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Homework Help: Green's Function

  1. May 17, 2005 #1

    a) Find two linearly independent solutions of [itex] t^2x''+tx' - x = 0[/itex]

    b) Calculate Green's Function for the equation [itex] t^2x''+tx' - x = 0[/itex], and use it to find a particular solution to the following inhomogeneous differential equation.

    [tex]t^2x''+tx'-x = t^4[/tex]

    c) Explain why the global existence and uniqueness theorem guarantees that, if [itex]f:(0,\infty)\rightarrow \mathbb{R}[/itex] is continuous, the the initial value problem

    [tex] t^2x''+tx' - x = f(t), \quad \quad x'(1) = x(1) = 0[/tex]

    has a unique solution on [itex](0,\infty)[/itex]. Find an example of a continuous function [itex]f:(0,\infty) \rightarrow \mathbb{R}[/itex] such that the solution of the above IVP satisfies [itex]|x(t)|\rightarrow \infty[/itex] as [itex]t\rightarrow 0+[/itex], so that the solution is not continuous on [itex][0,\infty)[/itex].
  2. jcsd
  3. May 17, 2005 #2


    Our differential equation is [itex]t^2x''+tx'-x=0[/itex] (1). A guess for a linearly independent solution to (1) is [itex]x(t) = t[/itex], so it makes sense that a second will be of the form [itex]x(t) = t - ty(t)[/itex].

    We can check this by taking [itex]x(t) = t[/itex] to be a solution to (1). Where [itex]x'(t) = 1[/itex], and [itex]x''(t) = 0[/itex].

    Substituting this into (1) we see that this is indeed a solution. Now we try [itex]x(t) = ty(t)[/itex], with [itex]x'(t) = y(t) + ty'(t)[/itex], and [itex]x''(t) = 2y'(t) + ty''(t)[/itex]. Note here that [itex]y(t)[/itex] is some function linearly independent to [itex]x(t)[/itex].

    Substituting this information into (1) gives

    [tex]t^2(2y'(t) + ty''(t)) + t(y(t)+ty'(t)) + ty(t) = 0[/tex]
    [tex]t^3y''(t) + 2t^2y'(t) + t^2y'(t) + ty(t) - ty(t) = 0[/tex]
    [tex]t^3y''(t) + 3t^2y'(t) = 0[/tex]

    Which implies that

    [tex]ty''(t) + 3y'(t) = 0[/tex]

    Now we write [itex]z = y'(t)[/itex], then [itex]tz' +3z = 0[/itex]. This is a separable differential equation and can be solved

    [tex]\int\frac{dz}{z} = -3\int\frac{1}{t}dt[/tex]
    [tex]\Rightarrow \ln|z| = -3\ln|t|+C[/tex]
    [tex]\Rightarrow z = \frac{c}{t^3}[/tex]
    [tex]\Rightarrow y = \int\frac{c}{y^3}dt[/tex]
    [tex]\Rightarrow y = -\frac{3}{2t^2} + C[/tex]

    Now take [itex]C = 0[/itex] and we have

    [tex]x_1(t) = -\frac{3}{2t^2}t = -\frac{3}{2t}[/tex]

    is another solution. If we let [itex]C = -3/2[/itex], then this can be written as [itex]x_1(t) = \frac{C}{t}[/itex]. And so the second linearly independent solution is

    [tex]x(t) = t + \frac{C}{t}[/tex]
  4. May 17, 2005 #3

    We can verify these two solutions are linearly independent by solving the corresponding Wronskian matrix

    [tex]W(x_1(t),x_2(t),t) = t\left(1-\frac{C}{t^2}\right) - \left(t+\frac{C}{t}\right) = -\frac{2C}{t} \neq 0[/tex]

    Now we have the 2 Linearly independent solutions needed to generate Green's Function. We define Green's Function as

    [tex]G(s,t) = -\frac{t}{2}\det \left( \begin{array}{cc}
    t & t+\frac{1}{t} \\
    s & s + \frac{1}{s}
    \end{array} \right)[/tex]

    [tex] = -\frac{t}{2}\left(t\left(s+\frac{1}{s}\right)-s\left(t+\frac{1}{t}\right)\right)[/tex]

    [tex]-\frac{t}{2}\left(ts+\frac{t}{s}-st -\frac{s}{t}\right)[/tex]

    [tex]-\frac{t^2}{2} + s[/tex]

    The solution should be

    [tex]x(t) = \int_0^s G(s,t)t^4dt[/tex]

    [tex]= \int_0^s -\frac{t^6}{2}ds + \int_0^s st^4ds [/tex]

    [tex]= \left.-\frac{st^6}{2}\right|_0^s + \left.\frac{s^2t^4}{2}\right|_0^2 [/tex]

    Ok, as you can probably guess, I am way off track!!! I stopped here because I was leading nowhere.

    Can anyone check all my working and find what I am doing wrong and how to proceed.

  5. May 17, 2005 #4


    User Avatar
    Homework Helper

    Why don't you just use t and 1/t as your solutions? if t is a solution and f(t) is a solution, then f(t)-t is also a solution, so you might as well subtract off t to make your life easier.
  6. May 17, 2005 #5
    [tex]W(t,1/t) = \det\left(
    t & 1/t \\
    s & 1/s

    [tex]= \frac{t}{s} - \frac{s}{t}[/tex]

    And so

    [tex]G(s,t) = \int_0^s G(s,t)t^4dt[/tex]

    [tex]= \int_0^s\left(\frac{t}{s} - \frac{s}{t}\right)t^4dt[/tex]

    [tex]= \int_0^s\frac{t^5}{s}dt - \int_0^s st^3dt[/tex]

    [tex]= \left.\frac{t^6}{6s}\right|_0^s - \left.\frac{st^4}{4}\right|_0^s[/tex]

    [tex]= \frac{s^5}{6} - \frac{s^5}{4}[/tex]

    [tex] = -\frac{s^5}{12}[/tex]
  7. May 18, 2005 #6

    Both solutions [itex]x_1(t) = t[/itex] and [itex]x_2(t) = 1/t[/itex] have continuous first-order partial derivatives with respect to [itex]x[/itex] on an open interval containing [itex](t_0,x_0)[/itex].
  8. May 18, 2005 #7


    User Avatar
    Homework Helper

    Don't you need boundary conditions to specify a Green's function?
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