Answer to Question #257882 in Differential Equations for JaytheCreator

Question

Solve the following B.V.P.

𝑢𝑡𝑡 = 𝑐 2𝑢𝑥𝑥; { 𝑢(𝑥, 0) = cos 𝑥 − 1, 𝑢𝑡 (𝑥, 0) = 0 𝑢(0,𝑡) = 0, 𝑢(2𝜋,𝑡) = 0 0 ≤ 𝑥 ≤ 2�

Answer 1

Given

$U_{tt}=2c\space U{xx}\\\frac{\delta ^2U}{\delta x^2}=\frac{1}{2c}\frac{\delta ^2 u}{\delta t^2}\\U(x,0)=f(x)=cos \space x-1\\U(x,0)=g(x)=0\\and\\U(0,t)=0,U(2\pi,t)=0\\0 ≤ X ≤ 2$

We know that

$U(x,t)=\frac{1}{2}[f(x+ct)-f(x-ct)]+\frac{1}{2c}\int^{x+ct}_{x-ct}g(s)ds$

$U(x,t)=\frac{1}{2}[cos(x+\sqrt{2ct})-1-cos (x-\sqrt{2ct})+1]+\frac{1}{2\sqrt{2c}}\int ^{x+\sqrt{2ct}}_{x-\sqrt{2ct}}0\space ds$

$U(x,t)=\frac{1}{2}[cos(x+\sqrt{2ct})-cos (x-\sqrt{2ct})+1]$

and

$U(0,t)=0\\and\\U(2\pi,t)=0\\0=0$

we know that $cos\theta=cos \theta$

since verified

and $U(2\pi,t)=0$

$0=\frac{1}{2}[cos(2\pi+\sqrt{2ct})-cos(2\pi-\sqrt{2ct})]\\=\frac{1}{2}[cos(\sqrt{2ct})-cos(-\sqrt{2ct})]\\0=0$

verified

hence solution is

$U(x,t)=\frac{1}{2}[cos(x+\sqrt{2ct})-cos (x-\sqrt{2ct})]$