Partial differentiation

**Deadstar** · November 30th 2007, 02:34 PM

Show that, if we change the independent variables (x,y) to (s,t), where $x = e^s cos t$ and $y = e^s sin t$ , then
$\frac{\partial^2 u}{\partial s^2} + \frac{\partial^2 u}{\partial t^2} = (x^2 + y^2)(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2})$

Now... Whats u? do you not need u to be equal to some function of x and y to be able to solve this?

**Opalg** · December 1st 2007, 12:10 AM

u is an unknown function. You don't need to know explicitly what it is (so long as it's smooth enough to be twice differentiable).

This is an exercise in using the chain rule. Start with $\frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s} = e^s\cos t\frac{\partial u}{\partial x} + e^s\sin t\frac{\partial u}{\partial y}$ . Differentiate this again, partially with respect to s, using the chain rule again, together with the product rule, to get an expression for $\frac{\partial^2 u}{\partial s^2}$ in terms of $\frac{\partial^2 u}{\partial x^2}$ , $\frac{\partial^2 u}{\partial x\partial y}$ and $\frac{\partial^2 u}{\partial y^2}$ .

Then do the same procedure, differentiating twice partially with respect to t this time, to get an expression for $\frac{\partial^2 u}{\partial t^2}$ . Add the two expressions together and hope that the $\frac{\partial^2 u}{\partial x\partial y}$ terms cancel out to give the result that you want.

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