Given a differential equation

$$ Lu(x) = f(x), $$

where $L$ is the differential operator, and $f$ is a given function of $x$, the general solution of $u(x)$ is

$$ u(x) = \int G(x, s) f(s)\, ds $$

$G(x, s)$ is the Green's function of $L$ and satisfies

$$ LG = \delta(x - s) $$

If we have Dirichlet boundary conditions like

$$ u(a) = \alpha, $$

$\alpha$ is a given number, this differential problem is called the Dirichlet problem.

My question is why $G(a, s) = 0$ with given a boundary condition $u(a) = \alpha$?

  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Commented Dec 25, 2022 at 11:37
  • $\begingroup$ Check again, may be field is curly. $\endgroup$ Commented Dec 26, 2022 at 11:34

1 Answer 1


The short answer is that your problem is linear non-homogeneous. You have two sources of inhomogeneity: the boundary condition ($\alpha$) and the sources ($f(x)$). Instead of solving the entire problem in one go, you want to treat the inhomogeneities one at a time and use superposition to reconstruct the original solution.

This is simply linear algebra. Take for example a $d$ dimensional vector space $E$ and $e^1,…,e^d$ basis vectors of the dual space. Say you want to solve the system of equations for $x\in E$: $$ \langle e^1,x\rangle =a^1\\ …\\ \langle e^d,x\rangle =a^d $$ and $a^1,…,a^d$ known constants. As you can see, you could try to solve the equation directly or construct the dual basis $e_1,…,e_d$ defined by: $$ \langle e^i,e_j\rangle=\delta^i_j $$ i.e. you normalize only one inhomogeneity and make the other constraints homogeneous. Your solution can this be directly solved to: $$ x=\sum_i a^ie_i $$

Formally, you can apply the same reasoning to your problem. The solution to your problem is therefore: $$ u(x)=v(x)+\int G(x,s)f(s)ds \tag{1} $$ with: $$ LG(x,s) = \delta(x-s), \quad G(a,s)=0 \\ Lv(x) = 0, \quad v(a)=\alpha $$ I think that part of your confusion comes from the fact that you forgot the first term.

Hope this helps.


Applying the method when $E$ is finite dimensional shows that it’s a pedantic way of introducing the inverse matrix. Take $E=\mathbb R^2$, and: $$ e^1 = (L^1_1, L^1_2) \\ e^2 = (L^2_1, L^2_2) \\ $$ then let: $$ e_1=\begin{pmatrix} G_1^1 \\ G_1^2 \end{pmatrix} \quad e_2=\begin{pmatrix} G_2^1 \\ G_2^2 \end{pmatrix} $$ The duality condition is equivalent to the definition of the inverse matrix: $$ LG=I_2 $$ In particular if you want to solve the system: $$ Lx=a $$ then you’ll use: $$ x=Ga $$

A more advanced example would be $L=-\frac{d^2}{dx^2}$ giving the bulk equation: $$ -\frac{d^2u}{dx^2}=f \tag{2} $$ and Dirichlet boundary conditions at $x=0,1$: $$ u(0)=\alpha_0 \\ u(1)=\alpha_1 \tag{3} $$ You then have: $$ G(x,s)= \min[(1-s)x,s(1-x)]\\ v(x) = \alpha_1x+\alpha_0(1-x) $$

Hope this helps.

  • $\begingroup$ I am not good at linear algebra. Can you give me an example above this line? $$x=\sum_i a^ie_i $$ $\endgroup$
    – IvanaGyro
    Commented Dec 25, 2022 at 13:56
  • $\begingroup$ Does $v(x)$ satisfy $Lv(x) = 0$? $\endgroup$
    – IvanaGyro
    Commented Dec 25, 2022 at 14:02
  • $\begingroup$ Yes it’s in the last equation. I’ll give some examples. Btw it’s best to be confident in linear algebra as most of physics is about extrapolating its results. $\endgroup$
    – LPZ
    Commented Dec 25, 2022 at 16:23
  • $\begingroup$ I am still confused by the last example. In your last example $Lv(x)$ is $-\alpha_1 + \alpha_0$, and $LG(x, s)$ is not $\delta(x -s)$. Do I misunderstand anything? $\endgroup$
    – IvanaGyro
    Commented Dec 25, 2022 at 17:34
  • $\begingroup$ My bad, I forgot the square… $\endgroup$
    – LPZ
    Commented Dec 25, 2022 at 19:02

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