Partial Differential Equation using D'Alembert's approach

**flaming** · February 23rd 2009, 01:02 PM

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

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

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

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

**ThePerfectHacker** · February 23rd 2009, 02:55 PM

Originally Posted by flaming

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

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

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

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

Since we have the boundary condition $u_x(0,t)=0$ we would consider even extensions. Let $f_1(x)\text{ and }g_1(x)$ 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 $u_tt = c^2u_{xx}$ for $(x,t) \in \mathbb{R}^2$ . Thus, the solution is given by:
$u(x,t) = \frac{1}{2}[f_1(x+ct) - f_1(x-ct)] + \frac{1}{2a}\int_{x-ct}^{x+ct}g_1(\xi) d\xi$ .
Notice that $u(-x,t) = u(x,t)$ therefore $u_x(0,t) = 0$ , this is precisely what we want.

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