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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.

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  • $\begingroup$ Hi and welcome to Math.SE. It would be preferable to use MathJax for mathematical expressions. You can get started here, and a more complete reference can be found here. $\endgroup$ May 13 at 9:42
  • $\begingroup$ @user3733558 Thank you so much $\endgroup$
    – Jeon JI No
    May 13 at 10:51
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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 $x=s$, $$ y_s'(x)=C(s)·[y_1'(\min(s,x))·u(s-x)·y_2(\max(s,x))+y_1(\min(s,x))·y_2'(\max(s,x))·u(x-s)],\\ 1=y_s'(s+0)-y_s'(s-0)=C(s)·[y_1(s)·y_2'(s)-y_1'(s)·y_2(s)]=C(s)·W[y_1,y_2](s). $$

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  • $\begingroup$ My Intuitions said that Dirac-Delta function(or distribution) is suited for construction of Green function since y(s) = 0, y'(s) = 1.(I watched Khan Dirac.) But right now I can't fully understand 𝛿∗𝑓=𝑓, why y_s = 𝑦1(min(𝑠,𝑥))·𝑦2(max(𝑠,𝑥)) etc. Thank to you, I could learned that Green function is orignated from Dirac Delta distribution to solve IVP-BVP or other O-PDEs. If I have a chance to ask one more chance, which book can I learn Delta-dirac distribution property? I didn't have enough upvotes, can't upvote your answer. Again, Thank you for sincere reply. $\endgroup$
    – Jeon JI No
    May 13 at 15:17
  • $\begingroup$ Mathematically, distribution theory is a rather heavy topic that comes long after ODEs. So in any strict mathematical treatment the Green function will be introduced without or with unsatisfying motivation. So you will find this approach more likely in books "calculus for physicists" and/or "engineers". $\endgroup$ May 13 at 15:25
  • $\begingroup$ Thank you so much. I keep that in mind. $\endgroup$
    – Jeon JI No
    May 13 at 15:34

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