Matematické Fórum

vlado_bb · 10. 07. 2020 07:25

↑↑ Dacu:Of course he is wrong, $3$ and $-3$ are not roots of $x^2-3=0$ .

Dacu · 10. 07. 2020 19:42 — Editoval Dacu (10. 07. 2020 19:46)

vlado_bb napsal(a):
↑↑ Dacu:Of course he is wrong, $3$ and $-3$ are not roots of $x^2-3=0$ .

Hello,

I made the same mistake and you saw that I corrected it ....Bati had to write $x=\mp \sqrt{3}$ ....

It is wrong what Bati says, namely:
"Let $c_1=c_2=0$ Then $f(x)-3g'(x)=0$ for all $x\in\mathbb{R}$ " referring to the functions $f(x) = \frac{1}{2} c_1 e^{-x} (e^{2 x} + 1) + \frac{3}{2} c_2 e^{-x} (e^{2 x} - 1) + x^2 + 2 x - 1$ and $g(x) = \frac{1}{6} c_1 e^{-x} (e^{2 x} - 1) + \frac{1}{2} c_2 e^{-x} (e^{2 x} + 1) + \frac{1}{3} (x^2 + 2 x - 1)$ ?

vlado_bb napsal(a):
↑↑ Dacu:Such functions do not exist (see the solution by ↑↑ jarrro:).

I repeat:
I think that one solution could be $f(x) = \frac{1}{2} c_1 e^{-x} (e^{2 x} + 1) + \frac{3}{2} c_2 e^{-x} (e^{2 x} - 1) + x^2 + 2 x - 1$ and $g(x) = \frac{1}{6} c_1 e^{-x} (e^{2 x} - 1) + \frac{1}{2} c_2 e^{-x} (e^{2 x} + 1) + \frac{1}{3} (x^2 + 2 x - 1)$ .
Please kindly prove to me that the above solution is not correct!Thank you very much!

All the best,

Dacu

vlado_bb

↑ Dacu:The proof by jarrro is very clear. Which part you do not understand?

Your functions $f,g$ can only exist on the set $\{-\sqrt{3},\sqrt{3}\}$ . However, they have not derivatives of this set, so they cannot be the solution of your system. So, there is no solution.

Bati · 28. 07. 2020 16:50

↑ Dacu:
I made an oversight, sorry for that. However, my conclusion remains the same. If $c_1=c_2=0$ , then your system becomes
$|x^2-3|=x^2-3$
$|x^2-3|=3-x^2$
The first equality is true iff $x\in (-\infty,-\sqrt3]\cup[\sqrt3,\infty)$ , while second is true iff $x\in[-\sqrt3,\sqrt3]$ . So your particular ``solution'' works only at points $\{\sqrt3,-\sqrt3\}$ (which could be achieved by a much simpler function if thats what you want)

vlado_bb · 28. 07. 2020 18:31

↑ Bati:Are you sure that we can speak about derivatives of functions defined at two points only? In my opinion no, hence there is no solution.

Bati · 28. 07. 2020 18:40

↑ vlado_bb:
Functions f,g from #21 are defined (and smooth) everywhere

vlado_bb · 28. 07. 2020 18:44

↑ Bati:Sure, but if they should be solutions, then they are defined only on $\{\sqrt3,-\sqrt3\}$ , correct?

Bati · 28. 07. 2020 23:15

↑ vlado_bb:
I would say no to that... those functions are defined in R and whether they solve something or not is irrelevant in that regard. But this debate is rather pointless, I just wanted to provide a direct argument which I believe Dacu was asking for. jarrro's answer is completely fine as well, of course

Dacu · 29. 07. 2020 07:24 — Editoval Dacu (29. 07. 2020 07:29)

Hello all,

From the "WolframAlpha" read:

Odkaz

Is it correct what "WolframAlpha" displays? I say yes!

All the best,

Dacu

Bati · 29. 07. 2020 09:29

Hi ↑ Dacu:

I say resolute NO. Lets find the minimal (un)working example:
$|f'(x)|=-1-x^2$
I hope that you agree that there exist no functions f solving this equation (as the left side is always >=0, but right side is always <0
But wolfram says:
$f(x) = c_1 - 1/3 x (x^2 + 3)$
$f(x) = c_1 + x^3/3 + x$
And it is also obvious what it does wrong: It writes $|f'(x)|=\sqrt{(f'(x))^2}$ , which is fine, but then it squares both sides of equation to convert it to a ''standard'' form, which however results in a very different ODE then the original one (simply because $x=1$ is not equivalent to $x^2=1$ etc.)

You should never trust blindly these computational engines. They are written (at least initially) by people and it is very difficult, if not impossible, to write it in a way that it processess any mathematically valid input. These tools, however, can be very useful for checking your intuition or your computations. That said, they can not think for you (yet).

Still, you may try to report this bug to WA. The desktop version of wolfram produces the same result:
DSolve[Abs[f'[x]] == -1 - x^2, f[x], x]
gives the same output as WA, but with a small warning:
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

Curiously enough, wolfram gives the right answer to $|f'(x)|=-1$ (again, no solution).

Dacu · 30. 07. 2020 06:30

Bati napsal(a):
Hi ↑ Dacu:

I say resolute NO. Lets find the minimal (un)working example:
$|f'(x)|=-1-x^2$
I hope that you agree that there exist no functions f solving this equation (as the left side is always >=0, but right side is always <0
But wolfram says:
$f(x) = c_1 - 1/3 x (x^2 + 3)$
$f(x) = c_1 + x^3/3 + x$
And it is also obvious what it does wrong: It writes $|f'(x)|=\sqrt{(f'(x))^2}$ , which is fine, but then it squares both sides of equation to convert it to a ''standard'' form, which however results in a very different ODE then the original one (simply because $x=1$ is not equivalent to $x^2=1$ etc.)

You should never trust blindly these computational engines. They are written (at least initially) by people and it is very difficult, if not impossible, to write it in a way that it processess any mathematically valid input. These tools, however, can be very useful for checking your intuition or your computations. That said, they can not think for you (yet).

Still, you may try to report this bug to WA. The desktop version of wolfram produces the same result:
DSolve[Abs[f'[x]] == -1 - x^2, f[x], x]
gives the same output as WA, but with a small warning:
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

Curiously enough, wolfram gives the right answer to $|f'(x)|=-1$ (again, no solution).

Hello,

I think that for $|f'(x)|=-1-x^2$ the answer https://www.wolframalpha.com/input/?i=% … 3D-1-x%5E2 is correct , because it considers that $f(x): \mathbb C \rightarrow \mathbb C$ , $x=a\cdot i$ , $a\in \mathbb R$ and $-1-x^2\geq 0$ ...

All the best,

Dacu

Bati · 30. 07. 2020 08:03 — Editoval Bati (30. 07. 2020 08:05)

↑ Dacu:
Even if this was true, this function $x^3/3+x$ cannot be a solution on the whole line $a\mapsto ai$ (it fails in $(-1,1)$ ). At any case, you really should specify this $f:\mathbb{C}\to\mathbb{C}$ first and then you should also explain what do you mean by a solution (for someone, it may be not enough that the solution works only pathwise). I think your system becomes a usual real ODE system after applying a simple substitution and so there is no reason to assume you haven't applied it already.

Matematické Fórum

#26 10. 07. 2020 07:25

Re: Systém diferenciálních rovnic

#27 10. 07. 2020 19:42 — Editoval Dacu (10. 07. 2020 19:46)

Re: Systém diferenciálních rovnic

vlado_bb napsal(a):

vlado_bb napsal(a):

#28 10. 07. 2020 20:30 — Editoval vlado_bb (10. 07. 2020 20:35)

Re: Systém diferenciálních rovnic

#29 28. 07. 2020 16:50

Re: Systém diferenciálních rovnic

#30 28. 07. 2020 18:31

Re: Systém diferenciálních rovnic

#31 28. 07. 2020 18:40

Re: Systém diferenciálních rovnic

#32 28. 07. 2020 18:44

Re: Systém diferenciálních rovnic

#33 28. 07. 2020 23:15

Re: Systém diferenciálních rovnic

#34 29. 07. 2020 07:24 — Editoval Dacu (29. 07. 2020 07:29)

Re: Systém diferenciálních rovnic

#35 29. 07. 2020 09:29

Re: Systém diferenciálních rovnic

#36 30. 07. 2020 06:30

Re: Systém diferenciálních rovnic

Bati napsal(a):

#37 30. 07. 2020 08:03 — Editoval Bati (30. 07. 2020 08:05)

Re: Systém diferenciálních rovnic

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