Timeline for A solution of a system of equations that involve directional derivatives
Current License: CC BY-SA 4.0
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| Apr 24, 2021 at 19:11 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 25, 2020 at 19:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 27, 2020 at 18:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 29, 2020 at 18:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 30, 2020 at 15:34 | answer | added | Igor Khavkine | timeline score: 1 | |
| Mar 30, 2020 at 6:06 | comment | added | Eilon | Thanks, Igor. The Poincare-Miranda Theorem allows one to derive conditions for the existence of a solution. Are there tools that allow one to derive conditions for uniqueness of solution? | |
| Mar 29, 2020 at 21:05 | comment | added | Igor Khavkine | If you are happy with using the intermediate value theorem in one dimension (in order to put desired hypotheses on $f(x)$ and $g(x)$), then you might also be happy with using its generalization in two or higher dimensions: the Poincaré-Miranda theorem. | |
| Mar 29, 2020 at 20:36 | comment | added | Igor Khavkine | Ah, I see. Sorry, I totally misread your question again! | |
| Mar 29, 2020 at 17:21 | history | edited | Eilon | CC BY-SA 4.0 |
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| Mar 29, 2020 at 15:53 | comment | added | Eilon | Note that $f$ and $g$ are given functions, and we look for a solution $(x,y) \in [0,1]^2$. We do NOT look for a set of functions $f$ and $g$ that satisfies the equations for every $x$ and $y$. | |
| Mar 29, 2020 at 15:52 | comment | added | Eilon | Since we consider the two-dimensional case, for every $y$ there is at most one $x$ that satisfies $$\frac{\partial f}{\partial x}(x,y) = \frac{\partial g}{\partial x}(1-x,1-y),$$ and for every $x$ there is at most one $y$ that satisfies $$\frac{\partial f}{\partial y}(x,y) = \frac{\partial g}{\partial y}(1-x,1-y).$$ It is not clear to me how we derive that there is a unique pair $(x,y)$ such that both equations hold for it. | |
| Mar 29, 2020 at 15:47 | comment | added | Eilon | Igor, you are a bit quick for me. In the one dimensional case, $x \mapsto f'(x)$ is decreasing and $x \mapsto g'(1-x)$ is increasing, and therefore, by imposing the appropriate inequalities on $f'(0)$, $f'(1`)$, $g'(0)$, and $g'(1)$ we deduce that there is exactly one point $x$ such that $f'(x) = g'(1-x)$. [continues in the next comment] | |
| Mar 29, 2020 at 10:46 | comment | added | Igor Khavkine | OK, now I understand! Thanks for explaining. But then the solution is pretty straightforward. In one variable, using the above notation, you want $f'(x) = g'(1-x) = -h'(x)$. The only possible solution is $h(x)=-f(x)$ or $g(x) = -f(1-x) + C$. Or with two variables, $g(x,y) = -f(1-x,1-y) + C$, for a constant $C$. | |
| Mar 29, 2020 at 5:35 | comment | added | Eilon | I guess the confusion is because of the way we interpret the expression $\frac{\partial g}{\partial x}(1-x)$: for me, $\frac{\partial g}{\partial x}(1-x)$ is the derivative of $g$ evaluated at the point $1-x$. For you, this is the derivative of the function $h(x) = g(1-x)$ evaluated at $x$. And these two derivatives are indeed of different signs. I apologize for the confusion. | |
| Mar 28, 2020 at 23:05 | comment | added | Igor Khavkine | Sorry, I still see a contradiction. For simplicity freeze $y$ and let $h(x) = g(1-x,1-y)$. According to what you are saying, $\frac{\partial}{\partial x} h(x) > 0$ but $h(x)$ is monotone decreasing. Is that not a contradiction? | |
| Mar 28, 2020 at 20:54 | comment | added | Eilon | The function $g(1-x,1-y)$ is indeed decreasing in each variable, but the directional derivative of $g$ in each variable at the point $(1-x,1-y)$ is still positive, or do I miss your question? | |
| Mar 28, 2020 at 13:28 | comment | added | Igor Khavkine | Isn't this impossible, since $g(1-x,1-y)$ is now strictly decreasing in each variable rather than increasing like $f(x,y)$? Or am I missing something? | |
| Mar 28, 2020 at 11:37 | history | asked | Eilon | CC BY-SA 4.0 |