4
$\begingroup$

Given the following BVP $$-(e^x Y'(x))'=\lambda e^x Y(x), \quad 0<x<\pi$$ $$Y(0)=0, Y(\pi)=0$$

I have found that the eigenfunctions are $Y_n(x)=e^{-\frac{x^2}{2}}\sin{nx}$

and the eigenvalues are $\lambda_n=\frac{1}{4}+n^2$

We check whether the eigenfunctions are orthogonal by calculating the inner product$\langle f,g\rangle_2=\int_{0}^{1} f(x)g(x) \,dx $

\[Lambda][n_] = 1/4 + n^2;
Y[x_, n_] = Exp[-1/2*x]*Sin[n*x]
Integrate[Y[x, 1]*Y[x, 2], {x, 0, Pi}]

It is not zero, so it is not orthogonal.

Normalizing eigenfunctions to the new inner product by weight

The norm $||f||_{\mu}=\int_{0}^{1} \mu(x)f(x)g(x) \,dx$ is now defined

norm[n_] = Sqrt[Integrate[Exp[x]*Y[x, n]^2, {x, 0, Pi}]]
Simplify[%, n \[Element] Integers && n > 0]
X[x_, n_] = Y[x, n]/norm[n];
Integrate[Exp[x]*X[x, 1]*X[x, 2], {x, 0, Pi}]

So the normalized eigenfunctions are orthogonal to the new inner product with weight

Expansion to the eigenfunctions in terms of the inner product by weight

We define the function

f[x_]=x*(Pi-x)
pic1=Plot[f[x],{x,0,1},PlotStyle->{Thick},PlotRange->All]

enter image description here

the coefficients of the expansion of g with respect to the normalized eigenfunctions are given

A[n_] = Integrate[Exp[x]*f[x]*X[x, n], {x, 0, Pi}]
Simplify[%, n \[Element] Integers && n > 0]

We define the partial sums

S[x_, N_] := Sum[A[n]*X[x, n], {n, 1, N}];
fig2 = Plot[S[x, 5], {x, 0, 1}, PlotStyle -> {Red, Dashed}]
fig3 = Plot[S[x, 10], {x, 0, 1}, PlotStyle -> {Green}]
fig4 = Plot[S[x, 20], {x, 0, 1}, PlotStyle -> {Black, Dashed}]
Show[fig1, fig2, fig3, fig4]

enter image description here enter image description here enter image description here enter image description here

Is the procedure I am following correct?

$\endgroup$

1 Answer 1

1
$\begingroup$

You have coded as :

\[Lambda][n_] = 1/4 + n^2; 

Y[x_, n_] = Exp[-1/2*x]*Sin[n*x]
Integrate[Y[x, 1]*Y[x, 2], {x, 0, Pi}]

But with the check for orthogonality, the integrand was not multiplied by the weight function. So change the last command to :

Integrate[Exp[x]*Y[x, 1]*Y[x, 2], {x, 0, Pi}]

There seems to be a typo in one of the commands for graphing

Show[fig1, fig2, fig3, fig4]

-- Actually , you mean

Show[pic1, fig2, fig3, fig4] 

--

Rest of the calculations are fine.

$\endgroup$
1
  • 1
    $\begingroup$ This may be more appropriate as a comment than as an answer. $\endgroup$
    – MarcoB
    Commented Aug 4, 2023 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.