一类系数含指数函数的二阶偏微分方程

青青子衿

这样的\(\,u=f\left(v\right)\,\)变量代换是怎么想到的？有相关理论吗？
\[ v(x,y)=\sqrt{2}\,e^{\frac{x+y}{2}}\sin\frac{x-y}{2} \]
\[ u(x,y)=f\left(\sqrt{2}\,e^{\frac{x+y}{2}}\sin\frac{x-y}{2}\right) \]
\begin{align*}
&&\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}&=u\cdot e^{x+y}\\
&&&\Downarrow&u=f\left(v\right)\\
&&f''\left(v\right)&=f\left(v\right)
\end{align*}

青青子衿

回复 1# 青青子衿
\begin{align*}  
&&\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}&=k\!\cdot\!u\cdot e^{ax+by}\\  
&&&\large\Downarrow&&\begin{cases}
\xi=ax+by\\
\eta=bx-ay
\end{cases}\\  
&&\dfrac{\partial^2u}{\partial\xi^2}+\dfrac{\partial^2u}{\partial\eta^2}&=\dfrac{k}{a^2+b^2}\!\cdot\!u\cdot e^\xi\\
&&&\large\Downarrow&&\begin{cases}
\varphi=\exp\left(\frac{\xi}{2}\right)\cos\left(\frac{\eta}{2}\right)\\
\psi=\exp\left(\frac{\xi}{2}\right)\sin\left(\frac{\eta}{2}\right)
\end{cases}\\  
&&\dfrac{\partial^2u}{\partial\varphi^2}+\dfrac{\partial^2u}{\partial\psi^2}&=\dfrac{4k}{a^2+b^2}\!\cdot\!u\\  
\end{align*}
\begin{align*}
\mathrm{d}^2u&=\dfrac{\partial^2u}{\partial\xi^2}\,\mathrm{d}\xi^2+2\dfrac{\partial^2u}{\partial\xi\partial\eta}\,\mathrm{d}\xi\mathrm{d}\eta+\dfrac{\partial^2u}{\partial\eta^2}\,\mathrm{d}\eta^2\\
&=\dfrac{\partial^2u}{\partial\xi^2}\left(a\mathrm{d}x+b\mathrm{d}y\right)^2+2\dfrac{\partial^2u}{\partial\xi\partial\eta}\,\left(a\mathrm{d}x+b\mathrm{d}y\right)\left(b\mathrm{d}x-a\mathrm{d}y\right)+\dfrac{\partial^2u}{\partial\eta^2}\,\left(b\mathrm{d}x-a\mathrm{d}y\right)^2\\
&=\left(a^2\dfrac{\partial^2u}{\partial\xi^2}+2ab\dfrac{\partial^2u}{\partial\xi\partial\eta}+b^2\dfrac{\partial^2u}{\partial\eta^2}\right)\mathrm{d}x^2+2\left(\cdots\right)\,\mathrm{d}x\mathrm{d}y+\left(b^2\dfrac{\partial^2u}{\partial\xi^2}-2ab\dfrac{\partial^2u}{\partial\xi\partial\eta}+a^2\dfrac{\partial^2u}{\partial\eta^2}\right)\,\mathrm{d}y^2\\
\end{align*}
\begin{align*}   
u&=\bigg(A_1\varphi+A_2\bigg)\bigg[B_1\cos\left(\alpha\psi\right)+B_2\sin\left(\alpha\psi\right)\bigg]&&-\dfrac{4k}{a^2+b^2}=\alpha^2\\   
u&=\bigg(A_1\varphi+A_2\bigg)\bigg[B_1\cosh\left(\alpha\psi\right)+B_2\sinh\left(\alpha\psi\right)\bigg]&&-\dfrac{4k}{a^2+b^2}=-\alpha^2\\   
u&=\bigg[A_1\cos\left(\alpha\varphi\right)+A_2\sin\left(\alpha \varphi\right)\bigg]\bigg(B_1\psi+B_2\bigg)&&-\dfrac{4k}{a^2+b^2}=\alpha^2\\   
u&=\bigg[A_1\cosh\left(\alpha\varphi\right)+A_2\sinh\left(\alpha\varphi\right)\bigg]\bigg(B_1\psi+B_2\bigg)&&-\dfrac{4k}{a^2+b^2}=-\alpha^2\\   
u&=\bigg[A_1\cos\left(\alpha\varphi\right)+A_2\sin\left(\alpha\varphi\right)\bigg]\bigg[B_1\cos\left(\beta\psi\right)+B_2\sin\left(\beta\psi\right)\bigg]&&-\dfrac{4k}{a^2+b^2}=\alpha^2+\beta^2\\   
u&=\bigg[A_1\cos\left(\alpha\varphi\right)+A_2\sin\left(\alpha\varphi\right)\bigg]\bigg[B_1\cosh\left(\beta\psi\right)+B_2\sinh\left(\beta\psi\right)\bigg]  
&&-\dfrac{4k}{a^2+b^2}=\alpha^2-\beta^2\\   
u&=\bigg[A_1\cosh\left(\alpha\varphi\right)+A_2\sinh\left(\alpha\varphi\right)\bigg]\bigg[B_1\cos\left(\beta\psi\right)+B_2\sin\left(\beta\psi\right)\bigg]  
&&-\dfrac{4k}{a^2+b^2}=-\alpha^2+\beta^2\\   
u&=\bigg[A_1\cosh\left(\alpha\varphi\right)+A_2\sinh\left(\alpha\varphi\right)\bigg]\bigg[B_1\cosh\left(\beta\psi\right)+B_2\sinh\left(\beta\psi\right)\bigg]  
&&-\dfrac{4k}{a^2+b^2}=-\alpha^2-\beta^2\\  
\end{align*}
？？？
吉米多维奇3493