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Show that the substitution $u=x^2-y^2$, $v=2xy$ converts the equation $\frac{\partial^2 W}{\partial x^2}+\frac{\partial^2 W}{\partial y^2}=0$ into $\frac{\partial^2 W}{\partial u^2}+\frac{\partial^2 W}{\partial v^2}=0$

I have $$\left\lbrace u=x^2-y^2 \atop v=2xy \right. $$ So $$\frac{\partial W}{\partial x} = \frac{\partial W}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial W}{\partial v}\frac{\partial v}{\partial y} $$ and $$\frac{\partial W}{\partial y} = \frac{\partial W}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial W}{\partial v}\frac{\partial v}{\partial y} $$ I can compute for example $\frac{\partial u}{\partial x}=2x$, $\frac{\partial u}{\partial y}=-2y $, $\frac{\partial v}{\partial x}=2y$ and $\frac{\partial v}{\partial y}=2x$ but how to find $\frac{\partial W}{\partial u}$ or $\frac{\partial W}{\partial v}$?

Valent
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1 Answers1

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Simply put, you don't!

You've correctly calculated that \begin{align*} W_{x} &= W_{u}\partial_{x}u + W_{v}\partial_{x}v = 2xW_{u} + 2yW_{v} \\ W_{y} &= W_{u}\partial_{y}u + W_{v}\partial_{y}v = -2yW_{u} + 2xW_{v}. \end{align*}

Next, (note: $\partial_{x}u = \frac{\partial u}{\partial x}$ and $\partial_{xx}^{2}u = \frac{\partial^{2}u}{\partial x^{2}}$), applying product/chain rule again, \begin{align*} W_{xx} &= W_{uu}(\partial_{x}u)^{2} + W_{uv}(\partial_{x}u)(\partial_{x}v) + W_{u}(\partial_{xx}^{2}u) \\ &\qquad + W_{vu}(\partial_{x}u)(\partial_{x}v) + W_{vv}(\partial_{x}v)^{2} + W_{v}(\partial_{xx}^{2}v) \\ &= 4x^{2}W_{uu} + 8xyW_{uv} + 4y^{2}W_{vv} + 2W_{u} \end{align*} as well as \begin{align*} W_{yy} &= W_{uu}(\partial_{y}u)^{2} + W_{uv}(\partial_{y}u)(\partial_{y}v) + W_{u}(\partial_{yy}^{2}u) \\ &\qquad + W_{vu}(\partial_{y}u)(\partial_{y}v) + W_{vv}(\partial_{y}v)^{2} + W_{v}(\partial_{yy}^{2}v) \\ &= 4y^{2}W_{uu} - 8xyW_{uv} + 4x^{2}W_{vv} - 2W_{u}. \end{align*} Now what can we deduce?

Christopher K
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