About Green function on ODE IVP, BVP

Question

I am sophomore student learning ODE.

While learning ODE, suddenly met Green function in IVP, BVP.

My 1st question is why it is introduced in IVP, BVP, such as: (Due to reduction of order & initial, boundary conditions??)

G(x,t)= \begin{cases} { y_1(t) y_2(x)\over W(t)} & \mbox{if } a\leq t\leq x \mbox{} \\ { y_1(x) y_2(t)\over W(t)}& \mbox{if } x\leq t\leq b\mbox{ } \end{cases}

Second, How can I understand that G(x,t) is dependent only on y₁ (x), y₂(x), but independent on f(x)? ( From the Initial or boundary values, get its Wronskian? ) ( From my book, A first course in diffrential equations 11th editions, Dennis G. Zill Chapter 4.8)

1st, 2nd questions might be related. Even though the answer is one, I wanted to ask whether it's the strongest, easiest way to use on IVP-BVP, compared to the others.

I'd really appreciate your sincere answer. My knowledge is just average sophomore early math major student.

Thank you so much for sparing your time.

Answer 1

We have a linear differential operator $L$, $L[y]=y''+py'+qy$, and want to solve $L[y]=f$ with homogeneous boundary conditions. If all the right sides $f$ that are "admissible" in some sense (for instance piecewise linear over a fixed subdivision) can be represented as a linear combinations of "atom" or kernel functions $f(x)=\sum_{k=1}^Nc_kg_k(x)$ then one would only have to solve $L[y_k]=g_k$ to construct any solution as $y(x)=\sum_{k=1}^N c_ky_k(x)$. So one can solve a large number of problems with the solutions of a finite number of BVPs.

If one carries the atomic decomposition construction to its extreme, then one ends up with the "sifting" property $f=\delta*f=\int_a^bf(s)\delta_s\,ds$. This reduces the solution of $L[y]=f$ for a general right side to the solution of $L[y_s]=δ_s$ for any $s\in[a,b]$. Then the general solution can be reconstructed by "summation" over $s$, $y(x)=\int_a^b f(s)y_s(x)\,ds$. Making $s$ into a function argument, this gives the Green function $G(x,s)=y_s(x)$.

Because of the properties of the Dirac delta distribution, one has $L[y_s](x)=0$ for $x\ne s$ and $y_s$ satisfies the homogeneous boundary conditions. Thus the separation into parts over $[a,s]$ and $[s,b]$ that are homogeneous solutions, at first independent. The continuity of the solution at $x=s$ then requires that $y_s(x)=C(s)·y_1(\min(s,x))·y_2(\max(s,x))$ where $y_1$ satisfies the left and $y_2$ the right boundary condition. Note that only one factor is "active", the other two are constants. The value of $C(s)$ follows from the condition that the first derivative $y_s'$ needs to have a unit jump at $x=s$, $$ y_s'(x)=C(s)·[y_1'(\min(s,x))·u(s-x)·y_2(\max(s,x))+y_1(\min(s,x))·y_2'(\max(s,x))·u(x-s)],\\ 1=y_s'(s+0)-y_s'(s-0)=C(s)·[y_1(s)·y_2'(s)-y_1'(s)·y_2(s)]=C(s)·W[y_1,y_2](s). $$