Derive d'Alembert's formula for u_(xx)=(1)/(a^2)u_(tt) with…

Derive d'Alembert's formula for $u_{xx}=\frac{1}{a^2}u_{tt}$ with $u(x,0)=f(x)$, $u_t(x,0)=F(x)$.

Step 1 · General solution and initial conditions

$u=\phi(x+at)+\psi(x-at)$. At $t=0$: $\phi(x)+\psi(x)=f(x)$. Differentiating in $t$ and setting $t=0$: $\phi'(x)-\psi'(x)=F(x)/a$.

Step 2 · Solve for $\phi,\psi$

Integrating the second relation from a fixed $x_0$ and combining with the first: \[ \phi(x)=\frac{f(x)}{2}+\frac{1}{2a}\int_{x_0}^xF(s)\,ds+\frac C2,\qquad \psi(x)=\frac{f(x)}{2}-\frac{1}{2a}\int_{x_0}^xF(s)\,ds-\frac C2. \]

Step 3 · Recombine

Substituting $x\to x+at$ in $\phi$ and $x\to x-at$ in $\psi$ and adding, the constants $\pm C/2$ cancel and the two integrals combine into one over $[x-at,x+at]$: \[ u(x,t) = \frac{f(x-at)+f(x+at)}{2}+\frac{1}{2a}\int_{x-at}^{x+at}F(s)\,ds. \]

Verification

At $t=0$: $u=f(x)$. \checkmark Differentiating in $t$ (Leibniz rule on the integral) and setting $t=0$ gives $u_t=F(x)$. \checkmark Differentiating twice more in each variable and comparing confirms $u_{tt}=a^2u_{xx}$ term by term. \checkmark