1

**Home Assignment 4 / Re: Problem 1**

« **on:**October 24, 2012, 09:30:43 PM »

Solution to Problem 1(d)

To show that eigenfunctions corresponding to different eigenvalues are orthogonal, we evaluate the following:

$$(\lambda_{n}-\lambda_{m})\intop_{0}^{l}X_{n}(x)X_{m}(x)dx$$

Notice that we can make a simple substitution, apply the Fundamental Theorem of Calculus using the boundary conditions. Then,

$$(\lambda_{n}-\lambda_{m})(X_{n}(x)X_{m}(x))=X_{n}"(x)X_{m}-X_{n}(x)X"_{m}(x)=(X_{n}'(x)X_{m}(x)-X_{n}(x)X'_{m}(x))'$$

Plugging into the original integral, we obtain:

$$\intop_{0}^{l}(X_{n}'(x)X_{m}(x)-X_{n}(x)X'_{m}(x))'dx=X_{n}'(l)X_{m}(l)-X_{n}(l)X'_{m}(l)-X_{n}'(0)X_{m}(0)+X_{n}(0)X'_{m}(0)=0$$

Therefore, the eigenfunctions corresponding to different eigenvalues are orthogonal.

To show that eigenfunctions corresponding to different eigenvalues are orthogonal, we evaluate the following:

$$(\lambda_{n}-\lambda_{m})\intop_{0}^{l}X_{n}(x)X_{m}(x)dx$$

Notice that we can make a simple substitution, apply the Fundamental Theorem of Calculus using the boundary conditions. Then,

$$(\lambda_{n}-\lambda_{m})(X_{n}(x)X_{m}(x))=X_{n}"(x)X_{m}-X_{n}(x)X"_{m}(x)=(X_{n}'(x)X_{m}(x)-X_{n}(x)X'_{m}(x))'$$

Plugging into the original integral, we obtain:

$$\intop_{0}^{l}(X_{n}'(x)X_{m}(x)-X_{n}(x)X'_{m}(x))'dx=X_{n}'(l)X_{m}(l)-X_{n}(l)X'_{m}(l)-X_{n}'(0)X_{m}(0)+X_{n}(0)X'_{m}(0)=0$$

Therefore, the eigenfunctions corresponding to different eigenvalues are orthogonal.