Find d’Alembert’s solution to the problem:
$\frac{∂^2u}{∂t^2} =\frac{∂^2u}{∂x^2} ,x \in \mathbb{R}$, $t > 0$
$u(x, 0) = x^2 \in \mathbb{R}$
$\frac{∂u}{∂t} (x, 0) = 0 ,x \in \mathbb{R}$.
My attempt : I thin $u(x,t)$ will be $0 $ because $u(x,0)=u'(x,0)=0$
Any Hints/solution will be appreciated
It is a straightforward application of d'Alembert's formula: $$u(x,t) = \frac{(x+t)^2 +(x-t)^2}{2} + \frac{1}{2}\int_{x-t}^{x+t}0\,{\rm d}s = x^2+t^2.$$