# Solve $f(x) = \lambda \int\limits_{0}^1(\max(x,t)+xt)f(t)dt$

I need to solve this: $\ f(x) = \lambda \int\limits_{0}^1(\max(x,t)+xt)f(t)dt$.

Rewriting it as: $\ f(x) = \lambda(\int\limits_0^x x(t+1)f(t)dt + \int\limits_x^1 t(x+1) f(t)dt)$.

1st derivative: \begin{align*} f'(x) &= \lambda\left(\int\limits_0^x (1+t)f(t)dt + x(1+x)f(x) - xf(x) + \int\limits_x^1 tf(t)dt - x^2f(x)\right)\\ &=\lambda\left(\int\limits_0^x f(t)dt + \int\limits_0^x tf(t)dt + \int\limits_x^1 tf(t)dt\right)\\ &=\lambda\left(\int\limits_0^1 tf(t)dt + \int\limits_0^x f(t)dt\right) \end{align*}

2nd derivative: $\ f''(x) = \lambda f(x)$

so $f(x) = c_1 e^{\sqrt{\lambda}x} + c_2 e^{-\sqrt{\lambda}x}$

How can I find $c_1$ and $c_2$?

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Did you try substituting your answer back in to the original equation? It looks like $c_1$ and $c_2$ may be free parameters (assuming $0 \leq x \leq 1$). – dls Nov 30 '11 at 19:35
No, they are not free. In fact there are no nonzero solutions for all but one value of $\lambda$, and only a one-parameter family for that one. – Robert Israel Nov 30 '11 at 20:04
@DavideGiraudo done. – Philipp G. Sinicyn Nov 30 '11 at 20:24
@Robert can you explain? – Philipp G. Sinicyn Dec 1 '11 at 8:19
Substitute the answer back into the original equation with $x=0$ and $x=1$ to solve for $c_1$ and $c_2$. – Jeff Dec 1 '11 at 18:32

Your integral equation (IE, for short) is of the type: $$\tag{1} (I-\lambda T) f =0$$ with $I$ the identity operator and $T$ given by: $$Tf(x):= \int_0^1 \big( \max \{x,t\}+xt\big)\ f(t)\ \text{d} t\; ,$$ which is a bounded, compact linear operator on $C^0([0,1])$.

The Fredholm alternative applies to your IE, hence there are two possibilities: either A) (1) has only the trivial solution $f(t):=0$, or B) there exists a linear space of nontrivial solutions. In case A one says that $\lambda$ is a regular value for $T$; on the other hand, in case B one says that $\lambda$ is a singular value for $T$. It is also known that the singular values of a compact operator form a sequence, say $\{\lambda_n\}_{n\in \mathbb{N}_0}$, such that $|\lambda_n|\to \infty$ and that $\{\lambda_n\}$ has no finite accumulation point.

Your computations show that any $C^0$ solution of (1) is actually of class $C^2$ and solves the second order linear ODE $f^{\prime \prime} -\lambda f=0$; moreover, it is not hard to see that any solution of your IE satisfies also the Robin boundary conditions $f(0)-f^\prime (0)=0$, $f(1)-f^\prime (1)=0$ (to prove this, it suffices to plug $x=0,1$ into the IEs for $f$ and $f^\prime$); therefore each solution of (1) is also a solution of the BVP: $$\tag{2} \begin{cases} f^{\prime \prime} (t)-\lambda f(t)=0 &\text{, in } ]0,1[ \\ f(0)-f^\prime (0)=0 \\ f(1)-f^\prime (1)=0 \end{cases}$$ which is an eigenvalue problem for the differential operator $\tfrac{\text{d}^2}{\text{d} t^2}$ under the given boundary conditions.

On the other hand, I think one could prove that the eigenvalue problem (2) is in fact equivalent to (1), in the sense that any $C^2$ solution of (2) is also a solution of (1) (in order to prove this, I bet some integration by parts and other elementary tricks have to be used).

Thus if you have to solve (1), then you have to look for the eigenvalues of (2).

To find the eigenvalues of (2) one has to distinguish three cases:

• $\lambda >0$: in such a case one can plug $\lambda =\omega^2$ with $\omega >0$ into the ODE to get the general integral: $$f(t):=A e^{\omega t} +B e^{-\omega t}\; ;$$ then either (2) has only the trivial solution, or $\lambda =\omega^2$ is an eigenvalue of (2): the latter eventuality occurs iff the homogeneous linear system: $$\begin{cases} f(0)-f^\prime (0)=0 \\ f(1)-f^\prime (1)=0 \end{cases} \quad \Leftrightarrow \quad \begin{cases} (1-\omega) A+(1+\omega) B=0 \\ (1-\omega )e^\omega A+(1-\omega)e^{-\omega} B=0 \end{cases}$$ possesses nontrivial solutions, i.e. iff $\omega =1$ and $\lambda =1=:\lambda_0$. If $\lambda=\lambda_0$, all the eigenfunctions corresponding to $\lambda_0$, i.e. the nontrivial solutions of (2), are in the form: $$f_0(t):=Ae^t\; .$$

• $\lambda =0$: in this case the general integral of the ODE is $f(t):=A+Bt$ and it is easy to see that there are no nontrivial solutions of (2).

• $\lambda <0$: one can set $\lambda =-\omega^2$ (with $\omega >0$) into the ODE to get the general integral: $$f(t):= A\cos \omega t+B\sin \omega t\; ;$$ then either (2) has only the trivial solution, or $\lambda =-\omega^2$ is an eigenvalue of (2): the latter case occurs iff the homogeneous linear system: $$\begin{cases} A-\omega B =0 \\ (\cos \omega +\omega \sin \omega) A + (\sin \omega -\omega \cos \omega)B=0\end{cases}$$ has some nontrivial solutions, i.e. iff $\omega =n\pi$ and $\lambda =-n^2\pi^2=:\lambda_n$ with $n\in \mathbb{N}$. If $\lambda=\lambda_n$, all the eigenfunctions corresponding to $\lambda_n$ are in the form: $$f_n(t):=n\pi B \cos \omega t + B \sin \omega t\; .$$

Finally, summing it up, you have the following result:

The integral equation

$$\tag{1} f(x) = \lambda \int_0^1\big(\max \{x,t\} +xt\big)f(t)\ \text{d} t$$

has only the trivial solution $f(x):=0$ iff $\lambda \notin \{ \lambda_n\}_{n\in \mathbb{N}_0}=\{1,-\pi^2, -4\pi^2,\ldots ,-n^2\pi^2,\ldots \}$; it has also nontrivial solutions in the form:

$$f_0 (x) := Ae^x \qquad \text{(} A\text{ an arbitrary constant)}$$

iff $\lambda =\lambda_0 =1$; it has also nontrivial solutions in the form:

$$f_n(x) =n\pi B \cos (n\pi x)+ B \sin (n\pi x) \qquad \text{(} B\text{ an arbitrary constant)}$$

iff $\lambda =\lambda_n =-n^2\pi^2$ for some $n\in \mathbb{N}$.

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This is not a complete answer and too long for a comment. If $f$ is a so is a constant multiple of $f$, therefore you need at least one more condition on $f$ for a unique solution.

We get two constraints on $f$ by putting $x = 0$ and $x = 1$ in turn. that gives us $$f(0) = \lambda \int_0^1 tf(t) dt \mbox{ and } f(1) = \lambda\left\{\int_0^1 f(t)dt + \int_0^1 tf(t) dt\right\}.$$

Now, we break the problem into three cases.

If $\lambda = 0,$ then $f$ is identically zero.

If $\lambda = - k^2$ is negative then write $f = A \cos kx + B \sin kx.$ use $\int_0^1 \cos kt dt = \frac{\sin k}{k}, \int_0^1 \sin kt dt = \frac{1-\cos k}{k}$ and

$\int_0^1 t\cos kt dt = \frac{k\sin k + \cos k - 1}{k^2}, \int_0^1 t\sin kt dt = \frac{\sin k-k\cos k}{k^2}$. I hope I have not made any errors in evaluating these integrals.

Now put these back in the constraints and demanding non trivial solution for the pair $A, B$ will give a relation connecting $k$ and $\lambda.$ depending on the uniqueness of condition on $f$ should now provide the eigenvalues $k.$

I have not done the case of positive $\lambda$ which would involve hyperbolic sine and cosines.

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