考虑偏微分方程（PDE）：
{ u t + ( u x ) 2 = 0 , t > 0 , x ∈ R , u ∣ t = 0 = − x 2 . \begin{cases} u_t + (u_x)^2 = 0, & t > 0, \, x \in \mathbb{R}, \\ u|_{t=0} = -x^2. \end{cases} {ut+(ux)2=0,u∣t=0=−x2.t>0,x∈R,
通过特征线法求解该 PDE。设 p = u x p = u_x p=ux，则 Hamilton 函数为 H ( p ) = p 2 H(p) = p^2 H(p)=p2，其导数为 H ′ ( p ) = 2 p H'(p) = 2p H′(p)=2p。特征方程为：
d x d t = H ′ ( p ) = 2 p , d p d t = − H x = 0 , d u d t = p H ′ ( p ) − H ( p ) = p ⋅ 2 p − p 2 = p 2 . \frac{dx}{dt} = H'(p) = 2p, \quad \frac{dp}{dt} = -H_x = 0, \quad \frac{du}{dt} = p H'(p) - H(p) = p \cdot 2p - p^2 = p^2. dtdx=H′(p)=2p,dtdp=−Hx=0,dtdu=pH′(p)−H(p)=p⋅2p−p2=p2.
由 d p d t = 0 \frac{dp}{dt} = 0 dtdp=0 知 p p p 沿特征线为常数。初始条件为 u ( x , 0 ) = − x 2 u(x,0) = -x^2 u(x,0)=−x2，故在 t = 0 t = 0 t=0 时， p = u x = − 2 x p = u_x = -2x p=ux=−2x。设初始点为 x = x 0 x = x_0 x=x0，则 p = − 2 x 0 p = -2x_0 p=−2x0（常数）。
解 x x x 的方程：
d x d t = 2 p = 2 ( − 2 x 0 ) = − 4 x 0 , \frac{dx}{dt} = 2p = 2(-2x_0) = -4x_0, dtdx=2p=2(−2x0)=−4x0,
积分得 x = − 4 x 0 t + c x = -4x_0 t + c x=−4x0t+c。代入初始条件 x ( 0 ) = x 0 x(0) = x_0 x(0)=x0，有 c = x 0 c = x_0 c=x0，故：
x = x 0 ( 1 − 4 t ) . x = x_0 (1 - 4t). x=x0(1−4t).
解 u u u 的方程：
d u d t = p 2 = ( − 2 x 0 ) 2 = 4 x 0 2 , \frac{du}{dt} = p^2 = (-2x_0)^2 = 4x_0^2, dtdu=p2=(−2x0)2=4x02,
积分得 u = 4 x 0 2 t + d u = 4x_0^2 t + d u=4x02t+d。代入初始条件 u ( 0 ) = u ( x 0 , 0 ) = − x 0 2 u(0) = u(x_0, 0) = -x_0^2 u(0)=u(x0,0)=−x02，有 d = − x 0 2 d = -x_0^2 d=−x02，故：
u = 4 x 0 2 t − x 0 2 = x 0 2 ( 4 t − 1 ) . u = 4x_0^2 t - x_0^2 = x_0^2 (4t - 1). u=4x02t−x02=x02(4t−1).
由 x = x 0 ( 1 − 4 t ) x = x_0 (1 - 4t) x=x0(1−4t) 解出 x 0 = x 1 − 4 t x_0 = \frac{x}{1 - 4t} x0=1−4tx（当 1 − 4 t ≠ 0 1 - 4t \neq 0 1−4t=0 时），代入 u u u 的表达式：
u ( x , t ) = ( x 1 − 4 t ) 2 ( 4 t − 1 ) = x 2 ( 1 − 4 t ) 2 ⋅ [ − ( 1 − 4 t ) ] = − x 2 1 − 4 t . u(x,t) = \left( \frac{x}{1 - 4t} \right)^2 (4t - 1) = \frac{x^2}{(1 - 4t)^2} \cdot [-(1 - 4t)] = -\frac{x^2}{1 - 4t}. u(x,t)=(1−4tx)2(4t−1)=(1−4t)2x2⋅[−(1−4t)]=−1−4tx2.
显式解为：
u ( x , t ) = − x 2 1 − 4 t , t ≠ 1 4 . u(x,t) = -\frac{x^2}{1 - 4t}, \quad t \neq \frac{1}{4}. u(x,t)=−1−4tx2,t=41.
当 t → 1 4 − t \to \frac{1}{4}^- t→41− 时，分母 1 − 4 t → 0 + 1 - 4t \to 0^+ 1−4t→0+。对任意固定 x ≠ 0 x \neq 0 x=0，有：
lim t → 1 4 − u ( x , t ) = lim t → 1 4 − ( − x 2 1 − 4 t ) = − ∞ . \lim_{t \to \frac{1}{4}^-} u(x,t) = \lim_{t \to \frac{1}{4}^-} \left( -\frac{x^2}{1 - 4t} \right) = -\infty. t→41−limu(x,t)=t→41−lim(−1−4tx2)=−∞.
因此，当 t t t 趋近于 1 4 \frac{1}{4} 41 时，解 u ( x , t ) u(x,t) u(x,t) 在任意 x ≠ 0 x \neq 0 x=0 处发散至 − ∞ -\infty −∞。解在 t = 1 4 t = \frac{1}{4} t=41 处未定义，且在该时刻解趋于无穷。