so i'm really having a problem with Duhamel's principle, i'll explain what i know and if possible could someone help me fix my lack of understanding. Thanks!
so from my understanding of Duhamel's principle, if we have an inhomgenous PDE (for example the heat equation) such as
(Current case is finite domain $(0,a)$ ) $$u_t = Du_{xx}+f(x,t),~~~ u(x,0)=\phi(x) \\ u(0,t)=h(t), ~~~~~~ u(a,t)=g(t)$$
my first issue, in the above we have inhomogenous boundry conditions, if we had homogenous boundry conditions can i still use Duhamel's principle? (Despite the fact that i could solve it using Seperation of variables) or is the inhomogenous BC's the "signal" to use Duhamels principle?
First we find $v(x,t)$ which satisfies an auxiliary question
$$v_t=Dv_{xx},\\~~v(x,0) = 0 \\ v(0,t)= g(t), ~~~~~~~~~~~~~~~~ v(a,t) = h(t)$$
which basically comes down to finding an interpolation of h and g. (so long as its matchs the boundary conditions its all good).
Second: we solve the problem
$$w_t = Dw_{xx}+f(x,t), \\w(x,0)= \phi(x) \\ w(0,t)= w(a,t) = 0$$
with solution
$$w(x,t) = \sum w_n(t) \sin{\frac{\pi n x}{a}}$$ $$\text{ where } w_n(t) = w_n(0)e^{\frac{-D \pi^2 n^2 t}{a^2}} + e^{\frac{-D \pi^2 n^2 t}{a^2}} \int_{0}^{t} e^{\frac{-D \pi^2 n^2 s}{a^2}} f_n(s)ds$$ $$w_n(0) = \frac{2}{a}\int_0^{a} \phi(x) \sin{\frac{\pi n x}{a}} dx$$ and $$f_n(x) = \frac{2}{a}\int_0^a f(x,t)\sin{\frac{\pi n x}{a}}$$
then the solution to our original problem is
$$u(x,t) = v(x,t) + w(x,t)$$
step 1: Solve a shifted problem by finding $v(x,t)$ which satisfies the Boundary conditions from the general problem.
step 2: solve the inhomogenous problem with homogeneous boundary conditions, whilst keeping the initial condition
step 3: Sum the two.
i believe this is the method to use the principle. In this instance (in terms of the heat equation) we have an external source which is adding values to the PDE. so the problem with solution v(x,t) is akin to shifting the initial amount of energy in the system by $\phi(x)$ and eliminating the external source. (this makes sense because if we dont have the external source then we should have zero initial energy
also extending the principle to neumann conditions. are all we doing again defining v to solve the boundary conditions in terms of $v$ and $v_t$?
in the next scenario we have 1 dimensional infinite domains so our problem is $$u_t = Du_{xx}+f(x,t) \\ u(x,0) = \phi(x)$$
in this instance we dont have boundary conditions so we solve the following
$$v_t=Dv_{xx}, \\v(x,0) = \phi(x)$$ ie we eliminate the source again
and have a solution of
$$v(x,t) = \int_{\mathbb{R}}\Phi(x-y)\phi(y)~dy$$
next we solve
$$w_t = Dw_{xx}+f(x,t) \\ w(x,0) = 0$$ which again gives $$w(x,t) = \int_0^t \int_{\mathbb{R}}\Phi(x-y,t-s)f(y,s)~dyds$$
which are true due to the translational and diliational invariance of the heat equation.
finally $u = v +w$
so to me it seems that practically we eliminate our initial condition and source term from our problem. solve the equation, then solve a new equation with the source term but no initial condition, and sum the two.
intuitively i have no idea why. or whether my understanding of the procedure is correct (in fairness ive been going over this for the last few days, tried researching it and at this point just frazzled) any help guys and girls?
thanks for the help.
Update: after having let my brain deglaze a bit and then redoing some research on the subject i've built up a little bit more of an intuitive rationale why, again i would need people to help confirm. from my current understanding the arguement is that a non-homogenous system is akin to a homogenous one where at some time period s < t a source term starts adding in energy to the system, in that case what we're really doing is just kind of "Psuedo" scaling back time and since v and w both solve our "basic" version of the heat equation they can be considered particular solutions that correspond to the different set ups of the system, Ie one solves for the Boundary conditions and one solves for the source itself.
am i on the right lines?
update 2: ive found a pdf on the physics exchange which explains the procedure more accurately.