A Functional Differential Equation: $f^prime(x) =frac{f(2x)}{2f(x)}$











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I was having a play with some trig. identities and noticed the following:



$$cos{x}=frac{sin{2x}}{2sin{x}}.tag{1}$$



Now, $cos{x} = frac{d}{dx}sin{x}$ so I made the following analogous differential equation:




$$f^prime(x) =frac{f(2x)}{2f(x)} tag{2}$$




I have not seen a differential equation which relates a function's derivative to a change in its argument, so I was wondering whether anyone knew what these were called?



Somewhat predictably, $f_1(x)=sin{x}$ is not the only solution, I found that $f_2(x)=Asin(omega x)$ where $aomega=1$ is also a solution. I then guessed another solution, $e^{lambda x}$, and found that $f_3(x)=e^{frac{1}{2}x}$ is also a solution.



My main questions are:




What are these type of equations called? and are there any other solutions to this one?




Thanks for reading.










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  • 2




    Interesting Question. I think a while. Here is the same question: math.stackexchange.com/questions/197569/…
    – Pocho la pantera
    Sep 9 '13 at 15:54






  • 1




    If you make a change of variable, you can get this as a "delay differential equation", where $f'(x)$ is related to $f(x)$ and $f(x-h)$ for some constant $h$. en.wikipedia.org/wiki/Delay_differential_equation
    – GEdgar
    Sep 9 '13 at 16:24










  • We had this same question two days ago: math.stackexchange.com/questions/2999456/… . It seems to be a duplicate of this question from 2013 which has surfaced $8$ minutes ago.
    – Christian Blatter
    Nov 17 at 19:26















up vote
12
down vote

favorite
2












I was having a play with some trig. identities and noticed the following:



$$cos{x}=frac{sin{2x}}{2sin{x}}.tag{1}$$



Now, $cos{x} = frac{d}{dx}sin{x}$ so I made the following analogous differential equation:




$$f^prime(x) =frac{f(2x)}{2f(x)} tag{2}$$




I have not seen a differential equation which relates a function's derivative to a change in its argument, so I was wondering whether anyone knew what these were called?



Somewhat predictably, $f_1(x)=sin{x}$ is not the only solution, I found that $f_2(x)=Asin(omega x)$ where $aomega=1$ is also a solution. I then guessed another solution, $e^{lambda x}$, and found that $f_3(x)=e^{frac{1}{2}x}$ is also a solution.



My main questions are:




What are these type of equations called? and are there any other solutions to this one?




Thanks for reading.










share|cite|improve this question




















  • 2




    Interesting Question. I think a while. Here is the same question: math.stackexchange.com/questions/197569/…
    – Pocho la pantera
    Sep 9 '13 at 15:54






  • 1




    If you make a change of variable, you can get this as a "delay differential equation", where $f'(x)$ is related to $f(x)$ and $f(x-h)$ for some constant $h$. en.wikipedia.org/wiki/Delay_differential_equation
    – GEdgar
    Sep 9 '13 at 16:24










  • We had this same question two days ago: math.stackexchange.com/questions/2999456/… . It seems to be a duplicate of this question from 2013 which has surfaced $8$ minutes ago.
    – Christian Blatter
    Nov 17 at 19:26













up vote
12
down vote

favorite
2









up vote
12
down vote

favorite
2






2





I was having a play with some trig. identities and noticed the following:



$$cos{x}=frac{sin{2x}}{2sin{x}}.tag{1}$$



Now, $cos{x} = frac{d}{dx}sin{x}$ so I made the following analogous differential equation:




$$f^prime(x) =frac{f(2x)}{2f(x)} tag{2}$$




I have not seen a differential equation which relates a function's derivative to a change in its argument, so I was wondering whether anyone knew what these were called?



Somewhat predictably, $f_1(x)=sin{x}$ is not the only solution, I found that $f_2(x)=Asin(omega x)$ where $aomega=1$ is also a solution. I then guessed another solution, $e^{lambda x}$, and found that $f_3(x)=e^{frac{1}{2}x}$ is also a solution.



My main questions are:




What are these type of equations called? and are there any other solutions to this one?




Thanks for reading.










share|cite|improve this question















I was having a play with some trig. identities and noticed the following:



$$cos{x}=frac{sin{2x}}{2sin{x}}.tag{1}$$



Now, $cos{x} = frac{d}{dx}sin{x}$ so I made the following analogous differential equation:




$$f^prime(x) =frac{f(2x)}{2f(x)} tag{2}$$




I have not seen a differential equation which relates a function's derivative to a change in its argument, so I was wondering whether anyone knew what these were called?



Somewhat predictably, $f_1(x)=sin{x}$ is not the only solution, I found that $f_2(x)=Asin(omega x)$ where $aomega=1$ is also a solution. I then guessed another solution, $e^{lambda x}$, and found that $f_3(x)=e^{frac{1}{2}x}$ is also a solution.



My main questions are:




What are these type of equations called? and are there any other solutions to this one?




Thanks for reading.







calculus differential-equations trigonometry functional-equations






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edited Nov 17 at 19:15









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asked Sep 9 '13 at 15:41









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  • 2




    Interesting Question. I think a while. Here is the same question: math.stackexchange.com/questions/197569/…
    – Pocho la pantera
    Sep 9 '13 at 15:54






  • 1




    If you make a change of variable, you can get this as a "delay differential equation", where $f'(x)$ is related to $f(x)$ and $f(x-h)$ for some constant $h$. en.wikipedia.org/wiki/Delay_differential_equation
    – GEdgar
    Sep 9 '13 at 16:24










  • We had this same question two days ago: math.stackexchange.com/questions/2999456/… . It seems to be a duplicate of this question from 2013 which has surfaced $8$ minutes ago.
    – Christian Blatter
    Nov 17 at 19:26














  • 2




    Interesting Question. I think a while. Here is the same question: math.stackexchange.com/questions/197569/…
    – Pocho la pantera
    Sep 9 '13 at 15:54






  • 1




    If you make a change of variable, you can get this as a "delay differential equation", where $f'(x)$ is related to $f(x)$ and $f(x-h)$ for some constant $h$. en.wikipedia.org/wiki/Delay_differential_equation
    – GEdgar
    Sep 9 '13 at 16:24










  • We had this same question two days ago: math.stackexchange.com/questions/2999456/… . It seems to be a duplicate of this question from 2013 which has surfaced $8$ minutes ago.
    – Christian Blatter
    Nov 17 at 19:26








2




2




Interesting Question. I think a while. Here is the same question: math.stackexchange.com/questions/197569/…
– Pocho la pantera
Sep 9 '13 at 15:54




Interesting Question. I think a while. Here is the same question: math.stackexchange.com/questions/197569/…
– Pocho la pantera
Sep 9 '13 at 15:54




1




1




If you make a change of variable, you can get this as a "delay differential equation", where $f'(x)$ is related to $f(x)$ and $f(x-h)$ for some constant $h$. en.wikipedia.org/wiki/Delay_differential_equation
– GEdgar
Sep 9 '13 at 16:24




If you make a change of variable, you can get this as a "delay differential equation", where $f'(x)$ is related to $f(x)$ and $f(x-h)$ for some constant $h$. en.wikipedia.org/wiki/Delay_differential_equation
– GEdgar
Sep 9 '13 at 16:24












We had this same question two days ago: math.stackexchange.com/questions/2999456/… . It seems to be a duplicate of this question from 2013 which has surfaced $8$ minutes ago.
– Christian Blatter
Nov 17 at 19:26




We had this same question two days ago: math.stackexchange.com/questions/2999456/… . It seems to be a duplicate of this question from 2013 which has surfaced $8$ minutes ago.
– Christian Blatter
Nov 17 at 19:26










3 Answers
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CLAIM: The only solutions $f(x)$ to your FDE that are analytic at $x=0$ with $f(0)ne 0$ are in the form $f(x)=ae^{x/2a}$ with $ainmathbb R$.




It is easy to demonstrate that functions of the described form satisfy your FDE, so all that remains is to establish that they are the only solutions.



Let's rewrite your functional equation as
$$2f(x)f'(x)=f(2x)$$
...ignoring the possible issues that this may cause when $f(x)=0$. Now let's assume that $f$ is analytic and that it has the Maclaurin series
$$f(x)=sum_{n=0}^infty frac{x^n a_n}{n!}$$
for some coefficients $a_n$. Then this functional equation is the same as
$$2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)=sum_{k=0}^infty frac{x^k 2^k a_{k}}{k!}$$
If we multiply together the series on the left, we get
$$begin{align}
2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)
&=sum_{m,n=0}^infty frac{x^{m+n}cdot 2 a_m a_{n+1}}{m!n!}\
&=sum_{k=0}^infty x^ksum_{m+n=k}frac{2a_m a_{n+1}}{m!n!}\
&=sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}\
end{align}$$

So our initial equality becomes
$$sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=sum_{k=0}^infty x^k frac{2^k a_{k}}{k!}$$
Now, by equating coefficients, we have that
$$sum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=frac{2^k a_{k}}{k!}$$
...for all $kin mathbb Z$. This can be interpreted as a recurrence relation for $a_n$, and it demonstrates that given $a_0=f(0)$, all other coefficients, and thus the function $f$, are determined.



Thus, since for any $a_0$, there exists $ainmathbb R$ such that $f(x)=ae^{x/2a}$ satisfies $f(0)=a_0$, we have that the value of $f(0)$ determines the function $f(x)$. More specifically,
$$f(x)=f(0)e^{x/2f(0)}$$



NOTE. Because I began this strategy by multiplying both sides of your FDE by $f(x)$, this recurrence messes up if $f(0)=0$. Indeed, if this is the case, then the recurrence shows that $a_n=0$ for all $n$, and $f(x)=0$, even though you have found a function $f_2(x)=sin(xomega)/omega$ that also satisfies $f_2(0)=0$ and is not constant. However, if $f(0)ne 0$, the recurrence should work out just fine.






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    • Observe that if $f_1$ is a solution to the functional differential equation (FDE)




      $$ f(2z) ~=~ 2f(z)f^{prime}(z)~equiv~frac{d[f(z)^2]}{dz}, tag{A} $$




      then $$z~mapsto~ f_{lambda}(z)~:=~frac{f_1(lambda z)}{lambda}, qquad lambda~in~mathbb{C}^{times}~equiv~mathbb{C}backslash{0}, tag{B}$$
      is also a solution.







    • Claim: The FDE (A) has the following solutions $f(z)=sum_{ninmathbb{N}_0}a_nz^n$ that are analytic in a neighborhood of $z=0$:




      1. "The exp solution": $f(z)~=~frac{exp(lambda z)}{2lambda}, lambda~in~mathbb{C}^{times}.$


      2. "The null solution": $f(z)~=~0.$


      3. "The sinh solution": $f(z)~=~frac{sinh(lambda z)}{lambda}, lambda~in~mathbb{C}^{times}.$


      4. "The identity solution": $f(z)~=~z.$ (Can be viewed as the $lambdato 0$ limit of the sinh solution.)








    • Recursion relation:
      $$ 2^{n-1} a_n~=~sum_{r=0}^n (r+1)a_{r+1}a_{n-r}, qquad ninmathbb{N}_0. tag{C} $$



    • Sketched proof of claim:




      1. Case $a_0neq 0$: This leads to the exp solution.


      2. Case $a_0=0=a_1$: This leads to the null solution.


      3. Case $a_0=0neq a_1$: This leads to the (extended) sinh solution. Here $a_3inmathbb{C}$ is a free parameter. $Box$




    • See also this related Math.SE post.







    share|cite|improve this answer






























      up vote
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      down vote













      To use "Picard's method" might work. You have
      $$
      y(x)=y_0+int_{x_0}^x frac{y(2t)}{y(t)} , dt
      $$



      Define
      $$
      y_{k+1}=y_0+int_{x_0}^x frac{y_k(2t)}{y_k(t)},dt
      $$
      and calculate $lim_{krightarrow infty} y_k$.
      I do not know if the sequence converges.






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
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        up vote
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        CLAIM: The only solutions $f(x)$ to your FDE that are analytic at $x=0$ with $f(0)ne 0$ are in the form $f(x)=ae^{x/2a}$ with $ainmathbb R$.




        It is easy to demonstrate that functions of the described form satisfy your FDE, so all that remains is to establish that they are the only solutions.



        Let's rewrite your functional equation as
        $$2f(x)f'(x)=f(2x)$$
        ...ignoring the possible issues that this may cause when $f(x)=0$. Now let's assume that $f$ is analytic and that it has the Maclaurin series
        $$f(x)=sum_{n=0}^infty frac{x^n a_n}{n!}$$
        for some coefficients $a_n$. Then this functional equation is the same as
        $$2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)=sum_{k=0}^infty frac{x^k 2^k a_{k}}{k!}$$
        If we multiply together the series on the left, we get
        $$begin{align}
        2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)
        &=sum_{m,n=0}^infty frac{x^{m+n}cdot 2 a_m a_{n+1}}{m!n!}\
        &=sum_{k=0}^infty x^ksum_{m+n=k}frac{2a_m a_{n+1}}{m!n!}\
        &=sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}\
        end{align}$$

        So our initial equality becomes
        $$sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=sum_{k=0}^infty x^k frac{2^k a_{k}}{k!}$$
        Now, by equating coefficients, we have that
        $$sum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=frac{2^k a_{k}}{k!}$$
        ...for all $kin mathbb Z$. This can be interpreted as a recurrence relation for $a_n$, and it demonstrates that given $a_0=f(0)$, all other coefficients, and thus the function $f$, are determined.



        Thus, since for any $a_0$, there exists $ainmathbb R$ such that $f(x)=ae^{x/2a}$ satisfies $f(0)=a_0$, we have that the value of $f(0)$ determines the function $f(x)$. More specifically,
        $$f(x)=f(0)e^{x/2f(0)}$$



        NOTE. Because I began this strategy by multiplying both sides of your FDE by $f(x)$, this recurrence messes up if $f(0)=0$. Indeed, if this is the case, then the recurrence shows that $a_n=0$ for all $n$, and $f(x)=0$, even though you have found a function $f_2(x)=sin(xomega)/omega$ that also satisfies $f_2(0)=0$ and is not constant. However, if $f(0)ne 0$, the recurrence should work out just fine.






        share|cite|improve this answer



























          up vote
          3
          down vote














          CLAIM: The only solutions $f(x)$ to your FDE that are analytic at $x=0$ with $f(0)ne 0$ are in the form $f(x)=ae^{x/2a}$ with $ainmathbb R$.




          It is easy to demonstrate that functions of the described form satisfy your FDE, so all that remains is to establish that they are the only solutions.



          Let's rewrite your functional equation as
          $$2f(x)f'(x)=f(2x)$$
          ...ignoring the possible issues that this may cause when $f(x)=0$. Now let's assume that $f$ is analytic and that it has the Maclaurin series
          $$f(x)=sum_{n=0}^infty frac{x^n a_n}{n!}$$
          for some coefficients $a_n$. Then this functional equation is the same as
          $$2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)=sum_{k=0}^infty frac{x^k 2^k a_{k}}{k!}$$
          If we multiply together the series on the left, we get
          $$begin{align}
          2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)
          &=sum_{m,n=0}^infty frac{x^{m+n}cdot 2 a_m a_{n+1}}{m!n!}\
          &=sum_{k=0}^infty x^ksum_{m+n=k}frac{2a_m a_{n+1}}{m!n!}\
          &=sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}\
          end{align}$$

          So our initial equality becomes
          $$sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=sum_{k=0}^infty x^k frac{2^k a_{k}}{k!}$$
          Now, by equating coefficients, we have that
          $$sum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=frac{2^k a_{k}}{k!}$$
          ...for all $kin mathbb Z$. This can be interpreted as a recurrence relation for $a_n$, and it demonstrates that given $a_0=f(0)$, all other coefficients, and thus the function $f$, are determined.



          Thus, since for any $a_0$, there exists $ainmathbb R$ such that $f(x)=ae^{x/2a}$ satisfies $f(0)=a_0$, we have that the value of $f(0)$ determines the function $f(x)$. More specifically,
          $$f(x)=f(0)e^{x/2f(0)}$$



          NOTE. Because I began this strategy by multiplying both sides of your FDE by $f(x)$, this recurrence messes up if $f(0)=0$. Indeed, if this is the case, then the recurrence shows that $a_n=0$ for all $n$, and $f(x)=0$, even though you have found a function $f_2(x)=sin(xomega)/omega$ that also satisfies $f_2(0)=0$ and is not constant. However, if $f(0)ne 0$, the recurrence should work out just fine.






          share|cite|improve this answer

























            up vote
            3
            down vote










            up vote
            3
            down vote










            CLAIM: The only solutions $f(x)$ to your FDE that are analytic at $x=0$ with $f(0)ne 0$ are in the form $f(x)=ae^{x/2a}$ with $ainmathbb R$.




            It is easy to demonstrate that functions of the described form satisfy your FDE, so all that remains is to establish that they are the only solutions.



            Let's rewrite your functional equation as
            $$2f(x)f'(x)=f(2x)$$
            ...ignoring the possible issues that this may cause when $f(x)=0$. Now let's assume that $f$ is analytic and that it has the Maclaurin series
            $$f(x)=sum_{n=0}^infty frac{x^n a_n}{n!}$$
            for some coefficients $a_n$. Then this functional equation is the same as
            $$2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)=sum_{k=0}^infty frac{x^k 2^k a_{k}}{k!}$$
            If we multiply together the series on the left, we get
            $$begin{align}
            2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)
            &=sum_{m,n=0}^infty frac{x^{m+n}cdot 2 a_m a_{n+1}}{m!n!}\
            &=sum_{k=0}^infty x^ksum_{m+n=k}frac{2a_m a_{n+1}}{m!n!}\
            &=sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}\
            end{align}$$

            So our initial equality becomes
            $$sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=sum_{k=0}^infty x^k frac{2^k a_{k}}{k!}$$
            Now, by equating coefficients, we have that
            $$sum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=frac{2^k a_{k}}{k!}$$
            ...for all $kin mathbb Z$. This can be interpreted as a recurrence relation for $a_n$, and it demonstrates that given $a_0=f(0)$, all other coefficients, and thus the function $f$, are determined.



            Thus, since for any $a_0$, there exists $ainmathbb R$ such that $f(x)=ae^{x/2a}$ satisfies $f(0)=a_0$, we have that the value of $f(0)$ determines the function $f(x)$. More specifically,
            $$f(x)=f(0)e^{x/2f(0)}$$



            NOTE. Because I began this strategy by multiplying both sides of your FDE by $f(x)$, this recurrence messes up if $f(0)=0$. Indeed, if this is the case, then the recurrence shows that $a_n=0$ for all $n$, and $f(x)=0$, even though you have found a function $f_2(x)=sin(xomega)/omega$ that also satisfies $f_2(0)=0$ and is not constant. However, if $f(0)ne 0$, the recurrence should work out just fine.






            share|cite|improve this answer















            CLAIM: The only solutions $f(x)$ to your FDE that are analytic at $x=0$ with $f(0)ne 0$ are in the form $f(x)=ae^{x/2a}$ with $ainmathbb R$.




            It is easy to demonstrate that functions of the described form satisfy your FDE, so all that remains is to establish that they are the only solutions.



            Let's rewrite your functional equation as
            $$2f(x)f'(x)=f(2x)$$
            ...ignoring the possible issues that this may cause when $f(x)=0$. Now let's assume that $f$ is analytic and that it has the Maclaurin series
            $$f(x)=sum_{n=0}^infty frac{x^n a_n}{n!}$$
            for some coefficients $a_n$. Then this functional equation is the same as
            $$2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)=sum_{k=0}^infty frac{x^k 2^k a_{k}}{k!}$$
            If we multiply together the series on the left, we get
            $$begin{align}
            2bigg(sum_{m=0}^infty frac{x^m a_m}{m!}bigg)bigg(sum_{n=0}^infty frac{x^n a_{n+1}}{n!}bigg)
            &=sum_{m,n=0}^infty frac{x^{m+n}cdot 2 a_m a_{n+1}}{m!n!}\
            &=sum_{k=0}^infty x^ksum_{m+n=k}frac{2a_m a_{n+1}}{m!n!}\
            &=sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}\
            end{align}$$

            So our initial equality becomes
            $$sum_{k=0}^infty x^ksum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=sum_{k=0}^infty x^k frac{2^k a_{k}}{k!}$$
            Now, by equating coefficients, we have that
            $$sum_{m=0}^k frac{2a_m a_{k-m+1}}{m!(k-m)!}=frac{2^k a_{k}}{k!}$$
            ...for all $kin mathbb Z$. This can be interpreted as a recurrence relation for $a_n$, and it demonstrates that given $a_0=f(0)$, all other coefficients, and thus the function $f$, are determined.



            Thus, since for any $a_0$, there exists $ainmathbb R$ such that $f(x)=ae^{x/2a}$ satisfies $f(0)=a_0$, we have that the value of $f(0)$ determines the function $f(x)$. More specifically,
            $$f(x)=f(0)e^{x/2f(0)}$$



            NOTE. Because I began this strategy by multiplying both sides of your FDE by $f(x)$, this recurrence messes up if $f(0)=0$. Indeed, if this is the case, then the recurrence shows that $a_n=0$ for all $n$, and $f(x)=0$, even though you have found a function $f_2(x)=sin(xomega)/omega$ that also satisfies $f_2(0)=0$ and is not constant. However, if $f(0)ne 0$, the recurrence should work out just fine.







            share|cite|improve this answer














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            edited Nov 17 at 20:05

























            answered Nov 17 at 19:32









            Frpzzd

            20.2k638104




            20.2k638104






















                up vote
                1
                down vote















                • Observe that if $f_1$ is a solution to the functional differential equation (FDE)




                  $$ f(2z) ~=~ 2f(z)f^{prime}(z)~equiv~frac{d[f(z)^2]}{dz}, tag{A} $$




                  then $$z~mapsto~ f_{lambda}(z)~:=~frac{f_1(lambda z)}{lambda}, qquad lambda~in~mathbb{C}^{times}~equiv~mathbb{C}backslash{0}, tag{B}$$
                  is also a solution.







                • Claim: The FDE (A) has the following solutions $f(z)=sum_{ninmathbb{N}_0}a_nz^n$ that are analytic in a neighborhood of $z=0$:




                  1. "The exp solution": $f(z)~=~frac{exp(lambda z)}{2lambda}, lambda~in~mathbb{C}^{times}.$


                  2. "The null solution": $f(z)~=~0.$


                  3. "The sinh solution": $f(z)~=~frac{sinh(lambda z)}{lambda}, lambda~in~mathbb{C}^{times}.$


                  4. "The identity solution": $f(z)~=~z.$ (Can be viewed as the $lambdato 0$ limit of the sinh solution.)








                • Recursion relation:
                  $$ 2^{n-1} a_n~=~sum_{r=0}^n (r+1)a_{r+1}a_{n-r}, qquad ninmathbb{N}_0. tag{C} $$



                • Sketched proof of claim:




                  1. Case $a_0neq 0$: This leads to the exp solution.


                  2. Case $a_0=0=a_1$: This leads to the null solution.


                  3. Case $a_0=0neq a_1$: This leads to the (extended) sinh solution. Here $a_3inmathbb{C}$ is a free parameter. $Box$




                • See also this related Math.SE post.







                share|cite|improve this answer



























                  up vote
                  1
                  down vote















                  • Observe that if $f_1$ is a solution to the functional differential equation (FDE)




                    $$ f(2z) ~=~ 2f(z)f^{prime}(z)~equiv~frac{d[f(z)^2]}{dz}, tag{A} $$




                    then $$z~mapsto~ f_{lambda}(z)~:=~frac{f_1(lambda z)}{lambda}, qquad lambda~in~mathbb{C}^{times}~equiv~mathbb{C}backslash{0}, tag{B}$$
                    is also a solution.







                  • Claim: The FDE (A) has the following solutions $f(z)=sum_{ninmathbb{N}_0}a_nz^n$ that are analytic in a neighborhood of $z=0$:




                    1. "The exp solution": $f(z)~=~frac{exp(lambda z)}{2lambda}, lambda~in~mathbb{C}^{times}.$


                    2. "The null solution": $f(z)~=~0.$


                    3. "The sinh solution": $f(z)~=~frac{sinh(lambda z)}{lambda}, lambda~in~mathbb{C}^{times}.$


                    4. "The identity solution": $f(z)~=~z.$ (Can be viewed as the $lambdato 0$ limit of the sinh solution.)








                  • Recursion relation:
                    $$ 2^{n-1} a_n~=~sum_{r=0}^n (r+1)a_{r+1}a_{n-r}, qquad ninmathbb{N}_0. tag{C} $$



                  • Sketched proof of claim:




                    1. Case $a_0neq 0$: This leads to the exp solution.


                    2. Case $a_0=0=a_1$: This leads to the null solution.


                    3. Case $a_0=0neq a_1$: This leads to the (extended) sinh solution. Here $a_3inmathbb{C}$ is a free parameter. $Box$




                  • See also this related Math.SE post.







                  share|cite|improve this answer

























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote











                    • Observe that if $f_1$ is a solution to the functional differential equation (FDE)




                      $$ f(2z) ~=~ 2f(z)f^{prime}(z)~equiv~frac{d[f(z)^2]}{dz}, tag{A} $$




                      then $$z~mapsto~ f_{lambda}(z)~:=~frac{f_1(lambda z)}{lambda}, qquad lambda~in~mathbb{C}^{times}~equiv~mathbb{C}backslash{0}, tag{B}$$
                      is also a solution.







                    • Claim: The FDE (A) has the following solutions $f(z)=sum_{ninmathbb{N}_0}a_nz^n$ that are analytic in a neighborhood of $z=0$:




                      1. "The exp solution": $f(z)~=~frac{exp(lambda z)}{2lambda}, lambda~in~mathbb{C}^{times}.$


                      2. "The null solution": $f(z)~=~0.$


                      3. "The sinh solution": $f(z)~=~frac{sinh(lambda z)}{lambda}, lambda~in~mathbb{C}^{times}.$


                      4. "The identity solution": $f(z)~=~z.$ (Can be viewed as the $lambdato 0$ limit of the sinh solution.)








                    • Recursion relation:
                      $$ 2^{n-1} a_n~=~sum_{r=0}^n (r+1)a_{r+1}a_{n-r}, qquad ninmathbb{N}_0. tag{C} $$



                    • Sketched proof of claim:




                      1. Case $a_0neq 0$: This leads to the exp solution.


                      2. Case $a_0=0=a_1$: This leads to the null solution.


                      3. Case $a_0=0neq a_1$: This leads to the (extended) sinh solution. Here $a_3inmathbb{C}$ is a free parameter. $Box$




                    • See also this related Math.SE post.







                    share|cite|improve this answer
















                    • Observe that if $f_1$ is a solution to the functional differential equation (FDE)




                      $$ f(2z) ~=~ 2f(z)f^{prime}(z)~equiv~frac{d[f(z)^2]}{dz}, tag{A} $$




                      then $$z~mapsto~ f_{lambda}(z)~:=~frac{f_1(lambda z)}{lambda}, qquad lambda~in~mathbb{C}^{times}~equiv~mathbb{C}backslash{0}, tag{B}$$
                      is also a solution.







                    • Claim: The FDE (A) has the following solutions $f(z)=sum_{ninmathbb{N}_0}a_nz^n$ that are analytic in a neighborhood of $z=0$:




                      1. "The exp solution": $f(z)~=~frac{exp(lambda z)}{2lambda}, lambda~in~mathbb{C}^{times}.$


                      2. "The null solution": $f(z)~=~0.$


                      3. "The sinh solution": $f(z)~=~frac{sinh(lambda z)}{lambda}, lambda~in~mathbb{C}^{times}.$


                      4. "The identity solution": $f(z)~=~z.$ (Can be viewed as the $lambdato 0$ limit of the sinh solution.)








                    • Recursion relation:
                      $$ 2^{n-1} a_n~=~sum_{r=0}^n (r+1)a_{r+1}a_{n-r}, qquad ninmathbb{N}_0. tag{C} $$



                    • Sketched proof of claim:




                      1. Case $a_0neq 0$: This leads to the exp solution.


                      2. Case $a_0=0=a_1$: This leads to the null solution.


                      3. Case $a_0=0neq a_1$: This leads to the (extended) sinh solution. Here $a_3inmathbb{C}$ is a free parameter. $Box$




                    • See also this related Math.SE post.








                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 18 at 15:12

























                    answered Nov 18 at 12:50









                    Qmechanic

                    4,72811851




                    4,72811851






















                        up vote
                        -1
                        down vote













                        To use "Picard's method" might work. You have
                        $$
                        y(x)=y_0+int_{x_0}^x frac{y(2t)}{y(t)} , dt
                        $$



                        Define
                        $$
                        y_{k+1}=y_0+int_{x_0}^x frac{y_k(2t)}{y_k(t)},dt
                        $$
                        and calculate $lim_{krightarrow infty} y_k$.
                        I do not know if the sequence converges.






                        share|cite|improve this answer

























                          up vote
                          -1
                          down vote













                          To use "Picard's method" might work. You have
                          $$
                          y(x)=y_0+int_{x_0}^x frac{y(2t)}{y(t)} , dt
                          $$



                          Define
                          $$
                          y_{k+1}=y_0+int_{x_0}^x frac{y_k(2t)}{y_k(t)},dt
                          $$
                          and calculate $lim_{krightarrow infty} y_k$.
                          I do not know if the sequence converges.






                          share|cite|improve this answer























                            up vote
                            -1
                            down vote










                            up vote
                            -1
                            down vote









                            To use "Picard's method" might work. You have
                            $$
                            y(x)=y_0+int_{x_0}^x frac{y(2t)}{y(t)} , dt
                            $$



                            Define
                            $$
                            y_{k+1}=y_0+int_{x_0}^x frac{y_k(2t)}{y_k(t)},dt
                            $$
                            and calculate $lim_{krightarrow infty} y_k$.
                            I do not know if the sequence converges.






                            share|cite|improve this answer












                            To use "Picard's method" might work. You have
                            $$
                            y(x)=y_0+int_{x_0}^x frac{y(2t)}{y(t)} , dt
                            $$



                            Define
                            $$
                            y_{k+1}=y_0+int_{x_0}^x frac{y_k(2t)}{y_k(t)},dt
                            $$
                            and calculate $lim_{krightarrow infty} y_k$.
                            I do not know if the sequence converges.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 9 '13 at 16:10









                            Pocho la pantera

                            2,218617




                            2,218617






























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