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For $D,\gamma ,a>0$, apply the superposition principle to two translated solutions for Gaussian initial data for the heat equation. The displayed solution of $u_t=D{u}_{xx}$ has odd initial profile $A(e^{-\gamma (x-a{)}^2}-e^{-\gamma (x+a{)}^2})$. Its total spatial integral is zero. For $A>0$, it is positive on $x>0$ and negative on $x<0$ at every $t\ge 0$.

Writing it as $2A{s}^{-1/2}{e}^{-\gamma (x^2+a^2)/s}\sinh (2\gamma ax/s)$ shows how cancellation affects its late-time shape. On the diffusive scale $x=O(\sqrt{s})$, its leading form is $4A\gamma ax{s}^{-3/2}{e}^{-\gamma x^2/s}$. Thus the two lobe maxima asymptotically lie at $x=\pm \sqrt{s/(2\gamma )}$, while their absolute heights decay as $s^{-1}$. Their centers are not fixed at the initial Gaussian centers. The first spatial moment remains ${\int }_{\mathbb{R}}xu(x,t)dx=2Aa\sqrt{\pi /\gamma }$, as follows directly from the Gaussian integral and translation.

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 $\pm 2$.

The heat equation is linear and homogeneous in $u$: the operator $L={\partial }_t-\lambda {\partial }_x^2$ satisfies $L(A{u}_1+B{u}_2)=AL{u}_1+BL{u}_2$. Thus the superposition principle applies to its solutions. Spatial translations are also allowed because the coefficients are constant.

Choose $a=1/(4\lambda )$ in part (b) and multiply by $\sqrt{a}$ to make the initial amplitude one. Taking the two shifts $b=1$ and $b=-1$ then givesAt $t=0$ this reproduces both initial peaks. Each term is an instance of Gaussian initial data for the heat equation, so their sum satisfies the equation. Its extension to negative times is a real smooth solution precisely on the connected intervalAt the lower endpoint the peaks narrow without bound and their heights diverge; there is no finite classical continuation there. For positive times each peak spreads and decreases in height while preserving its integral.

Use the conventionThese are the Fourier transform and inverse Fourier transform for suitably integrable data. Assuming the required ordinary transforms exist and the lower derivatives vanish at both ends, integration by parts gives $\widehat{g^{\prime }}=[g{e}^{-ikx}{]}_{-\infty }^{\infty }+ik\hat{g}=ik\hat{g}$. Repeating this argument yieldsThe decay assumptions are used to discard the boundary terms at each step.

For positive diffusivity $\alpha$, transforming the heat equation gives ${\partial }_t\hat{f}=-\alpha k^2\hat{f}$. Thus $\hat{f}(k,t)=e^{-\alpha k^{2}t}{\hat{f}}_0(k)$, where $f_0(x)=f(x,0)$. The convolution theorem and the Gaussian heat kernel giveFor $f_0(x)=e^{-x^2}$, the Fourier transform of a Gaussian is $\sqrt{\pi }{e}^{-k^2/4}$. Multiplying by the heat factor and inverting gives the Gaussian initial data for the heat equation solutionIt recovers the initial profile as $t\downarrow 0$ and preserves its spatial integral $\sqrt{\pi }$.

For the Gaussian initial data for the heat equation, substitute the initial profile into the heat-kernel solution. Completing the square in $y$ givesThe remaining Gaussian integral yieldsThe centre stays at $x_0$ and the squared width grows linearly with time.

For the interval pulse, the convolution reduces to $B{\int }_a^{b}G(x-y,t)dy$. Using $u=(y-x)/(2\sqrt{t})$ and the error function givesThis is the translated and scaled heat equation with interval-indicator initial data. At its jump points the $t\downarrow 0$ limits are $B/2$; away from them the initial data are recovered pointwise.

For fixed $a<x<b$ at late time, both error function arguments are small. Since $\mathrm{erf}u=2u/\sqrt{\pi }+O(u^3)$,The estimate is uniform for $x\in [a,b]$. Equivalently, on this fixed interval the Gaussian in the convolution is $1+O((b-a{)}^2/t)$.

The spreading distance is more apparent on the growing spatial scale. For the Gaussian pulse the exponent is order one when $|x-x_0|=O(\sqrt{t})$. For the interval pulse, set $x=c+\xi \sqrt{t}$ with $c=(a+b)/2$ and fixed $\xi$. Uniformly for bounded $\xi$, its convolution givesThus both nonzero pulses spread over a distance $O(\sqrt{t})$, while their heights are $O(t^{-1/2})$ and their total integrals remain constant. More quantitatively, the variances of their normalized profiles are $1/(2a_0)+2t$ and $(b-a{)}^2/12+2t$, respectively, confirming the same width law.

Complete the square in the Fourier transform exponent:The supplied shifted Gaussian integral therefore givesThis is the Fourier transform of a Gaussian, with angular-wavenumber convention $e^{-ikx}$.

Take the spatial Fourier transform of the heat equation. Two integrations by parts, with decaying fields and first derivatives at infinity, give $\mathcal{F}[{\psi }_{xx}]=-k^2\tilde{\psi }$. Thus each mode satisfies the ordinary differential equationInserting the transformed initial condition and integrating the time equation givesFinally set $A=t+1/(4\lambda )>0$. The inverse Fourier transform, evaluated by the same Gaussian identity, givesEquivalently the Gaussian initial data for the heat equation evolve asThe profile broadens and its peak decreases, while its total integral remains $\sqrt{\pi /\lambda }$, as required by diffusion without loss on the whole line. Direct differentiation verifies the equation and the initial limit.