I am trying to understand Duhamel's Principle by applying it to a simple first order ODE and then generalizing. I am thinking of [imath]P(t)[/imath] as expressing a bank account balance at time [imath]t[/imath], as my economics intuition is better than my physics intuition. The [imath]\pi[/imath] forcing term is just chosen for algebraic clarity (it's a bit more concrete than a parameter, but stays separate from other numeric values in the problem).
First I considered [imath]P'(t) = \frac{1}{10}P(t) + \pi,\ P(0) = 0[/imath]. I started by solving [imath]Q'(\tau) = \frac{1}{10}Q(\tau),\ Q(0) = \pi[/imath] (swapping the forcing term and initial condition) and got the answer [imath]Q(\tau) = \pi e^{\frac{1}{10}\tau}[/imath]. It appears that I can simply integrate this and produce the correct solution as [imath]P(t) = \int_0^t \pi e^{\frac{1}{10}\tau} d\tau = 10\pi e^{\frac{1}{10}t} - 10\pi[/imath]. In English, this seems to say, "To calculate the balance of a bank account starting with [imath]\$0[/imath] and continuously receiving deposits of [imath]\$\pi[/imath] per time unit, integrate between [imath]0[/imath] and [imath]t[/imath] the balance of a hypothetical bank account starting with [imath]\$\pi[/imath] and continuously receiving deposits of [imath]\$0[/imath] per time unit." Of note is that both the real and hypothetical accounts have [imath]10\%[/imath] continuous compounding, but their time units are not necessarily the same ([imath]t[/imath] and [imath]\tau[/imath] are presumably two forms of time).
1) I expected this relationship to be more obvious to me once spelled out in English than it is. Is there a different way to state the above quote that makes its truth more clear?
2) In this example, I simply integrated [imath]Q[/imath] over [imath]\tau[/imath] to solve for [imath]P[/imath]. How can I generalize this integration step to work for differential equations in many variables and of arbitrary order? Wikipedia states, "The integrand is the retarded solution [imath]P^s f[/imath], evaluated at time [imath]t[/imath], representing the effect, at the later time [imath]t[/imath], of an infinitesimal force [imath]f(x, s)\ ds[/imath] applied at time [imath]s[/imath]," but this is not meaningful to me and I can't find further explanation.
First I considered [imath]P'(t) = \frac{1}{10}P(t) + \pi,\ P(0) = 0[/imath]. I started by solving [imath]Q'(\tau) = \frac{1}{10}Q(\tau),\ Q(0) = \pi[/imath] (swapping the forcing term and initial condition) and got the answer [imath]Q(\tau) = \pi e^{\frac{1}{10}\tau}[/imath]. It appears that I can simply integrate this and produce the correct solution as [imath]P(t) = \int_0^t \pi e^{\frac{1}{10}\tau} d\tau = 10\pi e^{\frac{1}{10}t} - 10\pi[/imath]. In English, this seems to say, "To calculate the balance of a bank account starting with [imath]\$0[/imath] and continuously receiving deposits of [imath]\$\pi[/imath] per time unit, integrate between [imath]0[/imath] and [imath]t[/imath] the balance of a hypothetical bank account starting with [imath]\$\pi[/imath] and continuously receiving deposits of [imath]\$0[/imath] per time unit." Of note is that both the real and hypothetical accounts have [imath]10\%[/imath] continuous compounding, but their time units are not necessarily the same ([imath]t[/imath] and [imath]\tau[/imath] are presumably two forms of time).
1) I expected this relationship to be more obvious to me once spelled out in English than it is. Is there a different way to state the above quote that makes its truth more clear?
2) In this example, I simply integrated [imath]Q[/imath] over [imath]\tau[/imath] to solve for [imath]P[/imath]. How can I generalize this integration step to work for differential equations in many variables and of arbitrary order? Wikipedia states, "The integrand is the retarded solution [imath]P^s f[/imath], evaluated at time [imath]t[/imath], representing the effect, at the later time [imath]t[/imath], of an infinitesimal force [imath]f(x, s)\ ds[/imath] applied at time [imath]s[/imath]," but this is not meaningful to me and I can't find further explanation.