# Eigenfunction again!

1. May 31, 2010

### samleemc

[urgent] Eigenfunction again!

Find the eigenvalue and eigenfunction of the following
$$\int$$cos(2X)+cox(2t)y(t) dt = ky , k=eigenvalue , intergrate from 0 to Pi

thx, urgent

this is the original question

Last edited: May 31, 2010
2. May 31, 2010

### elibj123

Re: [urgent] Eigenfunction again!

$$\int_{0}^{\pi} cos(2t)y(t)dt$$

Is a number, yet you compare it to a function ( y(t)).

Also

$$\int_{0}^{\pi} cos(2x)y(t)dt$$

is an expression of x, and again you compare it to an expression of t ( ky(t))

Are you sure you wrote it correctly?

3. May 31, 2010

### samleemc

Re: [urgent] Eigenfunction again!

4. May 31, 2010

### samleemc

Re: [urgent] Eigenfunction again!

5. Jun 1, 2010

### HallsofIvy

Staff Emeritus
Re: [urgent] Eigenfunction again!

What is found in your attachment is
$$\int_0^\pi (cos(2x)+ cos(2t))y(t)dt= ky(x)$$
not quite what you originally wrote.

We can write that as
$$cos(2x)\int_0^\pi y(t)dt+ \int_0^\pi cos(2t)y(t)dt= ky(x)$$

As elibj123 pointed out, both of those integrals are NUMBERS that, of course, depend on y(x).

Let $$Y_1= \int_0^\pi y(t)dt$$
and $$Y_2= \int_0^\pi cos(2t)y(t)$$.

Now the equation reads simply $ky(t)= Y_1 cos(2x)+ Y_2$ and we can solve for y(t) just by finding the two numbers $Y_1$ and $Y_2$.

Multiply both sides of $ky(x)= Y_1 cos(2x)+ Y_2$ by cos(2x) and integrate from 0 to $\pi$:
$$k\int_0^\pi cos(2x)y(x)dx= Y_1 \int_0^\pi cos^2(2x) dx+ Y_2\int_0^\pi cos(2x)dx$$

Of course, $\int_0^\pi cos(2x)y(x)dx$ is the same as $\int_0^\pi cos(2t)y(t)dt$ which we have called $Y_2$. The other two integrals do not involve y(x) and so can be integrated. Rather than do them for you I am going to write $\int_0^\pi cos^2(2x)dx= A$ and $\int_0^\pi cos(2x)= B$ (although that second one ought to be obvious!).

Now, our equation is $kY_2= AY_1+ BY_2$ or $AY_1+ (B- k)Y_2= 0$.

If we simply integrate $ky(x)= Y_1 cos(2x)+ Y_2$ itself from 0 to $\pi$ we get
$$k\int_0^\pi y(x)dx= Y_1 \int_0^\pi cos(2x)dx+ Y_2\int_0^\pi dx= BY_1+ \pi Y_2$$
That is the same as $kY_1= BY_1+ \pi Y_2$ or $(B-k)Y_1+ \pi Y_2$.

That is, we can solve for y(x) by solving the pair of numerical equations
$(B- k)Y_1+ \pi Y_2= 0$ and $AY_1+ (B- k)Y_2= 0$
where A and B are given by the integrals above.

Of course, like any eigenvalue equation, those are satified by the "trivial" solution $Y_1= Y_2= 0$. Eigenvalues are values of k for which there exist non-trivial solutions. Non-trivial solutions for homogeneous systems of equations occur when the determinant of the coefficient matrix, here $(B- k)^2- A\pi$, is 0.

Last edited: Jun 1, 2010