Solve the integral equation $f(x)-\lambda\int^{1}_{0}\min(x,t)f(t)dt=\sin(\frac{\pi N x}{2})$

Question

In my studies of fractional analysis I have been solving problems to familiarize with the concepts of fredholm theory, and I found the following problem which I have been having problems solving:

For the following integral equation in $L^{2}[0,1] $ $f(x)-\lambda\int^{1}_{0}\min(x,t)f(t)dt=\sin(\frac{\pi N x}{2})$.
a) For any $\lambda\in\mathbb{C}$ solve the homogeneous equation
b)Specify the number of solutions on $\lambda, N\in\mathbb{N}$.

I tried to solve the homogeneous equation, but I am having problems doing it, this is my attempt:

We need to solve $f(x)-\lambda\int^{1}_{0}\min(x,t)f(t)dt=0$, for this we use the following $\min(x,t)=\frac{x+t}{2}+\frac{|x-t|}{2}$:

$$f(x)=\frac{\lambda}{2}\int^{1}_{0}(x+t)f(t)dt+\frac{\lambda}{2}\int^{1}_{0}|x-t|f(t)dt.$$

Then we differentiate with respect to $x$ yielding the following: $$f'(x)=\frac{\lambda}{2}\int^{1}_{0}f(t)dt+\frac{\lambda}{2}\int^{1}_{0}\frac{x-t}{|x-t|}f(t)dt$$

And we differentiate one more time and we get $f''(x)=0$, so we then can see that $f(x)=k_{1}x^{2}+k_{2}x+k_{3}$.

Once we get this we should substitute this on the homogeneous equation, calculate the integral and obtain a system of equations for the coefficients, the problem is that, when I did this I found that $k_{1}=k_{2}=k_{3}=0$, so I gues I did something wrong.

Any help would be appreciated.

Answer 1

As a first step, let us determine the spectrum of the linear operator $A: L^2([0,1]) \to L^2([0,1])$ defined by $$ (Af)(x)=\int\limits_0^1 {\min}(x,t)f(t)dt=\int\limits_0^xt f(t) dt+ x\int\limits_x^1f(t) dt. $$ The operator $A$ is selfadjoint as the integral kernel satisfies the condition $\min(x,t) = \overline{\min(t,x)}$. The eigenvalue equation $$\int\limits_0^x tf(t)dt+ x \int\limits_x^1 f(t) dt = a f(x),$$ implies $f(0)=0$ for the eigenfunction $f$ associated with the real eigenvalue $a$. Differentiation leads to $$ \int\limits_x^1 f(t) dt = a f^\prime(x), $$ implying the further condition $f^\prime(1)=0$ on the eigenfunction. Differentiating once more, gives the second order differential equation $$ a f^{\prime\prime}(x)+f(x)=0, \quad f(0)=0, \, f^\prime(1)=1 $$ with the (normalized) solutions $$f_n(x) = \sqrt{2} \, \sin \frac{(2n+1) \pi x}{2}, \quad n=0,1,2,\ldots $$ with associated eigenvalues $$ a_n= \left(\frac{2}{(2n+1) \pi} \right)^2.$$

Coming back to the original problem, task (a) was to solve the equation $f-\lambda Af=0$ for arbitrary complex $\lambda$. The case $\lambda=0$ admits only the trivial solution $f=0$ and we shall assume $\lambda\ne 0$ in the following. The equation $Af=\frac{1}{\lambda}f$ has a nontrivial solution $f= c f_n$ (for some $n$) only if $\lambda \in \{1/a_0, 1/a_1,1/a_2,\ldots\}$. On the other hand, in the case $\lambda \notin \{1/a_0,1/a_1,1/a_2,\ldots \}$, $f=0$ remains as the only solution.

Task (b) was to study the inhomogeneous integral equation $f-\lambda Af = g$ for the special case $g(x)=\sin (\pi N x/2)$ with positive integer $N$. In the following, I shall discuss the general case, leaving the remaining steps for the special case as an exercise. For $\lambda=0$, the (unique) solution is given by $f=g$. Assuming $\lambda \ne 0$ and writing the equation in the form $(\mathbf{1}-\lambda A)f=g$, it is obvious that we have to distinguish two cases depending on whether the operator $\mathbf{1}-\lambda A$ has an inverse or not. In the first case, the complex parameter $\lambda \notin \{1/a_0,1/a_1,1/a_2,\ldots\}$ and there exists a uniquely determined solution $$f=(\mathbf{1}-\lambda A)^{-1} g =\sum\limits_{n=0}^\infty\frac{\langle f_n| g\rangle}{1-\lambda a_n} f_n, \quad \langle f_n | g \rangle =\int\limits_0^1 \overline{f_n(t)} g(t) dt,$$ where the spectral decomposition of the operator was used. In the second case (where $\mathbf{1}-\lambda A$ has no inverse), we have $\lambda =1/a_m$ for a certain $m \in \{0,1,2,\ldots\}$ and the original equation can be written in the form $$f-f_m \langle f_m | f \rangle -\frac{1}{a_m} \sum\limits_{n\ne m} a_n\langle f_n|f\rangle f_n=g. $$ As the left-hand-side is a vector in the orthogonal complement of $f_m$, the condition $\langle f_m |g \rangle =0$ is required for the solvability of this equation. The final steps are straightforward and one finds the general solution $$f= c f_m +\sum\limits_{n\ne m}\frac{\langle f_n |g\rangle}{1-a_n/a_m} f_n $$ with $c$ arbitrary.

Answer 2

My answer is similar to Hyperon's but doesn't use any functional analysis methods. One can actually write the solution of the equation in closed form. If $\lambda = 0$ the equation is trivial so without loss of generality assume $\lambda \neq 0$. Define the operator $A$ by $$Af(x) = \int_0^x t f(t) \, dt + x \int_x^1 f(t) \, dt$$ Hence our integral equation can be written $$f - \lambda Af = g$$ where $g = \sin\left( \frac{\pi N x}{2} \right)$. To guide what I'm about to do I should remark that integral equations and second order ODEs are intimately related since integral operators and differential operators are often inverses of one another. Hence let us try to "invert" $A$.

Notice that $$\frac{d^2}{dx^2} Af(x) = - f(x)$$ So in a certain sense $-\frac{d^2}{dx^2}$ and $A$ are inverse of one another (not to be taken too literally since $-\frac{d^2}{dx^2}$ is not strictly invertible). Motivated by this we find

$$f^{\prime \prime}+ \lambda f = g^{\prime \prime}$$

Again let us be guided by linear algebra intuition: we have here the equation $\left( \frac{d^2}{dx^2} + \lambda \right)f = g^{\prime \prime}$ hence we would like to find $\left( \frac{d^2}{dx^2} + \lambda \right)^{-1}$. This is done by the theory of Green's functions. Indeed we see that the values of $\lambda$ for which this inverse does not exist is exactly the spectrum of $-\frac{d^2}{dx^2}$.

If we replace $f(x) = e^{ix\sqrt{\lambda}} h(x)$ we find

$$\frac{d}{dx}\left( h^\prime(x)+ 2i \sqrt{\lambda}h(x) \right) = e^{-i\sqrt{\lambda}x}g^{\prime \prime}(x)$$ Thus $$h^\prime(x)+ 2i \sqrt{\lambda}h(x) = C + \int_0^x e^{-i\sqrt{\lambda}t}g^{\prime \prime}(t) \, dt = C_1 + e^{-i\sqrt{\lambda}x} g^\prime(x)+ i \sqrt{\lambda}e^{-i\sqrt{\lambda}x}g(x) - \lambda \int_0^x e^{-i\sqrt{\lambda}t} g(t) \, dt$$ This final equation can be solved by an integrating factor. $$\frac{d}{dx}\left( h(x) e^{2i \sqrt{\lambda} x}\right) = C_1 e^{2i \sqrt{\lambda} x} + e^{i\sqrt{\lambda}x} g^\prime(x)+ i \sqrt{\lambda}e^{i\sqrt{\lambda}x}g(x) - e^{2i \sqrt{\lambda} x} \lambda \int_0^x e^{-i\sqrt{\lambda}t} g(t) \, dt$$ Then, after integrating by parts, we find $$ f(x) e^{i \sqrt{\lambda} x} = C_2 + C_1 e^{2i \sqrt{\lambda} x} + g(x)e^{i \sqrt{\lambda} x}- \lambda \int_0^x ds \int_0^s dt \, e^{2i \sqrt{\lambda} s} e^{-i \sqrt{\lambda} t} g(t) $$ This final integral may be simplified further. $$\int_0^x ds \int_0^s dt \, e^{2i \sqrt{\lambda} s} e^{-i \sqrt{\lambda} t} g(t) = \int_{[0,1]^2} \chi_{0 \leq t \leq s \leq x} e^{2i \sqrt{\lambda} s} e^{-i \sqrt{\lambda} t} g(t) \, dt \, ds $$ Notice that we can now integrate over $s$ from $t$ to $x$. This gives the solution $$f(x) = C_1 e^{i \sqrt{\lambda} x} + C_2 e^{-i \sqrt{\lambda} x} + g(x)- \lambda e^{-i \sqrt{\lambda}x} \int_0^x \frac{e^{2i \sqrt{\lambda} x} - e^{2i \sqrt{\lambda} t}}{2 i \sqrt{\lambda}} g(t) \, dt $$ With boundary conditions we might be able to determine $C_1$ and $C_2$. Indeed, the original integral equation implies certain boundary conditions, namely $f(0) = g(0)$ and $f^\prime(1) = g^\prime(1)$. The first implies that $C_1 + C_2 = 0$. Thus let $C = C_1 = -C_2$. The second condition gives (if I've done the algebra correctly!) $$ C (e^{i\sqrt{\lambda}}-e^{-i\sqrt{\lambda}}) = \frac{\sqrt{\lambda}}{2i} \int_0^1 (e^{2i\sqrt{\lambda}}+e^{2i\sqrt{\lambda}t}) g(t) \, dt$$ If $e^{i\sqrt{\lambda}}-e^{-i\sqrt{\lambda}} \neq 0$ then we may uniquely determine $C$. If $e^{i\sqrt{\lambda}}-e^{-i\sqrt{\lambda}} = 0$ then our integral equation will only have a solution if the above integral on the right hand side vanishes. This is essentially the "Fredholm alternative."