What is wrong with my calculation about a wave equation here? Consider the following wave equation in "negative" time:

$$u_{tt}=\Delta u, \quad x\in {\Bbb R}^3, \ \color{red}{t<0}$$
  with initial conditions
  $$
u(x,0)=g(x),\quad u_t(x,0)=0,\quad x\in{\Bbb R}^3.
$$

For $t>0$, define $v:\Bbb{R}^3\times[0,\infty)\to{\Bbb R}$ as 
$$v(x,s):=u(x,-s).$$
Then we have
$$
v_{s}(x,s)=u_t(x,-s)\cdot(-1),\quad v_{ss}(x,s)=u_{tt}(x,-s)\cdot (-1)(-1)=u_{tt}(x,-s)
$$
and $$
\Delta v(x,s)=\Delta u(x,-s).
$$
Since $u_{tt}(x,-s)=\Delta u(x,-s)$ for all $s>0$, we thus have

$$v_{tt}=\Delta v, \quad v\in {\Bbb R}^3, \ \color{blue}{t>0}$$
  with initial conditions
  $$
v(x,0)=g(x),\quad v_t(x,0)=0,\quad x\in{\Bbb R}^3.
$$
  Then by the Kirchhoof's formula, we have
  $$
v(x,t)=\frac{\partial}{\partial t}\left(\frac{1}{4\pi t}\int_{B(x;|t|)}g(y)dS(y)
\right),
$$
  which implies that when $t<0$,
  $$
u(x,t)=v(x,-t)=\frac{\partial}{\partial t}\left(\frac{1}{4\pi (-t)}\int_{B(x;|t|)}g(y)dS(y)
\right).
$$

When $g(x)\equiv 1$, using the formula above, we have
$$
u(x,t)=\frac{\partial}{\partial t}\left(\frac{1}{4\pi (-t)}(4\pi t^2)
\right)=-1.
$$
It follows that 
$$
\lim_{t\to 0-}u(x,t)=-1\not= u(x,0).
$$
What did I do wrong?
 A: The line where you perform $u(x,t) = v(x,-t) = \ldots $ is not correct. You're not allowed to just replace the variable inside the differentiation. What you really want is to calculate the derivative, and then evaluate at a particular point.
To better illustrate the incorrect substitution you did, consider the function $f(x) = 1$ for all $x$. Then it is true that
$$f(x) = \frac{d}{dx} (x).$$
But
$$ f(-x) \neq \frac{d}{dx}(-x)$$.
A: Thanks to Christopher's answer, I found that I did the last step wrong. I would like to write down the correct formula here.
I got correctly
$$
u(x,-t)=v(x,t)=\frac{\partial}{\partial t}\left(\frac{1}{4\pi t}\int_{B(x;|t|)}g(y)dS(y)
\right)
$$
for $t>0$. What I want is the formula for $u(x,t)$ for $t<0$. Let $q=h(t)=-t$. Then by the chain rule, I should have
$$
\begin{align}
u(x,q)&=u(x,-t)=\frac{\partial}{\partial q}\left(\frac{1}{4\pi (-q)}\int_{B(x;|q|)}g(y)dS(y)
\right)\cdot\frac{dq}{dt}\notag\\
&= 
\frac{\partial}{\partial q}\left(\frac{1}{4\pi (-q)}\int_{B(x;|q|)}g(y)dS(y)
\right)\cdot(-1)\notag\\
&=\frac{\partial}{\partial q}\left(\frac{1}{4\pi q}\int_{B(x;|q|)}g(y)dS(y)
\right),\qquad q<0.
\end{align}
$$
Replacing the symbol $q$ with $t$, we are done.
