Green's function

Consider \(ℒy = P_2 y'' + P_1 y' + P_0 y = f (x)\) on \(a<x <b\), with homogeneous boundary conditions \(y (a) = 0, y (b) = 0\).

Using variation of parameters we found the solution
\(y (x) = \int_a^b g (x, ξ ) f ( ξ ) \mathrm{d} ξ \)
where\[\tag1 g (x, ξ ) =\begin{cases}\frac{y_1 ( ξ ) y_2 (x)}{P_2 ( ξ ) W ( ξ )} & a< ξ <x<b,\\ \frac{y_2 ( ξ ) y_1 (x)}{P_2 ( ξ ) W ( ξ )} & a<x< ξ <b. \end{cases}\]

Here \(y_1\) and \(y_2\) are linearly independent solutions of the homogeneous ODE \(ℒy = 0\) satisfying one boundary condition each, i.e. \(y_1 (a) = 0, y_2 (b) = 0\).

The function \(g (x, ξ )\) is called a Green's function. The operation of multiplying by \(g\) and integrating can be thought of the inverse of the operator \(ℒ\).

Delta function

The delta function \(δ(x)\) has the properties that \(δ(x) = 0\) for all \(x \neq 0\) but \(\int_{- ∞}^{∞} δ(x) \mathrm{d} x = 1\).

This is not strictly a function, but rather a distribution, and is defined by its action on smooth test functions \(ϕ(x)\):

\[\int_{- ∞}^{∞} δ(x - ξ ) ϕ( ξ ) \mathrm{d} ξ = ϕ(x)\]

This is also called the sifting property.

\(δ(x)\) can be thought of as the derivative of the Heaviside function \(H (x) =\begin{cases}0 & x<0\\ 1 & x⩾0\end{cases}\)

Aside: \(δ(x)\) as be thought of the limit of an approximating sequence \(δ_n (x)\) as \(n→∞\).

Eg. \(δ_n (x) =\begin{cases}0 & | x | > \frac{1}{n}\\ \frac{n}{2} & | x |⩽\frac{1}{n}\end{cases}\) or \(δ_n (x) = \frac{n}{\sqrtπ} \mathrm{e}^{- n^2 x^2}\)

Delta-function construction of the Green's function

If we are to write the solution of \(ℒy = f\) as \(y (x) = \int_a^b g (x, ξ ) f ( ξ ) \mathrm{d} ξ \), then we need \(ℒy = \int_a^b ℒ_x g (x, ξ ) f ( ξ ) \mathrm{d} ξ = f (x)\). So we need

\(ℒ_x g (x, ξ ) = δ(x - ξ )\)
(*)

For \(y (x)\) to satisfy the boundary conditions \(B_1 y = 0\) and \(B_2 y = 0\), we need \(g (x, ξ )\) to satisfy the same condition (as a function of \(x\))
\(B_1 g = 0, B_2 g = 0\)

The Green's function can be thought of as the solution for a point force at \(x = ξ \).

To actually solve (*), note that \(δ(x - ξ ) = 0\) except at \(x = ξ \), so we can use the solution of the homogeneous equation
\(ℒ_x g = 0\)
on each interval \(a<x< ξ \) and \( ξ <x<b\).

We must also ensure that \(g\) is continuous at \(x = ξ \) [else \(g_x\) would be a delta function, and \(g_{x x}\) would have a worse singularity that wouldn't balance in (*)]. But \(g_x\) would be discontinuous at \(x = ξ \). Integrating (*) over \([ ξ - ε, ξ + ε]\):

\[\underbrace{\int_{ ξ - ε}^{ ξ + ε} P_2 g_{x x} \mathrm{d} x}_{\xrightarrow[ε→0]{} P_2 [g_x]_{ ξ -}^{ ξ +}} + \underbrace{\int_{ ξ - ε}^{ ξ + ε} P_1 g_x + P_0 g \mathrm{d} x}_{\xrightarrow[ε→0]{} 0} = \underbrace{\int_{ ξ - ε}^{ ξ + ε} δ(x - ξ ) \mathrm{d} ξ }_1\]So we need
\([g_x]_{ ξ -}^{ ξ +} = \frac{1}{P_2}\)
and \([g]_{ ξ -}^{ ξ +} = 0\)

Check (1) satisfies those conditions. But why don't we find \(g\) by solving them directly?

Example. \(ℒg = - y''\) on \(0<x<1, y (0) = y (1) = 0\)

The Green's function satisfies \(- g_{x x} = δ(x - ξ )\) with \(g (0, ξ ) = 0, g (1, ξ ) = 0\).

So \(g (x, ξ ) =\begin{cases}Ax & x< ξ \\ B (1 - x) & x > ξ \end{cases}\)(\(A, B\) may depend on \( ξ \))

Continuity of \(g\) at \( ξ ⇒A ξ = B (1 - ξ )\) and we need \(- [g_x]_{ ξ -}^{ ξ +} = 1⇒B + A = 1\)

Hence \(B = ξ , A = 1 - ξ \). So \(g (x, ξ ) =\begin{cases}x (1 - ξ ) & x< ξ \\ ξ (1 - x) & x > ξ \end{cases}\)

Then \(ℒy = f\) has solution \(y (x) = \int_0^1 g (x, ξ ) f ( ξ ) \mathrm{d} ξ \)

Example. Find the Green's function for

\[y'' + y = f (x) \quad 0<x<\fracπ2 \quad y (0) = 0 \quad y \left( \fracπ2 \right) = 0\]

\(g (x, ξ )\) satisfies \(g_{x x} + g = δ(x - ξ )\) with \(g (0, ξ ) = 0\) and \(g \left( \fracπ2, ξ \right) = 0\). So \(g (x, ξ ) =\begin{cases}A \sin x & 0<x< ξ \\ B \cos x & ξ <x<\fracπ2\end{cases}\)

Continuity at \( ξ ⇒A \sin ξ = B \cos ξ \)

and we need \([g_x]_{ ξ -}^{ ξ +} = 1⇒- B \sin ξ - A \cos ξ = 1⇒B = - \sin ξ , A = - \cos ξ \)

So \(g (x, ξ ) =\begin{cases}- \cos ξ \sin x & 0<x < ξ <\fracπ2\\ - \sin ξ \cos x & 0< ξ <x< \fracπ2\end{cases}\)

which is the same as what we found using variation of parameters last week.