Задача №1813
Условие
\(u=\ln\frac{1}{\sqrt{x^2+y^2}}\); показать, что \(\frac{\partial^2{u}}{\partial{x}^2}+\frac{\partial^2{u}}{\partial{y}^2}=0\).
Решение
\[
u=\ln\frac{1}{\sqrt{x^2+y^2}}
=-\frac{1}{2}\ln\left(x^2+y^2\right)
\]
\[
\frac{\partial{u}}{\partial{x}}
=-\frac{1}{2}\cdot\frac{1}{x^2+y^2}\cdot{2x}
=-\frac{x}{x^2+y^2}.
\]
\[
\frac{\partial{u}}{\partial{y}}
=-\frac{1}{2}\cdot\frac{1}{x^2+y^2}\cdot{2y}
=-\frac{y}{x^2+y^2}.
\]
\[
\frac{\partial^2{u}}{\partial{x}^2}
=-\frac{x^2+y^2-x\cdot{2x}}{\left(x^2+y^2\right)^2}
=\frac{x^2-y^2}{\left(x^2+y^2\right)^2}.
\]
\[
\frac{\partial^2{u}}{\partial{y}^2}
=-\frac{x^2+y^2-x\cdot{2y}}{\left(x^2+y^2\right)^2}
=\frac{y^2-x^2}{\left(x^2+y^2\right)^2}.
\]
\[
\frac{\partial^2{u}}{\partial{x}^2}+\frac{\partial^2{u}}{\partial{y}^2}
=\frac{x^2-y^2}{\left(x^2+y^2\right)^2}+\frac{y^2-x^2}{\left(x^2+y^2\right)^2}
=0.
\]
Ответ:
Равенство доказано.