2 Find the function u(x, t), dened for 0 ≤ x ≤ π and t ≥ 0, which satises the
following conditions:
∂u
∂t =
∂
2
u
∂x2
, u(0, t) = u(π, t) = 0, u(x, 0) = sin(x) + 3sin(2x) − 5sin(3x).

This problem involves solving the **one-dimensional heat equation**:

\[
\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}
\]

with the following boundary and initial conditions:

\[
u(0, t) = 0, \quad u(\pi, t) = 0 \quad \text{(boundary conditions)}
\]

\[
u(x, 0) = \sin(x) + 3\sin(2x) - 5\sin(3x) \quad \text{(initial condition)}.
\]

### Method: Separation of Variables

We solve this problem using **separation of variables**, where we assume the solution can be written as:

\[
u(x, t) = X(x) T(t),
\]
where \( X(x) \) depends only on \( x \) and \( T(t) \) depends only on \( t \).

Substitute this into the heat equation:

\[
\frac{d}{dt}(XT) = \frac{d^2}{dx^2}(XT),
\]
which simplifies to:
\[
X(x) \frac{dT(t)}{dt} = T(t) \frac{d^2X(x)}{dx^2}.
\]

Dividing both sides by \( X(x)T(t) \) gives:

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

This equation splits into two ordinary differential equations (ODEs):

1. **For \( T(t) \):**
   \[
   \frac{dT(t)}{dt} + \lambda T(t) = 0 \quad \Rightarrow \quad T(t) = e^{-\lambda t}.
   \]

2. **For \( X(x) \):**
   \[
   \frac{d^2X(x)}{dx^2} + \lambda X(x) = 0.
   \]

Given the boundary conditions \( u(0, t) = u(\pi, t) = 0 \), we require that \( X(0) = X(\pi) = 0 \). This is a standard boundary value problem for \( X(x) \), whose solutions are:

\[
X_n(x) = \sin(nx), \quad \lambda_n = n^2, \quad n = 1, 2, 3, \dots
\]

Thus, the general solution for \( u(x, t) \) is:

\[
u(x, t) = \sum_{n=1}^{\infty} A_n \sin(nx) e^{-n^2 t},
\]
where the constants \( A_n \) are determined by the initial condition \( u(x, 0) = \sin(x) + 3\sin(2x) - 5\sin(3x) \).

### Step 2: Use the Initial Condition

From the initial condition:

\[
u(x, 0) = \sin(x) + 3\sin(2x) - 5\sin(3x),
\]

we match this with the general solution at \( t = 0 \):

\[
u(x, 0) = \sum_{n=1}^{\infty} A_n \sin(nx).
\]

By comparing coefficients, we see that:

\[
A_1 = 1, \quad A_2 = 3, \quad A_3 = -5, \quad A_n = 0 \quad \text{for} \quad n > 3.
\]

### Final Solution

Thus, the solution to the heat equation is:

\[
u(x, t) = \sin(x) e^{-t} + 3\sin(2x) e^{-4t} - 5\sin(3x) e^{-9t}.
\]

This function satisfies the heat equation, the boundary conditions, and the initial condition.

---

Would you like more details on any part of the solution or have any questions?

Here are 5 questions that explore related topics:
1. How do separation of variables apply to solving PDEs like the heat equation?
2. What are the physical interpretations of the boundary and initial conditions in this problem?
3. How would the solution change if the initial condition was different?
4. Can you solve a heat equation with non-homogeneous boundary conditions?
5. What is the significance of the exponential decay terms in the solution?

**Tip:** When solving boundary value problems, carefully analyze the behavior of the function at the boundaries to select appropriate solution forms.

Find the function u(x, t), defined for 0 ≤ x ≤ π and t ≥ 0, which satisfies the following conditions: ∂u/∂t = ∂²u/∂x², u(0, t) = u(π, t) = 0, u(x, 0) = sin(x) + 3sin(2x) − 5sin(3x).

This page provides a detailed solution to the one-dimensional heat equation with boundary conditions u(0, t) = u(π, t) = 0 and initial condition u(x, 0) = sin(x) + 3sin(2x) − 5sin(3x). Learn how to apply the separation of variables method and derive the final solution.

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