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Mathematical Methods for Physicists: A concise introduction - Site Map

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SIMPLE LINEAR INTEGRAL EQUATIONS<br />

A Hermitian kernel has at least one eigenvalue and it may have an in®nite number.<br />

The proof will be omitted and we refer interested readers to the book by<br />

Courant and Hibert mentioned earlier (Chapter 3).<br />

The eigenvalues of a Hermitian kernel are real, and eigenfunctions belonging to<br />

di€erent eigenvalues are orthogonal; two functions f …x† and g…x† are said to be<br />

orthogonal if<br />

Z<br />

f *…x†g…x†dx ˆ 0:<br />

To prove the reality of the eigenvalue, we multiply the homogeneous Fredholm<br />

equation by u*…x†, then integrating with respect to x, we obtain<br />

Z b<br />

a<br />

u*…x†u…x†dx ˆ <br />

Z b Z b<br />

a<br />

a<br />

K…x; t†u*…x†u…t†dtdx:<br />

…11:20†<br />

Now, multiplying the complex conjugate of the Fredholm equation by u…x† and<br />

then integrating with respect to x, weget<br />

Z b<br />

a<br />

u*…x†u…x†dx ˆ *<br />

Z b Z b<br />

a<br />

a<br />

K*…x; t†u*…t†u…x†dtdx:<br />

Interchanging x and t on the right hand side of the last equation and remembering<br />

that the kernel is Hermitian K*…t; x† ˆK…x; t†, we obtain<br />

Z b<br />

a<br />

u*…x†u…x†dx ˆ *<br />

Z b Z b<br />

a<br />

a<br />

K…x; t†u…t†u*…x†dtdx:<br />

Comparing this equation with Eq. (11.2), we see that ˆ *, that is, is real.<br />

We now prove the orthogonality. Let i , j be two di€erent eigenvalues and<br />

u i …x†; u j …x†, the corresponding eigenfunctions. Then we have<br />

Z b<br />

u i …x† ˆ i K…x; t†u i …t†dt;<br />

a<br />

Z b<br />

u j …x† ˆ j K…x; t†u j …t†dt:<br />

Now multiplying the ®rst equation by u j …x†, the second by i u i …x†, and then<br />

integrating with respect to x, we obtain<br />

Z b<br />

Z b<br />

j u i …x†u j …x†dx ˆ i j<br />

a<br />

Z b<br />

Z b<br />

i u i …x†u j …x†dx ˆ i j<br />

a<br />

a<br />

a<br />

Z b<br />

a<br />

Z b<br />

a<br />

K…x; t†u i …t†u j …x†dtdx;<br />

K…x; t†u j …t†u i …x†dtdx:<br />

a<br />

…11:21†<br />

Now we interchange x and t on the right hand side of the last integral and because<br />

of the symmetry of the kernel, we have<br />

Z b<br />

Z b<br />

i u i …x†u j …x†dx ˆ i j<br />

a<br />

a<br />

Z b<br />

a<br />

K…x; t†u i …t†u j …x†dtdx:<br />

…11:22†<br />

422

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