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Before characteristic crossing, the method of characteristics for the Inviscid Burgers equation transports each initial value without changing it:Here $\xi$ labels the starting position. Suppose $u_0$ is a bounded continuously differentiable function with bounded derivative. As long as $1+t{u}_0^{\prime }(\xi )>0$ for every label, the characteristic flow map is invertible and differentiating it gives the displayed formula for $u_x$.^[ref]

If the negative minimum $m={\min }_{\xi }{u}_0^{\prime }(\xi )$ is attained, the first breaking time is $t_{\ast }=-1/m$. Two labels ${\xi }_1<{\xi }_2$ with $u_0({\xi }_1)>u_0({\xi }_2)$ have formal intersection timeTheir neighboring-label limit explains why the most negative initial slope breaks first. As $t\uparrow t_{\ast }$, the amplitude remains bounded but $u_x$ diverges: this is gradient blow-up. Continuing intersecting lines would assign incompatible values at one position; the post-breaking entropy solution instead uses shock waves. Nondecreasing data have no forward characteristic crossing by this mechanism.

For a concrete calculation in dimensionless quantities on the infinite line, choose $u_0(\xi )=1-\tanh \xi$, with limiting states $2$ and $0$. Then $u_0^{\prime }(\xi )=-(1-{\tanh }^2\xi )$ has minimum $-1$ at $\xi =0$, so $t_{\ast }=1$ and the breaking point is $(x,t)=(1,1)$. Along the central characteristic curve $x=t$, $u=1$ and $u_x=-1/(1-t)$ before breaking. The labels $\xi =\pm 1$ give intersecting formal lines at $t=1/\tanh 1\approx 1.313$, after the classical solution has already broken.

For Noncharacteristic Cauchy data parameterized by $t\ge 0$, the method of characteristics produces a surface $\Phi (s,t)=(x,y,z)$. The part uniquely fixed as a classical graph through the data must retain both the allowed footpoint range and a nonsingular branch of the projected map. On a branch where inversion gives $(s(x,y),t(x,y))$, require $t(x,y)\ge 0$, together with the branch condition and a nonzero Jacobian determinant. The strict inequality $t>0$ describes its open interior; the endpoint characteristic $t=0$ supplies a boundary trace. A fold caustic can supply another boundary.

An algebraic formula obtained by eliminating $s,t$ can extend beyond this domain of dependence. Its values there need not be selected by the original data, because their characteristic footpoints can lie on the unprescribed half of the data curve. Tracking the parameter range is therefore part of the uniqueness argument, separately from selecting a real square-root branch.

For the partial differential equation $u_x+u{u}_y=x$ with initial profile $u(0,\eta )=1+\sin \eta$, the method of characteristics gives the parametrized surfaceThe projection to the $(x,y)$ plane has Jacobian determinant $1+x\cos \eta$. It is positive for $|x|<1$, so the inverse function theorem makes the surface a unique local solution graph there. For $\eta \ge 0$, strict monotonicity and divergence to infinity give a global inverse onto $y\ge x+x^3/6$ at each such $x$.

The derivative $u_y=\cos \eta /(1+x\cos \eta )$ becomes singular when that projection first degenerates, although $u$ itself stays finite. At $x=1$ the degenerate labels are odd multiples of $\pi$, and at $x=-1$ they are even multiples. At either boundary the limiting map remains strictly increasing, with isolated zero derivatives, so fixed image values determine convergent labels. Consequently the gradient diverges at fixed $y=(2k+1)\pi +7/6$ as $x\uparrow 1$, and at fixed $y=2k\pi -7/6$ as $x\downarrow -1$, using $k\ge 0$ and $k\ge 1$, respectively, to stay in the half-line initial domain. At these first singularities $y_{\eta \eta }=0$ and $y_{\eta \eta \eta }\ne 0$; this is the onset of loss of invertibility, rather than an ordinary nondegenerate fold at the initial singular time.

For a first-order nonlinear partial differential equation $F(x,y,u,p,q)=0$, with $p=u_x,q=u_y$, Charpit's equations extend the method of characteristics to the five-dimensional space of position, value and gradient:The chain rule gives $dF/dt=0$ by direct cancellation, so a trajectory starting on $F=0$ remains there. Initial data $u_0(s)$ on $(x_0(s),y_0(s))$ require both $F(x_0,y_0,u_0,p_0,q_0)=0$ and $p_0{x}_0^{\prime }+q_0{y}_0^{\prime }=u_0^{\prime }$. The non-characteristic condition is $x_0^{\prime }{F}_q-y_0^{\prime }{F}_p\ne 0$. It makes the projection onto $(x,y)$ locally invertible. Compatible initial contact data then propagate as $du=pdx+qdy$, reconstructing a classical solution. Multiple permitted initial normal derivatives can give different local branches; loss of the projection Jacobian determinant marks a characteristic caustic.

For the linear partial differential equation $2y{u}_x-x{u}_y=0$, the method of characteristics gives $\dot{x}=2y$, $\dot{y}=-x$, $\dot{u}=0$. Thus $\varphi =x^2+2y^2$ is constant on each characteristic curve, whose nonzero level sets are ellipses. On a domain containing entire such ellipses, the solutions have the form $u=F(\varphi )$, subject to the desired regularity at the origin.

For the inhomogeneous equation $2y{u}_x-x{u}_y=xy(2y^2-x^2)$, a particular solution is $u_p=x^2{y}^2/2$, so the general solution is $u=x^2{y}^2/2+F(x^2+2y^2)$. Prescribing $u(x,1)=x^2$ fixes $F(s)=(s-2)/2$ only for $s\ge 2$. The inner ellipses $s<2$ never intersect that data line, so the boundary values leave their solution values undetermined. Smooth extensions must match the prescribed $F$ at $s=2$; choosing the same linear formula everywhere gives one polynomial extension.

A partial differential equation whose highest partial derivatives have order one. For $a(x,y)u_x+b(x,y)u_y=c(x,y,u)$, the method of characteristics follows the ordinary differential equations $\dot{x}=a$, $\dot{y}=b$, $\dot{u}=c$.

Particles keep their velocity and move along straight trajectories. The method of characteristics gives $f(t,x,v)=f_0(x-tv,v)$. The shear preserves Lebesgue measure and every Lp norm. Compact initial support remains compact at each finite time, but need not remain in one compact set for all times.

For $H_t+H^{n+1}{H}_x=1-2x$ with $H(\xi ,0)=0$, method of characteristics conserves $H^{n+2}/(n+2)-(x-x^2)$. A positive downstream branch therefore requires $0\le \xi <x<1-\xi$. Initial points in the negative-accumulation region cannot create nonnegative ice from a bare bed. Characteristic intersections and missing dry-bed physics prevent using the formal initial/boundary branches as an unrestricted global solution.

The equations of the method of characteristics give $(x,y,u)=(s+t^2/2,t,s{e}^t)$. Their parameter map to the plane has Jacobian determinant one, so it is globally invertible and $u=(x-y^2/2)e^y$ is uniquely determined exactly where $1\le x-y^2/2\le 2$. Both positive and negative $y$ occur. Outside this closed parabolic strip, $u=e^{y}F(x-y^2/2)$ allows arbitrary smooth changes of $F$ outside the initial parameter interval. The boundary characteristic curves carry the data from its endpoints.

Choose characteristic parameter $t$ with initial value at $t=0$. The characteristic equations areThe first two integrate immediately; substitution into the third givesEliminating $t$ gives the projected parabolaThe sign restriction selects the half-parabola actually reached from the initial point.

The projection map $(s,t)\mapsto (s{e}^t,s{e}^{2t})$ has Jacobian determinant $s{e}^{3t}\ne 0$, so the initial curve is noncharacteristic for every allowed $s$. Its global inverse isdefined precisely when $x,y$ have the same nonzero sign. Every such point is reached by exactly one characteristic. Hence the region of unique determination from these initial data projects ontothe first and third quadrants. The axes are not reached for finite parameter from $s\ne 0$, and the other two quadrants have no method of characteristics issued from this initial curve.

As a direct check, substitution of the inverse parameters gives the solutionIt has $z(s,s)=s^2$, and differentiating gives $x{z}_x+2y{z}_y=xy$. This is quadrant uniqueness for unequal dilation characteristics.

Introduce a characteristic parameter $s$ and label each point of the initial segment by $r\in [0,1]$. The method of characteristics givesIntegrating gives the parametric solutionThe projected characteristic curves satisfy $y=r(2-x)$, so each lies on a straight line through $(2,0)$. Its branch from the data always has $x<2$; the common vertex is approached only as $s\to -\infty$.

Eliminating $s,r$ using $e^s=2-x$ and $r=y/(2-x)$ yieldsIts derivatives give $(x-2)z_x+y{z}_y=y$, and substitution of $x=1$ gives the prescribed value one.

The Jacobian determinant of $(s,r)\mapsto (x,y)$ is $-e^{2s}\ne 0$, and the inverse just displayed is unique. Hence the data determine the solution exactly onThis is the domain determined by a finite segment under a radial characteristic flow. Its interior is $x<2$, $0<y<2-x$; the two boundary rays come from the endpoints of the initial segment. Every point there lies on exactly one characteristic from the data. Outside this sector, a characteristic either misses the initial segment or lies on the branch $x>2$, which cannot reach it. Thus the formula may be extended beyond this sector, but such an extension is not uniquely specified by the given data.

The method of characteristics usesFor a graph parametrized by $(x,y,u(x,y))$, a normal vector is $N=(-u_x,-u_y,1)$. The PDE says $N\cdot (a,b,c)=0$, so the characteristic vector is tangent to the solution surface. Flowing a suitable initial curve along these directions constructs a surface; where its projection to the $(x,y)$ plane has nonzero Jacobian determinant, the inverse function theorem makes it a graph. Local uniqueness of this general flow needs the usual regularity assumptions beyond mere continuity. The particular characteristic system below is smooth and can be solved explicitly.

Label the initial points by $\eta \ge 0$. Since $dx/ds=1$, take $x=s$. The other two equations integrate toEliminating $\eta =y-xu+x^3/3$ gives the implicit solutionFor each fixed $|x|<1$, the derivative of the displayed $y$ with respect to $\eta$ is $1+x\cos \eta \ge 1-|x|>0$. It increases from its value at $\eta =0$ to infinity, so it is a bijection from $[0,\infty )$ onto $[y_L(x),\infty )$, whereThus there is exactly one characteristic label, and hence one solution value, throughout the requested region, including the initial-label boundary as a trace.

The characteristic fold for sinusoidal data in a forced Burgers equation is seen by differentiating with respect to $y$:For a fixed integer $k\ge 0$, let $y_{\ast }=(2k+1)\pi +7/6$. As $x\to 1^{-}$ at this fixed $y_{\ast }$, the label tends to $(2k+1)\pi$, giving $u_y\to -\infty$. To justify the label limit, the limiting map at $x=1$ is strictly increasing despite isolated zero derivatives, so its inverse has a unique continuous limit. Similarly, at any fixed $y_{\ast }=2k\pi -7/6$ with $k\ge 1$, the limit as $x\to -1^{+}$ has $\eta \to 2k\pi$ and $u_y\to +\infty$. These fixed values lie within the region near the respective boundary, proving the requested gradient blow-up.

To find the characteristic domain from data on a half-line, retain the allowed range of its initial parameter. The portion connected to the initial data before the fold caustic has $s>-1$ and $t\ge 0$. Inverting the positive branch givesThe square root must be positive, and $t\ge 0$ requires $x+1\ge \sqrt{1+x+y}>0$. Squaring is now legitimate and gives $y\le x^2+x$. Thus the full characteristic-covered set on this branch isIts open interior, where both the data parameter and the projection are interior and nonsingular, is the classical domain of dependence

Each point of $D$ has exactly one backward characteristic meeting the interior $t>0$ of the prescribed data, so the Picard-Lindelöf theorem fixes its value. The upper boundary is the characteristic issuing from the endpoint $t=0$ and is fixed by the endpoint datum, consistently with the continuous trace from $D$. The lower boundary is excluded: $s=-1$ there, the Jacobian determinant vanishes, and the explicit solution derivatives diverge as $W\downarrow 0$.

Points of the half-plane $W>0$ outside this set have $t<0$ on the branch through the data. Their characteristic footpoints would lie on the unprescribed continuation of the initial line, so these values are not determined by the given half-line alone. For example, extend the initial value to negative $t$ as $1+\epsilon h(t)$, where $h$ is a smooth function supported on the negative half-line. Such extensions agree with all the prescribed data and, where noncharacteristic, produce different local extensions by the method of characteristics. The displayed formula supplies one possible extension there, but the data do not select it uniquely. This is why retaining the restriction on the initial half-line is essential. Continuing the parameterized surface past $s=-1$ instead produces the other sheet of the fold; it does not give a differentiable graph continuation through the caustic.

The method of characteristics replaces the first-order partial differential equation by the ordinary differential equationsAn integrating factor for the second equation gives $d(e^{-2s}y)/ds=\beta C{e}^{-s}$, so their general solution of a differential equation isHere $C,D,E$ are real constants. For $C\ne 0$, eliminating $s$ gives the characteristic projectionseach restricted to either $x>0$ or $x<0$. These are arcs of parabolas, with the straight-line exception $K=0$. Along an oriented characteristic curve, $|x|$ increases with $s$.

For $C=0$, the characteristic projections lie on $x=0$: $y=D{e}^{2s}$ describes either vertical half-line when $D\ne 0$. For $C=D=0$ the projection is the single point $(0,0)$, although $u=2s+E$ still changes. This last lifted characteristic curve cannot be contained in the function graph of a classical solution: at $(0,0)$ the original partial differential equation would require $0=2$.

Write the parameter along a characteristic curve as $r$. Charpit's equations areAt $r=0$, prescribe $x=x_0(s)$, $y=y_0(s)$, $u=u_0(s)$ and choose $p_0,q_0$ satisfying both $F(x_0,y_0,u_0,p_0,q_0)=0$ and the compatibility relation $p_0{x}_0^{\prime }+q_0{y}_0^{\prime }=u_0^{\prime }$. Direct differentiation givesThus the initial constraint propagates. A chosen initial-gradient branch is locally unique whenThis is simultaneously the implicit-function determinant for the initial gradients and the projection map determinant for the surface generated by the method of characteristics. Uniqueness continues while that projection map is invertible, the flow remains regular, and distinct characteristic curves do not give competing values at the same point. Multiple initial-gradient branches, characteristic curve initial points or a caustic require separate analysis.

Here the initial coordinate is $s=x$, so $p_0=3s^2/4$ and the equation forces $q_0^3=-s^3$, hence $q_0=-s$. The projection map determinant on the initial line is $3s^2$, nonzero for $0<s\le 1$. Solving the characteristic curve ordinary differential equations giveswith $p=3s^2{e}^{-2r}/4$ and $q=-s{e}^{-r}$. These satisfy both the equation and all the noncharacteristic initial values.

The projection map determinant is $J=3s^2{e}^r(2e^{-2r}-1)$. Starting from $r=0$, the first caustic occurs at $r_c=\frac{1}{2}\log 2$. At the initial endpoint $s=0$ the non-characteristic condition fails, and its entire characteristic curve remains the single point $(0,0)$; it does not generate an open region of determination.

The maximal regular sheet connected to the initial line uses $r<r_c$. Its interior image is exactlyIndeed the inverse on this sheet isand the inequalities say $0<s<1$ and $s>x/\sqrt{2}$. The upper boundary is the caustic; the lower boundary is the characteristic curve from $s=1$. Endpoints of that latter boundary can be retained as one-sided data. For forward characteristic curves alone, also impose $r\ge 0$: this retains $y\ge 0$, with lower boundary zero for $x\le 1$ and the displayed endpoint characteristic curve for $1<x<\sqrt{2}$. The other square-root branch belongs to the folded sheet after the caustic and does not provide a unique regular continuation from the initial data.

For zero viscosity, the method of characteristics gives $u=u_0(a)e^{-t}$ and $x=a+u_0(a)(1-e^{-t})$. On the sloping part of the initial profile, $u_0(a)=-2a$, so $x=a(2e^{-t}-1)$. While this map is increasing, inversion givesThe sloping interval shrinks from width one to zero while its height decreases from two to one. Its spatial derivative is $-2e^{-t}/(2e^{-t}-1)$ and diverges as $2e^{-t}-1\downarrow 0$. Thus the first shock formation by characteristic intersection occurs atFor $0\le t<t_{\ast }$ the profile consists of a left plateau, a straight descending ramp ending at zero, and a right zero plateau. At $t_{\ast }$ the ramp becomes a vertical jump. This is a damped Burgers equation; viscosity smooths that jump into a thin transition.

Write $F_p,F_q$ evaluated at the fixed pair $(p(s),q(s))$. Differentiating in $\tau$ gives $x_{\tau }=F_p$, $y_{\tau }=F_q$, and $u_{\tau }=p{F}_p+q{F}_q$, proving $u_{\tau }=p{x}_{\tau }+q{y}_{\tau }$.

Differentiate the other parameter. The difference of the required sides isThe first bracket is zero by the prescribed initial compatibility condition, and the second is zero by differentiating $F(p(s),q(s))=0$. Thus $u_s=p{x}_s+q{y}_s$ as well. These calculations use the differentiability of the supplied parametric functions.

If the coordinate Jacobian determinant $J=x_s{y}_{\tau }-y_s{x}_{\tau }$ is nonzero, the inverse function theorem gives local coordinates $(x,y)$. The chain rule then makes $(u_x-p,u_y-q)$ solve a homogeneous two-by-two linear system with determinant $J$. ThereforeThis is the method of characteristics for a first-order equation depending only on the gradient.

Write $A(x)=x-x^2$, $q=1/(n+2)$ and $p=(n+1)/(n+2)$. Integrating the steady mass conservation equation from the head givesFor the time-dependent first-order equation $H_t+H^{n+1}{H}_x=1-2x$, method of characteristics gives $\dot{x}=H^{n+1}$ and $\dot{H}=1-2x$. Consequently $d[H^{n+2}/(n+2)-A(x)]/dt=0$. On a characteristic starting at $(\xi ,0)$ with zero thickness,On a characteristic emitted by the boundary $x=0$, the invariant is zero and $H=H_0(x)$. The travel time of the one emitted at time zero isThe incomplete beta function is finite at both endpoints because $0<p<1$ for the physical exponents $n>0$. Before characteristic crossings, the initial-data branch and boundary-data branch have the displayed forms in the source, with the appropriate emission time $t-f(x)$ for the latter.

Glacier characteristic admissibility gives a substantive domain restriction. Factoring the invariant gives $A(x)-A(\xi )=(x-\xi )(1-x-\xi )$. A positive forward branch requiresEquivalently it is the image of this parameter domain under $(\xi ,x)\mapsto (x,t(\xi ,x))$, restricted further to regions where that map is single-valued. Characteristics cannot start with positive thickness growth at $\xi >1/2$, where the prescribed initial rate $H_t=1-2\xi$ is negative. This is why the printed piecewise expression is a formal characteristic construction, not a global nonnegative solution from a bare bed. For example for $x>1/2$, putting $L_{\xi }=1-2\xi$ givesBoth factors increase with $\xi$ there, so no positive initial-data branch exists at that $x$ for $0<t<f(x)$, contrary to a literal global reading of the first case. Compression can also make characteristics intersect; entropy solutions then require shock waves.

One can nevertheless give the exact zero-thickness endpoint locus of these formal characteristics. Let $K=f(1)=(n+2{)}^{-p}B(q,q)$. At $x=1-\xi$ the change of variable $z=\xi +(1-2\xi )y$ givesThis is the formal snout locus inferred from the given characteristic family; it retreats from $1$ towards $1/2$. It is not justified as the physical terminus of an accumulating glacier formed from zero initial ice. In particular it says nothing about a positive advancing snout at $t<K$. A physical free boundary problem must specify no ablation on a bare bed, maintain $H\ge 0$, and select an entropy solution or retain the small slope regularization. Those additional conditions are absent from the printed first-order construction. Thus a unique global physical snout trajectory cannot be claimed from that construction alone. The steady endpoint remains $x=1$. The degenerate $n=0$ limit would instead have the same endpoint time $K$ for every initial characteristic, so the inverted formula presumes $n>0$.

For a differentiable scalar function and a differentiable curve, the multivariable chain rule isLet $L=x{\partial }_x-2y{\partial }_y$. Choose $u=\alpha (x)y$ to be an invariant of the method of characteristics. Then $Lu=y\{x{\alpha }^{\prime }(x)-2\alpha (x)\}$, so $x{\alpha }^{\prime }=2\alpha$ and a convenient choice is $\alpha (x)=x^2$. For $v=y/x$, the chain rule gives $Lv=-3v$. Thus, writing the unknown in the new coordinates,Integrating with $u$ fixed yieldswhere $F$ is an arbitrary differentiable function on the relevant interval. This gives all local solutions on each patch where the specified change of variables is invertible: its Jacobian determinant is $\partial (u,v)/\partial (x,y)=3y$.

The axes need care because $v$ is undefined at $x=0$ and the Jacobian vanishes at $y=0$. Reparameterizing $F(u)=u^{2}G(u)$ gives the simpler chart expressionIt is the general solution on either component $x>0$ or $x<0$, including $y=0$ when $G$ is differentiable there. Indeed $(x,u=x^{2}y)$ is a nonsingular coordinate chart for $x\ne 0$, and the weighted Euler first-order equation reduces to $x{\partial }_{x}f{|}_u=6f$.

For a patch crossing $x=0$ with $y\ne 0$, the characteristic invariant $w=x\sqrt{|y|}$ instead gives $f=|y{|}^{-3}K(w)$, with an arbitrary differentiable function on each sign component of $y$. These chart descriptions must agree on overlaps. On the axes the original equation gives $f(x,0)=C_{\pm }{x}^6$ and $f(0,y)=D_{\pm }|y{|}^{-3}$ on their separate components; a solution defined smoothly through the origin must have $f(0,0)=0$ and compatible smooth limits, in particular $D_{\pm }=0$. No boundary data select a particular arbitrary function.

On a region with smooth positive area, set $q=G(Z)Q(\theta ,\zeta )$, $\zeta =h(Z)$. Substitution into the simple wave equation givesRemove the last term by $G^{\prime }/G=-A^{\prime }/(2A)$, and choose $h^{\prime }=G$. Starting at a regular section $Z_0$, one convenient normalized area transformation of a nonlinear acoustic simple wave isIn the notation $g(\zeta )$, set $g(\zeta )=G(h^{-1}(\zeta ))$. The remaining equation is $Q_{\zeta }-Q{Q}_{\theta }=0$. With initial value $f$ at $Z_0$, the method of characteristics givesThe map from the initial label loses invertibility when $1-f^{\prime }(\xi )\zeta =0$. For smooth data with bounded derivative and $M=\sup f^{\prime }>0$, the first forward characteristic crossing is at ${\zeta }_s=1/M$; if $M\le 0$, no forward shock occurs.

For spherical spreading, write physical radius as $R>0$ and specify data at an emitting radius $R_0>0$. Since $A(R)=R^2$,The shock distance for a spherical simple wave launched at finite radius isIf the axial distance is measured from the emitting surface, $Z=R-R_0$, then $A(Z)=(R_0+Z{)}^2$ andNo finite shock distance exists in this model when $M\le 0$.

The printed combination $A=Z^2$ with general bounded data at $Z=0$ is singular and does not define the intended outgoing spherical initial-value problem. Along a spherical characteristic, $d\theta /dR=-q$ and $dq/dR=-q/R$, so $Rq$ is constant. A bounded trace at $R=0$ forces that constant to be zero for every characteristic curve emanating from the origin; arbitrary nonzero bounded $f$ cannot be launched there. For example, the ostensibly admissible data $f\equiv 1$ would demand both $Rq=0$ and a nonzero trace, a contradiction. The logarithmic transformed distance is also divergent at the origin. Thus an unqualified numerical shock formula from those literal data would be unjustified. The finite-radius formulas above give the physical interpretation once the launching radius, or the shifted area law, is supplied.