# Partial Differential Equation using D'Alembert's approach

• Feb 23rd 2009, 01:02 PM
flaming
Partial Differential Equation using D'Alembert's approach
Solve by using D'Alembert's solution with the even extensions of f(x) and g(x)

$\displaystyle \mu_{tt} = c^2\mu_{xx}$ where $\displaystyle 0\leq x <\infty$

$\displaystyle \mu(x,0) = f(x)$, $\displaystyle \mu_t(x,0) = g(x)$ where $\displaystyle 0\leq x <\infty$

$\displaystyle \mu_x(0,t) = 0$ where $\displaystyle t\ge 0$
• Feb 23rd 2009, 02:55 PM
ThePerfectHacker
Quote:

Originally Posted by flaming
Solve by using D'Alembert's solution with the even extensions of f(x) and g(x)

$\displaystyle \mu_{tt} = c^2\mu_{xx}$ where $\displaystyle 0\leq x <\infty$

$\displaystyle \mu(x,0) = f(x)$, $\displaystyle \mu_t(x,0) = g(x)$ where $\displaystyle 0\leq x <\infty$

$\displaystyle \mu_x(0,t) = 0$ where $\displaystyle t\ge 0$

Since we have the boundary condition $\displaystyle u_x(0,t)=0$ we would consider even extensions. Let $\displaystyle f_1(x)\text{ and }g_1(x)$ be even extensions, and we will also assume that these extensions are well-behaved to satisfy the condition of D'Alembert's solution. Thus, we have that $\displaystyle u_tt = c^2u_{xx}$ for $\displaystyle (x,t) \in \mathbb{R}^2$. Thus, the solution is given by:
$\displaystyle u(x,t) = \frac{1}{2}[f_1(x+ct) - f_1(x-ct)] + \frac{1}{2a}\int_{x-ct}^{x+ct}g_1(\xi) d\xi$.
Notice that $\displaystyle u(-x,t) = u(x,t)$ therefore $\displaystyle u_x(0,t) = 0$, this is precisely what we want.