First kind Volterra integral equation regularity
0
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Let $K(x,y) in {L^2}({(0,1)^2})$ and $g in {L^2}(0,1)$. We consider the following integral equation
$$intlimits_0^x {K(x,t)f(t)dt = g(x)} $$
My question: what can we say about the regularity of $f$ if it exists? Thanks.
real-analysis functional-analysis ordinary-differential-equations integral-equations
asked Dec 15 '18 at 14:52
GustaveGustave
734211
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0
$begingroup$
Let $K(x,y) in {L^2}({(0,1)^2})$ and $g in {L^2}(0,1)$. We consider the following integral equation
$$intlimits_0^x {K(x,t)f(t)dt = g(x)} $$
My question: what can we say about the regularity of $f$ if it exists? Thanks.
real-analysis functional-analysis ordinary-differential-equations integral-equations
asked Dec 15 '18 at 14:52
GustaveGustave
734211
$endgroup$
1
$begingroup$
There is nothing we can say without knowing more about $g$, since any $f in L^2$ may be the solution of such an equation with a suitable $g$.
$endgroup$
– Hans Engler
Dec 15 '18 at 15:40
$begingroup$
Thank you sir. What if wa add $f$ to tge second member which will make the equation of second kind. Can we ensure that $f$ is $L^2$? thank you.
$endgroup$
– Gustave
Dec 16 '18 at 10:26
add a comment |
0
0
0
$begingroup$
Let $K(x,y) in {L^2}({(0,1)^2})$ and $g in {L^2}(0,1)$. We consider the following integral equation
$$intlimits_0^x {K(x,t)f(t)dt = g(x)} $$
My question: what can we say about the regularity of $f$ if it exists? Thanks.
real-analysis functional-analysis ordinary-differential-equations integral-equations
asked Dec 15 '18 at 14:52
GustaveGustave
734211
$endgroup$
Let $K(x,y) in {L^2}({(0,1)^2})$ and $g in {L^2}(0,1)$. We consider the following integral equation
$$intlimits_0^x {K(x,t)f(t)dt = g(x)} $$
My question: what can we say about the regularity of $f$ if it exists? Thanks.
real-analysis functional-analysis ordinary-differential-equations integral-equations
real-analysis functional-analysis ordinary-differential-equations integral-equations
asked Dec 15 '18 at 14:52
GustaveGustave
734211
asked Dec 15 '18 at 14:52
GustaveGustave
734211
asked Dec 15 '18 at 14:52
GustaveGustave
734211
asked Dec 15 '18 at 14:52
GustaveGustave
734211
asked Dec 15 '18 at 14:52
GustaveGustave
734211
734211
1
$begingroup$
There is nothing we can say without knowing more about $g$, since any $f in L^2$ may be the solution of such an equation with a suitable $g$.
$endgroup$
– Hans Engler
Dec 15 '18 at 15:40
$begingroup$
Thank you sir. What if wa add $f$ to tge second member which will make the equation of second kind. Can we ensure that $f$ is $L^2$? thank you.
$endgroup$
– Gustave
Dec 16 '18 at 10:26
add a comment |
1
$begingroup$
There is nothing we can say without knowing more about $g$, since any $f in L^2$ may be the solution of such an equation with a suitable $g$.
$endgroup$
– Hans Engler
Dec 15 '18 at 15:40
$begingroup$
Thank you sir. What if wa add $f$ to tge second member which will make the equation of second kind. Can we ensure that $f$ is $L^2$? thank you.
$endgroup$
– Gustave
Dec 16 '18 at 10:26
1
1
$begingroup$
There is nothing we can say without knowing more about $g$, since any $f in L^2$ may be the solution of such an equation with a suitable $g$.
$endgroup$
– Hans Engler
Dec 15 '18 at 15:40
$begingroup$
There is nothing we can say without knowing more about $g$, since any $f in L^2$ may be the solution of such an equation with a suitable $g$.
$endgroup$
– Hans Engler
Dec 15 '18 at 15:40
$begingroup$
Thank you sir. What if wa add $f$ to tge second member which will make the equation of second kind. Can we ensure that $f$ is $L^2$? thank you.
$endgroup$
– Gustave
Dec 16 '18 at 10:26
$begingroup$
Thank you sir. What if wa add $f$ to tge second member which will make the equation of second kind. Can we ensure that $f$ is $L^2$? thank you.
$endgroup$
– Gustave
Dec 16 '18 at 10:26
add a comment |
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1
$begingroup$
There is nothing we can say without knowing more about $g$, since any $f in L^2$ may be the solution of such an equation with a suitable $g$.
$endgroup$
– Hans Engler
Dec 15 '18 at 15:40
$begingroup$
Thank you sir. What if wa add $f$ to tge second member which will make the equation of second kind. Can we ensure that $f$ is $L^2$? thank you.
$endgroup$
– Gustave
Dec 16 '18 at 10:26