Solution $y(x)$ for $sin(x) = int_0^{2pi} max(y(t), y(x+t)) dt$












6












$begingroup$


I've been banging my head against a wall for a few weeks to find a feasible solution for $y(x)$.



$$sin(x) = int_0^{2pi} max(y(t), y(x+t)) dt$$



I don't think there is an unique solution, but I will accept any solution. I've tried various numerical minimization methods, convolution, guess the solution, etc, but nothing seems to work.



Does anybody have any ideas?






integration analysis integral-equations







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$endgroup$












  • $begingroup$
    try to use the fact $max(y(t),y(x+t))=|frac{y(t)+y(x+t)}{2}|+|frac{y(t)-y(x+t)}{2}|$
    $endgroup$
    – vidyarthi
    Dec 6 '18 at 17:22


















6












$begingroup$


I've been banging my head against a wall for a few weeks to find a feasible solution for $y(x)$.



$$sin(x) = int_0^{2pi} max(y(t), y(x+t)) dt$$



I don't think there is an unique solution, but I will accept any solution. I've tried various numerical minimization methods, convolution, guess the solution, etc, but nothing seems to work.



Does anybody have any ideas?






integration analysis integral-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    try to use the fact $max(y(t),y(x+t))=|frac{y(t)+y(x+t)}{2}|+|frac{y(t)-y(x+t)}{2}|$
    $endgroup$
    – vidyarthi
    Dec 6 '18 at 17:22
















6












6








6


5



$begingroup$


I've been banging my head against a wall for a few weeks to find a feasible solution for $y(x)$.



$$sin(x) = int_0^{2pi} max(y(t), y(x+t)) dt$$



I don't think there is an unique solution, but I will accept any solution. I've tried various numerical minimization methods, convolution, guess the solution, etc, but nothing seems to work.



Does anybody have any ideas?






integration analysis integral-equations







share|cite|improve this question











$endgroup$




I've been banging my head against a wall for a few weeks to find a feasible solution for $y(x)$.



$$sin(x) = int_0^{2pi} max(y(t), y(x+t)) dt$$



I don't think there is an unique solution, but I will accept any solution. I've tried various numerical minimization methods, convolution, guess the solution, etc, but nothing seems to work.



Does anybody have any ideas?






integration analysis integral-equations




integration analysis integral-equations






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 21:25







Jeffery Stout

















asked Dec 6 '18 at 16:49









Jeffery StoutJeffery Stout

343




343












  • $begingroup$
    try to use the fact $max(y(t),y(x+t))=|frac{y(t)+y(x+t)}{2}|+|frac{y(t)-y(x+t)}{2}|$
    $endgroup$
    – vidyarthi
    Dec 6 '18 at 17:22




















  • $begingroup$
    try to use the fact $max(y(t),y(x+t))=|frac{y(t)+y(x+t)}{2}|+|frac{y(t)-y(x+t)}{2}|$
    $endgroup$
    – vidyarthi
    Dec 6 '18 at 17:22


















$begingroup$
try to use the fact $max(y(t),y(x+t))=|frac{y(t)+y(x+t)}{2}|+|frac{y(t)-y(x+t)}{2}|$
$endgroup$
– vidyarthi
Dec 6 '18 at 17:22






$begingroup$
try to use the fact $max(y(t),y(x+t))=|frac{y(t)+y(x+t)}{2}|+|frac{y(t)-y(x+t)}{2}|$
$endgroup$
– vidyarthi
Dec 6 '18 at 17:22












1 Answer
1






active

oldest

votes


















3












$begingroup$

If $x=0$ then
begin{equation*}
0= sin(0)=int_{0}^{2pi}max(y(t),y(t)), mathrm{d}t = int_{0}^{2pi}y(t), mathrm{d}t .
end{equation*}
If $x = frac{3pi}{2}$ then
begin{equation*}
-1 = sinleft(frac{3pi}{2}right) = int_{0}^{2pi}maxleft(y(t),yleft(frac{3pi}{2}+tright)right), mathrm{d}t ge int_{0}^{2pi}y(t), mathrm{d}t = 0.
end{equation*}
Consequently, it does not exist any solution $y(x)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you obtain that inequality?
    $endgroup$
    – Szeto
    Dec 8 '18 at 11:50










  • $begingroup$
    I used that $max(a,b) ge a$.
    $endgroup$
    – JanG
    Dec 8 '18 at 14:11











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

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3












$begingroup$

If $x=0$ then
begin{equation*}
0= sin(0)=int_{0}^{2pi}max(y(t),y(t)), mathrm{d}t = int_{0}^{2pi}y(t), mathrm{d}t .
end{equation*}
If $x = frac{3pi}{2}$ then
begin{equation*}
-1 = sinleft(frac{3pi}{2}right) = int_{0}^{2pi}maxleft(y(t),yleft(frac{3pi}{2}+tright)right), mathrm{d}t ge int_{0}^{2pi}y(t), mathrm{d}t = 0.
end{equation*}
Consequently, it does not exist any solution $y(x)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you obtain that inequality?
    $endgroup$
    – Szeto
    Dec 8 '18 at 11:50










  • $begingroup$
    I used that $max(a,b) ge a$.
    $endgroup$
    – JanG
    Dec 8 '18 at 14:11
















3












$begingroup$

If $x=0$ then
begin{equation*}
0= sin(0)=int_{0}^{2pi}max(y(t),y(t)), mathrm{d}t = int_{0}^{2pi}y(t), mathrm{d}t .
end{equation*}
If $x = frac{3pi}{2}$ then
begin{equation*}
-1 = sinleft(frac{3pi}{2}right) = int_{0}^{2pi}maxleft(y(t),yleft(frac{3pi}{2}+tright)right), mathrm{d}t ge int_{0}^{2pi}y(t), mathrm{d}t = 0.
end{equation*}
Consequently, it does not exist any solution $y(x)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you obtain that inequality?
    $endgroup$
    – Szeto
    Dec 8 '18 at 11:50










  • $begingroup$
    I used that $max(a,b) ge a$.
    $endgroup$
    – JanG
    Dec 8 '18 at 14:11














3












3








3





$begingroup$

If $x=0$ then
begin{equation*}
0= sin(0)=int_{0}^{2pi}max(y(t),y(t)), mathrm{d}t = int_{0}^{2pi}y(t), mathrm{d}t .
end{equation*}
If $x = frac{3pi}{2}$ then
begin{equation*}
-1 = sinleft(frac{3pi}{2}right) = int_{0}^{2pi}maxleft(y(t),yleft(frac{3pi}{2}+tright)right), mathrm{d}t ge int_{0}^{2pi}y(t), mathrm{d}t = 0.
end{equation*}
Consequently, it does not exist any solution $y(x)$.






share|cite|improve this answer









$endgroup$



If $x=0$ then
begin{equation*}
0= sin(0)=int_{0}^{2pi}max(y(t),y(t)), mathrm{d}t = int_{0}^{2pi}y(t), mathrm{d}t .
end{equation*}
If $x = frac{3pi}{2}$ then
begin{equation*}
-1 = sinleft(frac{3pi}{2}right) = int_{0}^{2pi}maxleft(y(t),yleft(frac{3pi}{2}+tright)right), mathrm{d}t ge int_{0}^{2pi}y(t), mathrm{d}t = 0.
end{equation*}
Consequently, it does not exist any solution $y(x)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 9:33









JanGJanG

2,802514




2,802514












  • $begingroup$
    How did you obtain that inequality?
    $endgroup$
    – Szeto
    Dec 8 '18 at 11:50










  • $begingroup$
    I used that $max(a,b) ge a$.
    $endgroup$
    – JanG
    Dec 8 '18 at 14:11


















  • $begingroup$
    How did you obtain that inequality?
    $endgroup$
    – Szeto
    Dec 8 '18 at 11:50










  • $begingroup$
    I used that $max(a,b) ge a$.
    $endgroup$
    – JanG
    Dec 8 '18 at 14:11
















$begingroup$
How did you obtain that inequality?
$endgroup$
– Szeto
Dec 8 '18 at 11:50




$begingroup$
How did you obtain that inequality?
$endgroup$
– Szeto
Dec 8 '18 at 11:50












$begingroup$
I used that $max(a,b) ge a$.
$endgroup$
– JanG
Dec 8 '18 at 14:11




$begingroup$
I used that $max(a,b) ge a$.
$endgroup$
– JanG
Dec 8 '18 at 14:11


















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