Green Function - 3

[logistic_guy]

Solve.

$\displaystyle u'' + \frac{1}{x}u' - \frac{1}{x^2}u = x, \ \ \ u(1) = 3, \ \ \ u'(2) = -1$

 

[logistic_guy]

First, you have to notice that this is not an initial-value problem as the previous exercises, but rather a boundary-value problem. In other words, each condition starts at different point.

First condition starts at $\displaystyle x = 1$ while second condition starts at $\displaystyle x = 2$.

We can solve this problem as before, but we leave the starting limit of integration empty. This is the general solution if we wanna solve it as the previous exercises:

$\displaystyle u(x) = c_1x + \frac{c_2}{x} + \int_{}^{x}\frac{x^2 - s^2}{2x}s \ ds$

Here, the green function $\displaystyle g(x,s) = \frac{x^2 - s^2}{2x}$ is called One-Sided Green function.

Boundary-value problems are better solved by Two-Sided Green function.

With Two-Sided Green function, the general solution will look like this:

$\displaystyle u(x) = c_1x + \frac{c_2}{x} + \int_{1}^{2}G(x,s)s \ ds$

where $\displaystyle G(x,s)$ is Two-Sided Green function.

Our goal in this exercise is to learn how to solve boundary-value problems by Two-Sided Green function.

 

[logistic_guy]

It is not a bad idea to get a little help from some one who is faster than you. We start with Mrs. Alpha

$\displaystyle u'' + \frac{1}{x}u' - \frac{1}{x^2}u = x, \ \ \ u(1) = 3, \ \ \ u'(2) = -1$

If we let $\displaystyle \text{WolframAlpha}$ solve this problem, we get:


$\displaystyle u(x) = \frac{5x^4-57x^2+172}{40x}$


We will try to solve this problem by green function and see if our solution matches the solution of $\displaystyle \text{Alpha}$.

 

[logistic_guy]

I did some calculations and I have found that the green function for this problem is:

$\displaystyle \large G(x,s) =\begin{cases}-\frac{(s^2 - 1)(x^2 + 4)}{10x} & 1 \leq s \leq x\\[10pt]-\frac{(x^2 - 1)(s^2 + 4)}{10x} & x \leq s \leq 2\end{cases}$


In the next posts, we will show in detail how we got this green function. Tighten your seatbelt and get ready!

 

[logistic_guy]

First let us enjoy seeing the structure of green function.

$\displaystyle \large G(x,s) =\begin{cases}\frac{u_1(s)u_2(x)}{W(s)} & 1 \leq s \leq x\\[10pt]\frac{u_1(x)u_2(s)}{W(s)} & x \leq s \leq 2\end{cases}$

where $\displaystyle u_1(x)$ is a solution depends on the homogeneous boundary condition $\displaystyle u(1) = 0$ and $\displaystyle u_2(x)$ depends on $\displaystyle u(2) = 0$. And $\displaystyle W(s)$ is the Wronskian of $\displaystyle u_1(s)$ and $\displaystyle u_2(s)$.

 

[logistic_guy]

$\displaystyle u_2(x)$ depends on $\displaystyle u(2) = 0$.

Oops, $\displaystyle u_2(x)$ depends on $\displaystyle u'(2) = 0$.

We all know that the homogeneous solution (complementary solution) to the OP differential equation is:

$\displaystyle u(x) = c_1x + \frac{c_2}{x}$

Apply $\displaystyle u(1) = 0$. (Just remember that we are applying homogeneous conditions and not the actual conditions because green function depends on homogeneous conditions.)

$\displaystyle 0 = c_1 + c_2$

If we choose $\displaystyle c_1 = 1$ and $\displaystyle c_2 = -1$, the first condition will be satisfied.

Then,

$\displaystyle u_1(x) = x - \frac{1}{x}$

Apply $\displaystyle u'(2) = 0$ again to $\displaystyle u(x)$.

$\displaystyle 0 = c_1 - \frac{c_2}{4}$

If we choose $\displaystyle c_1 = 1$ and $\displaystyle c_2 = 4$, the second condition will be satisfied.

Then,

$\displaystyle u_2(x) = x + \frac{4}{x}$

 

[logistic_guy]

It's time to find the Wronskian.

$\large W(s) = W(u_1,u_2)\displaystyle = \left| \begin{matrix} u_1(s) & u_2(s) \\[10pt] u'_1(s) & u'_2(s) \end{matrix} \right| = \left| \begin{matrix} s - \frac{1}{s} & s + \frac{4}{s} \\[10pt] 1 + \frac{1}{s^2} & 1 - \frac{4}{s^2} \end{matrix} \right| = -\frac{10}{s}$

 

[logistic_guy]

Let us fill the gaps of green function.


$\displaystyle \large G(x,s) =\begin{cases}\frac{\left(s - \frac{1}{s}\right)\left(x + \frac{4}{x}\right)}{\left(-\frac{10}{s}\right)} & 1 \leq s \leq x\\[10pt]\frac{\left(x - \frac{1}{x}\right)\left(s + \frac{4}{s}\right)}{\left(-\frac{10}{s}\right)} & x \leq s \leq 2\end{cases}$

 

[logistic_guy]

Let us fill the gaps of green function.


$\displaystyle \large G(x,s) =\begin{cases}\frac{\left(s - \frac{1}{s}\right)\left(x + \frac{4}{x}\right)}{\left(-\frac{10}{s}\right)} & 1 \leq s \leq x\\[10pt]\frac{\left(x - \frac{1}{x}\right)\left(s + \frac{4}{s}\right)}{\left(-\frac{10}{s}\right)} & x \leq s \leq 2\end{cases}$

With a little simplification, green function becomes:


$\displaystyle \large G(x,s) =\begin{cases}-\frac{(s^2 - 1)(x^2 + 4)}{10x} & 1 \leq s \leq x\\[10pt]-\frac{(x^2 - 1)(s^2 + 4)}{10x} & x \leq s \leq 2\end{cases}$

 

[logistic_guy]

Our solution so far is:

$\displaystyle u(x) = c_1x + \frac{c_2}{x} + \int_{1}^{2}G(x,s)s \ ds$

Or

$\displaystyle u(x) = c_1x + \frac{c_2}{x} + \int_{1}^{x}-\frac{(s^2 - 1)(x^2 + 4)}{10x}s \ ds + \int_{x}^{2}-\frac{(x^2 - 1)(s^2 + 4)}{10x}s \ ds$

Or

$\displaystyle u(x) = c_1x + \frac{c_2}{x} - \frac{x^2 + 4}{10x}\int_{1}^{x}(s^3 - s) \ ds - \frac{x^2 - 1}{10x}\int_{x}^{2}(s^3 + 4s) \ ds$

 

[logistic_guy]

To find the constants $\displaystyle c_1$ and $\displaystyle c_2$, we apply the boundary conditions $\displaystyle u(1) = 3$ and $\displaystyle u'(2) = -1$ on the solution $\displaystyle u(x) = c_1x + \frac{c_2}{x}$.

Apply $\displaystyle u(1) = 3$.

$\displaystyle 3 = c_1 + c_2$

Apply $\displaystyle u'(2) = -1$.

$\displaystyle -1 = c_1 - \frac{c_2}{4} = c_1 - \frac{3 - c_1}{4}$

$\displaystyle -4 = 4c_1 - 3 + c_1$

This gives:

$\displaystyle c_1 = -\frac{1}{5}$

Then,

$\displaystyle 3 = -\frac{1}{5} + c_2$

This gives:

$\displaystyle c_2 = 3 + \frac{1}{5} = \frac{15}{5} + \frac{1}{5} = \frac{16}{5}$


Our solution so far is:

$\displaystyle u(x) = -\frac{x}{5} + \frac{16}{5x} - \frac{x^2 + 4}{10x}\int_{1}^{x}(s^3 - s) \ ds - \frac{x^2 - 1}{10x}\int_{x}^{2}(s^3 + 4s) \ ds$

 

[logistic_guy]

Let us try to solve the first integral.

$\displaystyle \int_{1}^{x}(s^3 - s) \ ds = \frac{s^4}{4} - \frac{s^2}{2}\bigg |_{1}^{x} = \left(\frac{x^4}{4} - \frac{x^2}{2}\right) - \left(\frac{1}{4} - \frac{1}{2}\right) = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{4}$


Our solution so far is:

$\displaystyle u(x) = -\frac{x}{5} + \frac{16}{5x} - \frac{x^2 + 4}{10x}\left(\frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{4}\right) - \frac{x^2 - 1}{10x}\int_{x}^{2}(s^3 + 4s) \ ds$

 

[logistic_guy]

Let us solve the second integral.

$\displaystyle \int_{x}^{2}(s^3 + 4s) \ ds = \frac{s^4}{4} + \frac{4s^2}{2}\bigg |_{x}^{2} = (4 + 8) - \left(\frac{x^4}{4} + 2x^2\right) = 12 - \frac{x^4}{4} - 2x^2$


Our solution so far is:

$\displaystyle u(x) = -\frac{x}{5} + \frac{16}{5x} - \frac{x^2 + 4}{10x}\left(\frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{4}\right) - \frac{x^2 - 1}{10x}\left(12 - \frac{x^4}{4} - 2x^2\right)$

 

[logistic_guy]

Simplify the third term.

$\displaystyle - \frac{x^2 + 4}{10x}\left(\frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{4}\right) = -\frac{x^5}{40} - \frac{x^3}{20} + \frac{7x}{40} - \frac{1}{10 x}$


Our solution so far is:

$\displaystyle u(x) = -\frac{x}{5} + \frac{16}{5x} -\frac{x^5}{40} - \frac{x^3}{20} + \frac{7x}{40} - \frac{1}{10 x}- \frac{x^2 - 1}{10x}\left(12 - \frac{x^4}{4} - 2x^2\right)$

 

[logistic_guy]

Simplify the last term.

$\displaystyle -\frac{x^2 - 1}{10x}\left(12 - \frac{x^4}{4} - 2x^2\right) = \frac{x^5}{40} + \frac{7x^3}{40} - \frac{7x}{5} + \frac{6}{5x}$


Our solution so far is:

$\displaystyle u(x) = -\frac{x}{5} + \frac{16}{5x} -\frac{x^5}{40} - \frac{x^3}{20} + \frac{7x}{40} - \frac{1}{10 x} + \frac{x^5}{40} + \frac{7x^3}{40} - \frac{7x}{5} + \frac{6}{5x}$

 

[logistic_guy]

Simplify further.

$\displaystyle -\frac{x}{5} + \frac{16}{5x} -\frac{x^5}{40} - \frac{x^3}{20} + \frac{7x}{40} - \frac{1}{10 x} + \frac{x^5}{40} + \frac{7x^3}{40} - \frac{7x}{5} + \frac{6}{5x}$


$\displaystyle = \frac{x^3}{8} - \frac{57x}{40} + \frac{43}{10 x}$


Our solution so far is:

$\displaystyle u(x) = \frac{x^3}{8} - \frac{57x}{40} + \frac{43}{10 x}$

 

[logistic_guy]

My solution is:

$\displaystyle \large u(x) = \frac{x^3}{8} - \frac{57x}{40} + \frac{43}{10 x}$


$\displaystyle \text{WolframAlpha}$'s solution is:

$\displaystyle \large u(x) = \frac{5x^4-57x^2+172}{40x}$


I will leave it for the $\displaystyle \text{audience}$ as an exercise to decide if they are equal.