(二阶线性椭圆型PDE) Tricomi方程

hbghlyj

DE1 notes 2022 page65
Example: Classify and reduce to normal form the PDE
$$
y u_{x x}+u_{y y}=0, \text { for } y>0
$$
$a c-b^2=y$ so the equation is elliptic when $y>0$.
The characteristic equation is
$$
y\left(y^{\prime}\right)^2+1=0
$$
that is
$$
y^{1 / 2} y^{\prime}= \pm i
$$
so integrating
$$
2 y^{3 / 2} \mp 3 i x=\text { const. }
$$
So take as variables $\zeta=2 y^{3 / 2} ; \eta=3 x$. Making the substitution, we find that
$$
3 \zeta\left(u_{\zeta \zeta}+u_{\eta \eta}\right)+u_\zeta=0,
$$but $\zeta \neq 0$ so the normal form is
$$
u_{\zeta \zeta}+u_{\eta \eta}=-\frac{u_\zeta}{3 \zeta}
$$

详细计算一下红色部分:
使用DSolveChangeVariables将$x,y$换为$\zeta,\eta$

1. DSolveChangeVariables[Inactive[DSolve][y (u^(2,0))[x,y]+(u^(0,2))[x,y]==0,u,{x,y}],u,{ζ,η},{ζ==2y^(3/2),η==3x}]//FullSimplify

复制代码

Mathematica输出
$\text{DSolve}\left[\frac{u^{(1,0)}(\zeta ,\eta )+3 \zeta  \left(u^{(0,2)}(\zeta ,\eta )+u^{(2,0)}(\zeta ,\eta )\right)}{\sqrt[3]{\zeta }}=0,u,\{\zeta ,\eta \}\right]$
分子,用普通数学符号写出:$u_\zeta+3 \zeta\left(u_{\zeta \zeta}+u_{\eta \eta}\right)=0$
又见
Tricomi equation canonical form and solution
Characteristic curves for second-order Tricomi equation

hbghlyj

\[\begin{cases}
\zeta=2y^{3/2}\\
\eta=3x
\end{cases}\implies
\begin{cases}
u_x=3u_\eta\\
u_y=u_{\zeta}3y^{1/2}
\end{cases}
\implies
\begin{cases}
u_{xx}=9u_{\eta\eta}\\
u_{yy}=u_{\zeta\zeta}9y+\frac32u_\zeta y^{-1/2}
\end{cases}
\]
代入方程$y u_{x x}+u_{y y}=0\implies 9yu_{\eta\eta}+u_{\zeta\zeta}9y+\frac32u_\zeta y^{-1/2}=0\implies$\[u_{\eta\eta}+u_{\zeta\zeta}=-\frac16u_\zeta y^{-3/2}=-\frac16u_\zeta\left(\zeta\over2\right)^{-1}=-\frac1{3\zeta}u_\zeta\]

Czhang271828

正好试一下 $y u_{x x}+u_{y y}=0$ 的量纲分析. 例如可取 $u$ 量纲为 $\mathrm{kg}$, $x$ 量纲为 $\mathrm m^3$, $y$ 量纲为 $\mathrm m^2$. 记 $T=y^3/x^2$ 为无量纲量. 计算得
$$
\begin{align*}
u_{xx}&=\left(u_T\cdot \dfrac{-2y^3}{x^3}\right)_x=u_T\cdot \dfrac{6y^3}{x^4}+u_{TT}\cdot \dfrac{4y^6}{x^6},\\
u_{yy}&=\left(u_T\cdot \dfrac{3y^2}{x^2}\right)_y=u_T\cdot \dfrac{6y}{x^2}+u_{TT}\cdot \dfrac{9y^4}{x^4}.
\end{align*}
$$
回代原方程, 得
$$
0=(6+6T)\cdot u_T+(4T^2+9T)\cdot u_{TT}.
$$
从而 $-\mathrm du_T=\dfrac{6(1+T)}{4T^2+9T}\cdot \mathrm dT$. 解得
$$
-u_T=\dfrac{2}{3}\ln T+\dfrac{5}{6}\ln (4T+9)+C.
$$
从而 $u(T)=\cdots$.

hbghlyj

'$u_\eta$' should be '$u_\zeta$'