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Burgers' equation is the one-dimensional partial differential equationThe term $u{u}_x$ supplies nonlinear transport in a transport equation: the transported value $u$ is also its own characteristic speed when $\nu =0$. Larger values catch up with smaller values ahead of them, steepening a decreasing profile. The diffusion term $\nu u_{xx}$ opposes this steepening. Positive $\nu$ gives the viscous Burgers equation; setting $\nu =0$ gives the Inviscid Burgers equation.

This small equation makes three different behaviors accessible in one model: Burgers characteristic breaking from continuous decreasing data, an entropy shock for a downward step, and a rarefaction wave for an upward step. The Burgers Riemann problem with positive flux and the Viscous Burgers travelling shock below give exact examples. These are mathematical calculations with chosen data; the figures contain no experimental measurements.

Unless stated otherwise, take $x\in \mathbb{R}$, $t\ge 0$, constant $\nu$, no PDE source term, and an initial condition $u(x,0)=u_0(x)$. There are no finite endpoints requiring boundary conditions. Where a profile has different limiting states at $x=\pm \infty$, those states are specified explicitly. On a finite interval, boundary conditions become part of the problem: positive diffusion normally requires a condition at each endpoint, whereas the Inviscid Burgers equation requires appropriate inflow transport boundary conditions. With dimensional variables, $u$ has units of velocity and $\nu$ of kinematic viscosity; all numerical examples below use dimensionless quantities.

For a pure rightgoing acoustic wave, retarded time $\theta =t-x$ and slow distance $X=Mx$ separate fast oscillations from nonlinear steepening. Expanding the equation of state gives $p=\rho +M(\gamma -1){\rho }^2/2+O(M^2)$. The leading perturbations satisfy $u_0={\rho }_0=p_0$. Adding the first-order mass conservation and momentum conservation equations cancels the unknown first corrections and yields the displayed viscous Burgers equation. Inviscid characteristics satisfy $\theta =s-(\gamma +1)F(s)X/2$, $p=F(s)$; loss of invertibility marks shock formation and the end of the smooth inviscid approximation.

For the negative-flux viscous Burgers equation, the Cole-Hopf transformation uses $f=2\alpha {\partial }_{\theta }\log \psi$. Initial N-wave data give ${\psi }_0=\exp [U(L^2-{\theta }^2)/(4\alpha )]$ inside $[-L,L]$ and one outside. Completing the square in the heat kernel convolution gives, with $a=1+UZ$,The Gaussian interval masses and their exponentially weighted tails must be treated together when taking the small-diffusion limit.

The Cole-Hopf transformation $u=-2\nu {\varphi }_x/\varphi$ converts the viscous Burgers equation into the heat equation ${\varphi }_t=\nu {\varphi }_{xx}$. Initial data $u(x,0)=g(x)$ correspond to $\varphi (x,0)=C\exp [-(2\nu {)}^{-1}{\int }^{x}g(s)ds]$.

The Fay solution is a periodic solution of the negative-flux viscous Burgers equation for $Z>0$. Its Fourier series has exponentially decreasing coefficients at each positive $\delta Z$. Expanding $1/\sinh (ns)=2{\sum }_{j\ge 0}{e}^{-(2j+1)ns}$ and summing the sine series givesFor $s\gg 1$, the first harmonic gives $q\sim 4\delta e^{-s}\sin \theta$. The Fay shock-layer asymptotics describe its opposite, small-$s$ regime.

For the negative-flux viscous Burgers equation $q_Z-q{q}_{\theta }=ϵq_{\theta \theta }$, set $q=2ϵ{\partial }_{\theta }\log \psi$. A sum of exponentials in the initial heat data evolves into $\psi ={\sum }_j{e}^{A_j}$, giving the displayed softmax function average. Two terms produce a traveling tanh shock from the lower to the higher state, with speed minus their mean.

The Rankine-Hugoniot conditions follow by integrating conservation laws across a thin moving layer, then letting its macroscopic thickness tend to zero. For example, normal momentum conservation in one dimension has the conservative formFor a shock wave moving at speed $V$, integrate over a control interval whose endpoints move with the discontinuity. The Reynolds transport theorem giveswhere $[f]=f_R-f_L$ denotes the difference of limiting states. Mass conservation similarly gives $V[\rho ]=[\rho u]$. If $v=u-V$ is the normal velocity in the shock frame, combining the two relations yieldsThe same construction applied to conservation of energy, for a non-radiating normal shock wave with nonzero mass flux, gives $[h+v^2/2]=0$, where $h$ is the specific enthalpy. An equation of state closes these relations, and the physical compression branch has increasing entropy.

Finite dynamic viscosity replaces the ideal discontinuity by a thin smooth transition. A viscous stress contributes a derivative to the momentum flux inside the layer and converts directed kinetic energy into internal energy. If its gradients vanish in the uniform far states, the same limiting Rankine-Hugoniot conditions still hold there. Balancing nonlinear transport with viscous diffusion gives a thickness of order $\nu /|u_L-u_R|$ in the viscous Burgers equation, where $\nu$ is the kinematic viscosity.

For $\nu >0$ and positive $\Psi$, set $f=\log \Psi$, so the Cole-Hopf transformation is $u=-2\nu f_x$. Substitution into the viscous Burgers equation givesHence the most general equation implied by this substitution isThe time normalization in the Cole-Hopf transformation accounts for the arbitrary function $C(t)$: multiplying $\Psi$ by a nonzero function of time leaves ${\partial }_x\log \Psi$ and therefore $u$ unchanged. Replacing $\Psi$ by $e^{-{\int }^{t}C(s)ds}\Psi$ sets $C=0$. With this conventional normalization,the heat equation.

For each $j=1,2$, the function ${\Psi }_j=e^{\nu {\kappa }_j^{2}t+{\kappa }_{j}x}$ obeys ${\partial }_t{\Psi }_j=\nu {\kappa }_j^2{\Psi }_j=\nu {\partial }_x^2{\Psi }_j$. By the superposition principle for this linear partial differential equation,is a solution, positive for all real $x,t,{\kappa }_1,{\kappa }_2$. Here ${\kappa }_j$ has units ${\mathrm{m}}^{-1}$ so both exponents are dimensionless. The positivity ensures that its logarithm and the resulting velocity are regular.

Use the Burgers reduction of weakly nonlinear rightgoing acoustics, with $\theta =t-x$, $X=Mx$, ${\partial }_t={\partial }_{\theta }$ and ${\partial }_x=-{\partial }_{\theta }+M{\partial }_X$. The equation of state expands asAt order one, mass conservation and momentum conservation give $({\rho }_0-u_0{)}_{\theta }=0$ and $(u_0-p_0{)}_{\theta }=0$. For the pure rightgoing acoustic wave with no independent leading mean density or velocity shift, ${\rho }_0=u_0=p_0$. This is the background choice implicit in the requested traveling-wave expansion; additional mean components would modify its phase convention. The source's duplicated phrase about expanding $p$ should refer to expanding $\rho$ as well as $u$.

At order $M$, the mass conservation equation becomeswhereas the momentum conservation equation becomesIn the latter equation, the two nonlinear products ${\rho }_0{u}_{0,\theta }$ and $-u_0{u}_{0,\theta }$ cancel. Substitute the equation of state relation into the former equation and add the two equations to eliminate the first corrections. The result isThe same calculation gives $u_1-p_1=-(\gamma +1)p_0^2/4+C(X)$, so the first corrections may be chosen without an explicit secular factor of $\theta$. This is why the retarded-time description can remain uniform for $|\theta |=O(M^{-1})$, provided the leading profile and its relevant derivatives remain bounded over that range and the slow propagation interval.

This Inviscid Burgers equation can develop a shock, so the source's uniformity assertion needs a pre-shock qualification. If $p_0(\theta ,0)=F(\theta )$ and $a=(\gamma +1)/2$, its method of characteristics givesThe smooth construction ends when this denominator vanishes; if $\max F^{\prime }>0$, the first such distance is $X_s=[a\max F^{\prime }{]}^{-1}$.

For viscosity write $\beta =M\nu$ with bounded positive $\nu$. Since $u_{0,xx}=p_{0,\theta \theta }+O(M)$, the order-$M$ momentum conservation equation instead reads $(u_1-p_1{)}_{\theta }+p_{0,X}=\nu p_{0,\theta \theta }$. Addition now yields the viscous Burgers equationIts positive diffusion resolves the inviscid steepening rather than supporting an unqualified smooth inviscid expansion through shock formation.

For this sign of the viscous Burgers equation, the Cole-Hopf transformation is $q=2\delta {\partial }_{\theta }\log \psi$. Integrating the initial sinusoid gives $\psi (\theta ,0)=C\exp [-\cos \theta /(2\delta )]$; the positive multiplicative constant cancels from $q$. Set $a=1/(2\delta )$ and $s=\delta Z$. The Fourier series of the initial exponential and the heat equation yield the exact sinusoidal Cole-Hopf solution for negative-flux Burgers flow:Each harmonic decays by its heat multiplier, which can also be obtained directly by convolving it with the Gaussian heat-kernel convolution. At $s\gg 1$ the first harmonic dominates:The error bounds use $|I_n(a)|\le I_0(a)$ for real positive $a$, also immediate from the Fourier-integral representation. Dividing, including the correction from the denominator, givesFor $\delta \ll 1$, the large-argument asymptotic expansion of a modified Bessel function gives $I_1(a)/I_0(a)\to 1$. HenceThe sign is positive because the first Fourier harmonic of $\psi$ is negative, so its angular derivative is positive. For a fixed nonzero $\sin \theta$, the relative corrections vanish as $\delta \to 0$ and $s\to \infty$; at the symmetry points both the exact solution and the leading expression vanish.

The Cole-Hopf transformation for the sign convention here is $Q=q_x/q$, with $q\ne 0$ and $q_t=q_{xx}$. Indeed $q_t/q=Q_x+Q^2$, soConversely a solution $Q$ determines a locally nonzero $q$ through $q_x=Qq$, $q_t=(Q_x+Q^2)q$; their compatibility is this Burgers equation. A spatially constant term can be removed by rescaling $q$ in time.

Set $\mu =q\psi$ in the heat-equation pair. Division by $q$ givesTheir compatibility can be checked without referring back to $q$: writing $A=ik-Q$, $B=-Q_x-Q^2-k^2$, we have ${\psi }_x=A\psi +1$, ${\psi }_t=B\psi +Q+ik$. The coefficient of $\psi$ in ${\psi }_{xt}-{\psi }_{tx}$ is $-Q_t+Q_{xx}+2Q{Q}_x$, and the constant term is $(ik-Q)(Q+ik)-Q_x-B=0$. Thus this linear auxiliary pair is compatible exactly for the required viscous Burgers equation. As in part (a), adjoining a constant component turns it into a homogeneous $2\times 2$ matrix Lax pair.

Set $u=-2\nu {\partial }_x\log \varphi$. Direct differentiation givesThus the Cole-Hopf transformation maps every nonvanishing solution of the heat equation to a solution of the viscous Burgers equation. Conversely the Burgers equation makes the ratio inside the derivative a function of time alone; a time-dependent rescaling of $\varphi$, which leaves $u$ unchanged, removes that function and gives the heat equation exactly.

Write $h=\log \psi$ and $q=2ϵh_{\theta }$, with $ϵ>0$. Direct differentiation givesThus the Cole-Hopf transformation with this negative-flux convention maps the heat equation to the stated viscous Burgers equation. Conversely its residual is independent of $\theta$ and can be removed by multiplying $\psi$ by a function of $Z$.

Convolving each initial exponential with the Gaussian heat kernel givesThe resulting finite-exponential Cole-Hopf solution isIt is a softmax function average of the constant states, so $q_1\le q\le q_N$ and $q_{\theta }={\mathrm{Var}}_w(q_j)/(2ϵ)\ge 0$.

For $N=2$, put $\Delta =q_2-q_1>0$ and $\overline{q}=(q_1+q_2)/2$. Equality of the two exponents occurs atThereforeThis is an increasing viscous shock from $q_1$ to $q_2$, moving at speed $-\overline{q}$ and with transition width of order $ϵ/\Delta$. That speed agrees with the Rankine-Hugoniot condition for flux $-q^2/2$; its concavity makes the increasing jump the entropy shock.

For general $N$, use the shock frame $\theta =-(q_1+q_N)Z/2+\xi$. The leading $Z$ coefficient in $A_j$ is $q_j(q_j-q_1-q_N)/(4ϵ)$. The endpoint coefficients coincide, and every interior coefficient is smaller by $(q_j-q_1)(q_N-q_j)/(4ϵ)>0$. The extremal-state asymptotics of exponential Burgers solutions therefore reduce the large-$Z$ profile to the same tanh shock with states $q_1,q_N$ and centreAt a fixed $\theta$, rather than in the moving frame, the endpoint of largest $|q_j|$ dominates. If $q_N=-q_1$, both endpoint rates coincide and the limiting shock is stationary. Intermediate plateaus disappear in the moving-frame limit.

For zero viscosity the Inviscid Burgers equation is $q_X-q{q}_{\theta }=0$. Along a characteristic curve, $dq/dX=0$ and $d\theta /dX=-q$. Labelling a characteristic by its initial position $\eta$ givesWhere $f$ is differentiable, $q_{\theta }=f^{\prime }(\eta )/(1-X{f}^{\prime }(\eta ))$. The first characteristic crossing occurs at $X=1/\max f^{\prime }$ if the maximum positive slope exists. A multivalued continuation is not a physical wave: rewrite the equation as the conservation law $q_X+{\partial }_{\theta }(-q^2/2)=0$ and replace the overturning part by an entropy-admissible shock wave. The Rankine-Hugoniot conditions give its trajectoryThe states on either side are supplied by the characteristics meeting the shock. Conservation supplies the equal-area construction, and the entropy condition requires characteristics to enter the shock. Since the flux is concave, a jump with $q_L<q_R$ is compressive.

For the rising step, $q_L=0$ and $q_R=U>0$, the left characteristics have slope zero and the right ones slope $-U$. They overlap immediately. The Burgers Riemann problem with negative flux is solved byIt is a shock, not a rarefaction wave; the minus sign in the nonlinear flux is decisive.

For positive viscosity, substitute $q=2ϵ{\partial }_{\theta }\log \psi$ into the viscous Burgers equation. Writing $w=\log \psi$ givesAn $X$-dependent multiplicative factor in $\psi$ removes the possible integration function. The Cole-Hopf transformation therefore gives ${\psi }_X=ϵ{\psi }_{\theta \theta }$. Normalize its initial value continuously at zero:Split the heat kernel integral at zero. Completing the square in the positive-half integral yieldsWhen differentiating, the Gaussian terms from the two complementary error functions cancel, becauseThus the exact viscous Burgers step solution with negative flux isFor $X>0$ its denominator is positive, its tails tend to $0$ and $U$, and it approaches the initial step as $X\downarrow 0$. At $\theta =-UX/2$ it equals $U/2$. Its vanishing-viscosity limit gives the entropy shock above. At large $U\sqrt{X/ϵ}$ near the moving shock center it approaches the travelling profile $U/[1+e^{-U(\theta +UX/2)/(2ϵ)}]$, with thickness of order $ϵ/U$.

Set $q=\log \psi$ and use the sign-appropriate Cole-Hopf transformation $f=2\epsilon q_{\theta }$. Direct differentiation givesThus the heat equation ${\psi }_z=\epsilon {\psi }_{\theta \theta }$ implies the viscous Burgers equation with the required negative nonlinear sign. Conversely, vanishing of the Burgers residual makes the final quotient depend only on $z$, and a multiplicative time-dependent factor in $\psi$ removes it. The algebraic substitution works for any nonzero $\epsilon$ where $\psi \ne 0$.

For a forward dissipative initial-value solution, take $\epsilon >0$ and $z>0$. This is essential for the supplied heat kernel: at $\epsilon <0$, $z>0$, its real Gaussian integral diverges even for $\psi (0,\varphi )=1$. More generally that kernel requires $\epsilon z>0$. The physical viscosity convention selects the positive case; the mere condition $\epsilon \ne 0$ does not justify a forward Gaussian function formula.

Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\epsilon )$ gives a positive continuous initial factor, normalized asIts heat kernel convolution converges for both signs of $U$, since a Gaussian function dominates the one-sided exponential. DefineThe left half-line contributes $J(\theta )$; completing the square on the right half-line gives $e^{S}J(-\theta -Uz)$. ThusWhen differentiating, the two moving-endpoint Gaussian function terms cancel, becauseOnly ${\partial }_{\theta }{e}^S=(U/(2\epsilon ))e^S$ remains in the logarithmic derivative. The viscous Burgers step solution with negative flux is thereforeThe denominator is strictly positive, so $f$ lies between $0$ and $U$, irrespective of the sign of $U$. Its initial limits away from $\theta =0$ are the required step.

For the spatial limits, the Gaussian function tails must be compared with $e^S$; inspecting that exponential alone gives the wrong inference when $U<0$. As $b\to +\infty$, the complementary error function asymptotic givesAt $\theta \to +\infty$, $J(\theta )e^{-S}$ tends to zero even for $U<0$, because its quadratic decay dominates any exponential linear in $\theta$. Hence $\alpha e^{-S}\to 0$. At $\theta \to -\infty$, use the identity above to obtainwhile $J(\theta )$ approaches its full Gaussian integral. Thusfor both signs of $U$.

Since $J$ is strictly decreasing, $\alpha =1$ holds exactly when $\theta =-\theta -Uz$. At this point $S=0$, so the midpoint symmetry of a viscous Burgers step givesMore generally the exact symmetry is $f(z,-Uz-\theta )=U-f(z,\theta )$. For $U>0$, the midpoint follows the inviscid shock trajectory. For $U<0$, it is instead the center of an expanding fan, not a shock.

The vanishing viscosity approximation makes the comparison precise. For fixed $z>0$ and $U>0$, near $\theta =-Uz/2$ both tail integrals tend to the full Gaussian integral, so $\alpha \to 1$ andThe layer has width $O(\epsilon /U)$ and tends to the entropy shock, with value $U/2$ at its center. Outside it the limits are $0$ and $U$.

For $U<0$, inside $0<\theta <-Uz$ both Gaussian function tails have large positive lower limits. Their asymptotics giveOutside this interval the limits are $0$ on the left and $U$ on the right. Thus positive viscosity selects the rarefaction wave, including the same midpoint value as the inviscid fan. The two sign cases agree with the preceding entropy solution construction, while a negative diffusivity would not furnish this dissipative selection.

Let $h=\log \psi$ and $f=2\alpha h_{\theta }$. Direct substitution into the viscous Burgers equation givesIf ${\psi }_Z=\alpha {\psi }_{\theta \theta }$, then $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, proving the Cole-Hopf transformation. A function of $Z$ inside the parentheses can be removed by rescaling $\psi$ by a time-dependent factor, which leaves $f$ unchanged.

For the forward heat equation and the displayed Gaussian kernel we require $\alpha >0$ and $Z>0$. The algebraic transformation works for nonzero $\alpha$, but the printed convolution is not a forward solution for negative $\alpha$: its Gaussian then grows and the step-data integral diverges.

Integrating the initial logarithmic derivative fixes a convenient positive initial heat datum,Split the heat kernel convolution at zero and complete the square in the positive half. With $h_{\ast }=U(2\theta +UZ)/(4\alpha )$, putThen $\psi =(A_0+e^{h_{\ast }}{C}_0)/\sqrt{4\pi \alpha Z}$. On differentiation, the moving-limit terms cancel because $e^{h_{\ast }}{e}^{-(\theta +UZ{)}^2/(4\alpha Z)}=e^{-{\theta }^2/(4\alpha Z)}$. HenceThis is the viscous Burgers step solution with negative flux. Both integrals can be written as $\sqrt{\pi \alpha Z}$ times a complementary error function.

For fixed $Z>0$, the Gaussian-tail asymptotics give $f\to U$ as $\theta \to +\infty$ and $f\to 0$ as $\theta \to -\infty$. For example $J{e}^{-h_{\ast }}$ tends to zero on the right with a Gaussian factor $e^{-(\theta +UZ{)}^2/(4\alpha Z)}$, while on the left it diverges with a Gaussian factor $e^{{\theta }^2/(4\alpha Z)}$. This holds for either sign of $U$; for $U=0$ the solution is already zero.

The Gaussian tail is strictly decreasing in its lower limit, so $J=1$ occurs exactly at $\theta =-UZ/2$. There $h_{\ast }=0$, and $f=U/2$. For $U>0$, the two lower limits are both far into the negative tail in the mature shock region, so $J\simeq 1$ throughout its thin transition. The solution is then approximately the traveling viscous frontcentred at the inviscid shock position, with thickness of order $\alpha /U$.

In the vanishing-viscosity limit, for $U>0$ the solution tends to the compressive entropy solution of part (a), away from its shock. For $U<0$, it tends instead to the rarefaction wave. To see the latter explicitly inside $0<\theta <-UZ$, both tails have positive lower limits, and their leading asymptotics give $J{e}^{-h_{\ast }}\to (-\theta -UZ)/\theta$. Thus $f\to -\theta /Z$ in the fan, with the constant states outside. Although $J=1$ still marks the fan midpoint, it is not approximately one throughout the expanding fan; replacing it by one there would create an inadmissible compressive-front approximation.

For the heat kerneldirect partial differentiation givesThus $K$ solves the heat equation. For bounded continuous $f$, differentiation under the integral sign givesWith $Y=(x-y)/\sqrt{4t}$,The Gaussian integral makes the weight have total mass one, and the dominated convergence theorem gives $u(x,t)\to f(x)$. This is the Gaussian approximate identity.

For the viscous Burgers equation with unit viscosity, the Cole-Hopf transformation $w=-2u_x/u$ reduces the equation to $u_t=u_{xx}$. To obtain initial value $g$, choose(up to an irrelevant positive constant), and set $u=K_t\ast f$. ThereforeEquivalently,and the approximate-identity limit gives $w(x,t)\to g(x)$.

Differentiate the equation for $f$ with respect to $x$, commuting mixed partial derivatives for a sufficiently smooth solution. With $y=f_x$ this givesThis is the viscous Burgers equation when $k>0$.

Substituting $u=-2\nu {\partial }_x\log \Psi$ into the viscous Burgers equation gives ${\partial }_x[({\Psi }_t-\nu {\Psi }_{xx})/\Psi ]=0$. Integration permits the arbitrary function $C(t)$ in the title. Multiplication of $\Psi$ by $e^{-\int C(t)dt}$ preserves $u$ and reduces its equation to the heat equation. Thus the heat-equation representation uses a permissible time normalization rather than a condition uniquely imposed on every representative $\Psi$.

The viscous Burgers equation combines a nonlinear transport equation with diffusion with constant $\nu >0$. On the infinite line, a bounded initial condition gives a smooth function solving the equation for every positive time; a jump in the initial condition is smoothed rather than retained as a discontinuity. Small $\nu$ can nevertheless produce a steep layer, as the Viscous Burgers travelling shock illustrates. The Cole-Hopf transformation $u=-2\nu {\partial }_x\log \varphi$, with positive $\varphi$, converts this equation to the heat equation ${\varphi }_t=\nu {\varphi }_{xx}$.^[ref]

Positive diffusion also explains the admissibility of the inviscid limit. For the entropy flux for a scalar conservation law associated with $\eta (u)=u^2/2$, direct multiplication by $u$ givesThe last term is nonpositive. With convergence sufficient to pass to the limit, the vanishing viscosity approximation therefore retains an entropy inequality rather than just the weak solution equation. For general bounded data, the full Kruzhkov entropy inequalities specify the entropy solution; the quadratic identity is a useful illustration of dissipation.^[ref]