# Help in partial derivative

## Homework Statement

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

$$\nabla^2u=0 \;\Rightarrow\; u_{xx}+u_{yy} = 0$$

From the book:

For $u(x,y)=v(x^2-y^2,2xy)$

$$u_x=2xv_x + 2yv_y$$

$$u_{xx} = 4x^2v_{xx} + 8 xyv_{xy} + 4y^2v_{yy} + 2v_x$$

$$u_y=2yv_x + 2xv_y$$

$$u_{yy} = 4y^2v_{xx} - 8 xyv_{xy} + 4x^2v_{yy} - 2v_x$$

$$\nabla^2u = u_{xx} + u_{yy} = (4x^2+4y^2)(v_{xx}+v_{yy}) = 0$$

This is my work:

I don't understand the solution the book gave.

$$\nabla^2u = \nabla \cdot \nabla u = \nabla \cdot \nabla v(x^2-y^2,2xy)$$

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

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

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

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

I don't even understand how $u_x=2xv_x + 2yv_y$

And $$u_{xx} = 4x^2v_{xx} + 8 xyv_{xy} + 4y^2v_{yy} + 2v_x$$

here stand for?

What is $v_x,\; v_{xx},\; v_y \hbox { and } v_{yy}$

Thanks

Alan

Last edited: