91Ó°ÊÓ

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 \(-\lambda\): $$\frac{C}{a^2}\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda.$$

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: $$X_n(x) = B_n\cos(k_nx).$$

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: $$T_n(t) = A_n e^{-k_n^2 \frac{a^2}{C}t}.$$

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: $$A_n = \frac{2}{L} \int_0^{\frac{L}{2}} \cos\left(\frac{n\pi x}{L}\right) dx.$$

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

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.