First-order PDE: Lagrange, Charpit, characteristics

In one sentence

A first-order PDE is an ODE in disguise: find the curves along which it becomes one, carry the initial data along them, and the only way it fails is when those curves do not reach where you need them.

Why the exam asks it

Because the method is mechanical once seen, and because two things go wrong in predictable places — the initial curve running tangent to a characteristic, and characteristics crossing each other, which is a shock. Both are questions with clean answers.

The idea

Quasi-linear equations and Lagrange

$$a(x,y,u)\,u_x + b(x,y,u)\,u_y = c(x,y,u).$$

Lagrange's method: solve the characteristic system

$$\frac{dx}{a} = \frac{dy}{b} = \frac{du}{c}.$$

Two independent first integrals $\varphi = C_1$ and $\psi = C_2$ give the general solution $F(\varphi,\psi) = 0$ for arbitrary $F$, or equivalently $\varphi = f(\psi)$.

The simplest case shows the shape of everything. For $u_x + 2u_y = 0$ the characteristics satisfy $dy/dx = 2$, so $y - 2x$ is constant along them, and $du = 0$ means $u$ is constant along each. Hence $u = f(y-2x)$ for any differentiable $f$ — the general solution is an arbitrary function of the characteristic variable, not an arbitrary constant. That is the structural difference from ODEs and it is where the arbitrariness lives.

Where the method fails

The initial curve tangent to a characteristic. Data is carried along characteristics, so an initial curve running tangent to one at a point carries no information away from it: either no solution exists or infinitely many do. The condition to check is that the initial curve is nowhere characteristic, and it is the hypothesis an option removes.

Characteristics crossing — shocks. For $u_x + u\,u_y = 0$ the speed depends on $u$ itself, so characteristics carrying different values can meet. With data $u(x,0) = g(x)$, the solution is implicit,

$$u = g(y - ux),$$

and the first crossing is at $t = -1/\min g'$. For $g(x) = -x$ that gives $t = 1$: a smooth initial condition producing a discontinuity in finite time, with no loss of smoothness in the equation. This is the PDE counterpart of $y' = y^2$ blowing up, and worth recognising as the same phenomenon.

Charpit for fully nonlinear equations

For $F(x,y,u,p,q) = 0$ with $p = u_x$, $q = u_y$, Charpit's equations are

$$\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{du}{pF_p + qF_q} = \frac{-dp}{F_x + pF_u} = \frac{-dq}{F_y + qF_u}.$$

Find one extra relation between $p$ and $q$, solve for both, and integrate $du = p\,dx + q\,dy$. In practice pick whichever ratio is simplest, which is usually the one where a variable is absent.

Complete, general and singular integrals are the three families: a complete integral has two arbitrary constants; the general integral comes from making one a function of the other and eliminating; the singular integral is the envelope, obtained by eliminating both constants via the partial derivatives. Options tend to confuse the general with the singular.

Where intuition breaks

"The general solution has arbitrary constants." It has an arbitrary function. $u = f(y-2x)$ for any $f$, which is infinitely more freedom than an ODE's constants.

"Any initial curve determines the solution." Not one tangent to a characteristic. Data must be given transverse to them, and the failure is either no solution or many.

"Smooth data and a smooth equation give a smooth solution." $u_x + u u_y = 0$ with $u(x,0) = -x$ develops a shock at $t=1$. Nonlinear first-order equations break smoothness in finite time.

"The singular integral is a special case of the general integral." It is not contained in it — it is the envelope of the complete integral and has to be found separately by eliminating both constants.

"Characteristics are always straight." Only when the coefficients are constant. For quasi-linear equations they depend on $u$, which is exactly why they can cross.

"$u = f(y-2x)$ needs $f$ differentiable, so weak solutions are irrelevant." Once shocks appear, the useful notion of solution is weaker than classical, and the Rankine–Hugoniot condition determines the shock speed. Worth knowing the phrase exists.

The exam's angle

1. Write the characteristic system immediately. $dx/a = dy/b = du/c$ is the whole method for quasi-linear equations.
2. Look for the easy pairing. One of the three ratios usually integrates on sight, giving the first integral cheaply.
3. Check transversality if the question gives an initial curve — tangency is the standard way to make "no solution" or "infinitely many" the right answer.
4. For shocks, compute $-1/\min g'$. If $g$ is decreasing anywhere, a shock forms and the time is that formula.
5. For fully nonlinear, use Charpit and name the integral asked for — complete, general or singular, since they are different objects.

The night before

• Lagrange: $\dfrac{dx}{a} = \dfrac{dy}{b} = \dfrac{du}{c}$; two first integrals give $F(\varphi,\psi)=0$.
• General solutions carry an arbitrary function, not constants. $u_x + 2u_y = 0$ gives $u = f(y-2x)$.
• The initial curve must be nowhere tangent to a characteristic, or the problem is ill-posed.
• $u_x + uu_y = 0$: characteristics carry $u$ at speed $u$, cross at $t = -1/\min g'$, and a shock forms. $g(x) = -x$ gives $t = 1$.
• Charpit for fully nonlinear $F(x,y,u,p,q)=0$; then integrate $du = p\,dx + q\,dy$.
• Complete integral: two constants. General: one made a function of the other. Singular: the envelope, found by eliminating both — not a case of the general.