First order partial differential equation

  • #1
161
1
How do I go about solving a differential equation of the form

[itex]\partial_{x}F_{x}(x,y) + \partial_{y}F_{y}(x,y) = g(x,y)[/itex]

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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Answers and Replies

  • #2
bigfooted
Gold Member
615
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How do I go about solving a differential equation of the form

[itex]\partial_{x}F_{x}(x,y) + \partial_{y}F_{y}(x,y) = g(x,y)[/itex]

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
[itex]\partial_{x}F(x,y) + \partial_{y}F(x,y) = g(x,y)[/itex]

the subscript x already implies a partial derivative with respect to x, which would mean that your equation is second order.
 
  • #3
pasmith
Homework Helper
2,016
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The derivatives are, of course, premultiplied by 1.

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

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 [itex]\partial_x^2 F + \partial_y^2 F = g[/itex] 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 [itex]F_x[/itex] and [itex]F_y[/itex] are cartesian components of a 2D vector field, in which case the PDE is [tex]\nabla \cdot \mathbf{F} = g,[/tex] which is one equation in two unknowns, and therefore does not have a unique solution.
 
  • #4
161
1
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?
 
  • #5
pasmith
Homework Helper
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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 [itex]\partial_x F_y - \partial_y F_x = h(x,y)[/itex] is also specified. If it happens that [itex]h = 0[/itex] then you can set [itex]F_x = \partial_x \phi[/itex], [itex]F_y = \partial_y \phi[/itex] to obtain Poisson's equation in the form [itex]\nabla^2\phi = \partial_x^2 \phi + \partial_y^2 \phi = g(x,y)[/itex]. In general you have to set [tex]
F_x = \partial_x \phi - \partial_y \psi \\
F_y = \partial_y \phi + \partial_x \psi
[/tex] which again yields Poisson's equation with[tex]
\nabla^2 \phi = g, \\
\nabla^2 \psi = h.
[/tex]
 

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