My Math Forum Differential equation proof

 Differential Equations Ordinary and Partial Differential Equations Math Forum

June 14th, 2017, 09:35 AM   #1
Math Team

Joined: Jul 2013
From: काठमाडौं, नेपाल

Posts: 831
Thanks: 60

Math Focus: सामान्य गणित
Differential equation proof

I got exam tomorrow!!
Attached Images
 20170614_230209.jpg (104.7 KB, 20 views)

 June 14th, 2017, 10:37 AM #2 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1383 $\displaystyle \frac{\partial z}{\partial r} = \frac{∂z}{∂x}\cdot\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial r} = 2r\frac{\partial z}{\partial x} + 2s\frac{\partial z}{\partial y}$, etc.
 June 14th, 2017, 10:40 AM #3 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित I know those rules but still can't prove. Could you give me some hint/help. Last edited by skipjack; June 14th, 2017 at 11:30 AM.
 June 14th, 2017, 11:31 AM #4 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1383 You'll also need to use the product rule. Can you post your attempt?
June 14th, 2017, 11:51 AM   #5
Math Team

Joined: Jul 2013
From: काठमाडौं, नेपाल

Posts: 831
Thanks: 60

Math Focus: सामान्य गणित
The solution for second order partial derivative I got:
Attached Images
 20170615_011932.jpg (96.1 KB, 5 views) 20170615_011921.jpg (95.8 KB, 4 views)

Last edited by skipjack; June 14th, 2017 at 09:41 PM.

 June 14th, 2017, 02:35 PM #6 Senior Member   Joined: Apr 2016 From: Australia Posts: 177 Thanks: 22 Math Focus: horses,cash me outside how bow dah, trash doves 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. Last edited by skipjack; June 14th, 2017 at 10:46 PM.
 June 14th, 2017, 10:39 PM #7 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1383 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*}
 June 15th, 2017, 03:38 AM #8 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित Are you not using product rule in your second time differentiation? Could you please explain what you are doing.
 June 15th, 2017, 04:57 AM #9 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,647 Thanks: 680 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 $2r\frac{\partial z}{\partial r}$ and gets $2\frac{\partial z}{\partial r}+ 2r\left(\frac{\partial^2 z}{\partial x^2}\frac{\partial x}{\partial r}+\frac{\partial^2 z}{\partial x\partial y}\frac{\partial y}{\partial r}\right)$, 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, $2s\frac{\partial z}{\partial y}$ has only one part because "2s" is independent of r (or, equivalently, the derivative of "2s" with respect to r, is 0).

 Tags differential, equation, proof

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post aletex Calculus 6 December 11th, 2015 07:16 PM Sonprelis Calculus 6 August 6th, 2014 10:07 AM PhizKid Differential Equations 0 February 24th, 2013 10:30 AM eric3353 Differential Equations 4 May 29th, 2008 04:46 AM eric3353 Differential Equations 2 May 28th, 2008 03:06 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top