Solve ∂u
∂t = α
2 ∂
2u
∂x2 subject to the conditions
(i) u is not infinite as t −→ ∞
(ii) u = 0 for x = 0 and x = π for all t
(iii) u = πx − x
2
for t = 0 in (0, π)

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.

Solve ∂u/∂t = α^2 ∂^2u/∂x^2 subject to the conditions (i) u is not infinite as t → ∞ (ii) u = 0 for x = 0 and x = π for all t (iii) u = πx − x^2 for t = 0 in (0, π).

Explore the solution to the heat equation ∂u/∂t = α^2 ∂^2u/∂x^2 with specific boundary and initial conditions. This solution involves separation of variables, Fourier series, and the calculation of Fourier coefficients. Suitable for advanced calculus or university-level students.

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