundary condition (ii): $u = 0$ for $x = 0$ and $x = \pi$ for all $t$.
3. Initial condition (iii): $u(x, 0) = \pi x - x^2$ for $0 < x < \pi$.

Solution Approach:

To solve this PDE, we use the method of separation of variables. Assume a solution of the form:

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

Substituting into the PDE, we get:

$X(x) \frac{dT(t)}{dt} = \alpha^2 T(t) \frac{d^2X(x)}{dx^2}$

Dividing both sides by $\alpha^2 X(x) T(t)$, we have:

$\frac{1}{\alpha^2 T(t)} \frac{dT(t)}{dt} = \frac{1}{X(x)} \frac{d^2X(x)}{dx^2} = -\lambda$

Here, $\lambda$ is a separation constant. This gives us two ordinary differential equations (ODEs):

1. For $T(t)$:

   $\frac{dT(t)}{dt} + \alpha^2 \lambda T(t) = 0$

   Solving this, we get:

   $T(t) = A e^{-\alpha^2 \lambda t}$

2. For $X(x)$:

   $\frac{d^2X(x)}{dx^2} + \lambda X(x) = 0$

   Solving this, we get:

   $X(x) = B \cos(\sqrt{\lambda} x) + C \sin(\sqrt{\lambda} x)$

Applying Boundary Conditions:

From the boundary condition $u = 0$ at $x = 0$ and $x = \pi$:

1. At $x = 0$:

   $X(0) = B = 0$

   Therefore, $X(x) = C \sin(\sqrt{\lambda} x)$.

2. At $x = \pi$:

   $X(\pi) = C \sin(\sqrt{\lambda} \pi) = 0$

   For non-trivial solutions ($C \neq 0$), $\sin(\sqrt{\lambda} \pi) = 0$. Thus:

   $\sqrt{\lambda} \pi = n\pi \implies \sqrt{\lambda} = n \implies \lambda = n^2$

where $n$ is a positive integer.

General Solution:

$u(x, t) = \sum_{n=1}^{\infty} C_n \sin(nx) e^{-\alpha^2 n^2 t}$

Applying Initial Condition:

Using $u(x, 0) = \pi x - x^2$:

$\pi x - x^2 = \sum_{n=1}^{\infty} C_n \sin(nx)$

This is a Fourier sine series, and we find coefficients $C_n$ using:

$C_n = \frac{2}{\pi} \int_0^{\pi} (\pi x - x^2) \sin(nx) \, dx$

Now, let's compute $C_n$:

$C_n = \frac{2}{\pi} \left[ \int_0^{\pi} \pi x \sin(nx) \, dx - \int_0^{\pi} x^2 \sin(nx) \, dx \right]$

Solving these integrals will give us the coefficients $C_n$.

Final Solution:

After finding $C_n$, the final solution is:

$u(x, t) = \sum_{n=1}^{\infty} C_n \sin(nx) e^{-\alpha^2 n^2 t}$

Would you like to see the step-by-step calculation of $C_n$, or do you have any other questions?

Related Questions:

1. How do you calculate Fourier coefficients for non-standard functions?
2. How does the solution change if the boundary conditions are altered?
3. What happens to the solution as $\alpha$ changes?
4. How would the solution differ for a different initial condition?
5. Can this method be used for non-homogeneous boundary conditions?

Tip: To solve PDEs with separation of variables, always carefully match initial and boundary conditions to determine constants correctly.