Paper 2, Section I, A

Differential Equations | Part IA, 2010

Let f(x,y)=g(u,v)f(x, y)=g(u, v) where the variables {x,y}\{x, y\} and {u,v}\{u, v\} are related by a smooth, invertible transformation. State the chain rule expressing the derivatives gu\frac{\partial g}{\partial u} and gv\frac{\partial g}{\partial v} in terms of fx\frac{\partial f}{\partial x} and fy\frac{\partial f}{\partial y} and use this to deduce that

2guv=xuxv2fx2+(xuyv+xvyu)2fxy+yuyv2fy2+Hfx+Kfy\frac{\partial^{2} g}{\partial u \partial v}=\frac{\partial x}{\partial u} \frac{\partial x}{\partial v} \frac{\partial^{2} f}{\partial x^{2}}+\left(\frac{\partial x}{\partial u} \frac{\partial y}{\partial v}+\frac{\partial x}{\partial v} \frac{\partial y}{\partial u}\right) \frac{\partial^{2} f}{\partial x \partial y}+\frac{\partial y}{\partial u} \frac{\partial y}{\partial v} \frac{\partial^{2} f}{\partial y^{2}}+H \frac{\partial f}{\partial x}+K \frac{\partial f}{\partial y}

where HH and KK are second-order partial derivatives, to be determined.

Using the transformation x=uvx=u v and y=u/vy=u / v in the above identity, or otherwise, find the general solution of

x2fx2y2x2fy2+fxyxfy=0x \frac{\partial^{2} f}{\partial x^{2}}-\frac{y^{2}}{x} \frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial f}{\partial x}-\frac{y}{x} \frac{\partial f}{\partial y}=0

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