Solutions to Laplace's equation

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    Laplace's equation
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Homework Help Overview

The discussion revolves around finding all solutions to Laplace's equation in two variables, specifically for functions of the form ax^3 + bx^2y + cxy^2 + dy^3. Participants are exploring the implications of the equation fxx + fyy = 0 and how it relates to the coefficients of the polynomial.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the derived equations (3a+c) and (3d+b) equating to zero, questioning the relationships between the coefficients a, b, c, and d. There is an exploration of the conditions under which the function satisfies Laplace's equation.

Discussion Status

Several participants are actively engaging with the problem, providing insights into the relationships between the coefficients. There is recognition that while certain conditions are necessary, they may not represent the most general solution. The conversation reflects a mix of attempts to clarify misunderstandings and to refine the conditions for the solutions.

Contextual Notes

Participants note that the function must satisfy Laplace's equation for all values of x and y, which raises questions about the generality of the proposed solutions and the assumptions made about the coefficients.

Miriverite
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[solved] solutions to Laplace's equation

Homework Statement


Find all solutions f(x,y) that satify Laplace's equation that are of the form:
ax^3 + bx^2y + cxy^2 + dy^3

Homework Equations


Laplace states that fxx + fyy = 0

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:
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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:
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?
 
Miriverite said:
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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