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Second order DE, long problem

  1. May 25, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    1)Calculate the general solution of [itex]y''+\frac{y'x}{1+x}-\frac{y}{1+x}=1+x[/itex].
    2)What behavior do the solutions have in [itex]x=-1[/itex]?
    3)Solve the boundary problem to the DE in the interval [0,1] with [itex]y(0)=0[/itex] and [itex]y(1)+y'(1)=0[/itex].
    4)Write down the DE under Sturm-Liouville's form and find a Green function for the DE with the boundary conditions in 3) over the interval [0,1]. Then solve the DE using Green's function.
    Hint:Use solutions [itex]y_1[/itex] and [itex]y_2[/itex] such that [itex]y_1(0)=0 \neq y_2 (0)[/itex] and [itex]y_2 (1)+y _2 '(1)=0 \neq y_1 (1)+y_1 ' (1)[/itex].

    2. Relevant equations

    For part 1, not sure.

    3. The attempt at a solution
    I tried to apply Frobenius method because of the singularity of the DE in [itex]x=-1[/itex].
    I reached [itex]\sum _{n=0}^ \infty (n+c)(n+c-1)a_n(x-1)^{n+c+2}+ \frac{x}{1+x}\sum _{n=0}^ \infty (n+c)a_n (x-1)^{n+c-1}- \frac{1}{1+x} \sum _{n=0}^ \infty a_n (x-1)^{n+c}=1+x[/itex].
    Noting really simplifies well, if I continue further I think I'll have problems in finding a recurrence relation that could be useful. I wonder if I'm not doing things wrong so far and if I should continue this way.

    Edit: this is not going well at all, my right hand side is different from 0.
     
    Last edited: May 25, 2012
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  3. May 25, 2012 #2

    sharks

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    For part (1), i think the forcing function or free term should be zero.

    The first step would be to write up the auxiliary equation and solve.

    I get: [tex]m=-1 \; or\; m=\frac{1}{x+1}[/tex]
     
    Last edited: May 25, 2012
  4. May 25, 2012 #3

    fluidistic

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    Thanks for helping me. I never met such terms before. If google helped me well, you mean that there's a function of only the variable x that should be 0?
    What I reached is [itex](x-1)^{c-2}a_0c(c-1)+(x-1)^{c-1}a_1 (c-1)c +\frac{x}{1+x} (x-1)^{c-1} c a_0 + \sum _{n=0}^{\infty } (x-1)^{n+c} [a_{n+2}(n+c+2)(n+1+c)+\frac{x}{1+x} (n+c+1)a_{n+1}- \frac{a_n}{x+1}] = 1+x[/itex]. I don't know where is the free term nor why a term should be 0.
     
  5. Jun 9, 2012 #4

    fluidistic

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    I haven't made much progress on this DE. I tried several ways, even with a friend. He told me apparently that the solution can only be kept as an integral though the integral is so horrible that even the basic special functions wouldn't help.
    On my own I've attempted to solve the homogeneous DE [itex]y''+\frac{y'x}{1+x}-\frac{y}{1+x}=0[/itex] via infinite series expansions.
    I reached that [itex]y_1 (z)= \sum _{n=0} ^{\infty } \frac{(-1)^nz^n}{n!}[/itex] and [itex]y_2 (z)= \sum _{n=0} ^{\infty } \frac{(-1)^n z^{n+2}}{(n+1)!}[/itex]. I do not recognize any function I know in these 2 linearly independent solutions. By the way here [itex]z=x+1[/itex].
    Now in order to find the particular solution to the general DE [itex]y''+\frac{y'x}{1+x}-\frac{y}{1+x}=1+x[/itex] I must calculate the wronskian of [itex]y_1[/itex] and [itex]y_2[/itex], which seems way too aweful to even be attempted.
    However according to wolfram alpha, the general solution is of the form [itex]y(x)=Ae^{-x}+Bx+Cx^2+D[/itex].
    Hmm now that I think about it, my solution [itex]y_1(z)=e^{-z}[/itex]. But I'm still unsure about my [itex]y_2(z)[/itex]. Any help is immensely appreciated on how to proceed further.
     
  6. Jun 10, 2012 #5

    sharks

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    For part (1): The general solution of the reduced equation is: [tex]y_c = Ae^{x/(1+x)}+Be^{-x}[/tex]
    The Wronskian is: [tex]\frac{x^2+2x+2}{(1+x)^2}[/tex]
     
  7. Jun 10, 2012 #6

    fluidistic

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    Ok thanks a lot, it means I've got one solution to the homogeneous problem right; the other wrong. I've spotted one error but even when corrected, the other solution to the DE doesn't match [itex]Ae^{x/(1+x)}[/itex].
    Thus I fear I have to show every single steps I've made.
    Let [itex]z=x+1[/itex] so that the orignal DE becomes [itex]y''+ \frac{(z-1)y'}{z}-\frac{y}{z}=0[/itex]. I propose a solution of the form [itex]y(z)= \sum _{n=0}^{\infty } a_n z^{n+c} \Rightarrow y'(z)=y''(z)=\sum _{n=0} ^{\infty } (n+c) a_n z^{n+c-1} \Rightarrow \sum _{n=0} ^{\infty } (n+c) (n+c-1) a_n z^{n+c-2}[/itex].
    Now I replace those series into the DE:
    [itex]\sum _{n=0} ^{\infty } (n+c) (n+c-1) a_n z^{n+c-2}+\frac{(z-1)}{z}\sum _{n=0} ^{\infty } (n+c) a_n z^{n+c-1}- \frac{1}{z} \sum _{n=0}^{\infty } a_n z^{n+c}=0[/itex].
    Developping a bit:
    [itex]\sum _{n=0} ^{\infty } (n+c) (n+c-1) a_n z^{n+c-2}+\sum _{n=0}^{\infty } (n+c) a_n z^{n+c-1} - \sum _{n=0}^{\infty } (n+c)a_nz^{n+c-2}- \sum _{n=0} ^{\infty } a_n z^{n+c-1}=0[/itex].
    Now I set every series in powers of [itex]z^{n+c-2}[/itex].
    [itex]\sum _{n=0} ^{\infty } z^{n+c-2} a_n [(n+c)(n+c-1)-(n+c)] + \sum _{n=1}^{\infty } z^{n+c-2} (n+c-1)a_{n-1}-\sum _{n=1}^{\infty } z^{n+c-2}a_{n-1}=0[/itex]
    So [itex]\sum _{n=0}^{\infty } z^{n+c-2}a_n (n+c)(n+c-2)+ \sum {n=1} ^{\infty } z^{n+c-2} a_{n-1}(n+c-2)=0[/itex].
    The secular equation is thus [itex]z^{c-2} a_0 c(c-2)=0[/itex] with roots [itex]c=0[/itex] or [itex]c=2[/itex].
    Continuing the work on the infinite series, I get that [itex]\sum _{n=1}^{\infty } z^{n+c-2} [a_n (n+c)(n+c-2)+a_{n-1}(n+c-2)]=0[/itex].
    Hence the reccurrence relation I get is [itex]a_n (n+c)(n+c-2)+a_{n-1}(n+c-2)=0[/itex].
    First case: [itex]c=0[/itex]:
    [itex]a_n=\frac{-a_{n-1}}{n}[/itex].
    Choosing [itex]a_0=1[/itex], I get [itex]a_1=-1[/itex], [itex]a_2=\frac{1}{2}[/itex], [itex]a_3=-\frac{1}{2 \cdot 3}[/itex]. In general I notice the pattern [itex]a_n = \frac{(-1)^n}{n!}[/itex] so that [itex]y_1 (z)= \frac{(-1)^n z^n}{n!}=e^{-z}[/itex]. This is a good sign because you obtained the same solution.
    Second case: [itex]c=2[/itex]:
    [itex]a_n=- \frac{a_{n-1}}{n+2}[/itex].
    Choosing [itex]a_0=1[/itex], I get [itex]a_1=- \frac{1}{3}[/itex], [itex]a_2 = \frac{1}{3 \cdot 4 }[/itex], [itex]a_3=- \frac{1}{3 \cdot 4 \cdot 5 }[/itex].
    In general I notice the pattern [itex]a_n = \frac{(-1)^n \cdot 2 }{(n+2)!} \Rightarrow y_2 (z) = \sum _{n=0} ^{\infty }\frac { 2 \cdot (-1)^n z^{n+2}}{(n+1)!}[/itex].
    Hmm now that I think about it, I don't really see how it's possible to reach a solution of the form [itex]e^{-\frac{1}{z}}[/itex] (this is what I should reach according to your solution) because this would mean a different form than powers of [itex]z^{n+c}[/itex], in fact this would require infinitely many negative powers of z. I'm at a loss at understanding what's going on.
     
  8. Jun 13, 2012 #7

    fluidistic

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    I just read some theory and I've "learned" something.
    If the indicial equation leads to two roots which differ by an integer then I cannot assume that (assuming that [itex]c_1<c_2[/itex]) [itex]y_2(z)= \sum _{n=0}^{\infty } a_n z^{n+c_2}[/itex]. This explains why my solution for [itex]y_1(z)[/itex] is right but not the one for [itex]y_2(z)[/itex].
    In order to find [itex]y_2(z)[/itex], I used the method which starts by assuming that [itex]y_2(z)=v(z)y_1(z)[/itex].
    After a lot of algebra I reached that [itex]y_2(z)=c_1(z-1)+c_2e^{-z}[/itex].
    Which implies that the complentary solution to the homogeneous DE is [itex]y_c (z)=c_1e^{-z}+c_2(z-1)[/itex].
    From this fact, I had some fun and checked out the validity of Abel's theorem.
    It only remains to find the particular solution to the DE [itex]y''+ \left ( \frac{z-1}{z} \right ) y' - \frac{y}{z}=z[/itex].
    I used variation of parameters but still haven't found out the solution (I think I should reach [itex]y_p(z)=z^2-2z+2[/itex] which is equivalent to [itex]x^2+1[/itex]). I reach an extremely complicated function given by wolfram; I'll recheck my algebra.
    I think your solution [itex]y_c = Ae^{x/(1+x)}+Be^{-x}[/itex] is equivalen to [itex]Ae^{-\frac{1}{z}}+Be^{-z}[/itex] which isn't equivalent to my solution. Is something I've said wrong?
     
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