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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 cubic-flux variant replaces the quadratic conservative flux of Burgers' equation by $-q^3/3$. Its inviscid characteristic speed is $-q^2$, while positive diffusivity regularizes shock waves. It is not linearized by the usual Cole-Hopf transformation.

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.

Write $L=\log \psi$, with $\psi >0$. Under the Cole-Hopf transformation $q=2ϵL_{\theta }$, direct differentiation givesSince $L_{\theta \theta }+L_{\theta }^2={\psi }_{\theta \theta }/\psi$, Burgers' equation is equivalent toThe ratio is therefore a function $a(Z)$ alone. Multiplying $\psi$ by $\exp [-\int a(Z)dZ]$ leaves $q$ unchanged and removes this scalar freedom. Thus one may choose the normalization for whichThis is the heat equation. The positive sign in $q=2ϵ{\partial }_{\theta }\log \psi$ corresponds to the negative conservative flux $-q^2/2$ used here; the more usual positive-flux convention has the opposite sign.

Integrate the initial Cole-Hopf transformation and choose a continuous positive normalization:Split the heat kernel convolution at zero, and complete the square in the positive half-line contribution. DefineThe resulting heat equation solution isDifferentiating the moving endpoints gives ${\partial }_{\theta }{I}_{-}=-e^{-{\theta }^2/(4ϵZ)}$ and $E{\partial }_{\theta }{I}_{+}=e^{-{\theta }^2/(4ϵZ)}$. These terms cancel. Hence ${\psi }_{\theta }=(4\pi ϵZ{)}^{-1/2}UE{I}_{+}/(2ϵ)$, and the viscous Burgers step solution with negative flux isThe lower limit of $I_{-}$ is $\theta$, as printed in the original PDF. The converted TeX's $-\theta$ does not give the correct convolution or endpoint cancellation.

For the vanishing-viscosity limit, the conservative flux is $F(q)=-q^2/2$ and the characteristic speed is $F^{\prime }(q)=-q$. If $U>0$, the characteristics from the right move left and collide with those on the left. The Rankine-Hugoniot condition gives shock speed $[F]/[q]=-U/2$. ThusThe viscous transition has width $O(ϵ/U)$, and its centre value is exactly $U/2$ by the midpoint symmetry of a viscous Burgers step.

If $U<0$, the right characteristics move right and separate from the left ones. The limit is a rarefaction wave,This follows either from the Burgers Riemann problem with negative flux or the large-argument Gaussian-tail asymptotics of the exact formula. The two signs therefore give a shock and a spreading rarefaction respectively, not two shocks.

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.

The initial data require $\Phi (\xi )=\exp [-(2\nu {)}^{-1}{\int }_{x_{\ast }}^{\xi }U(s)ds]$, up to a positive constant that cancels from $u$. Substituting this into the heat kernel formula gives the exponent $-G/(2\nu )$ withDifferentiate the kernel integral in $x$ and apply the Cole-Hopf transformation. Its constant prefactor cancels, leavingThe choice of $x_{\ast }$ adds an irrelevant constant to $G$; the usual convergence assumptions on the initial data justify the integrals and differentiation.

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 $h=\log \psi$, with $\psi >0$, and set $f=2\alpha h_{\theta }$. Direct differentiation givesThe heat equation ${\psi }_Z=\alpha {\psi }_{\theta \theta }$ implies $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, so the bracket vanishes. Conversely its being independent of $\theta$ can be absorbed into a $Z$-dependent multiplicative normalization of $\psi$, leaving $f$ unchanged. This proves the Cole-Hopf transformation with the positive sign appropriate to the negative-flux Burgers convention. Although the algebra works for nonzero $\alpha$, the given forward Gaussian function diffusion kernel and a physical vanishing-viscosity limit require $\alpha >0$.

Take $U,L>0$ for the stated Burgers N-wave. Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\alpha )$ and normalizing the exterior value to one givesThis function is continuous at $\pm L$; its logarithmic derivative has the specified jumps. Convolution with the heat kernel is positive and solves the heat equation for $Z>0$. Splitting the integral into the exterior baseline and the interior correction yieldsPut $a=1+UZ$. Completing the square givesChanging the interior integration variable to $\eta =a\phi$ changes its limits to $\pm La$ and the Gaussian function width to $Za$. The Jacobian and normalization leave the factor $a^{-1/2}$. Thus the Cole-Hopf solution for a Burgers N-wave isHere $I_{\alpha }$ is the normalized Gaussian function mass of its indicated interval. Its explicit error function representation isFor fixed $w>0$, the Gaussian function concentrates at $\phi =\theta$ as $\alpha \downarrow 0$. Consequentlyaway from the endpoints; at $\theta =\pm L$ the limit is $1/2$. The transition layer has width $O(\sqrt{\alpha w})$. This is an approximate identity argument, not a uniform step approximation across the endpoints.

For fixed $Z>0$ and $|\theta |\ll L$, both interval masses tend to one, and the exponentially large positive $\hat{\psi }$ dominates the exterior correction. Its logarithmic derivative therefore givesFor $|\theta |\gg La$, both interval masses are exponentially small and the weighted interior integral is also negligible, while the exterior contribution tends to one. Therefore $\boxed{f(Z,\theta )\simeq 0}$ in the specified far exterior.

An exponentially weighted Gaussian function tail should not be discarded solely because its unweighted interval mass tends to zero. In fact, comparing the order-one exterior term with $\hat{\psi }$ gives the sharper inviscid Burgers N-wave fronts $|\theta |=L\sqrt{1+UZ}$, not $La$. Away from these fronts, the vanishing-viscosity limit isInside $|\theta |<La$, this follows from the sign of $L^2-{\theta }^2/a$ in the exponential. Outside $La$, the constrained Gaussian function maximum lies at an interval endpoint and has negative exponent. The right shock wave's speed is $UL/(2\sqrt{a})$, equal both to the derivative of $L\sqrt{a}$ and to minus half the sum of its two limiting states; the left shock wave is its reflection. This checks consistency with the Rankine-Hugoniot condition and shows that the requested near-center and far-exterior approximations are compatible with the full entropy limit.

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 $w(x,t)\to g(x)$.

For $k\ne 0$, the chain rule gives $g_t=-f_{t}g/(2k)$ and $g_{xx}=(-f_{xx}/(2k)+f_x^2/(4k^2))g$. ConsequentlyThusThis is the potential form of the Cole-Hopf transformation, with $y=-2k{g}_x/g$. For $k>0$ it is a forward diffusion equation; at $k=0$ the exponential transformation is undefined, while negative $k$ gives a backward heat equation.

The real Gaussian function heat kernel $(4\pi \epsilon z{)}^{-1/2}\exp [-(\varphi -\theta {)}^2/(4\epsilon z)]$ is integrable only when $\epsilon z>0$. For forward time $z>0$, this requires positive diffusivity. A Cole-Hopf transformation is an algebraic substitution for any nonzero coefficient, but that does not make the real forward convolution valid for negative diffusivity. Constant initial heat data already produce a divergent integral with the wrong sign.

For $q_Z-q{q}_{\theta }=\delta q_{\theta \theta }$ and initial data $q=\sin \theta$, the Cole-Hopf transformation $q=2\delta {\psi }_{\theta }/\psi$ gives $\psi (\theta ,0)=\exp [-\cos \theta /(2\delta )]$. The Fourier series of this exponential and the heat equation give the displayed exact positive diffusion solution. Its first harmonic gives $q\sim 4\delta [I_1(a)/I_0(a)]e^{-\delta Z}\sin \theta$. The large-argument asymptotic expansion of a modified Bessel function makes the ratio tend to one as $\delta \to 0$.

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]

For positive diffusivity, the Cole-Hopf transformation with $f=2\alpha {\partial }_{\theta }\log \psi$ converts the negative-flux equation into the heat equation. Step data give a ratio of two Gaussian-tail integrals $J$. At $\theta =-UZ/2$, $J=1$ and $f=U/2$. Its vanishing-viscosity limit selects the shock or rarefaction wave in the Burgers Riemann problem with negative flux, depending on the sign of $U$.