Solving Green's Function for t^2x''+tx' - x = 0

[Oxymoron]

Question

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

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

t^2x''+tx'-x = t^4

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

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

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

 

[Oxymoron]

Solutions

a)

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

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

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

Substituting this information into (1) gives

t^2(2y'(t) + ty''(t)) + t(y(t)+ty'(t)) + ty(t) = 0
t^3y''(t) + 2t^2y'(t) + t^2y'(t) + ty(t) - ty(t) = 0
t^3y''(t) + 3t^2y'(t) = 0

Which implies that

ty''(t) + 3y'(t) = 0

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

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

Now take C = 0 and we have

x_1(t) = -\frac{3}{2t^2}t = -\frac{3}{2t}

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

x(t) = t + \frac{C}{t}

 

[Oxymoron]

b)

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

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

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

G(s,t) = -\frac{t}{2}\det \left( \begin{array}{cc}<br /> t &amp; t+\frac{1}{t} \\<br /> s &amp; s + \frac{1}{s}<br /> \end{array} \right)

= -\frac{t}{2}\left(t\left(s+\frac{1}{s}\right)-s\left(t+\frac{1}{t}\right)\right)

-\frac{t}{2}\left(ts+\frac{t}{s}-st -\frac{s}{t}\right)

-\frac{t^2}{2} + s

The solution should be

x(t) = \int_0^s G(s,t)t^4dt

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

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

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.

Cheers.

 

[StatusX]

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.

 

[Oxymoron]

W(t,1/t) = \det\left(<br /> \begin{array}{cc}<br /> t &amp; 1/t \\<br /> s &amp; 1/s<br /> \end{array}\right)

= \frac{t}{s} - \frac{s}{t}

And so

G(s,t) = \int_0^s G(s,t)t^4dt

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

= \int_0^s\frac{t^5}{s}dt - \int_0^s st^3dt

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

= \frac{s^5}{6} - \frac{s^5}{4}

= -\frac{s^5}{12}

 

[Oxymoron]

c)

Both solutions x_1(t) = t and x_2(t) = 1/t have continuous first-order partial derivatives with respect to x on an open interval containing (t_0,x_0).

 

[StatusX]

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