We have the following problem $$v_{tt}=c^2v_{xx}, \ x>0, t>0 \\ v(x,0)=\phi (x) , \ x\geq 0 \\ v_t(x,0)=\psi (x), \ x\geq 0\\ v_x(0,t)=0 , \ t\geq 0$$ I want to find a function $v$ that satifies the above problem. Then I want to show that the conditions $\phi '(0)=\psi '(0)=0$ are necessary for the existence of a smooth solution of the problem.
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The function that saatisfies the conditions (except the last one) is given by d'Alembert's formula, right?
So we have $$v(x,t)=\frac{1}{2}[\phi (x-ct]+\phi (x+ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi (\xi)d\xi$$ Then we have to take the partial derivative as for $x$ to use the last condition, right?
Then we have $$ v_x(x,t)=\frac{1}{2}[\phi' (x-ct)+\phi' (x+ct)]+\frac{1}{2c}[-\psi (x-ct)+\psi (x+ct)]$$ Is this correct?
Then we get $$v_x(0,t)=0\Rightarrow \frac{1}{2}[\phi' (-ct)+\phi' (ct)]+\frac{1}{2c}[-\psi (-ct)+\psi (ct)]=0$$ What do we get from here? Do we calculate that at $t=0$ ?