[偏微分方程教程习题参考解答]2.4完全非线性偏微分方程

在下列各题中, 设 $\dps{p=\frac{\p u}{\p x},\ q=\frac{\p u}{\p y}}$.

 

1. 求解下列初值问题:

 

(1). $\dps{\sedd{\ba{ll} pq=u,&x>0,\ y\in\bbR\\ u|_{x=0}=y^2 \ea}}$.

 

解答: 首先将初始条件写成参数形式 $$\bex x_0=0,\quad y_0=s,\quad u_0=s^2. \eex$$ 联合 ($F=pq-u$) $$\bex u_0'(s)-p_0(s)x_0'(s)-q_0(s)y_0'(s)=0,\quad F(0)=0 \eex$$ 知 $$\bex p_0=\frac{s}{2},\quad q_0=2s. \eex$$ 求解 $$\bex \sedd{\ba{ll} \cfrac{\rd x}{F_p}=\cfrac{\rd y}{F_q} =\cfrac{\rd u}{pF_p+qF_q} =-\cfrac{\rd p}{F_x+p F_u} =-\cfrac{\rd q}{F_y+qF_u}=\rd t\\ x(0)=y_0,\quad y(0)=y_0,\quad u(0)=u_0,\quad p(0)=p_0,\quad q(0)=q_0 \ea} \eex$$ 得 $$\bex x=2s(e^t-1),\quad y=\frac{s(e^t+1)}{2},\quad u=s^2e^{2t},\quad p=\frac{s}{2}e^t,\quad q=2se^t. \eex$$ 于是 $$\bex u=\sex{\frac{x+4y}{4}}^2. \eex$$

 

(2). $\dps{\sedd{\ba{ll} p^2+q^2=u^2\\ u(\cos s,\sin s)=1 \ea}}$.

 

解答: 首先将初始条件写成参数形式 $$\bex x_0=\cos s,\quad y_0=\sin s,\quad u_0=1. \eex$$ 联合 ($F=p^2+q^2-u^2$) $$\bex u_0'(s)-p_0(s)x_0'(s)-q_0(s)y_0'(s)=0,\quad F(0)=0 \eex$$ 知 $$\bex p_0=\pm \cos s,\quad q_0=\pm \sin s. \eex$$ 求解 $$\bex \sedd{\ba{ll} \cfrac{\rd x}{F_p}=\cfrac{\rd y}{F_q} =\cfrac{\rd u}{pF_p+qF_q} =-\cfrac{\rd p}{F_x+p F_u} =-\cfrac{\rd q}{F_y+qF_u}=\rd t\\ x(0)=y_0,\quad y(0)=y_0,\quad u(0)=u_0,\quad p(0)=p_0,\quad q(0)=q_0 \ea} \eex$$ 得 (注意 $pF_p+qF_q=2(p^2+q^2)=2u^2$) $$\bex x=\cos s[1\pm \ln(1+2t)],\quad y=\sin s[1\pm \ln(1+2t)],\quad u=-\frac{1}{1+2t}, \eex$$ $$\bex p=\pm \frac{\cos s}{1+2t},\quad q=\pm \frac{\sin s}{1+2t}. \eex$$ 于是 $$\bex u=e^{\pm (\sqrt{x^2+y^2}-1)}. \eex$$

 

(3). $\dps{\sedd{\ba{ll} q=p^3\\ u|_{x=0}=2y^\frac{3}{2} \ea}}$.

 

解答: 首先将初始条件写成参数形式 $$\bex x_0=0,\quad y_0=s,\quad u_0=2s^\frac{3}{2}. \eex$$ 联合 ($F=q-p^3$) $$\bex u_0'(s)-p_0(s)x_0'(s)-q_0(s)y_0'(s)=0,\quad F(0)=0 \eex$$ 知 $$\bex p_0=3^\frac{1}{3} s^\frac{1}{6},\quad q_0=3s^\frac{1}{2}. \eex$$ 求解 $$\bex \sedd{\ba{ll} \cfrac{\rd x}{F_p}=\cfrac{\rd y}{F_q} =\cfrac{\rd u}{pF_p+qF_q} =-\cfrac{\rd p}{F_x+p F_u} =-\cfrac{\rd q}{F_y+qF_u}=\rd t\\ x(0)=y_0,\quad y(0)=y_0,\quad u(0)=u_0,\quad p(0)=p_0,\quad q(0)=q_0 \ea} \eex$$ 得 $$\bex x=-3^\frac{5}{3}s^\frac{1}{3}t,\quad y=s+t,\quad u=2s^\frac{3}{2}-6s^\frac{1}{2}t,\quad p=3^\frac{1}{3}s^\frac{1}{6},\quad q=3s^\frac{1}{2}. \eex$$ 于是所求为 $$\bex x=-3^\frac{5}{3}s^\frac{1}{3}t,\quad y=s+t,\quad u=2s^\frac{3}{2}-6s^\frac{1}{2}t. \eex$$

 

(4). $\dps{\sedd{\ba{ll} u=xp+yq+\frac{1}{2}(p^2+q^2)\\ u|_{x=0}=\frac{1}{2}(1-y^2) \ea}}$.

 

解答: 首先将初始条件写成参数形式 $$\bex x_0=0,\quad y_0=s,\quad u_0=\frac{1}{2}(1-s^2). \eex$$ 联合 ($F=xp+yq+\frac{1}{2}(p^2+q^2)-u$) $$\bex u_0'(s)-p_0(s)x_0'(s)-q_0(s)y_0'(s)=0,\quad F(0)=0 \eex$$ 知 $$\bex p_0=\pm 1,\quad q_0=-s. \eex$$ 求解 $$\bex \sedd{\ba{ll} \cfrac{\rd x}{F_p}=\cfrac{\rd y}{F_q} =\cfrac{\rd u}{pF_p+qF_q} =-\cfrac{\rd p}{F_x+p F_u} =-\cfrac{\rd q}{F_y+qF_u}=\rd t\\ x(0)=y_0,\quad y(0)=y_0,\quad u(0)=u_0,\quad p(0)=p_0,\quad q(0)=q_0 \ea} \eex$$ 得 (注意由方程, $pF_p+qF_q=u+\frac{1}{2}(p^2+q^2)$) $$\bex x=\pm (e^t-1),\quad y=s,\quad u=e^t-\frac{s^2+1}{2},\quad p=\pm 1,\quad q=-s. \eex$$ 于是所求为 $$\bex u=\pm x+\frac{1-y^2}{2}. \eex$$

 

2. 求方程 $q=F(p)$ 满足 $u(x,0)=h(x)$ 的解, 其中 $F(\cdot)$ 是其变元的任意连续可微函数.

 

解答: 首先将初始条件写成参数形式 $$\bex x_0=s,\quad y_0=0,\quad u_0=h(s). \eex$$ 联合 ($G=q-F(p)$) $$\bex u_0'(s)-p_0(s)x_0'(s)-q_0(s)y_0'(s)=0,\quad G(0)=0 \eex$$ 知 $$\bex p_0=h'(s),\quad q_0=F(h'(s)). \eex$$ 求解 $$\bex \sedd{\ba{ll} \cfrac{\rd x}{G_p}=\cfrac{\rd y}{G_q} =\cfrac{\rd u}{pG_p+qG_q} =-\cfrac{\rd p}{G_x+p G_u} =-\cfrac{\rd q}{G_y+qG_u}=\rd t\\ x(0)=y_0,\quad y(0)=y_0,\quad u(0)=u_0,\quad p(0)=p_0,\quad q(0)=q_0 \ea} \eex$$ 得 $$\bex x=s-F'(h'(s)),\quad y=t,\quad u=h(s)+[F(h'(s))-pF'(h'(s))]t, \eex$$ $$\bex p=h'(s),\quad q=F(h'(s)). \eex$$ 于是所求为 $$\bex x=s-F'(h'(s)),\quad y=t,\quad u=h(s)+[F(h'(s))-pF'(h'(s))]t. \eex$$

 

3. 求方程 $xpq+yq^2=1$ 通过曲线 $u=x,y=0$ 的积分曲面.

 

解答: 首先将初始条件写成参数形式 $$\bex x_0=s,\quad y_0=0,\quad u_0=s. \eex$$ 联合 ($G=xpq+yq^2-1$) $$\bex u_0'(s)-p_0(s)x_0'(s)-q_0(s)y_0'(s)=0,\quad G(0)=0 \eex$$ 知 $$\bex p_0=1,\quad q_0=\frac{1}{s}. \eex$$ 求解 $$\bex \sedd{\ba{ll} \cfrac{\rd x}{G_p}=\cfrac{\rd y}{G_q} =\cfrac{\rd u}{pG_p+qG_q} =-\cfrac{\rd p}{G_x+p G_u} =-\cfrac{\rd q}{G_y+qG_u}=\rd t\\ x(0)=y_0,\quad y(0)=y_0,\quad u(0)=u_0,\quad p(0)=p_0,\quad q(0)=q_0 \ea} \eex$$ 得 (注意 $pF_p+qF_q=2$) $$\bex x=s+t,\quad y=t(s+t),\quad u=s+2t,\quad p=\frac{s}{s+t},\quad q=\frac{1}{s+t}. \eex$$ 于是所求为 $$\bex u=x+\frac{y}{x}. \eex$$

 

4. 求方程 $xp+yq=pq$ 的特征, 并导出通过曲线 $u=\frac{1}{2}x,\ y=0$ 的积分曲面.

 

解答: 设 $$\bex F=xp+yq-pq, \eex$$ 则特征方程为 $$\bex \frac{\rd x}{x-q} =\frac{\rd y}{y-p} =\frac{\rd u}{-pq} =\frac{\rd p}{-p} =\frac{\rd q}{-q}=\rd t. \eex$$ 解得特征为 $$\bex x=x_0e^t-\frac{q_0}{2}(e^t-e^{-t}),\quad y=y_0e^t-\frac{p_0}{2}(e^t-e^{-t}),\quad u=u_0+\frac{p_0q_0}{2}(e^{-2t}-1), \eex$$ $$\bex p=p_0e^{-t},\quad q=q_0e^{-t}. \eex$$ 将初始条件写成参数形式 (利用 $u_0'(s)-p_0(s)x_0'(s)-q_0(s)y_0'(s)=0, F(0)=0$) $$\bex x_0=s,\quad y_0=0,\quad u_0=\frac{1}{2}s,\quad p_0=\frac{1}{2},\quad q_0=s. \eex$$ 代入特征得到 $$\bex x=\frac{s}{2}(e^t+e^{-t}),\quad y=-\frac{1}{4}(e^t-e^{-t}),\quad u=\frac{s}{4}(1+e^{-2t}). \eex$$