Differentiation, change of variable

[thenewbosco]

The question:

if x=\rho cos \phi and y=\rho sin\phi

prove that if U is a twice differentiable function of x and y that

\frac{d^2U}{dx^2} + \frac{d^2U}{dy^2} = \frac{d^2U}{d\rho^2} + \frac{1}{\rho}\frac{dU}{d\rho} + \frac{1}{\rho^2}\frac{d^2U}{d\phi^2}

i am not sure how to approach this since the function is not given. Is there some assumption i am supposed to make about the function? like f(x,y) = xy or something?
any help on the first couple steps would be appreciated

 

[arildno]

With a rather more transparent notation, let u be a function of the polar coordinates, U the function in Cartesian coordinates, related by:
u(\rho(x,y),\theta(x,y))=U(x,y)

With this notation, you would have lower case u's on your right-hand side of the identity.

You need to compute the partial derivatives of the radial&angular variables with respect to x and y.

 

[NateTG]

i am not sure how to approach this since the function is not given. Is there some assumption i am supposed to make about the function?

Have you tried the chain rule?

 

[thenewbosco]

so \rho (x,y)= ((\frac{x}{cos \phi})^2 + (\frac{y}{cos \phi})^2)^\frac{1}{2}?

and \phi(x,y)=arctan(y/x) is this correct?

 

[thenewbosco]

to what would i apply the chain rule?

 

[arildno]

No, \rho(x,y)=\sqrt{x^{2}+y^{2}}

That ought to be obvious for anyone claiming to have understood Pythagoras' second most important theorem.

 

[thenewbosco]

so after i compute these derivatives of the radial and angular components with respect to x and y i will have everything and just have to put it all together so it looks like the identity i am trying to show?

 

[arildno]

No, you must also use the chain rule on u.
To give you a start on that:
\frac{\partial{U}}{\partial{x}}=\frac{\partial{u}}{\partial{r}}\frac{\partial{r}}{\partial{x}}+\frac{\partial{u}}{\partial{\theta}}\frac{\partial{\theta}}{\partial{x}}
and so on..

 

[thenewbosco]

where your r is actually a rho i assume then? if so thanks for the help i should be able to solve this one

 

[arildno]

Yes, r is rho.