I am dealing with a integral equation $$ f'(t) = -\int_0^t K(s) f(t-s)\quad t\in [0,t_\max] \tag{1}$$ in which $f(t)$ and $f'(t)$ are known, well-behaved functions of $t$ and $K(t)$ is the unknown. In addition, the physics of the problem dictate that $f(t)$ and $K(t)$ are even about $t=0$. I have discretized this equation on a uniform grid with $n$ intervals of width $\Delta t$. That is, I map $$ \begin{align} t &\mapsto i \Delta t \\ t_\max &\mapsto n\Delta t \\ \int_0^t ds &\mapsto \Delta t\sum_{j=0}^i w^{(i+1)}_j \end{align} $$ where $w^{(i+1)}$ are the weights for an $(i+1)$-point quadrature rule. With this, I arrive at a linear system for $K_1,\dots,K_n$: $$ f_i' + \Delta t w_0^{(i+1)}f_i K_0 = -\Delta t \sum_{j=1}^i w_j^{(i+1)} f_{i-j} K_j\quad i=1,\dots,n \tag{2} $$ with $K_0 = K(0)$ assumed known. In order to accommodate quadrature rules that require an odd number of points (e.g., Simpson's), I use the evenness of $f$ and $K$ to produce the equivalent discretization $$ f_i' + \frac{\Delta t}{2} w_i^{(2i+1)}f_i K_0 = -\Delta t \sum_{j=1}^i w_{i-j}^{(2i+1)} f_{i-j} K_j\quad i=1,\dots,n \tag{3} $$ based on extending the integral to the range $[-t,t]$.

I am testing my numerical solutions by comparing with analytic solutions to Eq. (1). Namely, $$ f(t) = \cos{(\omega t)} \implies K(t) = \omega^2 $$ and $$ f(t) = e^{-t/2} \left[\cos{(\alpha t/2)} + \frac{\sin{(\alpha t/2)}}{\alpha}\right] \implies K(t) = \frac{1+\alpha^2}{4}e^{-|t|} $$ In both cases, I find that the trapezoidal rule produces a numerical solution $K(t)$ that is in excellent agreement with the analytic one. On the other hand Simpson's rule appears to be unstable, producing numerical solutions that oscillate and rapidly grow in amplitude at $t$ increases. A sample for each test case is shown below: test cases

Evidently, there is something about Simpson's rule that makes the linear system in Eq. (3) ill-conditioned. Writing Eq. (3) in standard form, I have $$ b_i = \sum_{j}A_{ij}x_j $$ with $$ b_i=f'_i+\frac{\Delta t}{2}w^{(2i+1)}_if_iK_0\\ A_{ij}=-\Delta t w_{i-j}^{(2i+1)} f_{i-j}\\ x_j=K_j $$ Since the coefficient matrix $A_{ij}$ is lower-triangular, its eigenvalues are its diagonal entries: $$\lambda_i = A_{ii} = -\Delta t w_0^{(2i+1)} f_0\quad i=1,\dots,n$$ Whatever the quadrature rule, $w_0^{(2i+1)}$ is the same for all $i\ge 1$. For the trapezoidal rule $w_0=\frac{1}{2}$, and for Simpson's rule $w_0=\frac{1}{3}$. Then in both cases, there is only one unique eigenvalue of the matrix $A$.


  1. Can I use this information about the eigenvalues to ascertain why Simpson's rule is unstable, while trapezoidal is stable?
  2. Is there some analytic feature of Eq. (1) that could have led me to know a priori that Simpson's rule is unstable?
  3. If so, is there a more general result (or even rules of thumb) for 1st-kind Volterra equations that says certain types quadrature rules are to be preferred?
  • $\begingroup$ I find your notation slightly confusing because usually $K=K(t,s)$ is the kernel, a known function. Why does it matter that $K$ is even, when $K(s)$ only appears in the equation with $t>s>0$? $\endgroup$ – Kirill Jan 18 '18 at 23:07
  • $\begingroup$ I used $K$ being even en route to Eq. (3) after extending the integral to $[-t,t]$. W.r.t. notation, this really is an inverse problem. It just happens to also be a Volterra equation. $\endgroup$ – Endulum Jan 19 '18 at 3:08

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