Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is this differential equation? I'm going crazy

  1. Aug 11, 2015 #1
    I have been working on a math problem and I keep getting the some type of PDEs.

    x*dU/dx+y*dU/dy = 0
    x*dU/dx+y*dU/dy+z*dU/dz = 0 ...

    x1*dU/dx1+x2*dU/dx2+x3*dU/dx3 + ... + xn*dU/dxn= 0

    dU/dxi is the partial derivative with respect to the ith variable. Does anyone know about this type of PDE? It looks like the Euler-Cauchy ODE.
     
  2. jcsd
  3. Aug 11, 2015 #2

    jedishrfu

    Staff: Mentor

  4. Aug 12, 2015 #3

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In each case, we can write the differential operator as ##\vec{r}\cdot\nabla_r U##, so the operator is a directional derivative along the vector ##\vec{r}##. I'm not sure that the equation with this equal to zero has a special name, though there are equations named after both Euler and Cauchy in the theory of transport that involve a directional derivative.
     
  5. Aug 12, 2015 #4
    The equations can be reduced to a linear pde with constant coefficients with the substitution.

    xi = e^yi

    dU/dy1+dU/dy2 + ... + dU/dyn= 0

    Same substition you would use to solve Euler-Cauchy ODE.
     
  6. Aug 12, 2015 #5

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can actually do a bit more. On a patch where one coordinate does not vanish, ##x_n\neq 0##, we can show that the solutions ##U(x_i)## must only depend on the ratios ##x_i/x_n## (called homogenous or projective coordinates), i.e., ##U## is a function ##U(x_i/x_n)##.
     
  7. Aug 12, 2015 #6
    Wow your right... So U(xi/xn) could be written as U(e^(y1-yn),e^(y2-yn),...,e^(yn-1-yn)).
     
  8. Aug 12, 2015 #7
    The solution U(x1,x2,...,xn)=U(a*x1,a*x2,...,a*xn)

    Where a is constant. So it doesn't scale. I was just making substitutions to look for properties.

    ... Just realized the ratio of the two variables would cancel any constant coefficients anyway ... A lot of work for the same result.

    Now I need a condition to find a solution.
     
  9. Aug 12, 2015 #8

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Any (first-differentiable) function of the homogenous variables is a solution. You would need additional differential equations or boundary conditions to narrow the solutions.
     
  10. Aug 12, 2015 #9
    I know there has to be n-1 solutions for the solution I am looking for if they exist. So one less than the number of variables. Which I think is possible because you can rewrite the function as the ratio of one of the variables. Which makes one less than n inputs. Thanks for the help/enlightenment. I like the vector approach. I am going to write out the Fourier series for the solution, sub back in the ln(yi) for xi and see if I can find some conditions
     
  11. Aug 13, 2015 #10

    pasmith

    User Avatar
    Homework Helper

    Those are first-order linear PDEs which can in principle be solved by the method of characteristics.

    As an example, if [tex]
    x \frac{\partial U}{\partial x} + y\frac{\partial U}{\partial y} = 0
    [/tex] then you can make a change of independent variables from [itex](x,y)[/itex] to [itex](t,s)[/itex] where [itex]\frac{\partial x}{\partial t} = x[/itex], [itex]\frac{\partial y}{\partial t} = y[/itex] to obtain [tex]
    \frac{\partial U}{\partial t} = 0.[/tex] Thus [itex]U = f(s)[/itex] for some function [itex]f[/itex].
    Now we know that [itex](x,y) = (A(s),B(s))e^t[/itex] so all we need is to choose the functions [itex]A[/itex] and [itex]B[/itex] appropriately and we can then invert the relationship to give [itex](t,s)[/itex] in terms of [itex](x,y)[/itex]. Here we can take [itex](A(s),B(s)) = (\cos s, \sin s)[/itex] which gives a problem at [itex](x,y) = (0,0)[/itex], but if you look at the PDE you see that [itex]\nabla U[/itex] is not uniquely determined at (0,0) anyway.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is this differential equation? I'm going crazy
Loading...