I have difficulty about the explanation of Duhamel's principle on my book. Here is what's written on my book:
Take wave equation as an example. Consider the equation: \begin{cases} \frac{\partial^2u}{\partial t^2} = a^2\frac{\partial^2u}{\partial x^2} + f(t,x) &t>0, -\infty < x < +\infty\\ u|_{t=0} = 0, \frac{\partial u}{\partial t}|_{t=0} = 0 \end{cases} where $f(t,x)$ is the external force over a unit of mass of the string. $u(t,x)$ is the displacement at time $t$. By independent action principle, $u(t,x)$ is the summation of displacement $w(t,x;\tau)d\tau$ caused by the impulse $f(\tau, x)d\tau\ (0\leq\tau\leq t)$, thus $u(t,x) = \int_{0}^{t}w(t,x;\tau)d\tau$. Clearly, when $t<\tau$, $w(t,x;\tau) \equiv 0$. When $t>\tau$, the instantaneous impulse at time $\tau$ is the increase of momentum from $t=\tau-0$ to $t=\tau+0$. So $w(t,x;\tau)$ should satisfy the equation: \begin{cases} \frac{\partial^2w}{\partial t^2} = a^2\frac{\partial^2w}{\partial x^2} &t>\tau, -\infty < x < +\infty\\ w|_{t=\tau} = 0, \frac{\partial w}{\partial t}|_{t=\tau} = f(\tau,x) \end{cases} and $$ u(t,x) = \int_{0}^{t}w(t,x;\tau)d\tau $$
I checked that such $u$ is the solution to the inhomogeneous equation. But I just cannot understand the derivation of the Duhamel's principle. In the explanation, $w$ is also the displacement, its dimension is length. $\tau$ is the time. Why $u(t,x) = \int_{0}^{t}w(t,x;\tau)d\tau$? The dimension of $u$ becomes length times the time?
Actually I didn't quite understand other parts. Could anyone help me understand Duhamel's principle by explaining it in more detail?
Any help is appreciated!
First, I'll clear up the dimensional analysis (or equivalently scaling invariance). Say, $u$ has dimension $U$ (in general it does not need to be linked with length or time) and setting $a=1$, $x,t$ have dimension $T$ (length or time). $f$ has dimension $UT^{-2}$, and $w$ has dimension $[f]T = UT^{-1}$ (from the initial condition $\partial_tw(t=0)=f$). This makes sense since to get back $u$, you integrate in time which adds the dimension $T$.
Duhamel's principle allows you to convert inhomogeneous PDE's to integrals of homogeneous PDE's with the specific initial conditions.
The physicist's approach for your problem is to interpret the force as many different impulses at different time slices: $$ f(x,t) = \int f(x,\tau)\delta(t-\tau)d\tau $$ By superposition, you just look at the field cased by such an impulse for fixed $s$: $$ \partial_t^2w = \partial_x^2w+f(x,\tau)\delta(t-\tau) $$ Physically, you are applying an impulse of the system, so you'd expect a jump of velocity. Integrating over a small interval of time around $s$, you get, assuming continuity of $u$ with no time derivatives: $$ [\partial_t w]_{\tau^-}^{\tau^+} = f(x,\tau) $$ Taking the causal solution, $\partial_t u(\tau^-)=0$ (start from rest), so you get: $$ \partial_t w(\tau^+) = f(x,\tau) $$ and the continuity condition gives (still starting from rest): $$ [w]_{\tau^-}^{\tau^+} = 0\\ w(\tau^+) = 0 $$ which is the Cauchy problem you presented. You just need to superpose the solutions to get: $$ u = \int w d\tau $$
Actually, you can see this already in the harmonic oscillator: $$ \ddot u + u = f(t) $$ You just integrate the impulse responses of different times of forcing. To get the causal impulse response, you just solve after the impulse for a unit velocity given by the impulse.
Hope this helps.