Given : $$u(x)=x+2 \int_0^x e^{x-t}u(t)dt$$

Solve the Volterra Equation numerically using Trapezoidal Rule in $(0,5)$ choosing $n=8$ and compare with the exact values.

The Exact Solution I have found is : $$u(x)=x+\dfrac{2}{9}e^{3x}[1-e^{-3x}(3x+1)]$$

Numerically solving I have got : enter image description here

Both look disastrous. So can anyone show me where it is wrong or if possible provide solution.


The equation includes an exponential which produces very important variation of $u(x)$. So, is not surprising to observe "disastrous" results (as you say) if the number $n$ of intervals is too low.

Probably, the very small number $n=8$ was chosen in order to have only a few numbers to compare , respectively those given in the book and those computed by the student.

May be, one aim of the exercice is to make aware of the importance of the deviations due to a too low number of intervals.

The table below shows how the accuracy of the results becomes better, just by increassing $n$. That way, one can obtain results closed to the theoretical values.

enter image description here

  • $\begingroup$ If n increases then $\triangle t=\dfrac{b-a}{n}$ will decrease. Then the step size will go down. So how do we get $x=0.625$ $1.25$ and so on. These are in steps $0.625$, correct ? $\endgroup$ – creative Feb 20 '15 at 4:02
  • $\begingroup$ I supposed that $n$ is the number of intervals for $x$, from $x=0$ to $x=5$. So, $\Delta x=5/n=5/8=0.625$. I do not use $n=8$ for the number of steps in the integral. $\endgroup$ – JJacquelin Feb 20 '15 at 7:46

Trapezoidal rule of $\int^b_a f(x) dx$ with $n$ subintervals:


To get your textbook answer, notice that the length of your subinterval is $\frac{5}{8}$:

$$u(0)=0$$ $$u(5/8)=5/8+2\cdot \frac{5/8}{2}(e^{5/8-0}u(0)+e^{5/8-5/8}u(5/8))$$

This gives you $u(5/8)==1.666667$

$$u(5/4)=5/4+2\cdot \frac{5/4}{4}(e^{5/4-0}u(0)+2e^{5/4-5/8}u(5/8)+e^{5/4-5/4}u(5/4))$$

This would give you $u(5/4)=13.7124$. The following ones are similar.

I think the purpose of this problem is to show you it does not work. It will need some regularization to numerically solve the volterra equation.

  • $\begingroup$ Can you provide a method which is very close to exact ? I do not have much idea. $\endgroup$ – creative Feb 19 '15 at 14:42
  • $\begingroup$ I am not sure who to do that. Doesn't the book provide some insights on this topic in the future sections? I think if they show you by this example how it does not work, it should then later show how to make it work. What's the title of the book? $\endgroup$ – KittyL Feb 19 '15 at 14:54
  • $\begingroup$ Introduction to Integral Equations and applications by Abdul Jerri $\endgroup$ – creative Feb 20 '15 at 3:59

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