I Determining the coefficient of the legendre polynomial

Mayan Fung
Messages
131
Reaction score
14
We know that the solution to the Legendre equation:
$$ (1-x^2)\frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} + n(n+1) = 0 $$
is the Legendre polynomial $$ y(x) = a_n P_n (x)$$

However, this is a second order differential equation. I am wondering why there is only one leading coefficient. We need two conditions to determine the unique solution to a 2nd order ODE, don't we?
 
Physics news on Phys.org
In addition to the Legendre polynomials, there is a second linearly independent solution to the differential equation, usually denoted ##Q_n(x)##. These are usually thrown away as they are singular as ##x\to\pm 1## and a typical requirement is that the functions are regular in these limits.
 
  • Like
Likes Mayan Fung
Orodruin said:
In addition to the Legendre polynomials, there is a second linearly independent solution to the differential equation, usually denoted ##Q_n(x)##. These are usually thrown away as they are singular as ##x\to\pm 1## and a typical requirement is that the functions are regular in these limits.
That solved my puzzle! Thank you!
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top