Question

Laplace Equation,Question Attached

Answer 1

How do i go about doing them?

Answer 2

I know i have to prove them equal to 0 but it's not working for me :/ I don't know where I'm going wrong.

Answer 3

Do you have to prove that \(grad^2 f=0\)?

Answer 4

Yes.

Answer 5

fxx(x,y)+fyy(x,y)=0

Answer 6

so take some partial derivatives (using implicit differentiation)

Answer 7

Where are you at, or where do you get stuck?

Answer 8

I tried solving it but i don't get it equal to 0,if you could solve it i can check where I'm going wrong :/

Answer 9

You can post what you've go, perhaps is something simple.

Answer 10

*got

Answer 11

For example, what do you have for fx(x,y).

Answer 12

Well for 67

i did fxx which was equal to -x^2/root(x^2+y^2+z^2) and similarly fyy and fzz -y^2/root(x^2+y^2+z^2) and -z^2/root(x^2+y^2+z^2) . which surely doesn't equal to 0.I know I'm making a big mess

Answer 13

Have you done 66?
It is a good idea to start with fx, i.e. differentiate once wrt one variable, check it, and then proceed.

Answer 14

You don't really have to worry about the other variables because you can get them by symmetry.

Answer 15

Yeah 66 too same story :/

Answer 16

What did you get for fx(x,y) for 66?

Answer 17

For 66

fx(x,y)=x/x^2+y^2

Answer 18

That looks good to me, except for the parentheses, but I assume that you mean this:
\(\large f_x(x,y)=\frac{x}{x^2+y^2}\)
So can you proceed to find fxx?

Answer 19

Yeah i meant this

Answer 20

use the quotient rule!

Answer 21

I got fxx as -x^2+y^2/(x^2+y^2)^2

Answer 22

Something is wrong, because fxx should not be symmetrical with respect to x and y.
Let's do the quotient rule step by step:
d(u/v)=(vdu-udv)/v^2
so can you first do vdu/v^2?

Answer 23

That's what i did @mathmate

Answer 24

Oh...

Answer 25

I'm doing it again,made a mistake.

Answer 26

I guess

Answer 27

Well it's the same

Answer 28

Okay i got this one equal to 0

Answer 29

Can we get to the other one?

Answer 30

did you get
\(\large f_{xx}=\frac{-x^2+y^2}{(x^2+y^2)^2}\)
or did you get
\(\large f_{xx}=\frac{x^2-y^2}{(x^2+y^2)^2}\)

Answer 31

Parentheses count a lot in fractions!

Answer 32

The first one

Answer 33

and the second one in fyy

Answer 34

67 is similar.
Do fx, check it thoroughly, and then do fxx.
The rest is by symmetry.

Answer 35

I got fx as -x/root(x^2+y^2+z^2)

Answer 36

To give you a check, I got \(\Large f_x=\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\)

Answer 37

don't forget -1/2 becomes -3/2.

Answer 38

Stupid me,Iv'e been adding one to it the whole time .-. That's what was wrong. Thankyouuu

Answer 39

Great! I think you should be on your way!
If anything, post again!

Answer 40

Sure mate! :)

Answer 41

:)

Answer 42

Can you show what you've got so far, for 67, I guess?

Answer 43

I got fxx as -(x^2+y^2)^3/2 + 3y^2(x^2+y^2)^[1/2] / (x^2+y^2)^[3/2]

Answer 44

.-.

Answer 45

you are still working on 66?

Answer 46

No no 66 is done,I am talking about 67.

Answer 47

well then where did z in the denominator go?

Answer 48

fxx=(x^2+y^2+z^2)^3/2 + 3x^2(x^2+y^2+z^2)^[1/2] / (x^2+y^2+z^2)^[3/2]

Now?

Answer 49

well no, that's not what I got
first of all, what did you get for fx ?

Answer 50

-x/(x^2+y^2+z^2)^[3/2]

Answer 51

ok good, that is what I got, which I got leading to
-(x^2+y^2+z^2)^3/2 + 3x^2(x^2+y^2+z^2)^[1/2] / (x^2+y^2+z^2)^[3/2]

Answer 52

actually the denominator is squared, so
-(x^2+y^2+z^2)^3/2 + 3x^2(x^2+y^2+z^2)^[1/2] / (x^2+y^2+z^2)^3
but that doesn't matter
do you agree?

Answer 53

Agreed.That's exactly what i got.

Answer 54

I didn't see the negative sign when you wrote it
now factor out (x^2+^2+z^2)^1/2 from the numerator

Answer 55

I would say u=x^2+y^2+z^2 to avoid making such a mess of my paper lol

Answer 56

Haha well

Answer 57

Well but when i add those two factors i.e fxx and fyy i don't get 0

Answer 58

you need to add fzz too

Answer 59

Yeah fzz too :p

Answer 60

well what did you get for fxx after factoring out (x^2+y^2+z^2)^1/2 ?

Answer 61

u^[1/2] [-u^3+3x^2] / u^3

Answer 62

-u^3 is wrong
the exponent was 3/2, and you factored out 1/2, leaving...?

Answer 63

(u^3/2)/(u^1/2)=?

Answer 64

-u*

Answer 65

ah, not turn u back into x^2+y^2+z^2 and simplify

Answer 66

now*

Answer 67

Um

(2x^2+y^2+z^2) / (x^2+y^2+z^2)^[5/2]

Answer 68

you had u^[1/2] [-u^3+3x^2] / u^3
where did the 3x^2 go?

Answer 69

oh

Answer 70

(2x^2-y^2-z^2) / (x^2+y^2+z^2)^[5/2]

Answer 71

yes yes, that's it :)

Answer 72

just look at the numerator and consider what the numerators of fyy and fzz will be by symmetry

Answer 73

Omg here it is.Thanks alot man! It worked :p

Answer 74

:)
happy to help!

Answer 75

:)