Hi so I have the following question:
(ii) Simplify the equation $$ (x+y)\frac{\partial f}{\partial x} - (x-y)\frac{\partial f}{\partial y} =0 $$ using the change of variables $u = \ln\sqrt{x^2+y^2}$ and $v = \arctan (y/x)$.
This is part 2 of a question, part 1 asked me to find the total derivative and jacobian of $f(x,y,z)$, so I'm a bit confused as to how this leads on from that and I just don't really know where to begin so help would be greatly appreciated
On
This substitution is equivalent to log-polar coordinates.
First, compute the partial derivatives
$$ \frac{\partial u}{\partial x} = \frac{x}{x^2+y^2}, \quad \frac{\partial u}{\partial y} = \frac{y}{x^2+y^2} $$ $$ \frac{\partial v}{\partial x} = -\frac{y}{x^2+y^2}, \quad \frac{\partial v}{\partial y} = \frac{x}{x^2+y^2} $$
Then use the multi-variable chain rule $$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} = \dots $$ $$ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y} = \dots $$
Now all you have to do is substitute, and make everything be in terms of $u$ and $v$
The characteristic curves satisfy the differential equations $$ \dot x=x+y\\ \dot y=-x+y $$ which can be transformed to $$ \frac{d}{ds}(e^{-s}x)=e^{-s}y\\ \frac{d}{ds}(e^{-s}y)=-e^{-s}x $$ with the obvious solutions $$ x(s)=e^s(x_0\cos(s)-y_0\sin(s))\\ y(s)=e^s(y_0\cos(s)+x_0\sin(s)) $$ The change of variables corresponds to transforming these formulas into polar coordinates, which somewhat decouples these formulas.
To solve the given problem more along the given guidelines consider that the inverse transformation is $$ x=e^u\cos(v),\;y=e^u\sin(v). $$ Consider the composite function $$ g(u,v)=f\bigl(e^u\cos(v),\;e^u\sin(v)\bigr) $$ and compute its partial derivatives and how they are related in view of the PDE for $f$.