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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


    Staff: Mentor

  4. Aug 12, 2015 #3


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    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


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    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


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    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


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    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.
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