Solving Differential Equations for Finding Solutions

[neelakash]

Homework Statement

I am to show that the differential equataion D^2f+[A+V(x)]f=0
{A is a constant and V(x+m)=V(x)}
has the solutions of the form f=exp[ikx]U(x)where U(x+m)=U(X)

Homework Equations

The Attempt at a Solution

I tried to differentiate the given solution and put it in the equation...but the method is not working.Can you please help?

 

[siddharth]

I tried to differentiate the given solution and put it in the equation...but the method is not working.Can you please help?

What's U(x)? If the solution is given to you, then it must satisfy the differential equation.

It'd be easier to spot errors if you post your work. Have you seen the https://www.physicsforums.com/showthread.php?t=8997" for this forum yet? It's an easy way to post equations.

 

[tnho]

i think you may think of the Fourier Transformation method
and see it works or not~

 

[neelakash]

actually in the question paper it was given as v(x) instead U(x).
Note,in the equation it was V(x),and in solution it is v(x) and is said that v(x) is also periodic as v(x+m)=v(x).

I will try to adopt Latex notations.

whose Fourier transform should I consider?

 

[tnho]

actually in the question paper it was given as v(x) instead U(x).
Note,in the equation it was V(x),and in solution it is v(x) and is said that v(x) is also periodic as v(x+m)=v(x).

I will try to adopt Latex notations.

whose Fourier transform should I consider?

um...i don't quite understand you question, sorry. (mainly confused with V(x), v(x) and U(x)... )
would you mind telling me the in whole question once again..?

 

[neelakash]

OK,the question is I am to prove that=

D^2f+[A+V(x)]f=0
{A is a constant and V(x+m)=V(x),i.e.V(x) is periodic with period m}
has the solutions of the form f=exp[ikx]v(x)where v(x+m)=v(x)
i.e.v(x) is also periodic with period m

 

[neelakash]

After crude differentiation,I got this:
f"=[v"(x)+2ikv'(x)]*exp[ikx]-k^2*f

I write it as:
f"(x)=-V(x)f(x)-k^2*f(x)
and this can be written in desired form.

Here I assume V(x)=-{U"+2ikU'}/U and A=-k^2.Right?

Now I think we may check that V(x+m)=-{U"(x+m)+2ikU'(x+m)}/{U(x+m)}
=-{U"+2ikU'}/U

since after differentiation the period of the periodic function does not gets changed.Right?