Answer to Question #227031 in Differential Equations for smi

Question

Determine the Green function and express the solution as a definite integral

-(y^''+y)=f(x), y^'(0)=0, y(1)=0

Answer 1

We remind that the Green function is a function $G(x,s)$ such that $y(x)=\int_{x_0}^xG(x,s)f(s)ds$.

We rewrite the equation as $z=y'$ , $-z'-y=f(x)$. It can be rewritten as: $\left\{\right.\begin{matrix} y'=z \\ z'=-y-f(x) \end{matrix}$

It can be rewritten as: $\left(\begin{matrix} 1&0 \end{matrix}\right)\left(\begin{matrix} y'&z'\\ 0&0 \end{matrix}\right)=\left(\begin{matrix} 1&0 \end{matrix}\right)\left(\begin{matrix} y&z\\ 0&0 \end{matrix}\right)\left(\begin{matrix} 0&-1\\ 1&0 \end{matrix}\right)+\left(\begin{matrix} 1&0 \end{matrix}\right)\left(\begin{matrix} 0&-f(x)\\ 0&0 \end{matrix}\right)$

It is enough to solve the problem: $A'=A\sigma+F$, where $A=\left(\begin{matrix} y&z\\ 0&0 \end{matrix}\right)$, $\sigma=\left(\begin{matrix} 0&-1\\ 1&0 \end{matrix}\right)$ $F=\left(\begin{matrix} 0&-f(x)\\ 0&0 \end{matrix}\right)$. The solution of equation $A'=A\sigma$ is $A=Ce^{\sigma t}$ with matrix $C$. We suppose that $C=C(t)$. We substitute and get: $C'e^{\sigma t}+Ce^{\sigma t}{\sigma}=Ce^{\sigma t}{\sigma}+F$. We get: $C=\int Fe^{-\sigma t}dt+K$ with a constant matrix $K=\left(\begin{array}{cc}k_{11}&k_{12}\\k_{21}&k_{22}\end{array}\right).$ Thus, $A=(\int Fe^{-\sigma t}dt+K)e^{\sigma t}$. We notice that $y=\left(\begin{matrix} 1&0 \end{matrix}\right)\left(\begin{matrix} y&z\\ 0&0 \end{matrix}\right)\left(\begin{matrix} 1\\ 0 \end{matrix}\right)$ . Thus, $y=\left(\begin{matrix} 1&0 \end{matrix}\right)(\int Fe^{-\sigma t}dt+K)e^{\sigma t}\left(\begin{matrix} 1\\ 0 \end{matrix}\right)$. From the latter we can find the Green function.

Remind that $e^{\sigma t}=1+\sigma t+\frac{(\sigma t)^2}{2!}+\frac{(\sigma t)^3}{3!}+...$ Compute powers of matrix $\sigma$: $\sigma^2=\sigma=\left(\begin{matrix} -1&0\\ 0&-1 \end{matrix}\right)=-I$. Thus, $e^{-\sigma t}=I(1-\frac{(-t)^2}{2!}+\frac{(-t)^4}{4!}).-+...+\sigma(-t-\frac{(-t)^3}{3!}+\frac{(-t)^5}{5!})+...=Icos(\,-t)+\sigma\,sin(-\,t)$

As a result, we get:$y=\left(\begin{matrix} 1&0 \end{matrix}\right)(\int Fe^{-\sigma t}dt+K)e^{\sigma x}\left(\begin{matrix} 1\\ 0 \end{matrix}\right)=\left(\begin{matrix} 1&0 \end{matrix}\right)(\int \left(\begin{matrix} 0&-f(t)\\ 0&0 \end{matrix}\right)\left(\begin{matrix} cos\,t&sin\,t\\ -sin\,t&cos\,t \end{matrix}\right)dt+\left(\begin{array}{cc}k_{11}&k_{12}\\k_{21}&k_{22}\end{array}\right))e^{\sigma x}\left(\begin{matrix} 1\\ 0 \end{matrix}\right)=\left(\begin{matrix} 1&0 \end{matrix}\right) \left(\begin{matrix} \int f(t)\,sin\,t \,dt+k_{11}&-\int f(t)\,cos\,t\,dt+k_{12}\\ k_{21}&k_{22}\end{matrix}\right)\left(\begin{matrix} cos\,x&-sin\,x\\ sin\,x&cos\,x \end{matrix}\right)\left(\begin{matrix} 1\\ 0 \end{matrix}\right)=\left(\begin{matrix} 1&0 \end{matrix}\right)A\left(\begin{matrix} 1\\ 0 \end{matrix}\right)=\left(\begin{matrix} 1&0 \end{matrix}\right)\left(\begin{matrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{matrix}\right)\left(\begin{matrix} 1\\ 0 \end{matrix}\right)=a_{11}$.

$a_{11}= cos\,x(\int f(t)\,sin\,t \,dt+k_{11})+sin\,x(-\int f(t)\,cos\,t\,dt+k_{12})=k_{11} cos\,x+k_{12}sin\,x+\int f(t)(cos\,x\,sin\,t-sin\,xcos\,t) \,dt$

As we can see, $k_{11} cos\,x+k_{12}sin\,x$ is a solution of the homogenous problem $y''+y=0$. Therefore, we set $k_{11}=k_{12}=0$ and receive that the Green function is: $G(x,t)=cos\,x\,sin\,t-sin\,xcos\,t$