考虑偏微分方程(PDE):
{ut+(ux)2=0,t>0, x∈R,u∣t=0=−x2. \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=ux p = u_x p=ux,则 Hamilton 函数为 H(p)=p2 H(p) = p^2 H(p)=p2,其导数为 H′(p)=2p H'(p) = 2p H′(p)=2p。特征方程为:
dxdt=H′(p)=2p,dpdt=−Hx=0,dudt=pH′(p)−H(p)=p⋅2p−p2=p2. \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 dpdt=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)=−x2 u(x,0) = -x^2 u(x,0)=−x2,故在 t=0 t = 0 t=0 时,p=ux=−2x p = u_x = -2x p=ux=−2x。设初始点为 x=x0 x = x_0 x=x0,则 p=−2x0 p = -2x_0 p=−2x0(常数)。
解 x x x 的方程:
dxdt=2p=2(−2x0)=−4x0, \frac{dx}{dt} = 2p = 2(-2x_0) = -4x_0, dtdx=2p=2(−2x0)=−4x0,
积分得 x=−4x0t+c x = -4x_0 t + c x=−4x0t+c。代入初始条件 x(0)=x0 x(0) = x_0 x(0)=x0,有 c=x0 c = x_0 c=x0,故:
x=x0(1−4t). x = x_0 (1 - 4t). x=x0(1−4t).
解 u u u 的方程:
dudt=p2=(−2x0)2=4x02, \frac{du}{dt} = p^2 = (-2x_0)^2 = 4x_0^2, dtdu=p2=(−2x0)2=4x02,
积分得 u=4x02t+d u = 4x_0^2 t + d u=4x02t+d。代入初始条件 u(0)=u(x0,0)=−x02 u(0) = u(x_0, 0) = -x_0^2 u(0)=u(x0,0)=−x02,有 d=−x02 d = -x_0^2 d=−x02,故:
u=4x02t−x02=x02(4t−1). u = 4x_0^2 t - x_0^2 = x_0^2 (4t - 1). u=4x02t−x02=x02(4t−1).
由 x=x0(1−4t) x = x_0 (1 - 4t) x=x0(1−4t) 解出 x0=x1−4t x_0 = \frac{x}{1 - 4t} x0=1−4tx(当 1−4t≠0 1 - 4t \neq 0 1−4t=0 时),代入 u u u 的表达式:
u(x,t)=(x1−4t)2(4t−1)=x2(1−4t)2⋅[−(1−4t)]=−x21−4t. 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)=−x21−4t,t≠14. u(x,t) = -\frac{x^2}{1 - 4t}, \quad t \neq \frac{1}{4}. u(x,t)=−1−4tx2,t=41.
当 t→14− t \to \frac{1}{4}^- t→41− 时,分母 1−4t→0+ 1 - 4t \to 0^+ 1−4t→0+。对任意固定 x≠0 x \neq 0 x=0,有:
limt→14−u(x,t)=limt→14−(−x21−4t)=−∞. \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 14 \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=14 t = \frac{1}{4} t=41 处未定义,且在该时刻解趋于无穷。