Green Function - 2

[logistic_guy]

Solve.

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

 

[logistic_guy]

From the previous exercise, we know that the general solution to this type of differential equation is:

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

The main idea is to find the green function one time for a specific type of differential equation and then use it to solve all the differential equations of the same type. The most beautiful thing about green function is that it does not care much if the forcing function $\displaystyle f(x)$ or the conditions are changed.

The green function for this type of differential equation is always $\displaystyle g(x,s) = \frac{x^2 - s^2}{2x}$.

 

[logistic_guy]

Solve.

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

If we solve this Euler's differential equation by $\displaystyle \text{WolframAlpha}$, we get:

$\displaystyle u(x) = \frac{x^3}{8} - \frac{9x}{4} + \frac{9}{x}$

Let us solve it by green function and see if we can get the same result!

 

[logistic_guy]

Applying the initial conditions, we find that:

$\displaystyle c_1 = -\frac{5}{4} \ \$ and $\displaystyle \ \ c_2 = 7$

Then,

$\displaystyle u(x) = -\frac{5x}{4} + \frac{7}{x} + \int_{2}^{x}\frac{x^2 - s^2}{2x}s \ ds$

 

[logistic_guy]

Let us try to solve this integral.

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


$\displaystyle =\frac{x}{2}\left(\frac{x^2}{2} - \frac{2^2}{2}\right) - \frac{1}{2x}\left(\frac{x^4}{4} - \frac{2^4}{4}\right)$


$\displaystyle =\frac{x}{2}\left(\frac{x^2}{2} - 2\right) - \frac{1}{2x}\left(\frac{x^4}{4} - 4\right)$


$\displaystyle =\left(\frac{x^3}{4} - x\right) - \left(\frac{x^3}{8} - \frac{2}{x}\right)$


$\displaystyle = \frac{x^3}{4} - x - \frac{x^3}{8} + \frac{2}{x}$


$\displaystyle = \frac{x^3}{8} - x + \frac{2}{x}$

 

[logistic_guy]

$\displaystyle u(x) = -\frac{5x}{4} + \frac{7}{x} + \int_{2}^{x}\frac{x^2 - s^2}{2x}s \ ds$

Substitute the result back to the solution.

$\displaystyle u(x) = -\frac{5x}{4} + \frac{7}{x} + \frac{x^3}{8} - x + \frac{2}{x}$

Simplify.

$\displaystyle u(x) = -\frac{5x}{4} + \frac{9}{x} + \frac{x^3}{8} - \frac{4x}{4}$


$\displaystyle = -\frac{9x}{4} + \frac{9}{x} + \frac{x^3}{8}$

How to check if our solution is correct?

It matches the solution of $\displaystyle \text{WolframAlpha}$. This is one way to check that we have the correct green function.

$\displaystyle \rightarrow$ this is the beauty of chess!