First order partial differential equation

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SUMMARY

This discussion focuses on solving the first-order partial differential equation (PDE) of the form \partial_{x}F(x,y) + \partial_{y}F(x,y) = g(x,y). The method of characteristics is not applicable due to the absence of coefficients in front of the derivatives. Participants clarify that if the equation represents a two-dimensional vector field, it leads to a divergence equation \nabla \cdot \mathbf{F} = g, which lacks a unique solution unless additional conditions are specified. The discussion also highlights that if h(x,y) is defined, the problem can be transformed into Poisson's equation, solvable via Green's function or eigenfunction methods.

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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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