Recipe

Solve a quasilinear first-order PDE

Use when: the equation is $a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u)$ , with a coefficient depending on $u$ itself

Goal: $u(x,y)$ , built from the three coupled characteristic ODEs plus the prescribed initial data

Taught in Quasilinear first-order PDEs

Quick recipe

Write the three coupled characteristic ODEs
Decide whether $u$ can be solved for first
Substitute the known $u$ and solve the remaining ODEs
Use the initial data to eliminate the parameters
Check the solution by substitution

1
Write $x'(t)=a(x,y,u)$ , $y'(t)=b(x,y,u)$ , and $u'(t)=c(x,y,u)$ , using the given data $u(x_0,y_0)=u_0$ as the third initial condition.

Example

For $u_t+uu_x=0$ : $x'(t)=u$ , $y'(t)=1$ (here $y$ plays the role of $t$ ), and $u'(t)=0$ .

Why: the curve $x(t),y(t)$ a value of $u$ travels along now depends on $u$ itself, so all three unknowns are genuinely coupled together.
2
Look at $c(x,y,u)$ before solving anything else, since it decides whether $u(t)$ can be isolated from the other two equations.

Example

For $u_t+uu_x=0$ , $u'(t)=0$ gives $u(t)=u_0$ immediately, before $x(t)$ is touched.
- $c\equiv0$ — $u'(t)=0$ means $u=u_0$ stays fixed along the curve, so substitute this known constant into the remaining equations.
- $c$ genuinely depends on $u$ — all three equations must be solved together as one coupled system, since $u(t)$ keeps changing along the curve.
⚡ Shortcut

When the initial data itself is constant ( $u(x,0)=3$ for every $x$ ), every characteristic carries the same $u_0$ , so the whole solution is that same constant everywhere, with no further algebra needed.

Why: freezing $u$ at $u_0$ for the whole curve is only correct when $c\equiv0$ ; otherwise it silently drops the feedback that makes the equation quasilinear.
3
With $u$ known (constant, from the case above, or already solved), solve $x'(t)=a$ and $y'(t)=b$ for $x(t)$ and $y(t)$ in terms of $x_0,y_0,t$ .

Example

With $u=u_0$ : $x'(t)=u_0\Rightarrow x=x_0+u_0t$ .

Why: this is an ordinary ODE once $u$ is no longer an unknown sitting inside the coefficient.
4
Substitute the prescribed $u_0=u(x_0,y_0)$ into the solved curve, then solve for $x_0$ in terms of $x,y$ (or $x,t$ ).

Example

With $u_0=u(x_0,0)=x_0$ and $x=x_0(1+t)$ : $x_0=x/(1+t)$ , so $u(x,t)=x_0=x/(1+t)$ .

Why: the characteristic system alone only describes one curve at a time; the initial data is what ties every curve back to a single formula for $u(x,y)$ .
5
Differentiate the candidate $u$ and confirm it satisfies both the PDE and the initial condition exactly.

Example

$u_x=1/(1+t)$ , $u_t=-x/(1+t)^2$ , so $uu_x+u_t=x/(1+t)^2-x/(1+t)^2=0$ , and $u(x,0)=x$ .

Why: this catches an algebra slip in solving the coupled system before it's reported as the final answer.

Common mistakes

Freezing $u$ at its starting value $u_0$ for the whole curve when $c(x,y,u)$ is not identically zero.
Solving $x'(t)=a$ and $y'(t)=b$ before checking whether $u(t)$ needs to be found first or solved jointly with them.
Forgetting that constant initial data is a special case, not a hint that the equation itself has become linear.
Reporting the curve's equation in terms of $x_0,y_0,t$ instead of eliminating those parameters down to a formula in $x$ and $y$ alone.