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June 14th, 2017, 10:35 AM   #1
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Differential equation proof

I got exam tomorrow!!
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 June 14th, 2017, 11:37 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,271 Thanks: 1959 $\displaystyle \frac{\partial z}{\partial r} = \frac{∂z}{∂x}\cdot\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial r} = 2r\frac{\partial z}{\partial x} + 2s\frac{\partial z}{\partial y}$, etc.
 June 14th, 2017, 11:40 AM #3 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 879 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 12:30 PM.
 June 14th, 2017, 12:31 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,271 Thanks: 1959 You'll also need to use the product rule. Can you post your attempt?
June 14th, 2017, 12:51 PM   #5
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The solution for second order partial derivative I got:
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Last edited by skipjack; June 14th, 2017 at 10:41 PM.

 June 14th, 2017, 03:35 PM #6 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 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 11:46 PM.
 June 14th, 2017, 11:39 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,271 Thanks: 1959 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, 04:38 AM #8 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 879 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, 05:57 AM #9 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 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).

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