# Why the Green's function is zero on the Dirichlet boundary surface?

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

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

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.

Examples

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.

• I am not good at linear algebra. Can you give me an example above this line? $$x=\sum_i a^ie_i$$ Commented Dec 25, 2022 at 13:56
• Does $v(x)$ satisfy $Lv(x) = 0$? Commented Dec 25, 2022 at 14:02
• 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.
– LPZ
Commented Dec 25, 2022 at 16:23
• 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? Commented Dec 25, 2022 at 17:34
• My bad, I forgot the square…
– LPZ
Commented Dec 25, 2022 at 19:02