Question

In each exercise, use representation (8) to obtain the solution of the two- dimensional heat equation \(u_{t}(x, y, t)=\kappa\left[u_{x x}(x, y, t)+u_{y y}(x, y, t)\right], 0

Short answer

Question: Find the final solution for the two-dimensional heat equation given the initial conditions and the boundary values. Answer: The final solution for the two-dimensional heat equation is: $$ u(x, y, t) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{16}{mn\pi^2} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-\kappa(\lambda_{mn}^2) t} $$ for \(m\) and \(n\) odd.

Step-by-step solution

Using representation (8) to find the solution

According to representation (8), the solution can be found as the infinite series: $$ u(x, y, t) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} B_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-\kappa(\lambda_{mn}^2) t} $$ where \(\lambda_{mn}^2 = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2\) and \(B_{mn}\) is the coefficient, obtained by enforcing the initial condition.

Applying the initial condition

Using the initial condition \(u(x, y, 0) = f(x, y)\): $$ 8 \sin^2\left(\frac{\pi x}{a}\right) \sin^2\left(\frac{\pi y}{b}\right) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} B_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) $$ Since the sines are orthogonal in their intervals, we can find each term \(B_{mn}\) by multiplying both sides with \(\sin\left(\frac{m' \pi x}{a}\right) \sin\left(\frac{n' \pi y}{b}\right)\) and integrating over x and y: $$ B_{mn} = \frac{16}{ab} \int_0^a \int_0^b \sin^2\left(\frac{\pi x}{a}\right) \sin^2\left(\frac{\pi y}{b}\right) \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) dy dx $$

Compute B_mn

Calculate the terms and simplify when possible: $$ B_{mn} = \begin{cases} 0, & \text{if} \ m \ \text{or} \ n \ \text{is even} \\ \frac{16}{mn\pi^2}, & \text{if} \ m \ \text{and} \ n \ \text{are odd} \end{cases} $$ This result is because the integrals vanish when \(m\) or \(n\) is even. Now we have all the information needed to write the solution for the heat equation.

Write the final solution

Substitute \(B_{mn}\) back into the solution and simplify: $$ u(x, y, t) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{16}{mn\pi^2} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-\kappa(\lambda_{mn}^2) t} $$ for \(m\) and \(n\) odd. This is the final solution of the two-dimensional heat equation given the initial conditions and the boundary values.

Key concepts

Boundary Value Problems

When studying differential equations, particularly the heat equation, boundary value problems play a crucial role. In our exercise with the two-dimensional heat equation, boundary values are specified at the perimeter of a rectangular domain. Here, the solution must vanish on the boundary:- This means that at every point along the edges of the rectangle, the function representing temperature is zero.- Mathematically, this is expressed as \( u(x, y, t) = 0 \) when \( x = 0 \), \( x = a \), \( y = 0 \), and \( y = b \).In tackling these types of problems, we are interested in finding a function that satisfies both the heat equation and these specified conditions at the boundary edges. This approach helps ensure that the solution is appropriately bounded and behaves as expected under the physical constraints of the problem.

Fourier Series

Fourier series serve as a powerful tool to solve partial differential equations like the heat equation. In our context, they allow us to express the solution as an infinite series of sines and cosines: - The solution involves coefficients and terms as seen in the original representation (8) in the exercise. - Our task is to find these coefficients, which depend on initial conditions. This series expansion is beneficial because it transforms a complex differential equation into a simpler algebraic problem. The result is then a superposition of harmonic functions which satisfies the boundary and initial conditions inherent in the heat equation problem. Understanding Fourier series is essential for simplifying and solving these sophisticated mathematical models.

Orthogonal Functions

Orthogonal functions come into play when working with Fourier series since they provide a convenient basis for function decomposition. In our problem, the functions are sines with specific arguments:- These sine functions are orthogonal over the intervals \( [0, a] \) and \( [0, b] \), simplifying the process of finding coefficients in the solution series.- Orthogonality implies that the integral of the product of different sine functions over their defined interval is zero.When applying this property of orthogonality, we can separate the terms of our function, allowing each term in the series to be identified and solved independently. This simplifies solving complex boundary value problems by reducing many problems to separate easier ones.

Initial Conditions

Initial conditions are given values or functions that a solution must satisfy at the beginning of the observed process. For the heat equation, this means setting the temperature distribution at time \( t = 0 \):- Our specific initial condition is given by the function \( f(x, y) = 8 \sin^2\left(\frac{\pi x}{a}\right) \sin^2\left(\frac{\pi y}{b}\right) \).- This function delineates the initial temperature across the rectangular domain.By applying these conditions to our series representation of the solution, we can solve for the coefficients \( B_{mn} \), crucial for assembling the complete solution. Accurately incorporating initial conditions ensures that the final solution accurately reflects the process's starting state, which is fundamental for modeling real-world scenarios.