This is with regards to the first section of Wikipedia article Duhamel's principle (revision from July 2012). I want to see if I am understanding this.

Basically the inhomogeneous equation says that heat is being added at a rate of $f(x,t)$, so at each point $x$, $f(x,t)\,dt$ of heat is being added to the system.

And you can think of this as a set of systems $u_t(x,t)-\Delta u(x,t)=0$, $u(x,t_0)=f(x,t_0)$ for all $t_0\in(0,\infty)$, which says the initial heat distribution is $f(x,t)$, and no heat is being added, and if we integrate these solutions over $t_0$ we get the inhomogeneous solution...

It's the third paragraph I am a bit uncomfortable with. Can anyone offer some help?


3 Answers 3


Don't hurt your brain too much in trying to understand the physical interpretation. Start from the mathematics! Let me do the ODE version first:

Let $X: \mathbb{R}\to \mathbb{R}^N$ be a vector valued function of one (time) variable. Let $A:\mathbb{R}^N \to\mathbb{R}^N$ be a linear transformation that is independent of the variable $t$. Let $R_t$ denote the solution operator: that is $ X(t) = R_tX_0$ where $X_0$ is a fixed vector solves the equation $$ \dot{X} = AX $$ with initial data $X(0) = X_0$. Now let $Y(t)$ be another vector valued function. Consider the expression

$$ X(t) = \int_0^t R_{t-s}Y(s) \mathrm{d}s $$

Computing it explicitly we have that

$$ \left(\frac{d}{dt} - A\right)X(t) = \int_0^t\left(\frac{d}{dt} - A\right)\left[R_{t-s}Y(s)\right]~ \mathrm{d}s + R_{t-s}Y(s)\Big|_{s = t} $$

The second term comes from the fundamental theorem of calculus when the $\frac{d}{dt}$ derivative hits the integral sign.

By definition the term under the integral sign evaluates to 0, since $Z(t) = R_{t-s}Y(s)$ solves the homogeneous equation with initial data $Z(s) = Y(s)$. So we are left with

$$ (\frac{d}{dt} - A)X(t) = R_0Y(t) = Y(t) $$

i.e. that $X(t)$ solves the inhomogeneous problem.

For the PDE version, you just replace $\mathbb{R}^N$ with a Banach or Hilbert space, and the computation formally carries through in exactly the same way.

Coming back to the physical interpretation: you see that Wikipedia is being a bit imprecise: it is not the adding up of solutions that matters really; what drives Duhamel's principle is the fact that you are adding only a limited number of solutions.

Let me clarify: imagine you have a family of initial data $Y(s)$. And you write down the expression

$$ X(t) = \int_{a}^{b} R_{t-s}Y(s) \mathrm{d}s $$

which would be what we do if we were to just add (integrate) the contributions from all of the linear waves coming from the "inhomogeneity", by linearity it is clear that $X(t)$ will still solve the homogeneous equation, since it is a fixed (as in the limits of the integral) sum of may solutions.

The magic of Duhamel's principle is in that the upper-limit of the integral is time! That is, we defined

$$ X(t) = \int_0^t R_{t-s}Y(s) \mathrm{d}s $$

As you see from the derivation above, it is this upper-limit which, when acted on using the fundamental theorem of calculus, give you the inhomogeneous term. So what is the physical interpretation then? That the upper-limits also changes represents the fact that the solution at $X(t + \Delta t)$ consists of the forward time evolution of the solution at $X(t)$ plus a new contribution from the data in $(t,t+\Delta t)$ which was not included in the computation in $X(t)$. This "adding a new contribution" is precisely what we imagine the inhomogeneous term as, that is, a source term!

  • $\begingroup$ Thanks for this answer. In the equation after "So we are left with", shouldn't that be Y(t) instead of Y(s)? Before that we had (R(t - s) Y(s)) evaluated at s = t, which, I think, gives R(0) Y(t) = Y(t). Does that seem correct? $\endgroup$ Commented Jun 2, 2018 at 1:04
  • $\begingroup$ @RobertDodier: you are right. Let me fix that typo. $\endgroup$ Commented Jun 3, 2018 at 15:20

The intuition behind Duhamels principle is, in fact, quite simple. The input function f(x,t) determines, according to Newtons law, the velocity of the wave u(x.t)at all times. Instead of regarding f(x.t) as an input function, it might as well be regarded as a new “initial velocity function” f(x,s) of the wave, restarting repeatedly at all time s less than or equal to t. f(x,s) is, thus, regarded as a flow of velocities. To find the resulting velocity at any time t, you just add up (integrate) all these (repeting initial velocities) from s= zero to s=t. Thus, the time from zero to t is divided into infinitesimal small time-slice windows delta s, in each window the new velocity is calculated, and then f(s,t) is integrated from to zero to t. This may also be regarded as a convolution: https://www.youtube.com/watch?v=acAw5WGtzuk&ab_channel=PhysicsVideosbyEugeneKhutoryansky


The "General Considerations" section of the Wikipedia article makes the claims mentioned in the introduction a little more precise, but there is an intuitive argument that bears out the idea of integrating the homogeneous solutions, as long as they are sufficiently well-behaved:

Suppose that for each $s \in (0,\infty)$ one has a solution $u^{s}$ (here $s$ is a label, not an exponent), defined for times $t\geq s$, which satisfies

$$ (u^s)(x,s) = f(x,s)\text{, and, whenever }t>s\text{, }(u^s)_t(x,t_0) = (\Delta(u^s))(x,t_0)\,\,\, $$

If collection of functions $\{u^s\}_{s\in(0,\infty)}$ is nice enough, the integral

$$ I(x,t,s):=\int_0^s u^r(x,t)\,dr $$ is defined when $s\leq t$. If the collection $\{u^s\}_{s\in(0,\infty)}$ is very nice, then for $s\leq t$ one can differentiate under the integral sign to obtain $\Delta I$ and $I_t$, and apply the Fundamental Theorem of Calculus to obtain $I_s$: $$ (\Delta I)(x,t,s) = \int_0^s (\Delta (u^r))(x,t)\,dr = \int_0^s (u^r)_t(x,t)\,dr\ = I_t(x,t,s)\text{,}\qquad I_s(x,t,s) = u^s(x,t) $$

Assuming that the foregoing calculations were legitimate, the function $$ u(x,t):= \int_0^t (u^r)(x,t)\,ds = I(x,t,t) $$ is well-defined and satisfies $$ u_t(x,t) = I_t(x,t,t) + I_s(x,t,t) = (\Delta I)(x,t,t) + u^t(x,t) = (\Delta u)(x,t) + f(x,t) $$ for $t>0$ by the chain rule, and $$ u(x,0) = 0 $$ by definition.


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