Author Topic: Question 1 from TT2, 2015S  (Read 1184 times)

Tristan Fraser

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Question 1 from TT2, 2015S
« on: March 21, 2018, 08:17:23 PM »
Quote
Consider the eigenvalue problem
$$x^2 X″+2xX′+\lambda X=0,\ \  x \ \ \epsilon (\frac{2}{3},\frac{3}{2}), \ \  X′(\frac{2}{3})=0; \ \ \ X′(32)=0 \ \ \ \ \ (0)$$

Assume $ \lambda \geq 0 $. Find all the eigenvalues and the corresponding eigenfunctions.
 
Hint: as (1) is Euler equation, look   for elementary  solutions  in the form $x^m$).

I wrote the same trick and got the same characteristic of

$$ k(k-1) + 2k + \lambda = 0 \ \ \ \ \ (1) $$

What I do not understand is why, and how we were able to make the substitution of $ t = ln (\frac{3x}{2}) $ to arrive at   $$ \ddot{X} + \dot{X} + \lambda X =  0 \ \ \ \ \  (2) $$

Since my solution instead relied on examining the cases of $\lambda \geq 0 $, but even after plugging in $ x = \frac{3}{2} e^{t} $ I do not see how we would get to the above eigenvalue problem (2).

Jingxuan Zhang

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Re: Question 1 from TT2, 2015S
« Reply #1 on: March 21, 2018, 09:55:01 PM »
Tristan this is the trick used in 244 to solve Euler's equation:
$$\partial_t=x\partial_x;\,\partial^2_t=\partial_x+x^2\partial^2_x,\, x=e^t.$$

Victor Ivrii

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Re: Question 1 from TT2, 2015S
« Reply #2 on: March 22, 2018, 01:27:15 AM »
Actually it is a trick, used to explain, in what form one should look for a solution. After this is understood, you need to do it directly, without reductions