I am trying to solve the following IBVP:
\begin{cases} u_{tt} = 4u_{xx},\\ u_x(0,t) = 0, \\ u(x,0) = 1, u_t(x,0) = x \\ \end{cases}
with x > 0, t > 0. I have tried using D'Alambert's formula $u$(x,t) = $\frac{1}{2}\phi$(x - ct) + $\frac{1}{2}\phi$(x + ct) + $\frac{1}{2c}\int_{x-ct}^{x+ct}$ for x > ct, giving the solution u(x,t) = 1 + tx, and the evenly reflected version of D'alambert's Formula for $u$(x,t) $\frac{1}{2}\phi$(ct - x) + $\frac{1}{2}\phi$(x + ct) + $\frac{1}{2c}\int_{0}^{ct-x}\psi$(s)ds + $\frac{1}{2c}\int_0^{x+ct}\psi$(s)ds for x < ct giving the solution u(x,t) = 1 + $\frac{x^2}{4} + t^2$. However, the solution is apparently not supposed to be C$^2$ differentiable, but should have a jump discontinuity somewhere, which my solution doesn't seem to have. Am I making a mistake somewhere in my solution, or am I missing something?
