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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.