# Resolvent Kernel of Volterra Integral Equation

The resolvent kernal $R(x,t,\lambda)$ for the Volterra integral equation $$\phi(x)=x+\lambda\int\limits_a^x\phi(s)ds$$ is $\begin{array}1 1. e^{\lambda(x+t)} && 2. e^{\lambda(x-t)} && 3. \lambda e^{(x+t)} && 4. e^{\lambda xt} \end{array}$

From what I learned from here: Find iterative kernel and then try guess resolvent kernel, $R=\sum\limits_{i=1}^\infty K_i(x,t)$

Comparing with general form of Volterra equation of second kind, $$g(x)\phi(x)=f(x)+\lambda\int\limits_a^x K(x,t)\phi(t)dt$$

Kernel, $K(x,t)=1,K_1(x,t)=K(x,t)=1$

Now, $K_2(x,t)=\int\limits_t^x K(x,s)K_1(s,t)ds=\int\limits_t^x ds=x-t$

$K_3(x,t)=\int\limits_t^x K(x,s)K_2(s,t)ds \\=\int\limits_t^x (s-t)ds=\frac{s^2}2-ts\big|_{s=t}^x \\=\frac{x^2}2-tx-\frac{t^2}2+t^2=\frac{x^2}2-tx+\frac{t^2}2 \\=\frac1{\sqrt2}(x-t)^2$

And I am stuck, though I am tempted to choose option 2, since it has $x-t$.

I haven't done any coursework in Integral Equations, and I know only bits and pieces about it.

My doubts are:

1. Was I right in choosing $K(x,t)=1$? How does $\lambda$ find its place in resolvent kernel?
2. How to proceed further?
3. Are there any trial and error methods, where I can plug in the options to see if it is the answer?

It is better to start with the definition: $R(x,t,\lambda)$ is the resolvent kernel for $\phi(x)=f(x)+\lambda\int\limits_a^x\phi(s)ds$ if the solution can written as $$\phi(x) = f(x) + \lambda\int_a^x R(x,t,\lambda)f(t)dt.$$ Taking derivatives on both sides of the equation $\phi(x)=x+\lambda\int\limits_a^x\phi(s)ds$, we get $\phi'(x) = f'(x)+\lambda \phi(x)$. Therefore, $$\frac{d}{dx}e^{-\lambda x}\phi(x) = e^{-\lambda x} f'(x)$$ and \begin{align} e^{-\lambda x}\phi(x) &= e^{-\lambda a}\phi(a) + \int_a^x e^{-\lambda t}f'(t)dt\cr &=e^{-\lambda a} \phi(a)+\left.e^{-\lambda t}f(t)\right|_a^x -\int_a^x f(t)de^{-\lambda t} \cr &=e^{-\lambda x}f(x)+e^{-\lambda a}\big(\phi(a)-f(a)\big) +\lambda \int_a^x e^{-\lambda t}f(t)dt. \end{align} Using the fact $\phi(a)=f(a)$ (by evaluating the integral equation at $x=a$), we get $$\phi(x) = f(x)+\lambda\int_a^x e^{\lambda (x-t)}f(t)dt.$$ Therefore the resolvent kenrel is $R(x,t,\lambda)=e^{\lambda(x-t)}$.

If you continue your calculation, you will find (using induction) $K_n(x,t) = (x-t)^{n-1}/(n-1)!$, and the connection to the resolvent kernel is that $$R(x,t,\lambda) = K_1(x,t)+\lambda K_2(x,t) + \lambda^2 K_3(x,t)+\cdots =e^{\lambda(x-t)},$$ consistent with the above approach.

• Thanks a lot! That explains it! I made a small calculation mistake as well! And proper definition of resolvent kernel is what I missed! Will launch the bounty tomorrow! – Jesse P Francis Jan 23 '16 at 4:41
• Also, resolvent kernel, $R(x,t,\lambda)=\sum\limits_{i=1}^\infty \lambda^i K_i(x,t)$, I missed $\lambda$! – Jesse P Francis Jan 23 '16 at 4:51
• Thanks. That's very generous of you. – yhhuang Jan 26 '16 at 9:31
• not planning to cross the 500rep mark (again) any time soon, so if you find any posts worth a bounty/attracting attention to, do tell me! It's free give away! :) – Jesse P Francis Jan 26 '16 at 11:29

This is not a direct answer to the question which wording and notations seem not clear. Nevertheless, the analytic solution of the integral equation $\phi(x)=x+\lambda\int\limits_a^x\phi(s)ds$ is given below. Hoping this will help you to choose the right option. $$\frac{d\phi}{dx}=1+\lambda \phi(x)$$ The general solution of this linear ODE is : $$\phi(x)=-\frac{1}{\lambda}+c\:e^{\lambda x}$$ Bringing it back into the integral equation leads to :

$-\frac{1}{\lambda}+c\:e^{\lambda x}=x+\lambda\int\limits_a^x \left(-\frac{1}{\lambda}+c\:e^{\lambda s} \right)ds = a+c e^{\lambda x}-c\:e^{\lambda a}$

$c=\left(a+\frac{1}{\lambda}\right)e^{-\lambda a}$

$$\phi(x)=-\frac{1}{\lambda}+\left(a+\frac{1}{\lambda}\right)\:e^{\lambda (x-a)}$$

• This is the approach that comes naturally to me when I see this integral equation. I am curious how one would recover the form of the accepted solution from your last step -- surely it's not coincidental that the Resolvent kernel is in the solution, but I can't seem to reconcile the two equalities in a way that is natural. I realize this is about 2 years too late :). – David Kozak Dec 4 '17 at 2:16

Your aswer is also right. Instead of getting $(x-t)^2\over\sqrt2$, you will get $(x-t)^2\over2$. Simililarly $K_4$ can be find and when we generalize it , it will will be series of $exp(x-t)$

• You have contributed nothing new to an old question first having answers – Shailesh May 30 '16 at 9:07
• @Sailesh, what he mentioned is correct! I had mentioned it in once of the comments above. – Jesse P Francis Jun 6 '16 at 6:13

Your way of finding the Resolvent Kernel is also correct. There is a slight mistake in calculating $K_3(x,t)$. $$K_3(x,t)=\frac{(x-t)^2}{2!}$$$$K_4(x,t)=\frac{(x-t)^3}{3!}$$ and so on.

And about the final series to get the Resolvent Kernel, it should be $$R=\sum\limits_{i=1}^{\infty} \lambda^{i-1} K_i(x,t).$$ SO $R=1+ \lambda (x-t)+\frac{\lambda^2(x-t)^2}{2!}+\frac{\lambda^3(x-t)^3}{3!}+\dots=e^{\lambda(x-t)}.$