First order partial differential equation

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Discussion Overview

The discussion revolves around solving a first-order partial differential equation of the form ∂xFx(x,y) + ∂yFy(x,y) = g(x,y), where g(x,y) is a known function. Participants explore methods for obtaining an analytical solution and clarify the nature of the equation, including considerations of vector fields and potential connections to Poisson's equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant seeks guidance on solving the equation using the method of characteristics but notes the characteristic equation's dependence on coefficients that are zero.
  • Another participant points out that the original equation may be misinterpreted as second-order due to the notation used, suggesting it should be ∂xF(x,y) + ∂yF(x,y) = g(x,y).
  • A later reply suggests that if the original poster (OP) meant ∂x2F + ∂y2F = g, then the equation corresponds to Poisson's equation, which can be solved using Green's function methods or eigenfunction methods.
  • Another participant clarifies that if Fx and Fy represent components of a 2D vector field, the equation becomes ∇·F = g, which does not have a unique solution without additional constraints.
  • The OP confirms that the subscripts represent components of a 2D vector field and inquires about obtaining an analytical solution given that g is known.
  • Another participant responds that an analytical solution is not possible unless the quantity ∂xFy - ∂yFx is specified, indicating that if this quantity is zero, it leads to a form of Poisson's equation.

Areas of Agreement / Disagreement

Participants express differing interpretations of the original equation's order and form, with no consensus reached on a specific method for obtaining an analytical solution. Multiple competing views regarding the nature of the equation and the conditions for solvability remain unresolved.

Contextual Notes

Participants highlight the importance of specifying additional conditions or quantities to achieve a unique solution, indicating that the discussion is contingent on these factors.

matt_crouch
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How do I go about solving a differential equation of the form

\partial_{x}F_{x}(x,y) + \partial_{y}F_{y}(x,y) = g(x,y)

Where g(x,y) is a known function and I wish to solve for F. I thought i could apply the method of characteristics but the characteristic equation is dependent on coefficients in front of the derivatives which in this case are zero. If someone can point me in the right direction that could be great at least a way in which i can approach this. I am seeking an analytical solution
 
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matt_crouch said:
How do I go about solving a differential equation of the form

\partial_{x}F_{x}(x,y) + \partial_{y}F_{y}(x,y) = g(x,y)

Where g(x,y) is a known function and I wish to solve for F. I thought i could apply the method of characteristics but the characteristic equation is dependent on coefficients in front of the derivatives which in this case are zero. If someone can point me in the right direction that could be great at least a way in which i can approach this. I am seeking an analytical solution

The derivatives are, of course, premultiplied by 1.

Also, don't you mean the first order equation
\partial_{x}F(x,y) + \partial_{y}F(x,y) = g(x,y)

the subscript x already implies a partial derivative with respect to x, which would mean that your equation is second order.
 
bigfooted said:
The derivatives are, of course, premultiplied by 1.

Also, don't you mean the first order equation
\partial_{x}F(x,y) + \partial_{y}F(x,y) = g(x,y)

the subscript x already implies a partial derivative with respect to x, which would mean that your equation is second order.

If the OP really means \partial_x^2 F + \partial_y^2 F = g then his PDE is in fact Poisson's Equation, which can be solved by Green's function methods or eigenfunction methods as desired.

It is also possible that F_x and F_y are cartesian components of a 2D vector field, in which case the PDE is \nabla \cdot \mathbf{F} = g, which is one equation in two unknowns, and therefore does not have a unique solution.
 
Hi sorry the subscript on the function was to represent a two 2d vector field. Is there a way to obtain an analytical solution if the function g is known?
 
matt_crouch said:
Hi sorry the subscript on the function was to represent a two 2d vector field. Is there a way to obtain an analytical solution if the function g is known?

Not unless the quantity \partial_x F_y - \partial_y F_x = h(x,y) is also specified. If it happens that h = 0 then you can set F_x = \partial_x \phi, F_y = \partial_y \phi to obtain Poisson's equation in the form \nabla^2\phi = \partial_x^2 \phi + \partial_y^2 \phi = g(x,y). In general you have to set <br /> F_x = \partial_x \phi - \partial_y \psi \\<br /> F_y = \partial_y \phi + \partial_x \psi<br /> which again yields Poisson's equation with<br /> \nabla^2 \phi = g, \\<br /> \nabla^2 \psi = h.<br />
 

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