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I'm trying to find the original function from the Hessian defined as:

$Hf:= \begin{bmatrix}x-2y & x+2y\\x+2y & 2x+2y\end{bmatrix}$

Since the Hessian is symmetric, and the mixed order partials are equivalent, then there exists some $C^2$ function that can be differentiated twice to give this matrix. I'm not sure how I should go about solving this, though, and I should note it should not require integration.

Regardless, I did try to approach via integration by integrating $x-2y$ and $2x+2y$ to return the partials of x and y respectively, but these do not result in giving the mixed order derivatives.

Is there an alternative approach I should be taking, that does not involve integration? I guess you could logically deduce it...?

EDIT: Could I perhaps use a Taylor series expansion somehow? Though I don't think I have enough information...

TimelordViktorious
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  • Why do you think there should be a solution? You say that since the matrix is symmetric (which is certainly required) then it must be the Hessian of something. Is that a theorem? The fact that mixed partial derivatives are symmetric implies necessity not sufficiency – spaceisdarkgreen Apr 15 '17 at 17:50
  • Well the theorem does not talk about the existence of such a function, you are right, but it says that if $f: \mathbb{R^n} \to R$ is of class $C^2$ then the Hessian matrix of $f$ at $a \in \mathbb{R}$ is the symmetric $n x n$-matrix of second order partial derivatives. I guess that is an important distinction. – TimelordViktorious Apr 15 '17 at 17:52
  • Well then it's safe to say there's no reason to be perplexed. – spaceisdarkgreen Apr 15 '17 at 18:07
  • As noted in the answer below, it seems like there is no such $f$ in this case. We must have $f_x=x^2/2-2xy+G(y)$, but then $f_{xy}=-2x+G'(y)$, which doesn’t match the corresponding entry in the Hessian. – amd Apr 15 '17 at 19:18

1 Answers1

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Well, besides integration I don't see any other approach. You have: $f_{xx} = x-2y , f_{yy}=2x+2y , f_{xy} = x+2y$.

$$f(x,y)=xy^2+y^3/3+C(x)y+D(x)$$ $$f_{xx} = x-2y = yC''(x)+D''(x)$$

you have: $C''(x)=-2, D''(x)=x$. $C(x)=-x^2+Bx+A$, $D(x) = x^3/6 + B_1 x+C_1$.

In order to find the coefficients insert it to $f_{xy}$.

You get the following equations: $$2y+C'(x) = x+2y$$ plug and get: $$-2x+B=x$$ $$B=3x$$ This is impossible since $B$ is a constant with respect to $x$.

So this system doesn't seem to yield a solution not even a family of solutions.

MathematicalPhysicist
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