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I am looking for intuition for Green function in terms of linear algebra.

I was satisfied with the intuition explained in the answer to this question until recently.

The linear algebra intuition my PDE professor gave me goes like this:

Let $\{e_k\}_{k=1}^n$ be the standard basis for $\mathbb R^n$ and $\langle\cdot,\cdot\rangle$ be the standard Euclidean inner product.

Consider a linear system $Lu=f$ where $u$ and $f$ are $n$ dimensional column vectors and $L$ is a $n\times n$ matrix. We want to solve for $u$.

A green function $G=\begin{bmatrix} G_1 & G_2 & \cdots & G_n \end{bmatrix}$ (this is a matrix) is a solution to $L^TG_k=e_k$.

One can find $k$th component of $u$ as follows:

$$ \begin{align*} \langle Lu, G_k\rangle &= \langle f, G_k\rangle \\ \langle u, L^TG_k\rangle &= \langle f, G_k\rangle \\ \langle u, e_k\rangle &= \langle G_k, f\rangle \\ u_k &= G_k^Tf \end{align*} $$

The solution $u$ is thus $u=\begin{bmatrix} u_1 & u_2 & \cdots & u_n \end{bmatrix}^T$.

This is very different from the answer in the linked page. I still don't get what is going on here? Why the Green function is sometimes defined as $L^T_k=e_k$? What's the purpose of computing the adjoint $L^T$?

Jimmy Yang
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  • Devoid of the ODE context, the condition $L^TG_k=e_k$ for all $k$ simply means $L^TG=I$, i.e., $G^TL=(L^TG)^T=I$ or equivalently, $L^{-1}=G^T$. Hence the solution to $Lu=f$ is $u=L^{-1}f=G^Tf$. – user1551 Mar 04 '25 at 09:33
  • @user1551 Thanks, it addressed the crux of my question but if $L^{-1}=G^T$ then $LL^{-1}=I$ should hold as well? I can’t see why it should be true given that $L^TG_k=e_k$. – Jimmy Yang Mar 04 '25 at 09:53
  • I think this is true if we demand that $G$ (and $L$?) to be symmetric. In ODE context this is equivalent to saying that $G$ must satisfy Maxwell reciprocity and $L$ is self adjoint, I believe. – Jimmy Yang Mar 04 '25 at 10:07

1 Answers1

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Formally, you wan to solve for $u$ with given $v$: $$ Lu = v $$ The natural approach is to use a left inverse $\hat L$ so that $\hat LL = I$, so that: $$ u = \hat Lv $$ In practice, to find the left inverse, you need to solve an inhomogeneous PDE, so you want the differential operator acting on the left. This is why you naturally rather consider the dual problem: $$ L^T\hat L^T = I $$ This leads to the introduction of Green's function $G = \hat L^T = (L^T)^{-1}$, which is the elementary solution of the dual problem. You can solve the original problem by: $$ u = G^Tv $$

For example, consider the simplest ODE: $$ \dot x+x = y $$ with initial condition $x(t=-\infty)=0$. By variation of the constant, you know that the general solution is: $$ x(t) = \int_{—\infty}^te^{\tau-t}y(\tau)d\tau = \int G^T(t,\tau)y(\tau)d\tau $$ so you identify: $$ G(\tau,t) = e^{\tau-t}H(t-\tau) $$ Using: $$ L = \frac d{dt}+1 \quad L^T = -\frac d{d\tau}+1 $$ you can check that: $$ L^TG = \left(-\frac\partial{\partial\tau}+1\right)\left(e^{\tau-t}H(t-\tau)\right) = \delta(\tau-t) $$

Hope this helps.

LPZ
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