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MATHEMATICIAN June 14th, 2017 09:35 AM

Differential equation proof
1 Attachment(s)
Someone please prove this.
I got exam tomorrow!!

skipjack June 14th, 2017 10:37 AM

$\displaystyle \frac{\partial z}{\partial r} = \frac{∂z}{∂x}\cdot\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial r} = 2r\frac{\partial z}{\partial x} + 2s\frac{\partial z}{\partial y}$, etc.

MATHEMATICIAN June 14th, 2017 10:40 AM

I know those rules but still can't prove.
Could you give me some hint/help.

skipjack June 14th, 2017 11:31 AM

You'll also need to use the product rule. Can you post your attempt?

MATHEMATICIAN June 14th, 2017 11:51 AM

2 Attachment(s)
The solution for second order partial derivative I got:

Adam Ledger June 14th, 2017 02:35 PM

The very first thing I would be trying is what is the implied solution under the additional assumption imposed by making the substitution f(x,y)= u(x,y)+v(x,y)*I into the pde.

And then furthermore whether there are any initial values that can be dictated for u and v on (x,y).

But, don't pursue with any guarantee of the line of enquiry bearing fruit; I'm pretty sporadic with what pops into my head first.

skipjack June 14th, 2017 10:39 PM

You started off correctly, MATHEMATICIAN.

As $z$ has continuous second-order partial derivatives, $\displaystyle \frac{\partial^2 z}{\partial x\partial y} = \frac{\partial^2 z}{\partial y\partial x}$.

You went wrong towards the end, but you didn't show every detail, making it difficult for me to say what exactly you did that wasn't right.

For example, you correctly obtain $\displaystyle 2\frac{\partial z}{\partial x}$, but this appears twice in your final line for no apparent reason.

$\displaystyle \frac{\partial z}{\partial r} = 2r\frac{\partial z}{\partial x} + 2s\frac{\partial z}{\partial y}$
$\displaystyle \begin{align*}\partial^2z/\partial r^2 &= 2\frac{\partial z}{\partial x} + 2r\!\left(\frac{\partial^2z}{\partial x^2}\cdot\frac{\partial x}{\partial r} + \frac{\partial^2z}{\partial x\partial y}\cdot\frac{\partial y}{\partial r}\right) + 2s\!\left(\frac{\partial^2z}{\partial y\partial x}\cdot\frac{\partial x}{\partial r} + \frac{\partial^2z}{\partial y^2}\cdot\frac{\partial y}{\partial r}\right) \\
&= 2\frac{\partial z}{\partial x} + 2r\!\left(2r\frac{\partial^2z}{\partial x^2} + 2s\frac{\partial^2z}{\partial x\partial y}\right) + 2s\!\left(2r\frac{\partial^2z}{\partial y\partial x} + 2s\frac{\partial^2z}{\partial y^2}\right) \\
&= 4r^2\frac{\partial^2z}{\partial x^2} + 2\frac{\partial z}{\partial x} + 8rs\frac{\partial^2z}{\partial x\partial y} + 4s^2\frac{\partial^2z}{\partial y^2}\end{align*}$

MATHEMATICIAN June 15th, 2017 03:38 AM

Are you not using product rule in your second time differentiation?
Could you please explain what you are doing.

Country Boy June 15th, 2017 04:57 AM

Your reference to "second time differentiation" confused me because, at first, I thought you meant a second differentiation with respect to a "time" variable- but there is no such variable!

In fact, I think you mean just "differentiating the second time", taking the second derivative with respect to r.

skipjack does use the product rule when he differentiates and gets , the coefficient for the first term being "2" because, of course, the derivative of "2r" with respect to r, is "2".

But the second derivative with respect to r of the second term, has only one part because "2s" is independent of r (or, equivalently, the derivative of "2s" with respect to r, is 0).

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