Chain rule with partial derivitives

[JumboJam]

Hi all, was having difficulty with the following problem. Got an answer that was not the final answer but cannot find where I was going wrong. Help is appreciated!

Given:
\[x = \rho cos \phi\]
\[y = \rho sin \phi\]
Show that under the above transformation,
\[\frac{\partial ^{2}v }{\partial x^{2}} + \frac{\partial ^{2}v}{\partial x^{2}} = 0\]
becomes
\[\frac{\partial ^{2}v}{\partial \rho^{2}} + \frac{1}{\rho} \frac{\partial v}{\partial \phi} + \frac{1}{\rho^{2}}\frac{\partial ^{2}v}{\partial \phi^{2}} = 0\]

I decided to start from the second equation and work my way up. Thus, I solved for each partial and got

\[\frac{\partial ^{2}v}{\partial \rho^{2}} = cos^{2}\phi \frac{\partial ^{2}v}{\partial x^{2}} + 2sin\phi cos\phi \frac{\partial ^{2}v}{\partial x \partial y} + sin^{2}\phi \frac{\partial ^{2}}{\partial y^{2}}\]
\[\frac{\partial ^{2}v}{\partial \phi^{2}} = \rho ^{2}sin^{2}\phi \frac{\partial ^{2}v}{\partial x^{2}} - 2\rho ^{2}sin\phi cos\phi \frac{\partial ^{2}v}{\partial x \partial y} + \rho^{2}cos^{2}\phi \frac{\partial ^{2}v}{\partial y^{2}} - \rho cos \phi \frac{\partial v}{\partial x} - \rho sin \phi \frac{\partial v}{\partial y}\]
\[\frac{\partial v}{\partial \phi} = \rho cos \phi \frac{\partial v}{\partial y} - \rho sin \phi \frac{\partial v}{\partial x}\]

After plugging in terms, I get
\[v_{xx} + v_{yy} + cos \phi v_{y} - sin \phi v_x - \frac{sin \phi}{\rho}v_{y} - \frac{cos \phi}{\rho}v_{x}\]
I'm not sure quite what I'm doing wrong. After staring at my math for an hour, I decided to turn for help.

 

[stapel]

Given:

. . .\(\displaystyle x = \rho cos \phi\)
. . .\(\displaystyle y = \rho sin \phi\)

Show that under the above transformation,

. . .\(\displaystyle \dfrac{\partial ^{2}v }{\partial x^{2}} + \dfrac{\partial ^{2}v}{\partial x^{2}} = 0\)

becomes

. . .\(\displaystyle \dfrac{\partial ^{2}v}{\partial \rho^{2}} + \dfrac{1}{\rho} \dfrac{\partial v}{\partial \phi} + \dfrac{1}{\rho^{2}}\dfrac{\partial ^{2}v}{\partial \phi^{2}} = 0\)

How does \(\displaystyle v\) relate to \(\displaystyle x\) and \(\displaystyle y\)?

Thank you!

 

[Deleted member 4993]

Hi all, was having difficulty with the following problem. Got an answer that was not the final answer but cannot find where I was going wrong. Help is appreciated!

Given:
\[x = \rho cos \phi\]
\[y = \rho sin \phi\]
Show that under the above transformation,
\[\frac{\partial ^{2}v }{\partial x^{2}} + \frac{\partial ^{2}v}{\partial x^{2}} = 0\] ........................(1)
becomes
\[\frac{\partial ^{2}v}{\partial \rho^{2}} + \frac{1}{\rho} \frac{\partial v}{\partial \phi} + \frac{1}{\rho^{2}}\frac{\partial ^{2}v}{\partial \phi^{2}} = 0\]

I decided to start from the second equation and work my way up. Thus, I solved for each partial and got

\[\frac{\partial ^{2}v}{\partial \rho^{2}} = cos^{2}\phi \frac{\partial ^{2}v}{\partial x^{2}} + 2sin\phi cos\phi \frac{\partial ^{2}v}{\partial x \partial y} + sin^{2}\phi \frac{\partial ^{2}}{\partial y^{2}}\]
\[\frac{\partial ^{2}v}{\partial \phi^{2}} = \rho ^{2}sin^{2}\phi \frac{\partial ^{2}v}{\partial x^{2}} - 2\rho ^{2}sin\phi cos\phi \frac{\partial ^{2}v}{\partial x \partial y} + \rho^{2}cos^{2}\phi \frac{\partial ^{2}v}{\partial y^{2}} - \rho cos \phi \frac{\partial v}{\partial x} - \rho sin \phi \frac{\partial v}{\partial y}\]
\[\frac{\partial v}{\partial \phi} = \rho cos \phi \frac{\partial v}{\partial y} - \rho sin \phi \frac{\partial v}{\partial x}\]

After plugging in terms, I get
\[v_{xx} + v_{yy} + cos \phi v_{y} - sin \phi v_x - \frac{sin \phi}{\rho}v_{y} - \frac{cos \phi}{\rho}v_{x}\]
I'm not sure quite what I'm doing wrong. After staring at my math for an hour, I decided to turn for help.

I think the equation (1) should be:
\(\displaystyle [\frac{\partial ^{2}v }{\partial x^{2}} + \frac{\partial ^{2}v}{\partial y^{2}} = 0\)]

start from:

\(\displaystyle \rho^2 = x^2 + y^2 \) and

\(\displaystyle \tan(\phi) = \frac{y}{x}\)

\(\displaystyle \displaystyle{\frac{\partial v}{\partial x} \ = \ \frac{\partial v}{\partial \rho}*\frac{\partial \rho}{\partial x} + \frac{\partial v}{\partial \phi} * \frac{\partial \phi}{\partial x}}\)

and so on ... it is a lot of tedious algebra....

 

[JumboJam]

Thanks. I'll get on it and see if I can get the right answer.

 

[JumboJam]

\[\frac{1}{\rho} \frac{\partial v}{\partial \phi}\]

Just discovered that this must be a typo--similar problems elsewhere have this part as dv/dp, not dv/d(phi), which would make the problem work out.

I am currently studying for my quals and this is an old exam they gave a few years ago. I'm doomed if there are any typos in this year's exam.

 

[Ishuda]

Just discovered that this must be a typo--similar problems elsewhere have this part as dv/dp, not dv/d(phi), which would make the problem work out.

I am currently studying for my quals and this is an old exam they gave a few years ago. I'm doomed if there are any typos in this year's exam.

Yes, the standard wave/heat equation. See
https://www.math.ucdavis.edu/~saito/courses/21C.w11/polar-lap.pdf
for example.

As maybe a point of interest, the speed of sound in the ocean, c, is often modeled as c^-2 being piecewise linear because the cylindrically symmetric ocean with a harmonic source has an exact solution in the Airy (modified Bessel) functions.

 

[Deleted member 4993]

Just discovered that this must be a typo--similar problems elsewhere have this part as dv/dp, not dv/d(phi), which would make the problem work out.

I am currently studying for my quals and this is an old exam they gave a few years ago. I'm doomed if there are any typos in this year's exam.

One advice about quals - don't get stuck on one problem. Strictly divide your time among the problems - and attempt as many problems as necessary.

Remember that in a numerical (or symbolic) problem-solution situation the last ~30% of the effort generally garners ~10% of the credit. So don't waste time trying to finish a problem.

 

[JumboJam]

Thanks for the advice. Been stressing about it the last month or so, and you're right. It is definitely good to keep the problems in perspective.