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Put $\tau =t+b$ and $q=x+a$. For $\theta =f(t)e^{-q^2/(4\tau )}$, partial differentiation givesThe heat equation therefore requires $f^{\prime }=-f/(2\tau )$, giving $f=C/\sqrt{|t+b|}$ on an interval not containing $-b$. A decaying spatial Gaussian function uses $t+b>0$, so $f=C/\sqrt{t+b}$ there.

The initial profile is a difference of two translated Gaussian functions. Set $b=1/4$, $a=-2$ and $a=2$, with respective constants $C=1/2$ and $C=-1/2$. The superposition principle givesThis is an odd pair of Gaussian profiles under diffusion, obtained from Gaussian initial data for the heat equation. It is odd in $x$, vanishes at zero, is positive for $x>0$ and negative for $x<0$. The two lobes broaden and decrease in height, and $\theta (x,t)\to 0$ at every fixed $x$ as $t\to \infty$. Their extrema move; they do not remain exactly at $\pm 2$.

At an ordinary point, an order-$n$ linear ordinary differential equation has $n$ arbitrary constants in its general solution. Determining them uniquely requires $n$ independent, consistent scalar conditions. Prescribing $y,y^{\prime },\dots ,y^{(n-1)}$ at one ordinary point gives uniqueness under the standard continuous-coefficient assumptions. Conditions at different endpoints need an additional nonsingularity check: for example, $y^{\prime \prime }+y=0$, $y(0)=y(\pi )=0$ leaves every multiple of $\sin x$ possible. Thus counting boundary conditions alone does not guarantee uniqueness for a boundary value problem.

The generalized Leibniz rule follows by mathematical induction. The case $n=0$ is the identity $uv=uv$. Suppose the formula holds for $n$ and differentiate every term using the product rule:For an interior index $r$, the coefficient of $u^{(r)}{v}^{(n+1-r)}$ becomes $\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ by Pascal's identity. The two endpoint coefficients are one. This provesOnly finitely many derivatives are involved; it suffices that both functions are $n$ times differentiable.

For $u=x(x+1)$ and $v=e^{2x}$, the derivatives of $u$ beyond order two vanish. The three surviving terms of the generalized Leibniz rule giveThe simplified expression works for every integer $n\ge 0$, including the two cases where one or more of the three terms are absent.

For $u=x^2$ and $v=\sin x$, use $d^m(\sin x)/d{x}^m=\sin (x+m\pi /2)$. Again only three terms survive. With $T=x+n\pi /2$, they areSince $\sin (T-\pi /2)=-\cos T$ and $\sin (T-\pi )=-\sin T$, this becomesThe factors $n$ and $n(n-1)$ remove the nonexistent terms when $n=0$ or $1$.