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Find a 2-arguments function from six constraints on its derivatives

  1. Jul 5, 2013 #1
    I need to find a two-arguments function u(x,y) which satisfies six constraints on its derivatives. x and y are quantities so always positive.

    1&2: On the first derivatives:
    du/dx>0 for all x & du/dy>0 for all y (so u is increasing in x and y)

    3&4: On the second derivatives:
    d²u/dx²<0 for all x & d²u/dy²<0 for all y (so u is concave in x and y)

    5&6: On the crossed derivatives:
    d²u/dxdy<0 for all x+y<θ (or at least y<θ) & d²u/dxdy>0 for all x+y>θ (or at least y>θ) (θ is a threshold)

    I found one specific function that satisfies those conditions: u(x,y)=xy+1-exp(θ-x-y)

    But I don't think this is the only one. I would like to find the most general function that satisfies those six conditions. The best would be that this specific function that I found, belong to a pretty well-known category of functions. Don't know if it is possible. Maybe Weibull functions? Did not try yet.
    Could you help me please?

    Thanks a lot

    Last edited: Jul 5, 2013
  2. jcsd
  3. Jul 5, 2013 #2


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    Staff: Mentor

    u(x,y)=xy+1-exp(θ-x-y) is not a solution. The x-derivative at y=-5, x=5+θ is negative (-4).

    With "x+y<θ" and "x+y>θ":
    If u(x,y) is a solution for θ=0, then u(x-θ/2,y-θ/2) is a solution for θ. Therefore, it is sufficient to look for solutions with θ=0. In a similar way, if you want y=θ as threshold, it is sufficient to look for solutions for θ=0.

    Let f:R->R, and see if there is a solution for u(x,y)=f(x+y):
    The first four conditions are just f'>0, f''<0.
    The last conditions become f''<0 for x+y<0, and f''>0 for x+y>0. Obviously, that does not work together with f''<0. u has to be more complicated.
  4. Jul 5, 2013 #3
    I am doing a PhD in economics so I work with quantities and I forgot to tell you that x>=0 and y>=0 all the time. Sorry for this missing information and thanks for your answer
    As a matter of fact theta>0 too
  5. Jul 5, 2013 #4


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    Staff: Mentor

    Oh, that changes a lot.

    For every solution u(x,y), there is a set of solutions v(x,y)=u(x,y)+f(x)+g(y) if f'>0, f''<0, g'>0, g''<0. You can add everything which does not give a cross-derivative, and satisfies the first 4 conditions.
    Even better: For every function u(x,y) which satisfies the cross-solution conditions with bounded derivatives, you can find functions f,g such that the first four conditions are satisfied.

    How to construct many solutions:
    Let f:R->R be a function with f''(z)<0 for z<θ and bounded f''(z)>0 for z>θ
    Consider v(x,y)=f(x+y)+g(x)+h(y): It satisfies the cross-derivative conditions.
    d/dx v = (d/dx f(x+y))+g'(x) and d2/dx2 v = (d2/dx2 f(x+y))+g''(x): Find a function g such that the first is >0 and the second is <0.
    In the same way, find a function h (can be the same as g, if you like).
  6. Jul 5, 2013 #5
    Thanks again
    Last edited: Jul 5, 2013
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