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 18 hours ago
AD500712838
262
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 18 hours ago
AD500712838
262
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 18 hours ago
AD500712838
262
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 18 hours ago
AD500712838
262
asked 18 hours ago
AD500712838
262
asked 18 hours ago
AD500712838
262
asked 18 hours ago
AD500712838
262
asked 18 hours ago
AD500712838
262
262
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 17 hours ago
Julián Aguirre
67.6k24094
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
active
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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 17 hours ago
Julián Aguirre
67.6k24094
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 17 hours ago
Julián Aguirre
67.6k24094
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 17 hours ago
Julián Aguirre
67.6k24094
$fequiv0$ 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 17 hours ago
Julián Aguirre
67.6k24094
answered 17 hours ago
Julián Aguirre
67.6k24094
answered 17 hours ago
Julián Aguirre
67.6k24094
answered 17 hours ago
Julián Aguirre
67.6k24094
67.6k24094
add a comment |
add a comment |
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