Linear Volterra Integral Equation with deviating argument
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I am looking at a linear integral equation 2nd kind Volterra equation with deviating argument in the unknown $u(t) in L^2[a,b]$:
begin{equation}
u(x) = f(x)+int_a^{b-x+a} K(x,s) u(s) mathrm{d}s phantom{texttexttexttex}(1)
end{equation}
where $x in [a,b]$, $K(x,s)$ is absolutely continuous functions in $[a,b]times[a,b]$ and $f(x)in L^2[a,b]$. Does (1) have a unique solution?
functional-analysis ordinary-differential-equations
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add a comment |
$begingroup$
I am looking at a linear integral equation 2nd kind Volterra equation with deviating argument in the unknown $u(t) in L^2[a,b]$:
begin{equation}
u(x) = f(x)+int_a^{b-x+a} K(x,s) u(s) mathrm{d}s phantom{texttexttexttex}(1)
end{equation}
where $x in [a,b]$, $K(x,s)$ is absolutely continuous functions in $[a,b]times[a,b]$ and $f(x)in L^2[a,b]$. Does (1) have a unique solution?
functional-analysis ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
I am looking at a linear integral equation 2nd kind Volterra equation with deviating argument in the unknown $u(t) in L^2[a,b]$:
begin{equation}
u(x) = f(x)+int_a^{b-x+a} K(x,s) u(s) mathrm{d}s phantom{texttexttexttex}(1)
end{equation}
where $x in [a,b]$, $K(x,s)$ is absolutely continuous functions in $[a,b]times[a,b]$ and $f(x)in L^2[a,b]$. Does (1) have a unique solution?
functional-analysis ordinary-differential-equations
$endgroup$
I am looking at a linear integral equation 2nd kind Volterra equation with deviating argument in the unknown $u(t) in L^2[a,b]$:
begin{equation}
u(x) = f(x)+int_a^{b-x+a} K(x,s) u(s) mathrm{d}s phantom{texttexttexttex}(1)
end{equation}
where $x in [a,b]$, $K(x,s)$ is absolutely continuous functions in $[a,b]times[a,b]$ and $f(x)in L^2[a,b]$. Does (1) have a unique solution?
functional-analysis ordinary-differential-equations
functional-analysis ordinary-differential-equations
asked Jan 5 at 11:46
Nebojša ĐurićNebojša Đurić
1067
1067
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