4
$\begingroup$

After read http://math.tut.fi/~piche/pde/pde.pdf , do not know how to calculate eigenvector

How to find eigenvector of one second order differential equation?

why some people use sin as eigenvector? is it only sin can be eigenvector?

The problem is for eigenfunction expansion, first step is finding eigenvalue and eigenvector, but do not know how to calculate eigenvector for differential equation

for example Maple code

x*diff(f(x), x$2) + 2*x*diff(f(x),x) + f(x) = 0     x*diff(f(x), x$2) + 2*x*diff(f(x),x) + x = 0 

Updated

sol := dsolve(t*diff(phi(x),x$2)-x*diff(phi(x),x)+n*phi(x),phi(x)); phi := unapply(rhs(sol), x); BC := [phi(0)=0,phi(1)=0]; with(linalg): Ccoef := genmatrix(BC, [_C1,_C2]); CharEqn := det(Ccoef) = 0;  restart; sol := dsolve(t*diff(phi(x,t,n),x$2)-x*diff(phi(x,t,n),x)+n*phi(x,t,n),phi(x,t,n)); phi := unapply(rhs(sol), x); BC := [phi(0,0,0)=0,phi(1,1,1)=0]; with(linalg): Ccoef := genmatrix(BC, [_C1,_C2]); CharEqn := det(Ccoef) = 0; 

**sorry only Sunday have time to seriously read this file, i find the sin function coming from the step of calculating characteristic equation use pdf file's method to calculate above differential equation for eignvector,

this equation is Hermit after tried, characteristic equation is zero, it imply no eigenvector i guess this calculation maple code has something wrong

how to calculate this?**

Updated 2

Originally i expect to find Hermit H(x) and then use sum(H*z^m/m!, m=0..infinity) to find a A*exp(B) where B is in term of z and t and it is just a simple formula now following the steps, i guess the H is the solution of green function about the expansion

it become more compicated for H(x), and i find there is a D[2] but do not know where it come from. then do not know which step is H(x), i just guess vterm or vv

sol := dsolve(t*diff(phi(x),x$2)-x*diff(phi(x),x)+n*phi(x),phi(x)); phi := unapply(rhs(sol),x); odetest(sol,ode); eq1:=limit(rhs(sol),x=0,right)=0; eq2:=eval(rhs(sol),x=1)=0; Ccoef := LinearAlgebra:-GenerateMatrix([eq1,eq2],[_C1,_C2]); CharEqn:=LinearAlgebra:-Determinant(%[1])=0; solve(CharEqn,t); step1 := map(xi->simplify(subs(t=RootOf(KummerM(1/2-(1/2)*n, 3/2, 1/(2*_Z))),xi)),Ccoef); with(linalg): NN := nullspace(step1); subs(_C1=NN[1][1],_C2=NN[1][2],t=RootOf(KummerM(1/2-(1/2)*n, 3/2, 1/(2*_Z))),phi(x));  phi := (n,t,x) -> KummerM(1/2-(1/2)*n, 3/2, (1/2)*x^2/RootOf(KummerM(1/2-(1/2)*n, 3/2, 1/(2*_Z))))*x;  assume(j,posint): interface(showassumed=0): Gterm := unapply(-phi(n,t,x)*phi(n,t,x)*exp(-lambda(j)*t)/int(phi(n,t,x)^2,x=0..1),(j,n,x,y,t)): G:=Sum(Gterm(j,n,x,y,t),j=1..infinity); vterm := int(D[2](Gterm)(n,1,x,t-tau),tau=0..t); vv := sum(Sum(op(n,vterm),j=1..infinity),n=1..2); 
  • 0
    please explain your problem in more detail, the question is not clear2012-09-06
  • 0
    add some example2012-09-06
  • 0
    nice pdf file ... thanks for sharing2012-09-06
  • 0
    @M-Askman from the pdf, it says to read Zachmanoglou and Thoe to supplement the notes. So, you might get a copy of that to help better understand the pdf.2012-09-08
  • 0
    @M-Askman Minor nitpick. Operators have eigenvectors/values. Not equations.2012-09-08
  • 0
    What is the separation constant in here? $n$ or $t$ ?2012-09-09

3 Answers 3