Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 9

For a quasilinear partial differential equation, the characteristic equations for a quasilinear equation are the ordinary differential equationsA solution surface is locally a differentiable graph $z=u(x,y)$ whose tangent plane contains the vector $(P,Q,R)$ at every point. Its normal is $(-u_x,-u_y,1)$, so the tangency condition is precisely $P{u}_x+Q{u}_y=R$. Thus every characteristic lying on that graph follows the vector field without leaving the graph.

Parameterize the given initial curve by $\gamma (t)=(x_0(t),y_0(t),z_0(t))$. For each $t$, solve the characteristic equations for a quasilinear equation with initial value $\gamma (t)$ at $s=0$. The resulting map $(s,t)\mapsto (x(s,t),y(s,t),z(s,t))$ sweeps out a candidate solution surface. If the coefficients and initial data are continuously differentiable, the Picard-Lindelöf theorem supplies this local family. To make it a graph we also need the projection $(s,t)\mapsto (x,y)$ to be locally invertible. At $s=0$ its Jacobian determinant isThe Noncharacteristic Cauchy data condition is $J(0,t)\ne 0$. The inverse function theorem then gives $s=s(x,y)$ and $t=t(x,y)$, and hence $u(x,y)=z(s(x,y),t(x,y))$. Its derivatives obey the PDE by the chain rule. The same characteristic initial-value uniqueness gives local uniqueness of this solution wherever the projection remains invertible and the relevant characteristic meets the prescribed data.

The characteristic equations for a quasilinear equation here areSuccessive integration gives the general characteristicwhere $a,b,c$ are integration constants. Equivalently $z-2x$ and $y-z^2/4$ are constant along each curve: differentiating them with respect to $s$ gives zero. Three constants occur because shifting the origin of the parameter $s$ describes the same geometric curve again; there are two independent curve invariants.

Use $t\ge 0$ to label the initial half-line, with $(x,y,z)=(t,-t,1)$ at $s=0$. In the general characteristic from part (i), this sets $a=t$, $b=-t$, $c=1$. The parametric solution surface is thereforeAt $s=0$ this recovers exactly the given data. Moreover $(x_s,y_s,z_s)=(1,1+2s,2)=(1,z,2)$, so the surface follows the required characteristic vector field. The branch defining a classical graph through the data is determined below by restricting to the side of the projection fold containing $s=0$.

For fixed $t$, eliminate $s=x-t$ to obtain the projected parabolaAlong that curve,Equality occurs at $s=-1$, where the point is $(t-1,-t)$. Also $dy/dx=y_s/x_s=1+2s=-1$ there, so the tangent line has the same slope as the line $x+y=-1$ and passes through a point of that line. Thus every projected characteristic is tangent to that line, with no crossing of it. Because $t\ge 0$, the attained tangency points form only the half-line $x\ge -1$, $y=-x-1$.

Differentiating the projected parameterization givesIt vanishes exactly when $s=-1$. The image is the tangency half-line from part (iii), $x=t-1$, $y=-t$, $t\ge 0$. This is a fold caustic: the parameterized surface itself remains regular, but its projection ceases to be locally invertible, so it cannot there be expressed as a differentiable single-valued graph over the plane.

Put $W=1+x+y$. The identity $W=(s+1{)}^2$ gives $s=-1\pm \sqrt{W}$, and hence $z=-1\pm 2\sqrt{W}$. On the initial half-line $W=1$ and the prescribed value is $z=1$, which selects the positive sign. The branch containing the data has $s>-1$, soFor $W>0$, its derivatives are $z_x=z_y=W^{-1/2}$. Substitution checks the partial differential equation:On the data $y=-x$, the value is one. The other algebraic branch gives $z=-3$ there and does not meet these data. The positive-branch formula defines a solution on the larger half-plane $W>0$, but the part uniquely fixed by the prescribed half-line also requires a nonnegative characteristic footpoint $t$, as part (vi) explains.

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.