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Ecuaciones diferenciales en derivadas parciales

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Ecuaciones en derivadas parciales

Tomando u y v por nuevas variables independientes transformar la ecuación:
    \( \displaystyle x^2\frac{\partial^2 z}{\partial x^2} -(x^2+y^2)\frac{\partial^2 z}{\partial x \partial y} + y^2\frac{\partial^2 z}{\partial y^2} =0 \)
Siendo
    \( \displaystyle u = x+y \quad ; \quad v = \frac{1}{x} + \frac{1}{y} \)
- Respuesta 71

Obtenemos las derivadas parciales de z respecto a x e y:
    \( \displaystyle \begin{array}{l}
    \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x} = \frac{\partial z}{\partial u} - \frac{\partial z}{\partial v}·\left(\frac{1}{x^2}\right) \\
     \\
    \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} = \frac{\partial z}{\partial u} - \frac{\partial z}{\partial v}·\left(\frac{1}{y^2}\right)
    \end{array} \)
De igual modo, las derivadas segundas serán:
    \( \displaystyle \begin{array}{l}
    \frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial x} + \frac{\partial^2 z}{ \partial v \partial u}\frac{\partial v}{\partial x} - \left(\frac{1}{x^2}\right)\left(\frac{\partial^2 z}{\partial u \partial v}\frac{\partial u}{\partial x} + \frac{\partial^2 z}{\partial v^2}\frac{\partial v}{\partial x} \right) + \\
     \\
    + \left(\frac{2}{x^3}\right)\frac{\partial z}{\partial v} = \frac{\partial^2 z}{\partial u^2} - \left(\frac{2}{x^2}\right)\frac{\partial^2 z}{\partial v \partial u } + \left(\frac{1}{x^4}\right)\frac{\partial^2 z}{\partial v^2 } + \left(\frac{2}{x^3}\right)\frac{\partial z}{\partial v} \\
     \\
    \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial y} + \frac{\partial^2 z}{ \partial v \partial u}\frac{\partial v}{\partial y} - \left(\frac{1}{x^2}\right)\left(\frac{\partial^2 z}{\partial u \partial v}\frac{\partial u}{\partial y} + \frac{\partial^2 z}{\partial v^2}\frac{\partial v}{\partial y}\right) = \\
     \\
    = \frac{\partial^2 z}{\partial u^2} - \left(\frac{1}{y^2}\right)\frac{\partial^2 z}{\partial u \partial v } - \left(\frac{1}{x^2}\right)\frac{\partial^2 z}{\partial u \partial v } + \left(\frac{1}{x^2y^2}\right)\frac{\partial^2 z}{\partial v^2 } \\
     \\
    \frac{\partial^2 z}{\partial y^2} = \frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial y} + \frac{\partial^2 z}{ \partial v \partial u}\frac{\partial v}{\partial y} - \left(\frac{1}{y^2}\right)\left(\frac{\partial^2 z}{\partial u \partial v}\frac{\partial u}{\partial y} + \frac{\partial^2 z}{\partial v^2}\frac{\partial v}{\partial y}\right) + \\
     \\
    + \left(\frac{2}{y^3}\right)\frac{\partial z}{\partial v} = \frac{\partial^2 z}{\partial u^2} - \left(\frac{2}{y^2}\right)\frac{\partial^2 z}{\partial v \partial u } + \left(\frac{1}{y^4}\right)\frac{\partial^2 z}{\partial v^2 } + \left(\frac{2}{y^3}\right)\frac{\partial z}{\partial v}
    \end{array} \)
Sustituyendo en la expresión del enunciado resulta:
    \( \displaystyle\begin{array}{l}
    x^2\frac{\partial^2 z}{\partial x^2} -(x^2+y^2)\frac{\partial^2 z}{\partial x \partial y} + y^2\frac{\partial^2 z}{\partial y^2} = \\
     \\
    x^2\frac{\partial^2 z}{\partial u^2} - 2\frac{\partial^2 z}{\partial v \partial u } + \left(\frac{1}{x^2}\right)\frac{\partial^2 z}{\partial v^2 } + \left(\frac{2}{x}\right)\frac{\partial z}{\partial v} - \\
     \\ - (x^2+y^2) \frac{\partial^2 z}{\partial u^2} + \frac{x^2+y^2}{y^2}\frac{\partial^2 z}{\partial u \partial v } + \frac{x^2+y^2}{x^2}\frac{\partial^2 z}{\partial u \partial v } - \\
     \\ \frac{x^2+y^2}{x^2y^2}\frac{\partial^2 z}{\partial v^2 } + y^2 \frac{\partial^2 z}{\partial u^2} - 2\frac{\partial^2 z}{\partial v \partial u } + \frac{1}{y^2}\frac{\partial^2 z}{\partial v^2 } + \frac{2}{y}\frac{\partial z}{\partial v}=\\
     \\
    - 4 \frac{\partial^2 z}{\partial u \partial v} + \frac{2(x+y)}{xy}\frac{\partial z}{\partial v} + \frac{(x^2+y^2)^2}{x^2y^2}\frac{\partial^2 z}{\partial u \partial v}= 0
    \end{array}\)
El coeficiente del tercer término se puede transformar cómo sigue:
    \( \displaystyle \begin{array}{l}
    \left(\frac{x^2+y^2}{x·y}\right)^2 =\left(\frac{x^2+y^2 + 2xy - 2xy}{x·y}\right)^2 = \\
     \\
    = \left(\frac{(x+y)^2- 2xy}{x·y}\right)^2 = \left(\frac{(x+y)^2}{xy}- 2\right)^2 = (uv - 2)^2
    \end{array} \)
Por todo ello, la expresión final será:
    \( \displaystyle 4 \frac{\partial^2 z}{\partial u \partial v} + 2v\frac{\partial z}{\partial v} + (uv - 2)^2\frac{\partial^2 z}{\partial u \partial v}= 0 \)
Y simplificando:
    \( \displaystyle 2v\frac{\partial z}{\partial v} + \left[(uv)^2- 4·uv\right]\frac{\partial^2 z}{\partial u \partial v}= 0 \)
EJERCICIOS-ECUACIONES EN DERIVADAS PARCIALES
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