Apply d Alembert formula to solve the following problem
Question : Let u(x,t) be the solution of the initial value problem
\[u_{tt}-u_{xx}=0\]\[u\left(x,0\right)=x^3,\ u_t\left(x,0\right)=\sin(x)\]Then \[u\left(\pi,\pi\right)\] is
1. \[4\pi^3\]2. \[\pi^3\]3. \[0\]4. \[0\]
Solution :
Given u(x,t) be the solution of homogeneous wave equation
\[u_{tt}-c^2u_{xx}=0\]subject to general Cauchy initial conditions
\[u\left(x,0\right)=x^3=f\left(x\right),\]
\[u_t\left(x,0\right)=\sin{\left(x\right)}=g\left(x\right)\]Then D’ Alembert’s formula is given as
\[u\left(x,\ t\right)=\frac{1}{2}\ \left[f\left(x\ -\ ct\right)+\ f\left(x\ +\ ct\right)\right]\ +\frac{1}{2}\int_{x-ct}^{x+ct}{g(x)}dx\]
\[=\frac{1}{2}\ \left[\left(x\ -\ t\right)^3+\ \left(x\ +\ t\right)^3\right]\ +\frac{1}{2}\int_{x-t}^{x+t}\sin{\left(x\right)}dx\]
\[=\frac{1}{2}\ \left[\left(x\ -\ t\right)^3+\ \left(x\ +\ t\right)^3\right]-\frac{1}{2}\left[\cos{\left(x+t\right)}-\cos{\left(x-t\right)}\right]\]
\[u\left(x,\ t\right)=\frac{1}{2}\ \left[\left(x\ -\ t\right)^3+\ \left(x\ +\ t\right)^3\right]-\frac{1}{2}\left[\cos{\left(x+t\right)}-\cos{\left(x-t\right)}\right]\]Therefore
\[u\left(\pi,\ \pi\right)=\frac{1}{2}\ \left[0+\ 8\pi^3\right]-\frac{1}{2}\left[\cos{\left(2\pi\right)}-\cos{\left(0\right)}\right]\]
\[u\left(\pi,\ \pi\right)=4\pi^3\]
Therefore correct option is (1).