Question

Ifx+y=2e^θ cos ϕ,x-y=2ie^θ sin ϕ,& u is the function of x & y,then prove that
(∂^2 u)/(∂θ^2 )+(∂^2 u)/(∂ϕ^2 ) =4xy (∂^2 u)/∂x∂y

Answer 1

I can only thing of one thing

Answer 2

Are we thinking of the same thing?

Answer 3

my attempt

Answer 4

probably not :P

Answer 5

\(\Large \dfrac {∂^2 u}{∂θ^2 }+\dfrac{∂^2 u}{∂ϕ^2 } =4xy\dfrac{ ∂^2 u}{∂x∂y} \)

Answer 6

this is what i see http://puu.sh/byiYF/f8b94386cb.png

Answer 7

refresh the page

Answer 8

\(\Large x= e^{\theta +i\phi }\)
\(\Large y= e^{\theta -i\phi }\)

Answer 9

Is the first equation supposed to be \(x+y=2e^{i\theta}\) ?

Answer 10

no just
\(x+y = 2e^\theta \)

Answer 11

i mean
\(x+y = 2e^\theta \cos \phi \)

Answer 12

So given \(u=f(x,y)\) and \[\large\begin{cases}
x+y=2e^{\theta}\cos\varphi\\\\
x-y=2ie^{\theta}\sin\varphi
\end{cases}\]
you have to establish that
\[\large \dfrac {∂^2 u}{∂θ^2 }+\dfrac{∂^2 u}{∂ϕ^2 } =4xy\dfrac{ ∂^2 u}{∂x∂y}\]
just so I'm understanding correctly...

Answer 13

correct

Answer 14

typo: for that last equation I meant \(\dfrac{\partial^2u}{\partial\varphi^2}\)

Answer 15

yeah, i posted that too

Answer 16

got first partial derivate...how to get 2nd partial derivative ?

Answer 17

Are you given an expression for \(u(x,y)\)?

Answer 18

nopes

Answer 19

Second partial of \(u\) w.r.t. \(\theta\).
\[\begin{align*}\frac{\partial^2u}{\partial\theta^2}&=\frac{\partial}{\partial\theta}\left[\frac{\partial u}{\partial\theta}\right]\\\\
&=\frac{\partial}{\partial\theta}\left[\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial \theta}\right]\\\\
&=\frac{\partial }{\partial \theta}\left[\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial \theta}\right]+\frac{\partial }{\partial \theta}\left[\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial \theta}\right]\\\\
&=\left(\frac{\partial ^2u}{\partial \theta~\partial x }\cdot\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial x }\cdot\frac{\partial ^2x}{\partial \theta^2}\right)+\left(\frac{\partial ^2u}{\partial \theta~\partial y }\cdot\frac{\partial y}{\partial \theta}+\frac{\partial u}{\partial y }\cdot\frac{\partial ^2y}{\partial \theta^2}\right)
\end{align*}\]
It should be obvious that \(\dfrac{\partial x}{\partial \theta}=\dfrac{\partial x^2}{\partial \theta^2}=x\) and \(\dfrac{\partial y}{\partial \theta}=\dfrac{\partial^2 y}{\partial \theta^2}=y\), so you can reduce this a bit:
\[\begin{align*}
\frac{\partial^2u}{\partial\theta^2}&=\left(\frac{\partial ^2u}{\partial \theta~\partial x }\cdot\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial x }\cdot\frac{\partial ^2x}{\partial \theta^2}\right)+\left(\frac{\partial ^2u}{\partial \theta~\partial y }\cdot\frac{\partial y}{\partial \theta}+\frac{\partial u}{\partial y }\cdot\frac{\partial ^2y}{\partial \theta^2}\right)\\\\
&=x\left(\frac{\partial ^2u}{\partial \theta~\partial x }+\frac{\partial u}{\partial x }\right)+y\left(\frac{\partial ^2u}{\partial \theta~\partial y }+\frac{\partial u}{\partial y }\right)\end{align*}\]

Answer 20

Doing something similar with \(\varphi\) will yield
\[\begin{align*}
\frac{\partial^2u}{\partial\varphi^2}&=\left(\frac{\partial ^2u}{\partial \varphi~\partial x }\cdot\frac{\partial x}{\partial \varphi}+\frac{\partial u}{\partial x }\cdot\frac{\partial ^2x}{\partial \varphi^2}\right)+\left(\frac{\partial ^2u}{\partial \varphi~\partial y }\cdot\frac{\partial y}{\partial \varphi}+\frac{\partial u}{\partial y }\cdot\frac{\partial ^2y}{\partial \varphi^2}\right)\\\\

&=\left(\frac{\partial ^2u}{\partial \varphi~\partial x }\cdot ie^{\theta+i\varphi}-\frac{\partial u}{\partial x }\cdot e^{\theta+i\varphi}\right)+\left(\frac{\partial ^2u}{\partial \varphi~\partial y }\cdot (-ie^{\theta-i\varphi})-\frac{\partial u}{\partial y }\cdot e^{\theta-i\varphi}\right)\\\\

&=x\left(i\frac{\partial ^2u}{\partial \varphi~\partial x }-\frac{\partial u}{\partial x }\right)-y\left(i\frac{\partial ^2u}{\partial \varphi~\partial y }-\frac{\partial u}{\partial y }\right)\end{align*}\]

Answer 21

right...
how will we get
\(\large \dfrac{\partial^2 u}{\partial x \partial y}\)

from
\(\large \dfrac{\partial^2 u}{\partial \theta \partial x}\)
and other such terms...

Answer 22

\[\begin{align*}
\frac{\partial^2u}{\partial\varphi^2}
&=x\left(i\frac{\partial ^2u}{\partial \varphi~\partial x }-\frac{\partial u}{\partial x }\right)-y\left(i\frac{\partial ^2u}{\partial \varphi~\partial y }\color{red}+\frac{\partial u}{\partial y }\right)\end{align*}\]

Answer 23

i should add them now, right ?
i see few terms getting cancelled

Answer 24

Not sure about that just yet... Let's see what happens when we add the partials together:
\[\begin{align*}

\frac{\partial^2u}{\partial\theta^2}+\frac{\partial^2u}{\partial\varphi^2}&=
x\left(\frac{\partial ^2u}{\partial \theta~\partial x }+\frac{\partial u}{\partial x }\right)+y\left(\frac{\partial ^2u}{\partial \theta~\partial y }+\frac{\partial u}{\partial y }\right)\\\\
&~~~~~~~~+x\left(i\frac{\partial ^2u}{\partial \varphi~\partial x }-\frac{\partial u}{\partial x }\right)-y\left(i\frac{\partial ^2u}{\partial \varphi~\partial y }+\frac{\partial u}{\partial y }\right)\\\\
&=
x\left(\frac{\partial ^2u}{\partial \theta~\partial x }+i\frac{\partial ^2u}{\partial \varphi~\partial x }\right)+y\left(\frac{\partial ^2u}{\partial \theta~\partial y }-i\frac{\partial ^2u}{\partial \varphi~\partial y }\right)\\\\
\end{align*}\]

Answer 25

There's got to be a theorem relating second-order partial derivatives that lets you reduce terms like
\[\frac{\partial^2u}{\partial \theta~\partial x}+\frac{\partial^2u}{\partial \theta~\partial y}~...\]

Answer 26

\(\Large (\frac{\partial ^2u}{\partial \theta~\partial x }+i\frac{\partial ^2u}{\partial \phi~\partial x })\)

should = 2y \(\frac{\partial ^2u}{\partial x~\partial y}\)

and other = 2x \(\frac{\partial ^2u}{\partial x~\partial y}\)
need to search some theorem that gives this result...

Answer 27

**4 instead of 2

Answer 28

hey i'm getting
\[\frac{ \partial^2u }{ \partial \theta^2 }=x^2\frac{ \partial^2u }{ \partial x^2 }+2xy \frac{ \partial^2u }{ \partial y \partial x }+y^2\frac{ \partial^2u }{ \partial y^2}+x \frac{ \partial u }{ \partial x }+y \frac{ \partial u }{ \partial y }\]
and
\[\[\frac{ \partial^2u }{ \partial \phi^2 }=-x^2\frac{ \partial^2u }{ \partial x^2 }+2xy \frac{ \partial^2u }{ \partial y \partial x }-y^2\frac{ \partial^2u }{ \partial y^2}-x \frac{ \partial u }{ \partial x }-y \frac{ \partial u }{ \partial y }\\]\]
adding them
4

Answer 29

how ?

Answer 30

used any theorem ?

Answer 31

nope just a really lengthy calculation

Answer 32

am i on right track ?
or done something different from the beginning ?

Answer 33

ure approach is correct
\[\large x=e^{\theta+i \phi}\\y=e^{\theta-i \phi}\]

Answer 34

now from here first
\[\frac{ \partial u }{ \partial \theta }=\frac{ \partial u }{ \partial x }*\frac{ \partial x }{ \partial \theta }+\frac{ \partial u }{ \partial y }*\frac{ \partial y }{ \partial \theta }\\=e^{\theta+i \phi}*\frac{ \partial u }{ \partial x }+e^{\theta-i \phi}*\frac{ \partial u }{ \partial y }\\=x \frac{ \partial u }{ \partial x }+y \frac{ \partial u }{ \partial y }\]
ok so far?

Answer 35

yes

Answer 36

now next we need to find
\[\frac{ \partial^2u }{ \partial \theta^2 }\]

Answer 37

\[\frac{ d^2u }{ d \theta^2 }=\frac{ d }{ d \theta }(\frac{ du }{ d \theta })\]

Answer 38

thats what sith was doing, you're going in similar lines ?

Answer 39

\[=(x \frac{ \partial }{ \partial x }+y \frac{ \partial }{ \partial y })(x \frac{ \partial u }{ \partial x }+y \frac{ \partial u }{ \partial y })\\=x^2\frac{ \partial ^2 u }{ \partial x^2 }+2xy \frac{ \partial^2u }{ \partial y \partial x }+y^2\frac{ \partial ^2u }{\partial y^2 }+x \frac{ \partial u }{ \partial x }+y \frac{ \partial u }{ \partial y }\]

Answer 40

like this ok?

Answer 41

ohh1
thats new

Answer 42

and now
\[\frac{ \partial u }{ \partial \phi }=\frac{ \partial u }{ \partial x }*\frac{ \partial x }{ \partial \phi }+\frac{ \partial u }{ \partial y }\frac{ \partial y }{ \partial \phi }\]
same method--

Answer 43

u'll get--
\[\frac{ \partial u }{ \partial \phi }=i \frac{ \partial u }{ \partial x }x-iy \frac{ \partial u }{ \partial y }\]

Answer 44

again diffrentiating will give u
\[\frac{ \partial^2u }{ \partial \phi^2 }\]

Answer 45

thanks!

Answer 46

no problem
bohut tension mein hu campussing ke liye

Answer 47

:O
interview preparations should be started atleast 2-3 months before actual campus placements starts....

by campussing, you meant campus placements, right ?

Answer 48

yeah

Answer 49

clear concepts of C,C++ programming, data structures and algorithms,
these will help you in the interview, irrespective of which branch you're in

Answer 50

and excellent knowledge of 1 or 2 core subjects

Answer 51

thanks to @SithsAndGiggles also, for trying :)
have a look at sid's solution!

Answer 52

Yup never would have guessed that... Multivariable calc...