Solutions to Laplace's equation

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[solved] solutions to Laplace's equation

1. The problem statement, all variables and given/known data
Find all solutions f(x,y) that satify Laplace's equation that are of the form:
ax^3 + bx^2y + cxy^2 + dy^3

2. Relevant equations
Laplace states that fxx + fyy = 0

3. The attempt at a solution
fxx = 6ax + 2by
fyy = 6dy + 2cx
so 6ax + 2by + 6dy + 2cx = 0
(3a+c)x + (3d+b)y = 0

What do I do from here?
 
Last edited:

gabbagabbahey

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Well, if f(x,y) is a solution to Laplace's equation, then it satisfies it for all x and y....what does that tell you about (3a+c) and (3d+b)?
 
Well then:
(3a+c) = -(3d+b) = 0
a = -d
c = -b
c = -3a
b = 3a

Plugging in:
f(x,y) = ax^3 + 3ax^y - 3axy^2 - ay^3

But if you take the double partials:
fxx = 6x
fyy = 6y

and 6x-6y =/= 0 for all real numbers....

EDIT: Wow, I'm stupid. I took the double partials incorrectly so I thought I had done it wrongly.. only I hadn't. Thanks!
 
Last edited:

gabbagabbahey

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c=-3a and b=-3d are necessary, but I'm not sure why you think a=-b and b=-d are.

f(x,y)=ax^3-3dx^2y-3axy^2+dy^3 satisfies Laplace's equation for all a and d not just a=d
 
If a=-b adn b=-d, then my above function satisfies for all a - that's good enough, yes?
 

gabbagabbahey

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If a=-b adn b=-d, then my above function satisfies for all a - that's good enough, yes?
It does, but its not the most general solution...which is what you are looking for.

(3a+c) = -(3d+b) = 0 does not mean that a=-b and b=-d....the only requirement is that c=-3a and b=-3d.
 
I see. Thanks a bunch!
 

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