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$$ f(x)= \int_0^1 e^{|x-t|} f(t) \, dt+x-1 $$

I can't solve it, because I can't find the boundary conditions?

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Normally, one uses an iterative technique to solve this. Just start iterating using $x-1$. – Jon Dec 17 '11 at 19:17
Thank you for your comment, however in this exercise I have to use this method (using eigen values) to solve it. Do you know how I can do it? – eng Dec 17 '11 at 20:19
This is quite interesting as you can consider your integral a linear operator and compute $∫_0^1e^{|x-t|} u(t)dt=\lambda u(x)$. Looking at the answer by Didier, this just becomes $$\lambda u''(x)=3u(x).$$ Taking $\lambda=1$ you are back to the proposed answer. – Jon Dec 17 '11 at 20:47

What you call boundary conditions is not clear to me. Anyway, one can start from the identity $$ f(x)=\mathrm e^x\int_0^x\mathrm e^{-t} f(t)\mathrm dt+\mathrm e^{-x}\int_x^1\mathrm e^{t} f(t)\mathrm dt+x-1, $$ and differentiate it twice. The first differentiation yields $$ f'(x)=e^x\int_0^xe^{-t} f(t)dt-e^{-x}\int_x^1e^{t} f(t)dt+1, $$ and, unless I am mistaken, the second differentiation yields something like $$ f''(x)-3f(x)=1-x. $$ From this point, the usual methods apply.

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Thank you for your answer. however at this point how can i find characteristic values? – eng Dec 17 '11 at 20:38
@eng: I am not sure what you mean by characteristic values. The ODE is $(D+\sqrt{3})(D-\sqrt{3})f=1-x$, do you mean $\pm\sqrt{3}$? – robjohn Dec 17 '11 at 22:42
@robjohn: Indeed, it is completely useless to keep on reporting the right solution until OP does not give the text of his homework. So far, it is not clear at all what he is asking for. – Jon Dec 17 '11 at 23:01
@eng: Please tell me if my impression is right that (0) you cannot continue from the end of my post, (1) you do not know how to solve an ODE $y''-y=g(x)$, (2) you did not check the method explained on the page I linked to. – Did Dec 18 '11 at 0:14

Differentiating once yields $$ f'(x)=\int_0^1\operatorname{sgn}(x-t)e^{|x-t|}f(t)\;\mathrm{d}t+1 $$ Differentiating again, using the $\delta$ distribution when differentiating $\operatorname{sgn}$, yields $$ \begin{align} f''(x) &=\int_0^1\left(2\delta(x-t)+\operatorname{sgn}(x-t)^2\right)e^{|x-t|}f(t)\;\mathrm{d}t\\ &=2f(x)+\int_0^1e^{|x-t|}f(t)\;\mathrm{d}t\\ &=3f(x)-x+1 \end{align} $$ This gives the same ODE to solve as Didier got. To solve this type of equation, I tend to use an integrating factor.

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The operator: $$T:L^2(0,1)\ni f\mapsto \int_0^1 \exp(|x-t|)\ f(t)\ \text{d} t\in L^2(0,1)$$ is a linear, selfadjoint and compact operator on $L^2(0,1)$ (compactness and selfadjointness follows because the kernel $k(x,t):=\exp(|x-t|)$ is continuous and symmetric in $[0,1]^2$).

If you let $I$ be the identity operator on $L^2(0,1)$ and $g(x):=x-1$, your problem rewrites: $$\tag{1} (I-T)f =g\; .$$ It is known that an equation like (1) either has unique solution or it has some solution iff $g$ satisfies $g\bot\ \mathcal{N} (I-T)$ (i.e. $\langle g,u\rangle =0$ for each $u\in \mathcal{N}(I-T)$): this is the Fredholm alternative.

On the other hand, it is also known that an operator like $T$ possesses a discrete point spectrum and that $L^2(0,1)$ can be written as direct sum of the eigenspaces of $T$; moreover $L^2(0,1)$ has a Hilbert basis composed of eigenfunctions of $T$: this is the Hilbert-Schmidt theorem.

Therefore, let $\{e^n\}_{n\in \mathbb{N}}$ be a Hilbert basis of $L^2$ composed with the eigenfunction of $T$ and let $\{\lambda_n\}_{n\in \mathbb{N}}$ be the sequence of the eigenvalues of $T$ in which $\lambda_n$ corresponds to $e^n$; then you can write: $$f=\sum_{n=1}^\infty f_n\ e^n \qquad \text{and} \qquad g=\sum_{n=1}^\infty g_n\ e^n\; ,$$ with $g_n:=\langle g,e^n\rangle$, and: $$\sum_{n=1}^\infty g_n\ e^n =g=(I-T)f=\sum_{n=1}^\infty f_n\ (1-\lambda_n)\ e^n\; ;$$ if you multiply both leftmost and rightmost sides of the last formula by $e^N$ and you remember the orthogonality condition of the eigenfunctions, you gain: $$g_N =\sum_{n=1}^\infty g_n\ \delta_{n,N} =\sum_{n=1}^\infty g_n\ \langle e^n, e^N\rangle =\sum_{n=1}^\infty f_n\ (1-\lambda_n)\ \langle e^n, e^N\rangle = f_N\ (1-\lambda_N)$$ and: $$f_N=\frac{g_N}{1-\lambda_N}$$ if $\lambda_N\neq 1$; hence: $$\tag{S} f=\sum_{n=1}^\infty \frac{\langle g,e^n\rangle}{1-\lambda_n}\ e^n$$ is the solution of your problem.

It then follows that, when you want to use the eigenfunctions method to solve an integral equation, the first thing you have to do is finding the eigenvalues $\lambda_n$ and the eigenfunctions $e^n$ of $T$.

In order to do this, you can remember that $\lambda_n\neq 0$ is an eigenvalue iff $\mu_n :=1/\lambda_n$ is a singular value for the integral equation: $$f(x)=\mu\ \int_0^1 \exp (|x-t|)\ f(t)\ \text{d} t\; ,$$ i.e.: $$\tag{2} f(x)=\mu \bigg( e^x \underbrace{\int_0^x e^{-t}\ f(t)\ \text{d} t}_{A(x)} +e^{-x}\underbrace{\int_0^1 e^t\ f(t)\ \text{d} t}_{B(x)}\bigg)\; .$$ Therefore the problem of finding the eigenvalues of $T$ recasts as the problem of finding the singular values of the integral equation (2), i.e. the values of $\mu$ s.t. (2) has some nontrivial solutions.

Now assume $f\in L^2(0,1)$ to be a solution of (2). Since $e^{\pm t}\in C^\infty ([0,1])\subset L^2(0,1)$, then $e^{\pm t}\ f(t)\in L^1(0,1)$ and the functions $A,B$ are absolutely continuous; thus $f(x)=\mu e^x A(x)+\mu e^{-x}B(x)$ is continuous in $[0,1]$. But then $A,B$ are of class $C^1([0,1])$ and also $f$ is $C^1$; then again $A,B\in C^2([0,1])$ and $f\in C^2$... Bootstrapping you find $f\in C^\infty([0,1])$, hence you can differentiate (2) as many time you want.

Differentiating (2) twice you gain the ODE: $$f^{\prime \prime} = (1+2\mu) f$$ together with the Robin Bcs: $$f(0)=-f^\prime (0) \qquad \text{and} \qquad f(1)=f^\prime (1)$$ hence $f$ solves (2) iff it solves the BVP: $$\tag{R} \begin{cases} f^{\prime \prime} =(1+2\mu)\ f &\text{, in } ]0,1[ \\ f(0)=-f^\prime (0) \\ f(1)=f^\prime (1)\; .\end{cases}$$ Problem (R) possesses nontrivial solutions for some values of $\mu$ and you can compute all the singular values and the eigenfunctions using the method I used here.

Once you got $\{e^n\}$ and $\{\lambda_n\}$, you apply (S) to obtain the solution.

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Thank you for your useful and complete answer! – eng Dec 18 '11 at 7:13
could you please help me to solve the rest of this problem. if $$ \2λ-1=-ω^2\ $$ how can you find the roots (i.e. $$ ω =nπ $$)? – eng Dec 18 '11 at 7:20
While the function is:$${(A+ωB=0 <br/>(cosω+ωsinω)A+(sinω-ωcosω)B=0)$$ – eng Dec 18 '11 at 7:28

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