Assumptions:
$u(x,t)$ is solution of diffuse equation for $-\infty < x < \infty, t \geq 0$.
$\phi(x) = u(x,0)$ is even, resp. odd function.
I need to prove that for even: $u(x,t)=u(-x,t)$ and odd $u(x,t)=-u(-x,t)$.
I have:
EVEN:
$
\begin{align*}
u(x,t) &= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x-y)^2}{4kt}} \phi(y)dy \\
&= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x-y)^2}{4kt}} \phi(-y)dy \\
&= -\frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x+z)^2}{4kt}} \phi(z)dz = ??? = u(-x,t)
\end{align*}
$
I did the substitution of $z=-y$.
ODD:
$
\begin{align*}
u(x,t) &= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x-y)^2}{4kt}} \phi(y)dy \\
&= -\frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x-y)^2}{4kt}} \phi(-y)dy \\
&= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x+z)^2}{4kt}} \phi(z)dz =???= -u(-x,t)
\end{align*}
$
I don't know how to continue, or even worse if what I have done is right.
Best Answer
I figured it out. Here are answers if someone cares:
EVEN: $ \begin{align*} u(-x,t) &= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(-x-y)^2}{4kt}} \phi(y)dy \\ &= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x+y)^2}{4kt}} \phi(-y)dy \\ &= -\frac{1}{\sqrt{4\pi kt}} \int_{\infty}^{-\infty} e^{\frac{-(x-z)^2}{4kt}} \phi(z)dz \\ &= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x-z)^2}{4kt}} \phi(z)dz = u(x,t) \end{align*} $ substitution $z=-y$.
$ \begin{align*} u(-x,t) &= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(-x-y)^2}{4kt}} \phi(y)dy \\ &= \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x+y)^2}{4kt}} (-\phi(-y))dy \\ &= \frac{1}{\sqrt{4\pi kt}} \int_{\infty}^{-\infty} e^{\frac{-(x-z)^2}{4kt}} \phi(z)dz \\ &= -\frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty} e^{\frac{-(x-z)^2}{4kt}} \phi(z)dz = -u(x,t) \end{align*} $