# How to find eigenvector of one second order differential equation for Hermit?

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);  - please explain your problem in more detail, the question is not clear – James S. Cook Sep 6 '12 at 3:29 add some example – M-Askman Sep 6 '12 at 3:37 nice pdf file ... thanks for sharing – Santosh Linkha Sep 6 '12 at 4:10 @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. – James S. Cook Sep 8 '12 at 16:26 @M-Askman Minor nitpick. Operators have eigenvectors/values. Not equations. – Thiagarajan Sep 8 '12 at 23:47 ## 3 Answers Note that the principle of finding eigenvector of the second order linear ODE that arise from using separation of variables to a linear PDE is that finding the best form of the eigenvector so that we can get the most simplified form of the solution subjected to the B.C.s or I.C.s Theoretically, the form of the eigenvector can choose arbitrarily. However, since solving linear PDEs by using separation of variables subjected to the B.C.s or I.C.s should be unavoidable for performing kernel inversions. Choosing eigenvectors unwisely will face the too complicated kernel inversions and become trouble. So choosing eigenvectors wisely should be important when solving linear PDEs. Since solving a second order linear ODE will have two groups of linear independent solutions, so the best way is that making one of the linear independent solutions becomes zero when substituting most B.C.s or I.C.s, cause the last B.C. or I.C. we handling is remaining only one kernel. In fact finding the best eigenvector is mainly base on our personal experience. Note that$\sin$is only one of the common considerations but the only consideration, especially when the solution of the second order linear ODE fail to express in terms of$\sin$and$\cos$, because$\sin$and$\cos$have the important properties that for all integer$n$,$\sin n\pi=0$and$\cos n\pi=(-1)^n$. Think the following examples that why their eigenvectors are the best to be taken to those forms:$1$. Boundaries in heat equation:$-9\pi^2s^2-72$. Indication on how to solve the heat equations with nonconstant coefficients:$-\dfrac{4\pi^2s^2+1}{4}3$. Wave equation with initial and boundary conditions - is this function right?:$-\dfrac{(2m+1)^2\pi^2c^2}{4l^2}$- In the posted pdf on page 92 is mentions that$\{ \phi_i, \lambda_i \}$is a set of eigenfunctions and eigenvalues defined as solutions to $$\mathcal{L} \phi+\lambda \phi = 0 \qquad \mathcal{B}\phi =0$$ This notation is a quick way to denote a given PDE and associated boundary condition(BC). In the nice examples I teach in my DEqns course the values for$\phi$are forced from the given boundary conditions and the solutions$\phi$can be sine, cosine, hyperbolic sine or hyperbolic cosine or the constant function. The PDE+BC typically give a whole family of eigenvalues and eigenfunctions. In the separable case, the eigenvalues arise from proposing the solution is a product of univariate solutions then subtitution into the PDE and a little algebra brings you to an equation where the lhs and rhs are necessarily independent hence must be proportional to a constant. Customarily that constant is called the eigenvalue. For example, take a look at the explicit calculations of this in: (I have to run now,maybe there is a better thread to link here) Simplifying PDE The term eigenvector and eigenfunction are interchangeable in the context of function space. However, there are also some matrix calculations in that pdf and I'm not totally sure you are not reading something somewhere about a concrete column vector. The fact you mention "sin" gives me hope my interpretation of eigenvector=eigenfunction is correct. - sorry only Sunday have time to read this file seriously, updated an example which is Hermit differential equation, if success solve this example, then eigenvector can be known – M-Askman Sep 9 '12 at 3:39 Note sure that this is an answer, but hope that these points clarify some things for you. Any book on ordinary differential equations will cover the topic of eigenvalues/vectors. Googleing ODE and eigenvector will get you a lot of useful information. You should note that eigenvectors/values are only useful/relevant for linear differential equations. I am not real familiar with Maple, but the example that you posted appears to be non-linear so the idea of eigenvectors would not apply.$\sin(x)$is an eigenvector (more technically an eigenfunction) in differential equations because$\sin(x)$,$\cos(x)$, and$e^x$all maintain their basic form through the differentiation operator$\frac{d}{dx}$. So when you have a linear differential equation of the form$\dot{f}(x) + af(x) = 0$it is clear that$\dot{f}(x) = -af(x)$so$\dot{f}(x)$and$f(x)$are of the same form one is just a scaled version of the other. The only function with this property is$e^x\$.

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sorry only Sunday have time to read this file seriously, updated an example which is Hermit differential equation, if success solve this example, then eigenvector can be known – M-Askman Sep 9 '12 at 10:34
i have updated some code, after tried not success – M-Askman Sep 10 '12 at 13:41