Non trivial solution of Fredholm integral equation of second kind with constant kernel
Let us consider the following integral equation$$f(x) + lambda int_0^1 {K(s,x)f(s)ds = 0,{text{ x}} in {text{(0}}{text{,1)}}{text{.}}} $$
I'm looking of the values of $lambda$ so that the above equation has only $f=0$ as solution with a constant kernel.
Suppose that $K(s,x)=K$, we obtain
$$f(x) + lambda Kint_0^1 {f(s)ds = 0,{text{ x}} in {text{(0}}{text{,1)}}{text{.}}} $$
By taking the integral over $(0,1)$, we get $$(1 + lambda K)int_0^1 {f(s)ds = 0} $$. for all $f$. Now, if $lambda$ is different of $-1/K$, then $$int_0^1 {f(s)ds = 0} $$. I don't see how this can be helpful.
Any suggestions?. Thank you.
real-analysis integration functional-analysis differential-equations integral-equations
add a comment |
Let us consider the following integral equation$$f(x) + lambda int_0^1 {K(s,x)f(s)ds = 0,{text{ x}} in {text{(0}}{text{,1)}}{text{.}}} $$
I'm looking of the values of $lambda$ so that the above equation has only $f=0$ as solution with a constant kernel.
Suppose that $K(s,x)=K$, we obtain
$$f(x) + lambda Kint_0^1 {f(s)ds = 0,{text{ x}} in {text{(0}}{text{,1)}}{text{.}}} $$
By taking the integral over $(0,1)$, we get $$(1 + lambda K)int_0^1 {f(s)ds = 0} $$. for all $f$. Now, if $lambda$ is different of $-1/K$, then $$int_0^1 {f(s)ds = 0} $$. I don't see how this can be helpful.
Any suggestions?. Thank you.
real-analysis integration functional-analysis differential-equations integral-equations
1
The function has mean value 0 in integral sense over interval 0 to 1. It removes one degree of freedom. This means you have infinite set of solutions. Any function fulfilling the mean value equation $=0$ will do.
– mathreadler
Jan 4 at 15:48
So if $lambda= - 1/K$ we have only one solution?
– Gustave
Jan 4 at 15:54
If the other factor is $0$ then it does not matter what $f$ is, since the product will always be $0$ so then all functions $f$ will satisfy it.
– mathreadler
Jan 4 at 16:01
Thanks. I understand, but what I can say about the uniqueness of the trivial solution with respect to $lambda$?
– Gustave
Jan 4 at 16:09
add a comment |
Let us consider the following integral equation$$f(x) + lambda int_0^1 {K(s,x)f(s)ds = 0,{text{ x}} in {text{(0}}{text{,1)}}{text{.}}} $$
I'm looking of the values of $lambda$ so that the above equation has only $f=0$ as solution with a constant kernel.
Suppose that $K(s,x)=K$, we obtain
$$f(x) + lambda Kint_0^1 {f(s)ds = 0,{text{ x}} in {text{(0}}{text{,1)}}{text{.}}} $$
By taking the integral over $(0,1)$, we get $$(1 + lambda K)int_0^1 {f(s)ds = 0} $$. for all $f$. Now, if $lambda$ is different of $-1/K$, then $$int_0^1 {f(s)ds = 0} $$. I don't see how this can be helpful.
Any suggestions?. Thank you.
real-analysis integration functional-analysis differential-equations integral-equations
Let us consider the following integral equation$$f(x) + lambda int_0^1 {K(s,x)f(s)ds = 0,{text{ x}} in {text{(0}}{text{,1)}}{text{.}}} $$
I'm looking of the values of $lambda$ so that the above equation has only $f=0$ as solution with a constant kernel.
Suppose that $K(s,x)=K$, we obtain
$$f(x) + lambda Kint_0^1 {f(s)ds = 0,{text{ x}} in {text{(0}}{text{,1)}}{text{.}}} $$
By taking the integral over $(0,1)$, we get $$(1 + lambda K)int_0^1 {f(s)ds = 0} $$. for all $f$. Now, if $lambda$ is different of $-1/K$, then $$int_0^1 {f(s)ds = 0} $$. I don't see how this can be helpful.
Any suggestions?. Thank you.
real-analysis integration functional-analysis differential-equations integral-equations
real-analysis integration functional-analysis differential-equations integral-equations
asked Jan 4 at 15:23
GustaveGustave
720211
720211
1
The function has mean value 0 in integral sense over interval 0 to 1. It removes one degree of freedom. This means you have infinite set of solutions. Any function fulfilling the mean value equation $=0$ will do.
– mathreadler
Jan 4 at 15:48
So if $lambda= - 1/K$ we have only one solution?
– Gustave
Jan 4 at 15:54
If the other factor is $0$ then it does not matter what $f$ is, since the product will always be $0$ so then all functions $f$ will satisfy it.
– mathreadler
Jan 4 at 16:01
Thanks. I understand, but what I can say about the uniqueness of the trivial solution with respect to $lambda$?
– Gustave
Jan 4 at 16:09
add a comment |
1
The function has mean value 0 in integral sense over interval 0 to 1. It removes one degree of freedom. This means you have infinite set of solutions. Any function fulfilling the mean value equation $=0$ will do.
– mathreadler
Jan 4 at 15:48
So if $lambda= - 1/K$ we have only one solution?
– Gustave
Jan 4 at 15:54
If the other factor is $0$ then it does not matter what $f$ is, since the product will always be $0$ so then all functions $f$ will satisfy it.
– mathreadler
Jan 4 at 16:01
Thanks. I understand, but what I can say about the uniqueness of the trivial solution with respect to $lambda$?
– Gustave
Jan 4 at 16:09
1
1
The function has mean value 0 in integral sense over interval 0 to 1. It removes one degree of freedom. This means you have infinite set of solutions. Any function fulfilling the mean value equation $=0$ will do.
– mathreadler
Jan 4 at 15:48
The function has mean value 0 in integral sense over interval 0 to 1. It removes one degree of freedom. This means you have infinite set of solutions. Any function fulfilling the mean value equation $=0$ will do.
– mathreadler
Jan 4 at 15:48
So if $lambda= - 1/K$ we have only one solution?
– Gustave
Jan 4 at 15:54
So if $lambda= - 1/K$ we have only one solution?
– Gustave
Jan 4 at 15:54
If the other factor is $0$ then it does not matter what $f$ is, since the product will always be $0$ so then all functions $f$ will satisfy it.
– mathreadler
Jan 4 at 16:01
If the other factor is $0$ then it does not matter what $f$ is, since the product will always be $0$ so then all functions $f$ will satisfy it.
– mathreadler
Jan 4 at 16:01
Thanks. I understand, but what I can say about the uniqueness of the trivial solution with respect to $lambda$?
– Gustave
Jan 4 at 16:09
Thanks. I understand, but what I can say about the uniqueness of the trivial solution with respect to $lambda$?
– Gustave
Jan 4 at 16:09
add a comment |
1 Answer
1
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oldest
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Your step of taking the integral is too crude, at least initially. When you have
$$
f(x)+lambda Kint_0^1f(s),ds=0,
$$
you can write this as
$$
f(x)=-lambda Kint_0^1f(s),ds
$$
to conclude that $f$ is constant. If $lambda=0$, you get $f=0$. If $lambdane0$ and $lambdane-1/K$, your trick of integrating again gives you that $int_0^1 f=0$, so $f=0$.
When $lambda=-1/K$ the solution is not unique, as any constant $f$ will be a solution.
add a comment |
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1 Answer
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Your step of taking the integral is too crude, at least initially. When you have
$$
f(x)+lambda Kint_0^1f(s),ds=0,
$$
you can write this as
$$
f(x)=-lambda Kint_0^1f(s),ds
$$
to conclude that $f$ is constant. If $lambda=0$, you get $f=0$. If $lambdane0$ and $lambdane-1/K$, your trick of integrating again gives you that $int_0^1 f=0$, so $f=0$.
When $lambda=-1/K$ the solution is not unique, as any constant $f$ will be a solution.
add a comment |
Your step of taking the integral is too crude, at least initially. When you have
$$
f(x)+lambda Kint_0^1f(s),ds=0,
$$
you can write this as
$$
f(x)=-lambda Kint_0^1f(s),ds
$$
to conclude that $f$ is constant. If $lambda=0$, you get $f=0$. If $lambdane0$ and $lambdane-1/K$, your trick of integrating again gives you that $int_0^1 f=0$, so $f=0$.
When $lambda=-1/K$ the solution is not unique, as any constant $f$ will be a solution.
add a comment |
Your step of taking the integral is too crude, at least initially. When you have
$$
f(x)+lambda Kint_0^1f(s),ds=0,
$$
you can write this as
$$
f(x)=-lambda Kint_0^1f(s),ds
$$
to conclude that $f$ is constant. If $lambda=0$, you get $f=0$. If $lambdane0$ and $lambdane-1/K$, your trick of integrating again gives you that $int_0^1 f=0$, so $f=0$.
When $lambda=-1/K$ the solution is not unique, as any constant $f$ will be a solution.
Your step of taking the integral is too crude, at least initially. When you have
$$
f(x)+lambda Kint_0^1f(s),ds=0,
$$
you can write this as
$$
f(x)=-lambda Kint_0^1f(s),ds
$$
to conclude that $f$ is constant. If $lambda=0$, you get $f=0$. If $lambdane0$ and $lambdane-1/K$, your trick of integrating again gives you that $int_0^1 f=0$, so $f=0$.
When $lambda=-1/K$ the solution is not unique, as any constant $f$ will be a solution.
answered Jan 4 at 17:38
Martin ArgeramiMartin Argerami
124k1176175
124k1176175
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1
The function has mean value 0 in integral sense over interval 0 to 1. It removes one degree of freedom. This means you have infinite set of solutions. Any function fulfilling the mean value equation $=0$ will do.
– mathreadler
Jan 4 at 15:48
So if $lambda= - 1/K$ we have only one solution?
– Gustave
Jan 4 at 15:54
If the other factor is $0$ then it does not matter what $f$ is, since the product will always be $0$ so then all functions $f$ will satisfy it.
– mathreadler
Jan 4 at 16:01
Thanks. I understand, but what I can say about the uniqueness of the trivial solution with respect to $lambda$?
– Gustave
Jan 4 at 16:09