Mixing
Show that this second type Fredholm equation doesn't admit a solution using fredholm theorems
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Given the following fredholm integral equation $g(s) =f(s)+lambdaint_{0}^{2 pi}sin(t+s) g(t) dt$ defined from $C[0,2pi]$ over itself. Show that if $f(s) =s$ then the equation doesn't have solution ¿what about $f(s) =1$?.
Hi, I need to prove this problem using fredholm theorems (the ones that relate the primal no homogeneous equation with the dual homogeneous equation and viceversa) and I don't understand those theorems yet.
functional-analysis operator-theory spectral-theory integral-equations
asked Jan 25 at 12:45
NeuphoriaNeuphoria
283
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add a comment |
$begingroup$
Given the following fredholm integral equation $g(s) =f(s)+lambdaint_{0}^{2 pi}sin(t+s) g(t) dt$ defined from $C[0,2pi]$ over itself. Show that if $f(s) =s$ then the equation doesn't have solution ¿what about $f(s) =1$?.
Hi, I need to prove this problem using fredholm theorems (the ones that relate the primal no homogeneous equation with the dual homogeneous equation and viceversa) and I don't understand those theorems yet.
functional-analysis operator-theory spectral-theory integral-equations
asked Jan 25 at 12:45
NeuphoriaNeuphoria
283
$endgroup$
$begingroup$
If $lambda=0$, $g=f$ is the solution.
$endgroup$
– C.Ding
Jan 25 at 14:45
$begingroup$
I suppose that for $lambdaneq0$ then, otherwise the exercise would be pointless .
$endgroup$
– Neuphoria
Jan 25 at 15:13
add a comment |
$begingroup$
Given the following fredholm integral equation $g(s) =f(s)+lambdaint_{0}^{2 pi}sin(t+s) g(t) dt$ defined from $C[0,2pi]$ over itself. Show that if $f(s) =s$ then the equation doesn't have solution ¿what about $f(s) =1$?.
Hi, I need to prove this problem using fredholm theorems (the ones that relate the primal no homogeneous equation with the dual homogeneous equation and viceversa) and I don't understand those theorems yet.
functional-analysis operator-theory spectral-theory integral-equations
asked Jan 25 at 12:45
NeuphoriaNeuphoria
283
$endgroup$
Given the following fredholm integral equation $g(s) =f(s)+lambdaint_{0}^{2 pi}sin(t+s) g(t) dt$ defined from $C[0,2pi]$ over itself. Show that if $f(s) =s$ then the equation doesn't have solution ¿what about $f(s) =1$?.
Hi, I need to prove this problem using fredholm theorems (the ones that relate the primal no homogeneous equation with the dual homogeneous equation and viceversa) and I don't understand those theorems yet.
functional-analysis operator-theory spectral-theory integral-equations
functional-analysis operator-theory spectral-theory integral-equations
asked Jan 25 at 12:45
NeuphoriaNeuphoria
283
asked Jan 25 at 12:45
NeuphoriaNeuphoria
283
asked Jan 25 at 12:45
NeuphoriaNeuphoria
283
asked Jan 25 at 12:45
NeuphoriaNeuphoria
283
asked Jan 25 at 12:45
NeuphoriaNeuphoria
283
283
$begingroup$
If $lambda=0$, $g=f$ is the solution.
$endgroup$
– C.Ding
Jan 25 at 14:45
$begingroup$
I suppose that for $lambdaneq0$ then, otherwise the exercise would be pointless .
$endgroup$
– Neuphoria
Jan 25 at 15:13
add a comment |
$begingroup$
If $lambda=0$, $g=f$ is the solution.
$endgroup$
– C.Ding
Jan 25 at 14:45
$begingroup$
I suppose that for $lambdaneq0$ then, otherwise the exercise would be pointless .
$endgroup$
– Neuphoria
Jan 25 at 15:13
$begingroup$
If $lambda=0$, $g=f$ is the solution.
$endgroup$
– C.Ding
Jan 25 at 14:45
$begingroup$
If $lambda=0$, $g=f$ is the solution.
$endgroup$
– C.Ding
Jan 25 at 14:45
$begingroup$
I suppose that for $lambdaneq0$ then, otherwise the exercise would be pointless .
$endgroup$
– Neuphoria
Jan 25 at 15:13
$begingroup$
I suppose that for $lambdaneq0$ then, otherwise the exercise would be pointless .
$endgroup$
– Neuphoria
Jan 25 at 15:13
add a comment |
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$begingroup$
If $lambda=0$, $g=f$ is the solution.
$endgroup$
– C.Ding
Jan 25 at 14:45
$begingroup$
I suppose that for $lambdaneq0$ then, otherwise the exercise would be pointless .
$endgroup$
– Neuphoria
Jan 25 at 15:13