Solution $y(x)$ for $sin(x) = int_0^{2pi} max(y(t), y(x+t)) dt$
$begingroup$
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
asked Dec 6 '18 at 16:49
Jeffery StoutJeffery Stout
343
$endgroup$
add a comment |
$begingroup$
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
asked Dec 6 '18 at 16:49
Jeffery StoutJeffery Stout
343
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 6 '18 at 16:49
Jeffery StoutJeffery Stout
343
$endgroup$
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
asked Dec 6 '18 at 16:49
Jeffery StoutJeffery Stout
343
asked Dec 6 '18 at 16:49
Jeffery StoutJeffery Stout
343
edited Dec 6 '18 at 21:25
Jeffery Stout
asked Dec 6 '18 at 16:49
Jeffery StoutJeffery Stout
343
asked Dec 6 '18 at 16:49
Jeffery StoutJeffery Stout
343
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $y(x)$.
answered Dec 8 '18 at 9:33
JanGJanG
2,802514
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $y(x)$.
answered Dec 8 '18 at 9:33
JanGJanG
2,802514
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $y(x)$.
answered Dec 8 '18 at 9:33
JanGJanG
2,802514
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $y(x)$.
answered Dec 8 '18 at 9:33
JanGJanG
2,802514
$endgroup$
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 $y(x)$.
answered Dec 8 '18 at 9:33
JanGJanG
2,802514
answered Dec 8 '18 at 9:33
JanGJanG
2,802514
answered Dec 8 '18 at 9:33
JanGJanG
2,802514
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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