- #1

iainfs

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I have a couple of questions remaining on a differential equations example sheet that I can't seem to crack. They have a common theme -- changing variables in a PDE.

Here's the first one. I'm hoping that with a gentle nudge in the right direction the rest of it should fall into place :)

I think I've done the first two parts; I'd appreciate it if someone could check it over because I've probably ballsed up somewhere!

(i) Consider the change of variables

[tex]x=e^{-s} \sin t, \; y=e^{-s}\cos t[/tex] such that [tex]u(x,y) = v(s,t)[/tex].

Use the chain rule to express [tex]\partial v / \partial s[/tex] and [tex]\partial v / \partial t[/tex] in terms of [tex]x,y,\partial u/\partial x[/tex] and [tex]\partial u / \partial y[/tex].

[tex]\frac{\partial x}{\partial s} = -x, \; \frac{\partial x}{\partial t} = y, \; \frac{\partial y}{\partial s} = -y, \; \frac{\partial y}{\partial t} = -x[/tex]

Then we have

[tex]\frac{\partial v}{\partial s} = -x\frac{\partial u}{\partial x} - y\frac{\partial u}{\partial y}[/tex]

[tex]\frac{\partial v}{\partial t} = y\frac{\partial u}{\partial x} - x\frac{\partial u}{\partial y}[/tex].

(ii) Find, similarly, an expression for [tex]\partial^2 v/\partial t^2.[/tex]

Rewriting the operator [tex]\frac{\partial}{\partial t} = y\frac{\partial}{\partial x} - x\frac{\partial}{\partial y}[/tex], so we find [tex]\frac{\partial^2 v}{\partial t^2} = y\frac{\partial}{\partial x} \left( y\frac{\partial u}{\partial x} - x\frac{\partial u}{\partial y} \right) - x\frac{\partial}{\partial y} \left( y\frac{\partial u}{\partial x} - x\frac{\partial u}{\partial y} \right) [/tex]

[tex]=y \left[ \frac{\partial y}{\partial x}\frac{\partial u}{\partial x} + y \frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial y} - x \frac{\partial^2 u}{\partial x \partial y} \right] - x\left[ \frac{\partial u}{\partial x} + y\frac{\partial^2 u}{\partial y \partial x} - \frac{\partial x}{\partial y}\frac{\partial u}{\partial y} - x\frac{\partial^2 u}{\partial y^2}\right][/tex]

[tex]\Rightarrow \frac{\partial^2 v}{\partial t^2} = y^2 \frac{\partial^2 u}{\partial x^2} - 2xy\frac{\partial^2 u}{\partial x \partial y} + x^2 \frac{\partial^2 u}{\partial y^2} + y\frac{\partial y}{\partial x}\frac{\partial u}{\partial x} + x\frac{\partial x}{\partial y}\frac{\partial u}{\partial y} - y\frac{\partial u}{\partial y} - x \frac{\partial u}{\partial x} [/tex]

Further, [tex]\frac{x}{y}=\tan t[/tex], so [tex]\frac{\partial x}{\partial y} = \tan t = \frac{x}{y}[/tex], and similarly [tex]\frac{\partial y}{\partial x} = \frac{y}{x}[/tex].

Then we have

[tex]\frac{\partial^2 v}{\partial t^2} = y^2 \frac{\partial^2 u}{\partial x^2} - 2xy\frac{\partial^2 u}{\partial x \partial y} + x^2 \frac{\partial^2 u}{\partial y^2} + \frac{y^2}{x}\frac{\partial u}{\partial x} + \frac{x^2}{y}\frac{\partial u}{\partial y} - y\frac{\partial u}{\partial y} - x \frac{\partial u}{\partial x} [/tex]

(iii) Hence transform the equation [tex]y^2 \frac{\partial^2 u}{\partial x^2} - 2xy\frac{\partial^2 u}{\partial x \partial y} + x^2 \frac{\partial^2 u}{\partial y^2}=0[/tex] into a partial differential equation for v.

We have

[tex]\frac{\partial^2 v}{\partial t^2} - \frac{y^2}{x}\frac{\partial u}{\partial x} - \frac{x^2}{y}\frac{\partial u}{\partial y} + y\frac{\partial u}{\partial y} + x \frac{\partial u}{\partial x}=0[/tex]

[tex]\Rightarrow \frac{\partial^2 v}{\partial t^2} - \frac{\partial v}{\partial s} - \frac{y^2}{x}\frac{\partial u}{\partial x} - \frac{x^2}{y}\frac{\partial u}{\partial y}=0[/tex]

...but I can't see how to get rid of that last pair of terms.

Pointers much appreciated!

Many thanks,