Sturm–Liouville Form. A second-order equation is said to be in Sturm–Liouville form if it is expressed as [p(t)y'(t)]'+q(t)y(t)=0. Show that the substitutionsx1=y,x2=py' result in the normal form x'1=x2p,x2=-qx1. Ify(0)=a,y'(0)=b are the initial values for the Sturm–Liouville problem, what are x1(0)andx2(0)?

Short Answer

Expert verified

x1(0)=ax2(0)=p(0)b

Step by step solution

01

Express the equation in form of x

Here given.p(t)y'(t)'+q(t)y(t)=0

And

x1=y,x2=py'

The equation transforms as:

x'2+qx1=0x'2=-qx1y'=x2px'1=x2p

02

The initial conditions.

The given initial conditions are y(0)=a,y'(0)=b.

Initial conditions after transformations;

x1(0)=ax2(0)=p(0)y'(0)=p(0)b

This is the required result.

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