Fixed point of unusual integral equation
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I am a little rusty in this area so please forgive the slowness. I am trying to prove or disprove the existence of fixed points for the following integral equation. Throughout I am interested in the metric space $C[0,1]$ which is complete under the $Vert cdot Vert_infty$ norm.
Define $Phi: C[0,1] rightarrow C[0,1] $ by
$$Phi(x)(theta) = frac{c(theta) } {c(theta) + N int_0^1 x(t) c(t) dt }. $$ Surely some restrictions on $c(theta)$ will be needed, but I would rather make them as loose as possible. If it helps, for concreteness put $c(t) = t + t^p$. $N$ and $p$ are known positive parameters.
I am trying to show a solution to the equation $Phi(x) = x$ exists. I tried to show $Phi$ is a contraction mapping without success. Moreover, for the particular $c(cdot)$ above I can see that $Vert Phi(mathbf{1}) - Phi(mathbf{0})Vert = Vert mathbf{1} - mathbf{0} Vert = 1. $ Is this a correct counterexample to $Phi$ being a contraction mapping?
If so, do I have much hope of showing existence? I was hoping there was a clean fixed-point theorem I could quote, but I haven't found any so far.
functional-analysis functional-equations fixed-point-theorems integral-equations
asked Jan 15 at 16:18
user434180user434180
1108
$endgroup$
add a comment |
$begingroup$
I am a little rusty in this area so please forgive the slowness. I am trying to prove or disprove the existence of fixed points for the following integral equation. Throughout I am interested in the metric space $C[0,1]$ which is complete under the $Vert cdot Vert_infty$ norm.
Define $Phi: C[0,1] rightarrow C[0,1] $ by
$$Phi(x)(theta) = frac{c(theta) } {c(theta) + N int_0^1 x(t) c(t) dt }. $$ Surely some restrictions on $c(theta)$ will be needed, but I would rather make them as loose as possible. If it helps, for concreteness put $c(t) = t + t^p$. $N$ and $p$ are known positive parameters.
I am trying to show a solution to the equation $Phi(x) = x$ exists. I tried to show $Phi$ is a contraction mapping without success. Moreover, for the particular $c(cdot)$ above I can see that $Vert Phi(mathbf{1}) - Phi(mathbf{0})Vert = Vert mathbf{1} - mathbf{0} Vert = 1. $ Is this a correct counterexample to $Phi$ being a contraction mapping?
If so, do I have much hope of showing existence? I was hoping there was a clean fixed-point theorem I could quote, but I haven't found any so far.
functional-analysis functional-equations fixed-point-theorems integral-equations
asked Jan 15 at 16:18
user434180user434180
1108
$endgroup$
$begingroup$
I think this is confused. $Phi(x)(theta)$ is a function of $x$ and $theta$ which is well defined for a given $c(theta)$ by your equation. $x$ is not a function of $t$, which would allow it to be pulled out of the integral. To show that $Phi(x)(theta)=x$ is a solution for a specific $c(theta)$ you would just have to plug it into the equation. The contraction mapping theorem would apply if you somehow plugged $Phi$ into the right and iterated to convergence. You would have to define a metric on the space $Phi$ is in to show that the iteration is a contraction.
$endgroup$
– Ross Millikan
Jan 15 at 16:44
$begingroup$
Perhaps the extra (typo) 'exists' was confusing. I have removed it. But I don't see what you mean by '$x$ is not a function of $t$'. $x$ is some continuous function on [0,1] and $t$ just a dummy variable of integration. It can't be pulled out of the integral. The solution isn't $Phi(x)(theta)$, the solution would be a function $x(theta)$ that solves the functional equation $Phi(x) = x$. That is, it would solve $Phi(x)(theta) = x(theta)$ for all $theta in [0,1].$ Or perhaps I am indeed very confused. Please let me know if this clears it up.
$endgroup$
– user434180
Jan 15 at 16:53
add a comment |
$begingroup$
I am a little rusty in this area so please forgive the slowness. I am trying to prove or disprove the existence of fixed points for the following integral equation. Throughout I am interested in the metric space $C[0,1]$ which is complete under the $Vert cdot Vert_infty$ norm.
Define $Phi: C[0,1] rightarrow C[0,1] $ by
$$Phi(x)(theta) = frac{c(theta) } {c(theta) + N int_0^1 x(t) c(t) dt }. $$ Surely some restrictions on $c(theta)$ will be needed, but I would rather make them as loose as possible. If it helps, for concreteness put $c(t) = t + t^p$. $N$ and $p$ are known positive parameters.
I am trying to show a solution to the equation $Phi(x) = x$ exists. I tried to show $Phi$ is a contraction mapping without success. Moreover, for the particular $c(cdot)$ above I can see that $Vert Phi(mathbf{1}) - Phi(mathbf{0})Vert = Vert mathbf{1} - mathbf{0} Vert = 1. $ Is this a correct counterexample to $Phi$ being a contraction mapping?
If so, do I have much hope of showing existence? I was hoping there was a clean fixed-point theorem I could quote, but I haven't found any so far.
functional-analysis functional-equations fixed-point-theorems integral-equations
asked Jan 15 at 16:18
user434180user434180
1108
$endgroup$
I am a little rusty in this area so please forgive the slowness. I am trying to prove or disprove the existence of fixed points for the following integral equation. Throughout I am interested in the metric space $C[0,1]$ which is complete under the $Vert cdot Vert_infty$ norm.
Define $Phi: C[0,1] rightarrow C[0,1] $ by
$$Phi(x)(theta) = frac{c(theta) } {c(theta) + N int_0^1 x(t) c(t) dt }. $$ Surely some restrictions on $c(theta)$ will be needed, but I would rather make them as loose as possible. If it helps, for concreteness put $c(t) = t + t^p$. $N$ and $p$ are known positive parameters.
I am trying to show a solution to the equation $Phi(x) = x$ exists. I tried to show $Phi$ is a contraction mapping without success. Moreover, for the particular $c(cdot)$ above I can see that $Vert Phi(mathbf{1}) - Phi(mathbf{0})Vert = Vert mathbf{1} - mathbf{0} Vert = 1. $ Is this a correct counterexample to $Phi$ being a contraction mapping?
If so, do I have much hope of showing existence? I was hoping there was a clean fixed-point theorem I could quote, but I haven't found any so far.
functional-analysis functional-equations fixed-point-theorems integral-equations
functional-analysis functional-equations fixed-point-theorems integral-equations
asked Jan 15 at 16:18
user434180user434180
1108
asked Jan 15 at 16:18
user434180user434180
1108
edited Jan 15 at 16:50
user434180
asked Jan 15 at 16:18
user434180user434180
1108
asked Jan 15 at 16:18
user434180user434180
1108
asked Jan 15 at 16:18
user434180user434180
1108
1108
$begingroup$
I think this is confused. $Phi(x)(theta)$ is a function of $x$ and $theta$ which is well defined for a given $c(theta)$ by your equation. $x$ is not a function of $t$, which would allow it to be pulled out of the integral. To show that $Phi(x)(theta)=x$ is a solution for a specific $c(theta)$ you would just have to plug it into the equation. The contraction mapping theorem would apply if you somehow plugged $Phi$ into the right and iterated to convergence. You would have to define a metric on the space $Phi$ is in to show that the iteration is a contraction.
$endgroup$
– Ross Millikan
Jan 15 at 16:44
$begingroup$
Perhaps the extra (typo) 'exists' was confusing. I have removed it. But I don't see what you mean by '$x$ is not a function of $t$'. $x$ is some continuous function on [0,1] and $t$ just a dummy variable of integration. It can't be pulled out of the integral. The solution isn't $Phi(x)(theta)$, the solution would be a function $x(theta)$ that solves the functional equation $Phi(x) = x$. That is, it would solve $Phi(x)(theta) = x(theta)$ for all $theta in [0,1].$ Or perhaps I am indeed very confused. Please let me know if this clears it up.
$endgroup$
– user434180
Jan 15 at 16:53
add a comment |
$begingroup$
I think this is confused. $Phi(x)(theta)$ is a function of $x$ and $theta$ which is well defined for a given $c(theta)$ by your equation. $x$ is not a function of $t$, which would allow it to be pulled out of the integral. To show that $Phi(x)(theta)=x$ is a solution for a specific $c(theta)$ you would just have to plug it into the equation. The contraction mapping theorem would apply if you somehow plugged $Phi$ into the right and iterated to convergence. You would have to define a metric on the space $Phi$ is in to show that the iteration is a contraction.
$endgroup$
– Ross Millikan
Jan 15 at 16:44
$begingroup$
Perhaps the extra (typo) 'exists' was confusing. I have removed it. But I don't see what you mean by '$x$ is not a function of $t$'. $x$ is some continuous function on [0,1] and $t$ just a dummy variable of integration. It can't be pulled out of the integral. The solution isn't $Phi(x)(theta)$, the solution would be a function $x(theta)$ that solves the functional equation $Phi(x) = x$. That is, it would solve $Phi(x)(theta) = x(theta)$ for all $theta in [0,1].$ Or perhaps I am indeed very confused. Please let me know if this clears it up.
$endgroup$
– user434180
Jan 15 at 16:53
$begingroup$
I think this is confused. $Phi(x)(theta)$ is a function of $x$ and $theta$ which is well defined for a given $c(theta)$ by your equation. $x$ is not a function of $t$, which would allow it to be pulled out of the integral. To show that $Phi(x)(theta)=x$ is a solution for a specific $c(theta)$ you would just have to plug it into the equation. The contraction mapping theorem would apply if you somehow plugged $Phi$ into the right and iterated to convergence. You would have to define a metric on the space $Phi$ is in to show that the iteration is a contraction.
$endgroup$
– Ross Millikan
Jan 15 at 16:44
$begingroup$
I think this is confused. $Phi(x)(theta)$ is a function of $x$ and $theta$ which is well defined for a given $c(theta)$ by your equation. $x$ is not a function of $t$, which would allow it to be pulled out of the integral. To show that $Phi(x)(theta)=x$ is a solution for a specific $c(theta)$ you would just have to plug it into the equation. The contraction mapping theorem would apply if you somehow plugged $Phi$ into the right and iterated to convergence. You would have to define a metric on the space $Phi$ is in to show that the iteration is a contraction.
$endgroup$
– Ross Millikan
Jan 15 at 16:44
$begingroup$
Perhaps the extra (typo) 'exists' was confusing. I have removed it. But I don't see what you mean by '$x$ is not a function of $t$'. $x$ is some continuous function on [0,1] and $t$ just a dummy variable of integration. It can't be pulled out of the integral. The solution isn't $Phi(x)(theta)$, the solution would be a function $x(theta)$ that solves the functional equation $Phi(x) = x$. That is, it would solve $Phi(x)(theta) = x(theta)$ for all $theta in [0,1].$ Or perhaps I am indeed very confused. Please let me know if this clears it up.
$endgroup$
– user434180
Jan 15 at 16:53
$begingroup$
Perhaps the extra (typo) 'exists' was confusing. I have removed it. But I don't see what you mean by '$x$ is not a function of $t$'. $x$ is some continuous function on [0,1] and $t$ just a dummy variable of integration. It can't be pulled out of the integral. The solution isn't $Phi(x)(theta)$, the solution would be a function $x(theta)$ that solves the functional equation $Phi(x) = x$. That is, it would solve $Phi(x)(theta) = x(theta)$ for all $theta in [0,1].$ Or perhaps I am indeed very confused. Please let me know if this clears it up.
$endgroup$
– user434180
Jan 15 at 16:53
add a comment |
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$begingroup$
I think this is confused. $Phi(x)(theta)$ is a function of $x$ and $theta$ which is well defined for a given $c(theta)$ by your equation. $x$ is not a function of $t$, which would allow it to be pulled out of the integral. To show that $Phi(x)(theta)=x$ is a solution for a specific $c(theta)$ you would just have to plug it into the equation. The contraction mapping theorem would apply if you somehow plugged $Phi$ into the right and iterated to convergence. You would have to define a metric on the space $Phi$ is in to show that the iteration is a contraction.
$endgroup$
– Ross Millikan
Jan 15 at 16:44
$begingroup$
Perhaps the extra (typo) 'exists' was confusing. I have removed it. But I don't see what you mean by '$x$ is not a function of $t$'. $x$ is some continuous function on [0,1] and $t$ just a dummy variable of integration. It can't be pulled out of the integral. The solution isn't $Phi(x)(theta)$, the solution would be a function $x(theta)$ that solves the functional equation $Phi(x) = x$. That is, it would solve $Phi(x)(theta) = x(theta)$ for all $theta in [0,1].$ Or perhaps I am indeed very confused. Please let me know if this clears it up.
$endgroup$
– user434180
Jan 15 at 16:53