$d^2u/dx^2 - d^2u/dy^2 = f(x,y)$ $\rightarrow$ $d^2u/dξdη = \frac{1}{4}f(\frac{1}{2}(ξ+η),\frac{1}{2}(η-ξ))$

Question

$$\frac{\partial^2u}{\partial x^2} - \frac{\partial^2u}{\partial y^2} = f(x,y) \implies \frac{\partial^2u}{\partial \partial } = \frac{1}{4}f\left(\frac{1}{2}(+),\frac{1}{2}(-)\right)$$

Configuración $ = x y, = x + y$ Recibo $x = \frac{1}{2}(+)$ et $y=\frac{1}{2}(-)$

Parece que no puedo mostrar que $\frac{\partial^2u}{\partial x^2} - \frac{\partial^2u}{\partial y^2} = 4\times \frac{\partial^2u}{\partial \partial }$

Mi trabajo es:

$$\partial u/\partial x = \frac{1}{2}\frac{\partial u} {\partial \xi} + \frac{1}{2}\frac{\partial u}{\partial \eta}$$

$$\partial u/\partial x = -\frac{1}{2}\frac{\partial u} {\partial \xi} + \frac{1}{2}\frac{\partial u}{\partial \eta}$$

$$\frac{\partial^2u}{\partial x^2} = \frac{1}{4}\frac{\partial^2u}{\partial \xi^2} + \frac{1}{4}\frac{\partial^2u}{\partial \eta^2}+ \frac{1}{2}\frac{\partial^2u}{\partial \partial }$$

$$\frac{\partial^2u}{\partial y^2} = \frac{1}{4}\frac{\partial^2u}{\partial \xi^2} + \frac{1}{4}\frac{\partial^2u}{\partial \eta^2}- \frac{1}{2}\frac{\partial^2u}{\partial \partial }$$

¿En qué me equivoco? ¿Podría alguien ayudarme?

Answer 1

Has aplicado mal la regla de la cadena.

Aplicando la regla de la cadena dos veces, obtenemos \begin{align} \frac{\partial^2u}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x} \right) = \frac{\partial}{\partial x} \left( \frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \eta} \right)\\ &= \frac{\partial^2 u}{\partial \xi^2}\frac{\partial \xi}{\partial x} + \frac{\partial^2 u}{\partial \xi \partial \eta}\frac{\partial \eta}{\partial x} + \frac{\partial^2 u}{\partial \eta \partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial^2 u}{\partial \eta^2}\frac{\partial \eta}{\partial x}\\ &= \frac{\partial^2 u}{\partial \xi^2} + 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2}, \end{align} desde $$ \frac{\partial \xi}{\partial x} = \frac{\partial \eta}{\partial x} = 1. $$ Análogamente, \begin{align} \frac{\partial^2u}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial y}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial y} \right) = \frac{\partial}{\partial x} \left( -\frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \eta} \right)\\ &= -\frac{\partial^2 u}{\partial \xi^2}\frac{\partial \xi}{\partial y} - \frac{\partial^2 u}{\partial \xi \partial \eta}\frac{\partial \eta}{\partial y} + \frac{\partial^2 u}{\partial \eta \partial \xi}\frac{\partial \xi}{\partial y} + \frac{\partial^2 u}{\partial \eta^2}\frac{\partial \eta}{\partial y}\\ &= \frac{\partial^2 u}{\partial \xi^2} - 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2}, \end{align} desde $$ \frac{\partial \xi}{\partial y} = -1, \quad\text{and}\quad = \frac{\partial \eta}{\partial y} = 1. $$ Restando los dos se obtiene el resultado.