91Ó°ÊÓ

To start this part, recall the solution's form from Problem 13, which is: $$ u(x, t) = \phi(x+at) + \psi(x-at) $$ First, analyze the initial conditions. We know that \(u(x, 0) = 0\), so let's substitute t = 0 into the solution equation: $$ u(x, 0) = \phi(x) + \psi(x) = 0 $$ This gives us the equation for the sum of \(\phi\) and \(\psi\): $$ \phi(x) + \psi(x) = 0 $$ Next, let's consider the second initial condition, \(u_t(x, 0) = g(x)\). Firstly, compute \(u_t(x,t)\), the partial derivative of u with respect to time: $$ u_t(x, t) = a\phi'(x+at) - a\psi'(x-at) $$ Now, substitute t = 0 to obtain: $$ u_t(x, 0) = a\phi'(x) - a\psi'(x) = g(x) $$ This gives us the second equation: $$ a\phi'(x) - a\psi'(x) = g(x) $$ So, we have the following two equations: $$ \begin{array}{c}{\phi(x)+\psi(x)=0} \\\ {a \phi^{\prime}(x)-a\psi^{\prime}(x)=g(x)}\end{array} $$

First, let's use the first equation of part (a) to show that \(\psi^{\prime}(x)=-\phi^{\prime}(x)\). We have: $$ \phi(x) + \psi(x) = 0 $$ Take derivative with respect to x, $$ \phi'(x) + \psi'(x) = 0 $$ This gives us: $$ \psi'(x) = -\phi'(x) $$ Now, use the second equation from part (a) to show that \(-2 a \phi^{\prime}(x)=g(x)\): $$ a \phi^{\prime}(x)-a\psi^{\prime}(x)=g(x) $$ Substitute \(\psi^{\prime}(x)=-\phi^{\prime}(x)\): $$ a \phi^{\prime}(x)+ a\phi^{\prime}(x)=g(x) $$ $$ -2 a \phi^{\prime}(x) = g(x) $$ This implies: $$ \phi'(x) = -\frac{1}{2a}g(x) $$ Integrate by x from \(x_0\) to \(x\): $$ \phi(x) = -\frac{1}{2a}\int_{x_0}^x g(\xi)d\xi+\phi(x_0) $$ Now, we can determine \(\psi(x)\) using the relation we have between \(\phi(x)\) and \(\psi(x)\). We know that \(\psi(x) = - \phi(x)\), so: $$ \psi(x) = \frac{1}{2a}\int_{x_0}^x g(\xi)d\xi - \phi(x_0) $$

We're now ready to determine \(u(x, t)\). Recall the solution's form from Problem 13: $$ u(x, t) = \phi(x+at) + \psi(x-at) $$ Using our expressions for \(\phi(x)\) and \(\psi(x)\), we have: $$ u(x, t) = -\frac{1}{2a}\int_{x_0}^{x+at} g(\xi)d\xi + \phi(x_0) + \frac{1}{2a}\int_{x_0}^{x-at} g(\xi)d\xi - \phi(x_0) $$ Combine the integrals and simplify: $$ u(x, t) = \frac{1}{2a}\int_{x-at}^{x+at} g(\xi) d\xi $$ This is the final expression for \(u(x, t)\) based on the given initial conditions.