直标系下三维齐次亥姆霍兹方程的特解

青青子衿

\begin{align*}
\dfrac{\partial^2u}{\partial\,x^2}+\dfrac{\partial^2u}{\partial\,y^2}+\dfrac{\partial^2u}{\partial\,z^2}+\lambda\cdot u=0
\end{align*}
\begin{align*}
u&=\bigg[A_1\cos\left(\alpha x\right)+A_2\sin\left(\alpha x\right)\bigg]\bigg[B_1\cos\left(\beta y\right)+B_2\sin\left(\beta y\right)\bigg]\bigg(C_1z+C_2\bigg)&&\lambda=\alpha^2+\beta^2\\
u&=\bigg[A_1\cos\left(\alpha x\right)+A_2\sin\left(\alpha x\right)\bigg]\bigg[B_1\cosh\left(\beta y\right)+B_2\sinh\left(\beta y\right)\bigg]\bigg(C_1z+C_2\bigg)&&\lambda=\alpha^2-\beta^2\\
u&=\bigg[A_1\cos\left(\alpha x\right)+A_2\sin\left(\alpha x\right)\bigg]\bigg[B_1\cos\left(\beta y\right)+B_2\sin\left(\beta y\right)\bigg]\cos\left(\gamma z\right)&&\lambda=\alpha^2+\beta^2+\gamma^2\\
u&=\bigg[A_1\cos\left(\alpha x\right)+A_2\sin\left(\alpha x\right)\bigg]\bigg[B_1\cos\left(\beta y\right)+B_2\sin\left(\beta y\right)\bigg]\sin\left(\gamma z\right)&&\lambda=\alpha^2+\beta^2+\gamma^2\\
u&=\bigg[A_1\cosh\left(\alpha x\right)+A_2\sinh\left(\alpha x\right)\bigg]\bigg[B_1\cos\left(\beta y\right)+B_2\sin\left(\beta y\right)\bigg]\cos\left(\gamma z\right)&&\lambda=-\alpha^2+\beta^2+\gamma^2\\
u&=\bigg[A_1\cosh\left(\alpha x\right)+A_2\sinh\left(\alpha x\right)\bigg]\bigg[B_1\cos\left(\beta y\right)+B_2\sin\left(\beta y\right)\bigg]\sin\left(\gamma z\right)&&\lambda=-\alpha^2+\beta^2+\gamma^2\\
u&=\bigg[A_1\cosh\left(\alpha x\right)+A_2\sinh\left(\alpha x\right)\bigg]\bigg[B_1\cosh\left(\beta y\right)+B_2\sinh\left(\beta y\right)\bigg]\cos\left(\gamma z\right)&&\lambda=-\alpha^2-\beta^2+\gamma^2\\
u&=\bigg[A_1\cosh\left(\alpha x\right)+A_2\sinh\left(\alpha x\right)\bigg]\bigg[B_1\cosh\left(\beta y\right)+B_2\sinh\left(\beta y\right)\bigg]\sin\left(\gamma z\right)&&\lambda=-\alpha^2-\beta^2+\gamma^2\\
u&=\bigg[A_1\cosh\left(\alpha x\right)+A_2\sinh\left(\alpha x\right)\bigg]\bigg[B_1\cosh\left(\beta y\right)+B_2\sinh\left(\beta y\right)\bigg]\cosh\left(\gamma z\right)&&\lambda=-\alpha^2-\beta^2-\gamma^2\\
u&=\bigg[A_1\cosh\left(\alpha x\right)+A_2\sinh\left(\alpha x\right)\bigg]\bigg[B_1\cosh\left(\beta y\right)+B_2\sinh\left(\beta y\right)\bigg]\sinh\left(\gamma z\right)&&\lambda=-\alpha^2-\beta^2-\gamma^2\\
\end{align*}