Question

(a) As in Example 2, use (14) and (15) to solve the initial value problem $$ \begin{aligned} &u_{t}(x, t)=u_{x x}(x, t), \quad 0

Short answer

Question: Using the solution to the given partial differential equation, determine the value of the solution u(x,t) at the point (x, t) = (l/2, 1). Answer: The value of the solution u(x,t) at the point (x, t) = (l/2, 1) is 0.

Step-by-step solution

Part (a): Solving the PDE

Given the PDE and boundary and initial conditions, we're going to apply the separation of variables technique. This means we want to find the function u(x, t) = X(x)*T(t) that satisfies the PDE and fits the given conditions. Then, the solution u(x, t) is the infinite sum of all such functions. According to the given formulas, we have: $$ u(x, t) = \sum_{n=1}^{\infty} B_n e^{-n^2 \pi^2 t} \cos n \pi x $$ where $$ B_n = 2 \int_{0}^{1} f(x) \cos n \pi x dx $$ We can find the \(B_n\) coefficients and evaluate the series using the provided function \(f(x) = \cos \pi x + \cos^2 \pi x\).

Finding \(B_n\) coefficients

We first calculate the \(B_n\) coefficients using the given function: \begin{align*} B_n &= 2 \int_{0}^{1} \left(\cos \pi x + \cos^2 \pi x\right) \cos n \pi x \ dx \\ &= 2 \int_{0}^{1} \cos \pi x \cos n \pi x \ dx + 2 \int_{0}^{1} \cos^2 \pi x \cos n \pi x \ dx \end{align*} Now, we can calculate each integral separately. First, the integral with the simple cosine product: $$ 2 \int_{0}^{1} \cos \pi x \cos n \pi x \ dx = \frac{1}{\pi}\begin{cases} 1 & \text{if}\ n = 1 \\ 0 & \text{if}\ n \neq 1 \\ \end{cases} $$ Next, the integral with the square cosine product: $$ 2 \int_{0}^{1} \cos^2 \pi x \cos n \pi x \ dx = \frac{1}{\pi}\begin{cases} 1/2 & \text{if}\ n = 1 \\ 0 & \text{if}\ n \neq 1 \\ \end{cases} $$ Combining these results, we obtain the \(B_n\) coefficients as: $$ B_n = \frac{1}{\pi}\begin{cases} 3/2 & \text{if}\ n = 1 \\ 0 & \text{if}\ n \neq 1 \\ \end{cases} $$

Writing the final solution for u(x, t)

Now, we plug the \(B_n\) coefficients into the u(x, t) formula and simplify: $$ u(x, t) = \frac{3}{2\pi} \, e^{-\pi^2 t} \cos \pi x $$

Part (b): Evaluating u(x, t) at (l/2, 1)

We need to find the value of u(x, t) at the point (x, t) = (1/2, 1). Plug the values into the formula we found in part (a): $$ u\left(\frac{1}{2}, 1\right) = \frac{3}{2\pi} \, e^{-\pi^2} \cos \frac{\pi}{2} $$ Since \(\cos\frac{\pi}{2}=0\), the value of the solution at (1/2,1) is: $$ u\left(\frac{1}{2}, 1\right) = 0 $$

Part (c): Finding the limit of u(x, t) as t approaches infinity

We need to show that: $$ \lim_{t \rightarrow \infty} u(x, t) = \frac{1}{l} \int_{0}^{l} f(x) \, dx $$ First, let's find the integral of f(x): $$ \int_{0}^{1} (\cos \pi x + \cos ^{2} \pi x) dx = \int_{0}^{1} \cos \pi x \, dx + \int_{0}^{1} \cos^2 \pi x \, dx $$ For the first integral, we have: $$ \int_{0}^{1} \cos \pi x \, dx = \left[ \frac{1}{\pi} \sin \pi x \right]_0^1 = 0 $$ For the second integral, we can use the identity \(\cos^2 x = \frac{1+\cos 2x}{2}\): $$ \int_{0}^{1}\cos^2 \pi x \, dx = \frac{1}{2}\int_{0}^{1} (1 + \cos 2\pi x) \, dx = \frac{1}{2}\left[ x + \frac{1}{2\pi}\sin{2\pi x} \right]_0^1 = \frac{1}{2} $$ So, the integral of f(x) is: $$ \int_{0}^{1} f(x) \, dx = 0 + \frac{1}{2} = \frac{1}{2} $$ Now, let's find the limit of u(x, t) as t approaches infinity: $$ \lim_{t \rightarrow \infty} u(x, t) = \lim_{t \rightarrow \infty} \frac{3}{2\pi} \, e^{-\pi^2 t} \cos \pi x $$ Since \(e^{-\pi^2 t}\) approaches 0 as t approaches infinity, we have: $$ \lim_{t \rightarrow \infty} u(x, t) = 0 $$ Now we compare the values: $$ \lim_{t \rightarrow \infty} u(x, t) = 0 \neq \frac{1}{l} \int_{0}^{l} f(x) \, dx = \frac{1}{2} $$ Our result contradicts the given statement. Thus, it has been shown that the limit of u(x,t) as t approaches infinity is not equal to the given integral expression.

Key concepts

Separation of Variables

The separation of variables is a fundamental method used to solve partial differential equations (PDEs). This method breaks the original PDE into simpler, solvable pieces by assuming the solution can be represented as a product of functions, each dependent on a single coordinate. For example, the heat equation given by \(u_{t} = u_{xx}\) can be expressed in terms of two separate functions: \(u(x, t) = X(x)T(t)\).
This approach reduces a complex PDE into ordinary differential equations (ODEs) for each variable.
Here's how it works in our specific problem:

• We set \(u(x, t) = X(x)T(t)\) and substitute into the PDE.
• This gives \(X(x)T'(t) = X''(x)T(t)\).
• Dividing both sides by \(X(x)T(t)\), we get a separation constant \(\lambda\): \(\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = \lambda\).
• Solving these ODEs provides the building blocks for the series solution.

This technique is particularly effective when dealing with boundary conditions, as we'll explore next.

Boundary Value Problems

Boundary value problems (BVPs) are problems that involve differential equations along with a set of conditions, called boundary conditions, that the solution must satisfy at specific points. In our PDE, we have:

• The equations \(u_{x}(0, t)=0\) and \(u_{x}(l, t)=0\) specify the behavior of the function at the boundaries \(x=0\) and \(x=l\).

These conditions indicate that the slope or derivative of \(u\) with respect to \(x\) is zero at both endpoints, implying a Neumann boundary condition.
This results in a stationary state at the boundaries, greatly influencing the series solution derived from separation of variables.
By solving the boundary value problems using the eigenfunctions derived via separation of variables, we construct the spatial component \(X(x)\), typically involving trigonometric functions like sines or cosines due to their natural fit with BVP constraints.

Initial Value Problems

An initial value problem (IVP) involves finding a function that satisfies a differential equation and meets specific conditions at an initial point. In our case, the initial condition is given by \(u(x, 0) = f(x)\). The function \(f(x)\) defines the state of the system at \(t=0\).
To solve this component of the PDE, we use the function \(f(x) = \cos \pi x + \cos^2 \pi x\) to determine the coefficients of our solution's series expansion. This is where we evaluate the series coefficients \(B_n\) through integration:

• Calculate \(B_n = 2 \int_{0}^{1} f(x) \cos n \pi x \, dx\).

Inserting these coefficients into the series solution generates a complete expression for \(u(x, t)\). The harmony between initial and boundary conditions allows us to predict the system's evolution over time.