Integral eigenvectors and eigenvalues

I need to find the eigenvalues e eigenvectors of this integral.

$$\int_{0}^{2\pi}(\cos^2(x+y)+1/2)\phi (y)dy$$

b)- Solved thanks
$$\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy$$

Solutions Collecting From Web of "Integral eigenvectors and eigenvalues"

First, let me try to rephrase part of the question in a way that is likely to be more palatable for most folks on math.SE:

Edited question

I would like to find the eigenvalues and eigenvectors of the following integral:

$$\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy$$

I tried to find a matrix so I could compute the determinant, but I’m not sure what matrix I should use.

Discussion on the edit

Note that we are at least indicating some attempt to indicate our thoughts. Even if this is minimal, it demonstrates to folks who are donating there time to adress your question where you are. Also, never use URGENT in all caps in your title; I promise that it won’t go over well.

Finally, it’s generally considered polite to upvote an answer that gives reasonable help and to accept the best answer whenever appropriate. I notice that you have 14 questions with only one accept.

A partial answer

At any rate, after that, someone is likely to politely explain that this is not a problem involving matrices. Rather, you are looking for a function $\phi(x)$ and a non-zero, real number $\lambda$ such that

$$\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy = \lambda \phi(x).$$

My recommendation is that you guess and refine. If you let $\phi(x)=ax^2+bx+c$ and compute the integral, you should find that

$$\int_{0}^{1}(x^2y^2-2/45)(ay^2+by+c)dy = -\frac{2 a}{135}-\frac{2 c}{45}-\frac{2 b \
x}{45}+\left(\frac{a}{5}+\frac{c}{3}\right) x^2+\frac{b x^3}{3}.$$

From there, the question is simply, can you find $a$, $b$, and $c$ to make this work? It’s reasonably clear that $b=0$, right? I bet you can make some progress from here.

One more hint: My solutions for $\lambda$ have a 45 in the denominator. I also solved for $a$ in terms of $c$, which is to be expected in an eigenvalue problem.

Good luck!

Related problem:(I), (II). Here is a start and you have to work out the details. First, you need to find the eigen values $\lambda$ by solving the integral equations

$$ \phi(x) = \lambda\int_{0}^{2\pi}(\cos^2(x+y)+1/2)\phi (y)dy,$$

$$ \phi(x) = \lambda\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy .$$

Now, both of the integral equations have separable kernels which the following method can be used.


$$ \cos^2(x+y)=\frac{\cos(2x+2y)}{2}+\frac{1}{2} $$

$$ \cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y). $$

Added: It seems your book adopting the convention

$$ \lambda \phi(x) = \int_{0}^{2\pi}(\cos^2(x+y)+1/2)\phi (y)dy.$$

Now, using the above hints, we have

$$ \lambda \phi(x) = \int_{0}^{2\pi}\left(\frac{\cos(2x+2y)}{2} + 1/2 \right)\phi(y)dy$$

$$\implies \lambda \phi(x) = \cos(2x)\int_{0}^{2\pi} \frac{\cos(2y)}{2}\phi(y)dy – \sin(2 x) \int_{0}^{2\pi} \frac{\sin(2y)}{2}\phi(y)dy$$

$$ + \int_{0}^{2\pi}\phi(y) dy \longrightarrow (1). $$

$$ \implies \lambda \phi(x) = \cos(2x) c_1 – \sin(2 x) c_2 + c_3 $$

$$ \implies \phi(x) = \frac{1}{\lambda}\cos(2x) c_1 – \frac{1}{\lambda}\sin(2 x) c_2 + \frac{1}{\lambda}c_3. $$

Now, subs back in $(1)$ and simplify, you get

$$ \left( c_{{1}}-{\frac {c_{{1}}\pi }{\lambda}} \right) \cos \left( 2\,
x \right) + \left( -c_{{2}}-{\frac {c_{{2}}\pi }{\lambda}} \right)
\sin \left( 2\,x \right) +c_{{3}}-2\,{\frac {c_{{3}}\pi }{\lambda}}=0.$$

Now, by equating the coefficients, we have $\lambda=2\pi,\pi $ and $-\pi$. However, $\lambda=2\pi$ will give the solution $\phi(x)=1$. You can check this solution by substituting in $(1)$.