13 Green functions notes

\(\blacksquare \) Green function is what is called impulse response in control. But it is more general, and can be used for solving PDE also.

Given a differential equation with some forcing function on the right side. To solve this, we replace the forcing function with an impulse. The solution of the DE now is called the impulse response, which is the Green’s function of the differential equation.

Now to find the solution to the original problem with the original forcing function, we just convolve the Green function with the original forcing function. Here is an example. Suppose we want to solve   \(L\left [ y\left ( t\right ) \right ] =f\left ( t\right ) \) with zero initial conditions. Then we solve \(L\left [ g\left ( t\right ) \right ] =\delta \left ( t\right ) \). The solution is \(g\left ( t\right ) \). Now \(y\left ( t\right ) =g\left ( t\right ) \circledast f\left ( t\right ) \). This is for initial value problem.  For example. \(y^{\prime }\left ( t\right ) +kx=e^{at}\), with \(y\left ( 0\right ) =0\). Then we solve \(g^{\prime }\left ( t\right ) +kg=\delta \left ( t\right ) \). The solution is \(g\left ( t\right ) =\left \{ \begin {array} [c]{cc}e^{-kt} & t>0\\ 0 & t<0 \end {array} \right . \), this is for causal system. Hence \(y\left ( t\right ) =g\left ( t\right ) \circledast f\left ( t\right ) \). The nice thing here, is that once we find \(g\left ( t\right ) \), we can solve \(y^{\prime }\left ( t\right ) +kx=f\left ( t\right ) \) for any \(f\left ( t\right ) \) by just convolving the Green function (impulse response) with the new \(f\left ( t\right ) \).

\(\blacksquare \) We can think of Green function as an inverse operator. Given \(L\left [ y\left ( t\right ) \right ] =f\left ( t\right ) \), we want to find solution \(y\left ( t\right ) =\int _{-\infty }^{\infty }G\left ( t;\tau \right ) f\left ( \tau \right ) d\tau \). So in a sense, \(G\left ( t;\tau \right ) \) is like \(L^{-1}\left [ y\left ( t\right ) \right ] \).

\(\blacksquare \) Need to add notes for Green function for Sturm-Liouville boundary value ODE. Need to be clear on what boundary conditions to use. What is B.C. is not homogeneous?

\(\blacksquare \) Green function properties:

  1. \(G\left ( t;\tau \right ) \) is continuous at \(t=\tau \). This is where the impulse is located.
  2. The derivative \(G^{\prime }\left ( t\right ) \) just before \(t=\tau \) is not the same as \(G^{\prime }\left ( t\right ) \) just after \(t=\tau \). i.e. \(G^{\prime }\left ( t;t-\varepsilon \right ) -G^{\prime }\left ( t;t+\varepsilon \right ) \neq 0\). This means there is discontinuity in derivative.
  3. \(G\left ( t;\tau \right ) \) should satisfy same boundary conditions as original PDE or ODE (this is for Sturm-Liouville or boundary value problems).
  4. \(L\left [ G\left ( t;\tau \right ) \right ] =0\) for \(t\neq \tau \)
  5. \(G\left ( x;\tau \right ) \) is symmetric. i.e. \(G\left ( x;\tau \right ) =G\left ( \tau ;x\right ) \).

\(\blacksquare \) When solving for \(G\left ( t;\tau \right ) \), in context of 1D, hence two boundary conditions, one at each end, and second order ODE (Sturm-Liouville), we now get two solutions, one for \(t<\tau \) and one for \(t>\tau \).

So we have \(4\) constants of integrations to find (this is for second order ODE) not just two constants as normally one would get , since now we have 2 different solutions. Two of these constants from the two boundary conditions, and two more come from property of Green function as mentioned above. \(G\left ( t;\tau \right ) =\left \{ \begin {array} [c]{cc}A_{1}y_{1}+A_{2}y_{2} & 0<t<\tau \\ A_{3}y_{1}+A_{4}y_{2} & \tau <t<L \end {array} \right . \)