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

• Oxymoron
In summary: Also, don't you need the function F(t) to solve for x(t)?In summary, we found two linearly independent solutions for the given differential equation and used them to calculate Green's function. However, further analysis is needed to fully solve the inhomogeneous differential equation and guarantee the existence of a unique solution. Additionally, boundary conditions and a function F(t) are necessary to fully solve the problem.
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 $[0,\infty)$.

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

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

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

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

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

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

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

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.

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.

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

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

And so

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

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

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

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

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

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

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 $(t_0,x_0)$.

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

## 1. What is a Green's function?

A Green's function is a mathematical tool used to solve differential equations. It represents the response of a system to a delta function input, and can be used to find the solution to a wide variety of differential equations.

## 2. How is the Green's function used to solve differential equations?

The Green's function is used by convolving it with the input function in order to obtain the solution to the differential equation. This is known as the Green's function method and is commonly used in physics and engineering.

## 3. What is the significance of t^2x''+tx' - x = 0 in the Green's function?

The equation t^2x''+tx' - x = 0 represents a second-order linear differential equation with variable coefficients. This means that the Green's function solution can be used to solve a wide range of differential equations with similar form.

## 4. How do you find the Green's function for t^2x''+tx' - x = 0?

The Green's function for t^2x''+tx' - x = 0 can be found by using the method of variation of parameters or by applying the Laplace transform. Both methods yield the same result, which is the Green's function for the given differential equation.

## 5. What are the applications of solving the Green's function for t^2x''+tx' - x = 0?

The Green's function for t^2x''+tx' - x = 0 has various applications in physics and engineering. It can be used to solve problems related to heat transfer, wave propagation, and electrical circuits. It is also used in the study of quantum mechanics and quantum field theory.

• Introductory Physics Homework Help
Replies
9
Views
348
• Precalculus Mathematics Homework Help
Replies
7
Views
943
• Introductory Physics Homework Help
Replies
34
Views
997
• Introductory Physics Homework Help
Replies
29
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
884
• Introductory Physics Homework Help
Replies
9
Views
2K
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
165
• Differential Equations
Replies
1
Views
1K
• Calculus
Replies
19
Views
3K
• Calculus and Beyond Homework Help
Replies
10
Views
2K