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Help in partial derivative

  • Thread starter yungman
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  • #1
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Homework Statement



Show if v is harmonic ie. [itex]\; \nabla^2v=0 \;[/itex] , then [itex]\nabla^2u=0 \hbox { where } u(x,y)=v(x^2-y^2,2xy) [/itex]

[tex] \nabla^2u=0 \;\Rightarrow\; u_{xx}+u_{yy} = 0 [/tex]

From the book:

For [itex]u(x,y)=v(x^2-y^2,2xy)[/itex]

[tex]u_x=2xv_x + 2yv_y[/tex]

[tex] u_{xx} = 4x^2v_{xx} + 8 xyv_{xy} + 4y^2v_{yy} + 2v_x[/tex]

[tex]u_y=2yv_x + 2xv_y[/tex]

[tex] u_{yy} = 4y^2v_{xx} - 8 xyv_{xy} + 4x^2v_{yy} - 2v_x[/tex]

[tex] \nabla^2u = u_{xx} + u_{yy} = (4x^2+4y^2)(v_{xx}+v_{yy}) = 0 [/tex]




This is my work:

I don't understand the solution the book gave.

[tex]\nabla^2u = \nabla \cdot \nabla u = \nabla \cdot \nabla v(x^2-y^2,2xy)[/tex]

[tex]\nabla v(x^2-y^2,2xy) = [ \frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial x } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial x }]\hat{x} \;+\; [\frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial y } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial y }]\hat{y} [/tex]

[tex]\nabla v(x^2-y^2,2xy) = [2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) } ] \;\hat{x} \;\;+\;\; [-2y\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2x \frac{\partial v}{\partial (2xy) } ] \;\hat{y} [/tex]

[tex] \nabla^2 v = \nabla \cdot \nabla v = \frac{\partial}{\partial x} [2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) } ] \;\;+\;\; \frac{\partial}{\partial y} [-2y\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2x \frac{\partial v}{\partial (2xy) } ] [/tex]

[tex]\nabla^2 v = 4(x^2+y^2) [ \frac{\partial v}{\partial (x^2-y^2) } + \frac{\partial v}{\partial (2xy) }][/tex]



I don't even understand how [itex]u_x=2xv_x + 2yv_y[/itex]

And [tex] u_{xx} = 4x^2v_{xx} + 8 xyv_{xy} + 4y^2v_{yy} + 2v_x[/tex]

here stand for?


What is [itex]v_x,\; v_{xx},\; v_y \hbox { and } v_{yy} [/itex]


Please help explain to me.

Thanks

Alan
 
Last edited:

Answers and Replies

  • #2
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Anyone please?
 
  • #3
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Please!!!
 

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