Question

\(u=e^{(x+y)}+\log (x^3+y^3-x^2 y-xy^2) \)

is this an homogeneous function ?
how would i prove it ?
if it is, is the order 3 ?

Answer 1

what is the definition of homogenous function

Answer 2

the total exponent of each term should be same

Answer 3

for homogeneious function we prove

\(\large u(tx,ty) = t^ku(x,y)\)

Answer 4

\(u=e^{t(x+y)}+\log (x^3+y^3-x^2 y-xy^2)+3 \log t\)

Answer 5

the original qustion is

State and Prove Euler^' s Theorem for homogeneous functions in two variables and
hence find the value of

x^2 (∂^2 u)/(∂x^2 ) +2xy (∂^2 u)/∂x∂y+y^2 (∂^2 u)/(∂y^2 )+x ∂u/∂x+y ∂u/∂y
for
u=e^(x+y)+log (x^3+y^3-x^2 y-xy^2)


i did get
x^2 (∂^2 u)/(∂x^2 ) +2xy (∂^2 u)/∂x∂y+y^2 (∂^2 u)/(∂y^2 )+x ∂u/∂x+y ∂u/∂y = n^2 u
now how do i use this theorem for 'u'

Answer 6

is it even homogeneous ?

Answer 7

yeah it doesn't look homogeneous but we can't say anything unless we prove/disprove it. but from your partial derivatives expression gives the feeling that u is a homogeneous function right ?

Answer 8

right, the theorem is asked for homogeneous functions and we are asked to find that value for u ...so i guess they are implying u must be homogeneous

Answer 9

if not, should we do something like

u = e^a + log b where a ans b are homogeneous , for sure...

Answer 10

and use chainrule+euler theorem is it ? that looks like a good idea!

Answer 11

but euler's will be applicable for a and b only ...

Answer 12

\( \partial u/ \partial x = e^a \partial a/ \partial x + 1/b \partial b/ \partial x\)

something like this ?
won't it become much complicated ...espicially with the 2nd order partial derivative finding...

Answer 13

\(u_{xx} = e^a a_{xx} + e^a (a_x)^2 + 1/b b_{xx} -1/b^2 (b_x)^2\)

such steps will show up....

Answer 14

not sure what to do of
terms like
(a_x)^2 and (b_x)^2

i hope those get cancelled while finding
u_xy

Answer 15

\( u_x = e^a a_x + 1/b b_x \)
\(u_{xy} = e^a a_y a_x + e_a a_{xy} + 1/b b_{xy} -1/b^2 b_x b_y\)

Answer 16

but thats just working the partials, not using euler's theorem

Answer 17

will be using eurler's for a and b

won't calculate
a_x, b_x, a_xx, b_xx and so on...

Answer 18

still that doesn't work out

terms like a_x ^2 and a_x a_y will create problems...

Answer 19

any other approach....

Answer 20

homogeneous just means that it's set equal to zero, usually
I'm guessing the differential equation is what they mean

Answer 21

What I know about Euler's Theorem is that it's from Number Theory
must be some other Euler theorem than from the one I know
well, here's what I know about the Euler Theorem:

https://en.wikipedia.org/wiki/Euler's_theorem

Answer 22

but I'm guessing that we need a different one here

Answer 23

http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html

Answer 24

right,
this euler theorem is different

it states

x^2 (∂^2 u)/(∂x^2 ) +2xy (∂^2 u)/∂x∂y+y^2 (∂^2 u)/(∂y^2 )+x ∂u/∂x+y ∂u/∂y = n^2 u
for the homogeneous function u of order 'n'

that i have already proved

Answer 25

Oh my! Euler's theorem from number theory would have been fun :D

Answer 26

Yes, indeed.

Answer 27

I still wonder if that log is in the bottom or exponent

Answer 28

nvm either way its same level of difficulty >.<

Answer 29

the statement of the Theorem is that whole page that I linked to above ^
and the proof could take a while... let's see where to start

Answer 30

http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html

Answer 31

i did the proof already!

Answer 32

ok, so they're asking us to restrict it to two variables, that helps

Answer 33

Can I see it?

Answer 34

I don't even see it stated anywhere

Answer 35

sure
,just a sec

Answer 36

Wow, okay. Haha. That looks way longer that what I was about to do.

Answer 37

than*

Answer 38

I'm trying to see how the two compare, one sec...

Answer 39

That just seems like a way different approach, but it seems to work.
it just relies on a u substitution

I can put the other one there as well, just for reference
as another approach, I guess... it works for the general theorem, not just for two variables. This one seems specific to two variables - it's also shorter

Answer 40

true, lets think of how to use that theorem to find the required expression,,,

Answer 41

http://prntscr.com/4qpdey

Answer 42

Ok, so that's how I would state the theorem and the proof. A bit different, but it accomplishes the same thing I'm sure. ^

Answer 43

So, let's see...
both yours and mine wind up with the same formula actually
we just need to find what the degree is, the n
oh, maybe we just leave it as n... for now at least, but here's the thing
all we really need to do is take the second derivative of that statement in the proof.
and we'll have something close to what we need.

Answer 44

same formula?

i am getting n^2 f
with that formula, you will be getting
2^n f
right ?

Answer 45

ah, I see what they do, it's a bit different. That comes later ^
you're looking at the solution to the whole thing... I'm just talking about the first bold line - no squared terms there

Answer 46

:o I see... yours is better for what we're doing here
mine is time based, but yours is based more on x and y derivatives
Hmmmm

Answer 47

yeah, it's the same thing though lol
the notation just has me thinking too hard

Answer 48

So, at this step, we actually want to take x-derivatives and y-derivatives of both sides
that way the n's and t's are constants... now it makes more sense

Answer 49

Solution to the problem: Taking the partial derivative with respect to x on both sides of the previous equation gives us:

x d^2f /dx^2 + df/dx + y df/dydx = n df/dx

Answer 50

those should all be partial derivatives ^

Answer 51

now do the same thing but with a y-derivative

Answer 52

x d^2f /dxdy + df/dy + y d^2f/dydx = n df/dy

Answer 53

what are we trying to do there ?
Lets use 'u' please....are you trying to prove anything here ??

My aim as of now is th find the value of that big expression for 'u' using the theorem i proved...

Answer 54

yeah, ok... you can get lost in all these lines pretty easily

Answer 55

well, we should scrap all of my work then

Answer 56

and concentrate on 'u'

Answer 57

we've already done the proof, but I guess we need to do it a different way
to have the u in it

Answer 58

seems like you have the solution already
I'll just be following that to make sure I get the notation you want
otherwise, without that solution, I'd probably do it a slightly different way, on my own.

Answer 59

well, let me just ask you this then
what part of the solution that you have that you don't understand fully already?

Answer 60

seems this question is repelling people in MSE too, nobody is answering
http://math.stackexchange.com/questions/947807/if-u-exy-ln-x3y3-x2y-xy2-find-the-value-of

Answer 61

i understood everything i did in that proof, but look at the question, there's the 2nd part...

Answer 62

because all we have to do is to multiply n^2 by all of what u is and that's what they're after... that's what the theorem is telling us we can do
the proof and the work is long and drawn out
but the punchline is to just multiply u by n^2 to get what that entire expression is equal to

Answer 63

\(u\) is not homogeneous so thats where we're stuck

Answer 64

i did get

e^(x+y)+ln(x+y)+2ln(x−y)
but working out partials is not an option, i think!

the question clearly says,
"hence" find the value of ...

Answer 65

so i need to use the theorem i proved...but how is the question...

Answer 66

then the other stuff, up to the second bold line
that's all setup for the form they gave us...since we're dealing with second derivatives
they just too both partial derivatives of the form from the theorem and added them together... that's how they get all that stuff on the second bold line
which just happens to have the same form as what they are asking for

and, due to all the work on that page, winds up being equal to just n^2 times u
no matter what u is

Answer 67

see, it seems like this problem assumes a different definition for homogeneous as well
no mention of t at all
I even have that in my definition of homogeneous

Answer 68

but thats true only if u is homogeneous
if it is, and if i can prove, then the answer would just be n^2 u, where n is the order of u

Answer 69

right, so do we have to show that this u is homogeneous, and find the order?

Answer 70

that first line where they introduce u for the first time
that must be their definition of homogeneous; which is pretty different from mine
I don't doubt that they are similar

Answer 71

I think that's what has me thrown off

Answer 72

I see now, I can just take that as a definition (which I can't find anywhere else, by the way) and then the rest seems to make sense

now, let me pick it up where you left off by changing variables on u

Answer 73

ah, yeah, I knew something seemed off a bit
where did you get this solution? The notation seems a bit wrong, actually in a way
They treat x there the way all the other things I'm looking at treat a scalar
like the t I was using

let me keep digging around, this y/x business has me confused
from what I'm looking at, it should be y/t and x/t
with the scalar t on the bottom
but trying to make it with one variable over the other doesn't seem to agree with the other things I'm seeing

Answer 74

u = f(a,b)
a and b are functions of x and y

u_x in terms of a_x, b_x ?? (it will have a_y, b_y terms also na ?)

Answer 75

u_xy, u_xx, u_yy in terms of corresponding 'a's and 'b's ...doesn't seems easy...

Answer 76

I'm just scanning a doc real quick, to make sure i do this corresponding change of variables right

Answer 77

so, there's a change of variables where v = y/x