Non trivial solution of Fredholm integral equation of second kind with constant kernel












1














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







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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
















1














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







share|cite|improve this question


















  • 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








1







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







share|cite|improve this question













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






share|cite|improve this question













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asked Jan 4 at 15:23









GustaveGustave

720211




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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














  • 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










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.






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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.






    share|cite|improve this answer


























      1














      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.






      share|cite|improve this answer
























        1












        1








        1






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 17:38









        Martin ArgeramiMartin Argerami

        124k1176175




        124k1176175






























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