Homogeneous Fredholm integral equation of the first kind with positive symmetric kernel
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Given the equation
$$int^{1}_{-1}K(|x-t|)varphi(t)dt=0,$$
where the kernel is positive: $K(x)>0$; equation is satisfied for $xin [-1,1]$. $K(x)$ and $varphi(x)$ are real and continuous functions. I'm looking for nontrivial solutions $varphi(x)$.
Unfortunately, the explicit form of kernel $K(x)$ is unknown, but approximately K(x) is close to $exp(-lambda x)$.
I suppose, that if kernel $K(x)$ is not constant, then only trivial solutions exist. (If kernel is constant, then any odd function satisfies the equation.) Is this so? And how to prove it?
integral-equations
asked Nov 20 at 20:06
And111
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up vote
0
down vote
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Given the equation
$$int^{1}_{-1}K(|x-t|)varphi(t)dt=0,$$
where the kernel is positive: $K(x)>0$; equation is satisfied for $xin [-1,1]$. $K(x)$ and $varphi(x)$ are real and continuous functions. I'm looking for nontrivial solutions $varphi(x)$.
Unfortunately, the explicit form of kernel $K(x)$ is unknown, but approximately K(x) is close to $exp(-lambda x)$.
I suppose, that if kernel $K(x)$ is not constant, then only trivial solutions exist. (If kernel is constant, then any odd function satisfies the equation.) Is this so? And how to prove it?
integral-equations
asked Nov 20 at 20:06
And111
1
New contributor
And111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given the equation
$$int^{1}_{-1}K(|x-t|)varphi(t)dt=0,$$
where the kernel is positive: $K(x)>0$; equation is satisfied for $xin [-1,1]$. $K(x)$ and $varphi(x)$ are real and continuous functions. I'm looking for nontrivial solutions $varphi(x)$.
Unfortunately, the explicit form of kernel $K(x)$ is unknown, but approximately K(x) is close to $exp(-lambda x)$.
I suppose, that if kernel $K(x)$ is not constant, then only trivial solutions exist. (If kernel is constant, then any odd function satisfies the equation.) Is this so? And how to prove it?
integral-equations
asked Nov 20 at 20:06
And111
1
New contributor
And111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given the equation
$$int^{1}_{-1}K(|x-t|)varphi(t)dt=0,$$
where the kernel is positive: $K(x)>0$; equation is satisfied for $xin [-1,1]$. $K(x)$ and $varphi(x)$ are real and continuous functions. I'm looking for nontrivial solutions $varphi(x)$.
Unfortunately, the explicit form of kernel $K(x)$ is unknown, but approximately K(x) is close to $exp(-lambda x)$.
I suppose, that if kernel $K(x)$ is not constant, then only trivial solutions exist. (If kernel is constant, then any odd function satisfies the equation.) Is this so? And how to prove it?
integral-equations
integral-equations
asked Nov 20 at 20:06
And111
1
New contributor
And111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 20:06
And111
1
New contributor
And111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 20:06
And111
1
New contributor
And111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 20:06
And111
1
asked Nov 20 at 20:06
And111
1
1
New contributor
And111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
And111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
And111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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And111 is a new contributor. Be nice, and check out our Code of Conduct.
And111 is a new contributor. Be nice, and check out our Code of Conduct.
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