What is wrong with my calculation about a wave equation here?

Question

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?

Answer 1

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 $$ f(-x) \neq \frac{d}{dx}(-x)$$.

Answer 2

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.