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

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

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

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

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

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]

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

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

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?