Question

Solve the initial-boundary value problem. Perform numerical experiments for specific values of \(L\) and \(a\). $$ \begin{aligned} &\text { C } u_{t}=a^{2} u_{x x}, \quad 00\\\ &u_{x}(0, t)=0, \quad u(L, t)=0, \quad t>0\\\ &u(x, 0)=\left\\{\begin{array}{ll} 1, & 0 \leq x \leq \frac{L}{2}, \\ 0, & \frac{L}{2}

Short answer

Question: Determine the form of the solution to the initial-boundary value problem given by the heat equation \(C u_t = a^2 u_{xx}\), with given boundary and initial conditions. Answer: The solution to the initial-boundary value problem is given by: $$u(x, t) = \sum_{n=1}^\infty \frac{2}{n\pi}\left(1-(-1)^n\right)\cos\left(\frac{n\pi x}{L}\right)e^{-\left(\frac{n\pi a}{L}\right)^2 \frac{t}{C}}.$$

Step-by-step solution

Separate Variables

Assume that the solution \(u(x, t)\) is of the form \(u(x, t) = X(x)T(t)\). Substitute this into the heat equation: $$C(X(x)T'(t)) = a^2(X''(x) T(t)).$$ Now, divide both sides by \(a^2 X(x)T(t)\): $$\frac{C}{a^2}\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)}.$$ Since the left side is a function of \(t\) only, and the right side is a function of \(x\) only, both sides must be equal to a constant, say \(-\lambda\): $$\frac{C}{a^2}\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda.$$

Solve the ODE for \(X(x)\)

The \(\frac{X''(x)}{X(x)} = -\lambda\) equation can be rewritten as \(X''(x) + \lambda X(x) = 0\). This is a second-order homogeneous ordinary differential equation (ODE) with constant coefficients. The given boundary conditions are: $$X'(0) = 0 \text{ and } X(L) = 0.$$ To solve for \(X(x)\), we need to consider three cases: \(\lambda > 0\), \(\lambda = 0\), and \(\lambda < 0\). For this problem, the eigenvalues of the ODE will be positive and denoted as \(\lambda = k_n^2\), where \(n = 1, 2, 3, \ldots\). Therefore, the general solution of the ODE for \(X(x)\) is: $$X_n(x) = B_n\cos(k_n x) + C_n\sin(k_n x).$$ From \(X'(0) = 0\), we have: $$-B_nk_n\sin(k_n0) + C_nk_n\cos(k_n0) = 0.$$ Solving for \(C_n\), we get: $$C_n = 0.$$ Thus, the equation for \(X_n(x)\) becomes: $$X_n(x) = B_n\cos(k_nx).$$

Solve the ODE for \(T(t)\)

Now, consider the equation for \(T(t)\): $$\frac{C}{a^2}\frac{T'(t)}{T(t)} = -k_n^2.$$ This equation can be rewritten as: $$T'(t) = -k_n^2 \frac{a^2}{C} T(t).$$ The solution for this first-order ODE is: $$T_n(t) = A_n e^{-k_n^2 \frac{a^2}{C}t}.$$

Combine Solutions and Apply Initial Condition

Now that we have the solutions for \(X_n(x)\) and \(T_n(t)\), we can write the general solution for \(u(x, t)\) as follows: $$u(x, t) = \sum_{n=1}^\infty A_n \cos(k_n x)e^{-k_n^2 \frac{a^2}{C}t}.$$ To determine \(A_n\) and \(k_n\), we need to use the given initial condition \(u(x, 0) = 1\) if \(0 \leq x \leq \frac{L}{2}\), and \(0\) if \(\frac{L}{2} < x < L\). We first consider the \(u(x, 0)\) equation, which gives: $$u(x, 0) = \sum_{n=1}^\infty A_n \cos(k_n x) = \left\{\begin{array}{ll} 1, & 0 \leq x \leq \frac{L}{2}, \\ 0, & \frac{L}{2}<x<L \end{array}\right..$$ This equation describes a piecewise function, and we can use Fourier cosine series to represent it. The Fourier cosine series has the form: $$f(x) = \frac{A_0}{2} + \sum_{n=1}^{\infty} A_n \cos\left(\frac{n\pi x}{L}\right)$$ Comparing the given initial condition to the Fourier cosine series, we find that \(k_n = \frac{n\pi}{L}\) and the coefficients \(A_n\) can be determined by using Fourier's formula, which is given by: $$A_n = \frac{2}{L} \int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) dx.$$ For this specific problem, the formula for \(A_n\) is: $$A_n = \frac{2}{L} \int_0^{\frac{L}{2}} \cos\left(\frac{n\pi x}{L}\right) dx.$$

Calculate the Coefficients \(A_n\)

Compute the coefficients \(A_n\) using the formula: $$A_n = \frac{2}{L} \int_0^{\frac{L}{2}} \cos\left(\frac{n\pi x}{L}\right) dx = \frac{2}{n\pi}\left(1-(-1)^n\right).$$

Final Solution

Now that we have the coefficients \(A_n\) and the eigenvalues \(k_n\), the final solution of the initial-boundary value problem is: $$u(x, t) = \sum_{n=1}^\infty \frac{2}{n\pi}\left(1-(-1)^n\right)\cos\left(\frac{n\pi x}{L}\right)e^{-\left(\frac{n\pi a}{L}\right)^2 \frac{t}{C}}.$$ For specific values of \(L\) and \(a\), we can perform numerical experiments to check the convergence and stability of the solution.

Key concepts

Initial-Boundary Value Problem

An initial-boundary value problem (IBVP) is a mathematical challenge where we need to find a function that satisfies a particular differential equation, along with specific initial and boundary conditions. In the context of heat distribution, the heat equation often serves as the differential equation involved. Here, we explored the equation: \[Cu_t = a^2 u_{xx}\]This equation models how heat moves over time across a given rod or surface. The initial condition tells us how heat is distributed at the start, while the boundary conditions guide how heat behaves at the extremes of the spatial domain.

• Initial condition: Defined heat at the start, such as the piecewise function given.
• Boundary conditions: These involve constraints like a fixed or insulated boundary.

Understanding IBVPs allows us to predict heat flow in real-life objects by solving these equations under specific constraints.

Separation of Variables

The method of separation of variables is a popular technique in solving partial differential equations, like the heat equation. The idea is to assume the solution can be expressed as a product of functions, each depending on a single variable. For our problem, it takes the form:\[u(x, t) = X(x)T(t)\]By substituting this form into the heat equation, we separate the equation into two ordinary differential equations (ODEs), one in time and one in space. This step transforms a complex PDE into more manageable ODEs. Both sides of the derived equation should equate to a constant, which in our case is \(-\lambda\), to maintain the equality for all values of \(x\) and \(t\). This crucial step helps break down the PDE and effectively simplifies finding the solution.

Numerical Experiments

Numerical experiments involve using computations to approximate solutions to complex mathematical problems, such as differential equations, that might be difficult to solve analytically. After reaching the final formula for the solution, conducting numerical experiments with specific values of parameters, such as \(L\) and \(a\), helps to validate the solution. These experiments test:

• Convergence: Whether the solution approaches a stable state as time progresses.
• Stability: Ensure that small changes in initial conditions don't lead to significant deviations in the solution.

By varying these parameters, one can observe changes in heat distribution, ensuring that the analytical solution reflects real-world scenarios. This step is essential in demonstrating the reliability of mathematical models.

Fourier Series

Fourier series is a powerful method to represent functions as sums of sine and cosine terms, which are individual frequencies. This approach decomposes complex periodic functions into simple, calculable components. For our IBVP, the initial condition is expressed using a Fourier cosine series, owing to the fixed boundary at one end of the domain. The general form of a Fourier cosine series is:\[f(x) = \frac{A_0}{2} + \sum_{n=1}^{\infty} A_n \cos\left(\frac{n\pi x}{L}\right)\]This cosine representation facilitates handling the boundary conditions involving derivatives or zero terms. Computing the coefficients \(A_n\) allows assembly of a complete picture of the function over the interval. This makes Fourier series invaluable in solving problems with periodic boundary conditions, such as those encountered in thermal and wave equations.