Homogeneous Fredholm Integral Equation of Second Kind with Boundary Conditions
I have come across a claim that I am quite certain is true, but would be benefited from seeing a proof sketch.
Suppose we know the following equation:
$$
f(z)=int_0^T K(z,x)f(x),mathrm{d}x,
$$
which is a standard homogeneous Fredholm equation of the second kind. Suppose that $K>0$ on $[0,T]times [0,T]$ except $K(T,x)=0$ for all $x in [0,T].$ Therefore, any solution satisfies $f(T)=0.$
Assume $f(z)$ is a solution. Intuitively, it seems like if $f(0)=0,$ then $f(z)equiv 0.$ I say this because we have
begin{align*}
0&=int_0^T K(0,x)f(x),mathrm{d}x \
&=int_0^Tint_0^TK(0,x)K(x,x_1)f(x_1),mathrm{d}x_1mathrm{d}x \
&=int_0^Tint_0^Tint_0^T K(0,x)K(x,x_1)K(x_1,x_2)f(x_2),mathrm{d}x_2mathrm{d}x_1mathrm{d}x \
&vdots \
&=int_0^T...int_0^T K(0,x)...K(x_{n-1},x_n)f(x_n), mathrm{d}x_n ...mathrm{d}x.
end{align*}
In all of the integrals, $K>0$ a.e. It seems like by repeating this process, we must have $fequiv 0.$ Note that $K(z,x)neq K(x,z)$ and $K$ is not separable. Thank you.
functional-analysis reproducing-kernel-hilbert-space
asked Jan 4 at 13:46
AD500712838AD500712838
263
add a comment |
I have come across a claim that I am quite certain is true, but would be benefited from seeing a proof sketch.
Suppose we know the following equation:
$$
f(z)=int_0^T K(z,x)f(x),mathrm{d}x,
$$
which is a standard homogeneous Fredholm equation of the second kind. Suppose that $K>0$ on $[0,T]times [0,T]$ except $K(T,x)=0$ for all $x in [0,T].$ Therefore, any solution satisfies $f(T)=0.$
Assume $f(z)$ is a solution. Intuitively, it seems like if $f(0)=0,$ then $f(z)equiv 0.$ I say this because we have
begin{align*}
0&=int_0^T K(0,x)f(x),mathrm{d}x \
&=int_0^Tint_0^TK(0,x)K(x,x_1)f(x_1),mathrm{d}x_1mathrm{d}x \
&=int_0^Tint_0^Tint_0^T K(0,x)K(x,x_1)K(x_1,x_2)f(x_2),mathrm{d}x_2mathrm{d}x_1mathrm{d}x \
&vdots \
&=int_0^T...int_0^T K(0,x)...K(x_{n-1},x_n)f(x_n), mathrm{d}x_n ...mathrm{d}x.
end{align*}
In all of the integrals, $K>0$ a.e. It seems like by repeating this process, we must have $fequiv 0.$ Note that $K(z,x)neq K(x,z)$ and $K$ is not separable. Thank you.
functional-analysis reproducing-kernel-hilbert-space
asked Jan 4 at 13:46
AD500712838AD500712838
263
add a comment |
I have come across a claim that I am quite certain is true, but would be benefited from seeing a proof sketch.
Suppose we know the following equation:
$$
f(z)=int_0^T K(z,x)f(x),mathrm{d}x,
$$
which is a standard homogeneous Fredholm equation of the second kind. Suppose that $K>0$ on $[0,T]times [0,T]$ except $K(T,x)=0$ for all $x in [0,T].$ Therefore, any solution satisfies $f(T)=0.$
Assume $f(z)$ is a solution. Intuitively, it seems like if $f(0)=0,$ then $f(z)equiv 0.$ I say this because we have
begin{align*}
0&=int_0^T K(0,x)f(x),mathrm{d}x \
&=int_0^Tint_0^TK(0,x)K(x,x_1)f(x_1),mathrm{d}x_1mathrm{d}x \
&=int_0^Tint_0^Tint_0^T K(0,x)K(x,x_1)K(x_1,x_2)f(x_2),mathrm{d}x_2mathrm{d}x_1mathrm{d}x \
&vdots \
&=int_0^T...int_0^T K(0,x)...K(x_{n-1},x_n)f(x_n), mathrm{d}x_n ...mathrm{d}x.
end{align*}
In all of the integrals, $K>0$ a.e. It seems like by repeating this process, we must have $fequiv 0.$ Note that $K(z,x)neq K(x,z)$ and $K$ is not separable. Thank you.
functional-analysis reproducing-kernel-hilbert-space
asked Jan 4 at 13:46
AD500712838AD500712838
263
I have come across a claim that I am quite certain is true, but would be benefited from seeing a proof sketch.
Suppose we know the following equation:
$$
f(z)=int_0^T K(z,x)f(x),mathrm{d}x,
$$
which is a standard homogeneous Fredholm equation of the second kind. Suppose that $K>0$ on $[0,T]times [0,T]$ except $K(T,x)=0$ for all $x in [0,T].$ Therefore, any solution satisfies $f(T)=0.$
Assume $f(z)$ is a solution. Intuitively, it seems like if $f(0)=0,$ then $f(z)equiv 0.$ I say this because we have
begin{align*}
0&=int_0^T K(0,x)f(x),mathrm{d}x \
&=int_0^Tint_0^TK(0,x)K(x,x_1)f(x_1),mathrm{d}x_1mathrm{d}x \
&=int_0^Tint_0^Tint_0^T K(0,x)K(x,x_1)K(x_1,x_2)f(x_2),mathrm{d}x_2mathrm{d}x_1mathrm{d}x \
&vdots \
&=int_0^T...int_0^T K(0,x)...K(x_{n-1},x_n)f(x_n), mathrm{d}x_n ...mathrm{d}x.
end{align*}
In all of the integrals, $K>0$ a.e. It seems like by repeating this process, we must have $fequiv 0.$ Note that $K(z,x)neq K(x,z)$ and $K$ is not separable. Thank you.
functional-analysis reproducing-kernel-hilbert-space
functional-analysis reproducing-kernel-hilbert-space
asked Jan 4 at 13:46
AD500712838AD500712838
263
asked Jan 4 at 13:46
AD500712838AD500712838
263
asked Jan 4 at 13:46
AD500712838AD500712838
263
asked Jan 4 at 13:46
AD500712838AD500712838
263
asked Jan 4 at 13:46
AD500712838AD500712838
263
263
add a comment |
add a comment |
1 Answer
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$fequiv0$ is certainly a solution. The question is wether there are other nontrivial solurtions. Consider the case $K(z,x)=g(z),h(x)$ with $g(T)=0$, $gnotequiv0$. Then $g$ is a solution if $int_0^Th(x),g(x),dx=1$.
answered Jan 4 at 14:11
Julián AguirreJulián Aguirre
67.7k24094
Your counterexample is interesting. For what it's worth, I resolved my problem differently. Thank you.
– AD500712838
Jan 5 at 15:53
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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votes
$fequiv0$ is certainly a solution. The question is wether there are other nontrivial solurtions. Consider the case $K(z,x)=g(z),h(x)$ with $g(T)=0$, $gnotequiv0$. Then $g$ is a solution if $int_0^Th(x),g(x),dx=1$.
answered Jan 4 at 14:11
Julián AguirreJulián Aguirre
67.7k24094
Your counterexample is interesting. For what it's worth, I resolved my problem differently. Thank you.
– AD500712838
Jan 5 at 15:53
add a comment |
$fequiv0$ is certainly a solution. The question is wether there are other nontrivial solurtions. Consider the case $K(z,x)=g(z),h(x)$ with $g(T)=0$, $gnotequiv0$. Then $g$ is a solution if $int_0^Th(x),g(x),dx=1$.
answered Jan 4 at 14:11
Julián AguirreJulián Aguirre
67.7k24094
Your counterexample is interesting. For what it's worth, I resolved my problem differently. Thank you.
– AD500712838
Jan 5 at 15:53
add a comment |
$fequiv0$ is certainly a solution. The question is wether there are other nontrivial solurtions. Consider the case $K(z,x)=g(z),h(x)$ with $g(T)=0$, $gnotequiv0$. Then $g$ is a solution if $int_0^Th(x),g(x),dx=1$.
answered Jan 4 at 14:11
Julián AguirreJulián Aguirre
67.7k24094
$fequiv0$ is certainly a solution. The question is wether there are other nontrivial solurtions. Consider the case $K(z,x)=g(z),h(x)$ with $g(T)=0$, $gnotequiv0$. Then $g$ is a solution if $int_0^Th(x),g(x),dx=1$.
answered Jan 4 at 14:11
Julián AguirreJulián Aguirre
67.7k24094
answered Jan 4 at 14:11
Julián AguirreJulián Aguirre
67.7k24094
answered Jan 4 at 14:11
Julián AguirreJulián Aguirre
67.7k24094
answered Jan 4 at 14:11
Julián AguirreJulián Aguirre
67.7k24094
67.7k24094
Your counterexample is interesting. For what it's worth, I resolved my problem differently. Thank you.
– AD500712838
Jan 5 at 15:53
add a comment |
Your counterexample is interesting. For what it's worth, I resolved my problem differently. Thank you.
– AD500712838
Jan 5 at 15:53
Your counterexample is interesting. For what it's worth, I resolved my problem differently. Thank you.
– AD500712838
Jan 5 at 15:53
Your counterexample is interesting. For what it's worth, I resolved my problem differently. Thank you.
– AD500712838
Jan 5 at 15:53
add a comment |
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