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Solve differential equation using power series

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve ##y^{''}+zy=0## where ##y(0)=0## and ##y^{'}(0)=1##


    2. Relevant equations

    ##y(z)=z^r\sum _{k=0}^{\infty } C_kz^k##

    3. The attempt at a solution

    Well firstly:

    ##r(r-1)+p_0r+q_0=0## where obviously ##p_0=q_0=0## so ##r_1=0## and ##r_2=1##.

    In general ##y(z)=\sum _{k=0}^{\infty } C_kz^{k+r}##, meaning ##y{'}(z)=\sum _{k=0}^{\infty } (k+r)C_kz^{k+r-1}## and ##y{''}(z)=\sum _{k=0}^{\infty } (k+r)(k+r-1)C_kz^{k+r-2}##. Substituting this into original DE:

    ##y^{''}+zy=0##

    ##\sum _{k=0}^{\infty } (k+r)(k+r-1)C_kz^{k+r-2}+z\sum _{k=0}^{\infty } C_kz^{k+r}=0##

    Now looking at the lowest exponents:

    ##z^{r-2}##: ##(r-1)rC_0=0##
    ##z^{r-1}##: ##(r+1)rC_1=0##
    ##z^{r}##: ##(r+1)(r+2)C_2=0##
    ##z^{r+1}##: ##(r+2)(r+3)C_3=0## and
    ##\sum _{k=4}^{\infty }[(k+r-1)(k+r)C_k+C_{k-4}]z^{k+r-2}## so ##(k+r-1)(k+r)C_k+C_{k-4}=0##

    For ##r_1=0##:
    ##C_0\neq 0## also ##C_1\neq 0## while ##C_2=C_3=0##.

    using ##(k+r-1)(k+r)c_k+C_{k-4}=0## we get ##C_k=\frac{-C_{k-4}}{(k+r-1)(k+r)}=\frac{-C_{k-4}}{(k-1)k}##

    Now after some time you find out that there will be two sums:

    a) ##k=4m##:

    ##C_k=\frac{-C_{k-4}}{(k-1)k}##

    ##C_{4m}=\frac{(-1)^mC_{0}}{(4m)!}##

    b) ##k=4m+1##:

    ##C_k=\frac{-C_{k-4}}{(k-1)k}##

    ##C_{4m+1}=\frac{(-1)^mC_{1}}{(4m+1)!}##

    Hopefully so far everything is ok?

    If yes,than one solution should be: ##y(z)=\sum _{m=0}^{\infty }\frac{(-1)^mC_{0}}{(4m)!}z^{4m}+\sum _{m=0}^{\infty }\frac{(-1)^mC_{1}}{(4m+1)!}z^{4m+1}##

    While for ##r_2=1##:

    Only ##C_0\neq 0## while ##C_1=C_2=C_3=0##.

    so ##C_k=C_{4m}=\frac{(-1)^mC_{0}}{(4m)!}##

    and finally ##y(z)=\sum _{m=0}^{\infty }\frac{(-1)^mC_{0}}{(4m)!}z^{4m+1}##


    Ok, hopefully so far I haven't done any bigger mistakes. Now my problems start. Which ##y(z)## is now the solution of my DE? Or is it the linear combination of both? o_O I don't understand that part since we never did that.
     
  2. jcsd
  3. Apr 29, 2014 #2

    LCKurtz

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    That is the relevant equation if ##z=0## is a regular singular point. But it isn't. The relevant equation is $$y=\sum _{k=0}^{\infty } c_kz^k$$Since y(0) = 0 and$$
    y = c_0 +\sum _{k=1}^{\infty } c_kz^k$$ you have ##c_0=0##. Also since ##y'(0)=1## and$$
    y'=c_1+\sum _{k=2}^{\infty } c_kkz^{k-1}$$you have ##c_1=1##. Now calculate ##y''## and plug the series into your equation. You should be able to determine all the ##c_k## and get your solution.
     
    Last edited: Apr 29, 2014
  4. Apr 29, 2014 #3

    SammyS

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    I've only looked this far.

    The zr+1 term also has a contribution from the second sum, (for k = 0 in that sum).

    So you have: ##(r+2)(r+3)C_3+C_0=0##
     
  5. Apr 29, 2014 #4
    That seems to be something I should have known... With that in mind, here is version 2.0:

    ##y=C_0+\sum _{k=1}^{\infty }C_kz^k## and using ##y(0)=0## also ##C_0=0##.

    ##y^{'}=C_1+\sum _{k=2}^{\infty }kC_kz^{k-1}## and using ##y^{'}(0)=1## also ##C_1=1##.

    ##y^{''}=\sum _{k=2}^{\infty }k(k-1)C_kz^{k-2}##

    Plugging that into DE:

    ##\sum _{k=2}^{\infty }k(k-1)C_kz^{k-2}+z[C_0+\sum _{k=1}^{\infty }C_kz^k]=0##

    ##\sum _{k=2}^{\infty }k(k-1)C_kz^{k-2}+\sum _{k=1}^{\infty }C_kz^{k+1}=0##

    Again for different exponents:

    ##z^0##: ##2C_2=0##
    ##z^1##: ##6c_3=0## and

    ##\sum _{k=4}^{\infty }k(k-1)C_kz^{k-2}+\sum _{k=4}^{\infty }C_{k-3}z^{k-3+1}=0##

    ##\sum _{k=4}^{\infty }z^{k-2}[k(k-1)C_k+C_{k-3}]=0##

    So right now we know that ##C_0=C_2=C_3=0## and ##C_1=1## and ##C_k=\frac{-C_{k-3}}{k(k-1)}##

    Obviously only ##C_k=C_{2m+1}\neq 0## and If I am not mistaken: ##C_k=\frac{(-1)^mC_1}{(2m+1)!}##, which leads me to

    ##y(z)=\sum _{m=0}^{\infty }\frac{(-1)^m}{(2m+1)!}z^{2m+1}##

    Right?
     
  6. Apr 29, 2014 #5

    LCKurtz

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    That's good to here. But the problem begins with the "obviously" below :frown:

    You are mixing m and k together. What you need to do is write out some of the ##c_k## for ##k=1,4,7...##. You will find that ##2m+1## doesn't generate those powers, and you will discover the denominator isn't quite a factorial.
     
  7. Apr 29, 2014 #6
    Aham.., hmm, let's take a look.

    ##C_1=1##

    ##C_4=\frac{-C_1}{4\cdot 4}=-\frac{1}{4\cdot 3}##

    ##C_7=\frac{-C_4}{7\cdot 6}=\frac{1}{7\cdot 6 \cdot 4 \cdot 3}## and one more

    ##C_{10}=\frac{-C_7}{10\cdot 9}=-\frac{1}{10\cdot 9 \cdot 7\cdot 6 \cdot 4 \cdot 3}##

    Ok, ##3m+1## should generate those powers... ##(3m+1)=1,4,7,10,13,...##.

    Now... Denominator is almost factorial - only ##(2+3n)## numbers are missing, where ##n=0,1,2,3,...##. So If I could somehow include that sequence in numerator, that would be perfect.

    I'm lost here. Anyhow, wolfram alpha shoots out that ##3,4,6,7,9,10,..=\frac{1}{4}(6n-(-1)^n+5)##

    I am not really sure if those sequences are any help at all here o_O
     
  8. Apr 29, 2014 #7

    LCKurtz

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    Usually you just write out the product instead of some obscure formula like Wolfram gave. So, for example, your term with ##z^{3m+1}## would be written$$
    \frac {(-1)^m z^{3m+1}}{(3m+1)(3m)(3m-2)(3m-3)...(4)(3)}$$How many factors there are in the denominator depends on how large ##m## is. Notice that when ##m=0## which would give the ##z^1## term, the coefficient formula doesn't work. So when you write your final answer you would write ##y = z + \text{sum from 1 up}##, keeping the ##m=0## case separate.
     
    Last edited: Apr 29, 2014
  9. Apr 29, 2014 #8
    Aaaa, ok, I get it!

    Solution is than ##y(z)=z+\sum _{m=1}^{\infty }\frac {(-1)^m z^{3m+1}}{(3m+1)(3m)(3m-2)(3m-3)...(4)(3)}##


    Just to clear things out: Taking ##y(z)=z^r\sum _{k=0}^{\infty } C_kz^k## in this case where ##z=0## is not a regular singular point is WRONG or does it just complicate things? I am asking because in my original post I did find out that ##r_1=0## or ##r_2=1##. I can see no reason why would taking ##r=0## bring me to any different result than we just got.

    THANK YOU!
     
  10. Apr 29, 2014 #9

    LCKurtz

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    I suspect the answer is that it just complicates things. I didn't try working it all the way through the way you started, but since r=0 should give the answer, I'm guessing that the ##r=1## case probably gives the same answer. The reason I say that is that since ##c_0=0##, then ##x## is a factor of the solution and that's why ##r=1## works. But if ##c_0\ne 0##, I'm guessing ##r=1## wouldn't work. I haven't checked that all out though, so don't take this answer too seriously.
     
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