Question: (a) Given that ${\mathit{\psi }}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$and y₂(x)are solutions of (12.19)with f(x)=0, and that${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{=}\mathbf{0}$, find the Green function [as in (12.11) to (12.16)] and so obtain the solution (12.20). Then find the particular solution (12.21) as discussed for (12.18) and (12.21).

(b) The method of variation of parameters is an elementary way of finding a particular solution of (12.19) when you know the solutions of the homogeneous equation. Show as follows that this method leads to the same result (12.21) as the Green function method. Start with the known solution of the homogeneous equation, say$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}}$and allow the "constants" to be functions ofxto be determined so thatysatisfies (12.19). (The c's are the "parameters" which are to be "varied" in the expression "variation of parameters".) You want to find y'and y"to substitute into (12.19). First find y'and set the sum of the terms involving derivatives of the c's equal to zero. Differentiate the rest of y'again to get y". Now substitute y, y'and y"into (12.19)and use the fact that y₁and y₂both satisfy the homogeneous equation [that is, (12.19) with f(x) = 0 }. You should have the two equations:

$\begin{array}{l}{\mathbf{c}}_{\mathbf{1}}^{\mathbf{\text{'}}}{\mathbf{\xi }}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}^{\mathbf{\text{'}}}{\mathbf{y}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\\ {\mathbf{c}}_{\mathbf{1}}^{\mathbf{\text{'}}}{\mathbf{\xi }}_{\mathbf{1}}^{\mathbf{\text{'}}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{y}}_{\mathbf{2}}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\end{array}$

Solve this pair of equations for C₁and C₂'[say by determinants, and note that the denominator determinant is the Wronskian as in (12.20) and (12.21)]. Write the indefinite integrals for c₁and c₂, and write$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}}$to get (12.21).

(a) The given expressions are

$G\left(x,x^{\text{'}}\right)=\left\{\begin{array}{ll}A\left(x^{\text{'}}\right)y_1\left(x\right),& a<x<x^{\text{'}}<b\\ B\left(x^{\text{'}}\right)y_2\left(x\right),& a<x<x^{\text{'}}<b\end{array}\dots \dots \left(2\right)\right.$

(b) The given expressions are

$G\left(x,x^{\text{'}}\right)=\left\{\begin{array}{ll}A\left(x^{\text{'}}\right)y_1\left(x\right),& a<x<x^{\text{'}}<b\\ B\left(x^{\text{'}}\right)y_2\left(x\right),& a<x<x^{\text{'}}<b\end{array}\right.$

In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions.

(a) The Green's function satisfies:

$G^{\text{'}\text{'}}\left(x,x^{\text{'}}\right)+p\left(x\right)G^{\text{'}}\left(x,x^{\text{'}}\right)+q\left(x\right)G\left(x,x^{\text{'}}\right)=\delta \left(x-x^{\text{'}}\right).$

Now, for x ≠ x' we have:

$G^{\text{'}\text{'}}\left(x,x^{\text{'}}\right)+p\left(x\right)G^{\text{'}}\left(x,x^{\text{'}}\right)+q\left(x\right)G\left(x,x^{\text{'}}\right)=0$

and from the problem we have two solutions of this equations which we call y₁(x) and y₂(x). Therefore, we can write the solution of Green's function as:

$G\left(x,x^{\text{'}}\right)=\left\{\begin{array}{ll}A\left(x^{\text{'}}\right)y_1\left(x\right),& a<x<x^{\text{'}}<b\\ B\left(x^{\text{'}}\right)y_2\left(x\right),& a<x^{\text{'}}<x<b\end{array}\right.$

For x = x' we must have$G_1\left(x^{\text{'}},x^{\text{'}}\right)=G_2\left(x^{\text{'}},x^{\text{'}}\right)$ where 1 and 2 denote the two cases$\left(a<x<x^{\text{'}}<b\right.and\left.a<x^{\text{'}}<x<b\right)$. Therefore, we have:

$\left.a<x^{\text{'}}<x<b\right)$

The next thing we need to do is find the derivative:

$\frac{\text{d}G\left(x,x^{\text{'}}\right)}{\text{d}x}=\left\{\begin{array}{ll}A\left(x^{\text{'}}\right)y_1^{\text{'}}\left(x\right),& a<x<x^{\text{'}}<b\\ B\left(x^{\text{'}}\right)y_2^{\text{'}}\left(x\right),& a<x^{\text{'}}<x<b\end{array}\right.$

We have a jump in the derivative between these two cases and we can evaluate it by integrating the equation 1 from$x=x^{\text{'}}-ϵ$to$x=x^{\text{'}}+ϵ$and letting$ϵ\to 0$. We get:

${\left.\left(\frac{\text{d}G\left(x,x^{\text{'}}\right)}{\text{d}x}\right)\right|}_{x=x^{\text{'}}-ϵ}^{x=x^{\text{'}}+ϵ}+{\int }_{x^{\text{'}}-ϵ}^{x^{\text{'}}+ϵ}p\left(x\right)\frac{\text{d}G\left(x,x^{\text{'}}\right)}{\text{d}x}\text{d}x+{\int }_{x^{\text{'}}-ϵ}^{x^{\text{'}}+ϵ}q\left(x\right)G\left(x,x^{\text{'}}\right)\text{d}x={\int }_{x^{\text{'}}-ϵ}^{x^{\text{'}}+ϵ}\delta \left(x-x^{\text{'}}\right)\text{d}x=1.$

Third term is equal to 0 because of the property$G_1\left(x^{\text{'}},x^{\text{'}}\right)=G_2\left(x^{\text{'}},x^{\text{'}}\right)$and the fact that we are letting$ϵ\to 0$. For the second term we can use integration by parts:

$\begin{array}{l}{\int }_{x^{\text{'}}-ϵ}^{x^{\text{'}}+ϵ}p\left(x\right)\frac{\text{d}G\left(x,x^{\text{'}}\right)}{\text{d}x}\text{d}x=\mid u=p\left(x\right),\text{d}v=\frac{\text{d}G\left(x,x^{\text{'}}\right)}{\text{d}x}\text{d}x;\text{d}u=p^{\text{'}}\left(x\right)\text{d}x,\\ v=G\left(x,x^{\text{'}}\right)\mid ={\left.\left[p\left(x\right)G\left(x,x^{\text{'}}\right)\right]\right|}_{x=x^{\text{'}}-ϵ}^{x=x^{\text{'}}+ϵ}-{\int }_{x^{\text{'}}-ϵ}^{x^{\text{'}}+ϵ}{p}^{\text{'}}\left(x\right)G\left(x,x^{\text{'}}\right)\text{d}x.\end{array}$

The second term in the upper equation is also equal to 0 for the same reason as before. Therefore, if we now insert this into the previous equation we get:

${\left.\left(\frac{\text{d}G\left(x,x^{\text{'}}\right)}{\text{d}x}+p\left(x\right)G\left(x,x^{\text{'}}\right)\right)\right|}_{x=x^{\text{'}}-ϵ}^{x=x^{\text{'}}+ϵ}=1.$

We can now evaluate this equation by inserting

${\left.\left(\frac{\text{d}G\left(x,x^{\text{'}}\right)}{\text{d}x}+p\left(x\right)G\left(x,x^{\text{'}}\right)\right)G\left(x,x^{\text{'}}\right)\right|}_{x=x^{\text{'}}-ϵ}^{x=x^{\text{'}}+ϵ}=1.$and $G^{\text{'}}\left(x,x^{\text{'}}\right)$. We get:

$\begin{array}{l}B\left(x^{\text{'}}\right)\frac{\text{d}{y}_2}{\text{d}x}\left(x^{\text{'}}\right)-A\left(x^{\text{'}}\right)\frac{\text{d}{y}_1}{\text{d}x}\left(x^{\text{'}}\right)+p\left(x^{\text{'}}\right)B\left(x^{\text{'}}\right)y_2\left(x^{\text{'}}\right)-p\left(x^{\text{'}}\right)A\left(x^{\text{'}}\right)y_1\left(x^{\text{'}}\right)\\ =1-A\left(x^{\text{'}}\right)\left[p\left(x^{\text{'}}\right)y_1\left(x^{\text{'}}\right)+\frac{\text{d}{y}_1}{\text{d}x}\left(x^{\text{'}}\right)\right]+B\left(x^{\text{'}}\right)\left[p\left(x^{\text{'}}\right)y_2\left(x^{\text{'}}\right)+\frac{\text{d}{y}_2}{\text{d}x}\left(x^{\text{'}}\right)\right]=1\end{array}$

We need to solve equations 2 and 3 for functions A(x') and B(x') to get the final Green's function. We substitute B(x') = A(x')y₁(x') / y₂(x') from equation 2 into the equation 3 to get:

$\begin{array}{c}-A\left(x^{\text{'}}\right)\left[p\left(x^{\text{'}}\right)y_1\left(x^{\text{'}}\right)+\frac{\text{d}{y}_1}{\text{d}x}\left(x^{\text{'}}\right)-\frac{y_1\left(x^{\text{'}}\right)}{y_2\left(x^{\text{'}}\right)}p\left(x^{\text{'}}\right)y_2\left(x^{\text{'}}\right)-\frac{y_1\left(x^{\text{'}}\right)}{y_2\left(x^{\text{'}}\right)}\frac{\text{d}{y}_2}{\text{d}x}\left(x^{\text{'}}\right)\right]\\ =1A\left(x^{\text{'}}\right)\\ =\frac{y_2\left(x^{\text{'}}\right)}{y_1\left(x^{\text{'}}\right)\frac{\text{d}{y}_2}{\text{d}x}\left(x^{\text{'}}\right)-y_2\left(x^{\text{'}}\right)\frac{\text{d}{y}_1}{\text{d}x}\left(x^{\text{'}}\right)}B\left(x^{\text{'}}\right)\\ =A\left(x^{\text{'}}\right)\frac{y_1\left(x^{\text{'}}\right)}{y_2\left(x^{\text{'}}\right)}\\ =\frac{y_1\left(x^{\text{'}}\right)}{y_1\left(x^{\text{'}}\right)\frac{\text{d}{y}_2}{\text{d}x}\left(x^{\text{'}}\right)-y_2\left(x^{\text{'}}\right)\frac{\text{d}{y}_1}{\text{d}x}\left(x^{\text{'}}\right)}.\end{array}$

Therefore our Green's function is:

$G\left(x,x^{\text{'}}\right)=\left\{\begin{array}{ll}\frac{y_2\left(x^{\text{'}}\right)}{W\left(x^{\text{'}}\right)}{y}_1\left(x\right),& a<x<x^{\text{'}}<b\\ \frac{y_1\left(x^{\text{'}}\right)}{W\left(x^{\text{'}}\right)}{y}_2\left(x\right),& a<x^{\text{'}}<x<b\end{array}\right.$

The solution of the differential equation is therefore:

To find the particular solution we use the discussion above the equation $12.18$ from the chapter. In other words, we are subtracting the homogeneous solution by inverting the limits of the second integral and omitting the constant limits. Applying this we get the particular solution:

$y_p\left(x\right)=y_2\left(x\right)\int \frac{y_1\left(x\right)f\left(x\right)}{W\left(x\right)}\text{d}x-y_1\left(x\right)\int \frac{y_2\left(x\right)f\left(x\right)}{W\left(x\right)}\text{d}x$

(b) Let y₁(x) and y₂(x) be the homogeneous solutions of the differential equation. We assume our particular solution to be:

$y_p\left(x\right)=c_1\left(x\right)y_1\left(x\right)+c_2\left(x\right)y_2\left(x\right).$

To find the coefficients we need to insert this solution into the differential equation. Let's find the needed derivatives first:

$$\begin{gathered} y_p^{\text{'}}=c_1^{\text{'}}{y}_1+c_1{y}_1^{\text{'}}+c_2^{\text{'}}{y}_2+c_2{y}_2^{\text{'}} \\ {c}_1^{\text{'}}{y}_1+c_2^{\text{'}}{y}_2=0 \end{gathered}$$

and get the first derivative to be:

$y_p^{\text{'}}=c_1{y}_1^{\text{'}}+c_2{y}_2^{\text{'}}$

The second derivative is:

$y_p^{\text{'}\text{'}}=c_1^{\text{'}}{y}_1^{\text{'}}+c_1{y}_1^{\text{'}\text{'}}+c_2^{\text{'}}{y}_2^{\text{'}}+c_2{y}_2^{\text{'}\text{'}}.$

We can now insert this into the differential equation:

$$\begin{gathered} c_1^{\text{'}}{y}_1^{\text{'}}+c_1{y}_1^{\text{'}\text{'}}+c_2^{\text{'}}{y}_2^{\text{'}}+c_2{y}_2^{\text{'}\text{'}}+p\left(x\right)\left(c_1{y}_1^{\text{'}}+c_2{y}_2^{\text{'}}\right)+q\left(x\right)\left(c_1{y}_1+c_2{y}_2\right)=f\left(x\right) \\ {c}_1\left(y_1^{\text{'}\text{'}}+p\left(x\right)y_1^{\text{'}}+q\left(x\right)y_1\right)+c_2\left(y_2^{\text{'}\text{'}}+p\left(x\right)y_2^{\text{'}}+q\left(x\right)y_2\right)+c_1^{\text{'}}{y}_1^{\text{'}}+c_2^{\text{'}}{y}_2^{\text{'}}=f\left(x\right). \end{gathered}$$

We can use the fact that y₁ and y₂ satisfy the homogeneous equation and put the terms in the parentheses equal to 0. We get:

$c_1^{\text{'}}{y}_1^{\text{'}}+c_2^{\text{'}}{y}_2^{\text{'}}=f\left(x\right)$

To get the coefficients we need to solve the equations 4 and 5 for their derivatives. We use the Cramer's rule to do that and get:

To get the coefficients we need to solve the equations 4 and 5 for their derivatives. We use the Cramer's rule to do that and get:

$\begin{array}{l}{c}_1^{\text{'}}\left(x\right)=\frac{\left|\begin{array}{cc}0& y_2\\ f\left(x\right)& y_2^{\text{'}}\end{array}\right|}{\left|\begin{array}{ll}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right|}=\frac{-y_2\left(x\right)f\left(x\right)}{W\left(x\right)}\\ c_2^{\text{'}}\left(x\right)=\frac{\left|\begin{array}{ll}{y}_1& 0\\ y_1^{\text{'}}& f\left(x\right)\end{array}\right|}{\left|\begin{array}{ll}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right|}=\frac{y_1\left(x\right)f\left(x\right)}{W\left(x\right)}\end{array}$

where we recognized the Wronskian in the denominator. Integrating these equations we get the coefficients to be:

$\begin{array}{l}{c}_1\left(x\right)=\int \frac{-y_2\left(x\right)f\left(x\right)}{W\left(x\right)}\text{d}x,\\ c_2\left(x\right)=\int \frac{y_1\left(x\right)f\left(x\right)}{W\left(x\right)}\text{d}x.\end{array}$

We can now insert this into y_p and get the particular solution:

$y_p\left(x\right)=c_1\left(x\right)y_1\left(x\right)+c_2\left(x\right)y_2\left(x\right)$