The general idea of Green’s functions is to find the solution with forcing on the RHS
(ie if the forcing is applied at only one point),
and then superposition all of those solutions to get a complete function.

In practice:
Let be some linear operator only dependent on .
If solves for every fixed ,
then solves

Heat equation



We split the problem into two:


  1. Then the solution is given by superposition.
    First problem is easily solved by taking Fourier transform in and then using convolution theorem.
    Second problem has the same approach:


    Hence, by initial conditions, we can obtain

    And now apply convolution theorem UNDER the integral (so the final sol has a double integral).

Laplace equation

We can solve on some domain by finding the Green’s function for every . With some trouble we get:

up to some constants. Hence
If we further want on we first find for , and then for each find s.t. on . Think of this as placing charges s.t. the electrical potentials cancel out. Now inside because is always zero because .

Wave equation


Solved very similar to Heat equation, by separating the problem into bits with 0 forcing and 0 initial data. The inverse Fourier is UGLY to compute generally … but it involves some in 1 dimension or some in 3 dimensions.

Wave on halfline


Firstly, by appropriate substitution, transform this problem into

u(0,t)=0\\u(x,0)=f(x)\\u_t(x,0)=g(x)\end{cases}$$ Now consider the odd extensions $F_{odd}, f_{odd}, g_{odd}$ defined for $x\in R$. Because they are odd, they should cancel out at $x=0$ at give appropriate initial condition $u(0,t)=0$. Hence solve the wave on full line.