1. Reduction of order and Variation of Parameters. Define the operator
(a) Check that y(x) = x is a solution of ℒy = 0. Use reduction of order to find the general solution.
For $y(x)=x$, we have $ℒy(x)=x^2⋅0-x(x+2)⋅1+(x+2)⋅x=0$, so it is a solution. Let $y(x)=x⋅v(x)$, then $y'(x)=x⋅v'(x)+v(x),y''(x)=x⋅v''(x)+2v'(x)$\begin{align*}ℒy(x)&=x^3v''(x)+2x^2v'(x)-x^2(x+2)v'(x)-x(x+2)v(x)+x(x+2)v(x)\\&=x^3v''(x)-x^3v'(x)\end{align*}$⇒v''(x)=v'(x)⇒v(x)=C_1e^x+C_2⇒y(x)=C_1xe^x+C_2x$ is the general solution.
(b) Solve the following problem by variation of parameters:
$\left.\array{ℒ(y)=0\\y(1)=0}\right\}⇒y_1(x)=x(e^x-e)$ $\left.\array{ℒ(y)=0\\y(2)=0}\right\}⇒y_2(x)=x(e^x-e^2)$\begin{align*}W(ξ)&=y_1(ξ)y'_2(ξ)-y_2(ξ)y'_1(ξ)=(e-1)e^{ξ+1}ξ^2\\ c_1(x)&=\int_x^2\frac{f(ξ)y_2(ξ)}{P_2(ξ)W(ξ)}\mathrm{~d} ξ=-\frac{x+e^{2-x}-3}{(e-1) e}\\ c_2(x)&=\int_1^x\frac{f(ξ)y_1(ξ)}{P_2(ξ)W(ξ)}\mathrm{~d} ξ=\frac{x+e^{1-x}-2}{(e-1) e}\\ ⇒y(x)&=c_1(x)y_1(x)+c_2(x)y_2(x)=\frac{\left(e (x-2)+e^x-e^2 (x-1)\right) x}{(e-1) e}\end{align*}
2. Green’s function via variation of parameters.
(a) Use variation of parameters to solve the problem \[y''(x)-2y'(x)+2 y(x)=f(x), y(0)=0, y\left(\fracπ2\right)=0\tag∗\] where f is a given continuous function. Show that the solution can be written in the form \[y(x)=\int_0^{π/2} g(x,ξ) f(ξ) \mathrm{~d}ξ\] for a function g (the Green’s function) which you should determine.
$\left.\array{ℒ(y)=0\\y(0)=0}\right\}⇒y_1(x)=e^x\sin x$ $\left.\array{ℒ(y)=0\\y\left(\fracπ2\right)=0}\right\}⇒y_2(x)=e^x\cos x$\begin{align*}W(ξ)&=y_1(ξ)y'_2(ξ)-y_2(ξ)y'_1(ξ)=-e^{2ξ}\\ c_1(x)&=\int_x^{π/2}\frac{f(ξ)y_2(ξ)}{P_2(ξ)W(ξ)}\mathrm{~d} ξ=-\int_x^{π/2}f(ξ)e^{-ξ}\cosξ\mathrm{~d}ξ\\ c_2(x)&=\int_0^x\frac{f(ξ)y_1(ξ)}{P_2(ξ)W(ξ)}\mathrm{~d} ξ=-\int_0^xf(ξ)e^{-ξ}\sinξ\mathrm{~d}ξ\\y(x)&=c_1(x)y_1(x)+c_2(x)y_2(x)\\ ⇒g(x,ξ)&=\begin{cases}-e^{x-ξ}\sinξ\cos x&0<ξ<x\\-e^{x-ξ}\cosξ\sin x&x<ξ<π/2\end{cases} \end{align*}
(b) Evaluate the integral when f(x) = ex and check that the resulting expression for y does indeed satisfy (∗).
\begin{align*} c_1(x)&=\int_x^{π/2}{-\cosξ}\mathrm{~d}ξ=\sin x-1\\ c_2(x)&=\int_0^x{-\sinξ}\mathrm{~d}ξ=\cos x-1\\⇒y(x)&=c_1(x)y_1(x)+c_2(x)y_2(x)=e^x(1-\sin x-\cos x)\text{ does satisfy (∗)} \end{align*}
3. Variation of parameters for an Initial Value Problem. Consider the inhomogeneous ODE
ℒy(x) ≡ P2(x) y″(x) + P1(x) y′(x) + P0(x) y(x) = f(x)(∗)
for x > 0, with initial conditions y(0) = y′(0) = 0. Suppose that the homogeneous ODE ℒy = 0 has linearly independent solutions y1 and y2 satisfying y1(0) = 0 and y′2(0) = 0. Use variation of parameters to construct the solution to (∗), and determine the Green’s function g such that$$y(x)=\int_0^∞ g(x,ξ) f(ξ) \mathrm{~d}ξ$$
$y=c_1y_1+c_2y_2,c_1'y_1+c_2'y_2=0⇒y'=c_1y_1'+c_2y_2'$ $y'(0)=y_2'(0)=0≠y_1'(0)⇒c_1(0)=0$ $y(0)=y_1(0)=0≠y_2(0)⇒c_2(0)=0$ \begin{align*} c_1(x)&=-\int_0^x\frac{f(ξ)y_2(ξ)}{P_2(ξ)W(ξ)}\mathrm{~d} ξ\\ c_2(x)&=\int_0^x\frac{f(ξ)y_1(ξ)}{P_2(ξ)W(ξ)}\mathrm{~d} ξ\\ y(x)&=c_1(x)y_1(x)+c_2(x)y_2(x)\\ ⇒g(x,ξ)&=\begin{cases}\frac{-y_2(ξ)y_1(x)+y_1(ξ)y_2(x)}{P_2(ξ)W(ξ)}&0≤ξ≤x\\0&ξ ≥x\end{cases} \end{align*}
4. Adjoint problems. For each of the problems below, use the adjoint relation ⟨ℒy, w⟩ ≡ ⟨y, ℒ*w⟩ to determine the differential operator and boundary conditions for the adjoint problem. In each case state whether the operator and/or the full system is self-adjoint.
(a) ℒy = y″, 2y(0) + y′(0) = 0, y(1) + y′(1) = 0.
Integrating by parts\begin{aligned}⟨ℒy,w⟩=\int_0^1 y''(x) w(x) \d x & =-\int_0^1 y'(x) w'(x) \d x+\left[y'(x) w(x)\right]_0^1 \\ & =\int_0^1 y(x) w''(x) \d x+\left[y'(x) w(x)-y(x) w'(x)\right]_0^1≡\left<y, ℒ^* w\right>\end{aligned}To ensure that the boundary terms vanish\[\tag{23}y'(1)w(1)-y'(0)w(0)-y(1)w'(1)+y(0)w'(0)=0\]Substitute $y'(0)=-2y(0),y'(1)=-y(1)$ in (23)\[-y(1)[w(1)+w'(1)]+y(0)[2w(0)+w'(0)]=0\]The adjoint problem is\[ℒ^*w=w'',2w(0)+w'(0)=0,w(1)+w'(1)=0\]So the full system is self-adjoint.
(b) ℒy = y″, 2y(0) + y′(1) = 0, y(1) + y′(0) = 0.
Substitute $y'(1)=-2y(0),y'(0)=-y(1)$ in (23)\[y(0)[w'(0)-2w(1)]+y(1)[w(0)-w'(1)]=0\]The adjoint problem is\[ℒ^*w=w'',w'(0)-2w(1)=0,w(0)-w'(1)=0\]So the operator is self-adjoint but not the boundary condition.
(c) ℒy = y⁗− y′, y′(0) − y″(0) = 0, y‴(0) = 0, y(1) = 0, y′(1) − y‴(1) = 0.
Integrating by parts\begin{aligned}⟨ℒy,w⟩&=\int_0^1 [y''''(x)-y'(x)] w(x) \d x\\&=\int_0^1 y(x) w''''(x) \d x+\left[y'''(x) w(x)-y''(x) w'(x)+y'(x) w''(x)-y(x) w'''(x)\right]_0^1\\& +\int_0^1 y(x) w'(x) \d x-\left[y(x) w(x)\right]_0^1≡\left<y, ℒ^* w\right>\end{aligned}To ensure that the boundary terms vanish\begin{multline*}y'''(1) w(1)-y''(1) w'(1)+y'(1) w''(1)-y(1) w'''(1)-y(1) w(1)\\=y'''(0) w(0)-y''(0) w'(0)+y'(0) w''(0)-y(0) w'''(0)-y(0) w(0)\end{multline*}Substitute $y''(0)=y'(0),y'''(0)=0,y(1)=0,y'''(1)=y'(1)$\[y'(1)[w(1)+w''(1)]-y''(1) w'(1)=y'(0)[-w'(0)+w''(0)]-y(0)[w'''(0)+w(0)]\]The adjoint problem is\[ℒ^*w=w''''+w',w(1)+w''(1)=0,w'(1)=0,w'(0)-w''(0)=0,w'''(0)+w(0)=0\]So neither the operator nor the boundary condition is self-adjoint.
5. Sturm–Liouville form. Consider the general second order eigenvalue problem
ℒy(x) ≡ A(x)y″(x) + B(x)y′(x) + C(x)y(x) = λy(x), a < x < b (∗)
where A(x), B(x), C(x) are given functions with A(x) ≠ 0 for x ∈ [a, b].
(a) Use an integrating factor to show that (∗) can always be put into Sturm–Liouville form,$$\hatℒy(x)≡-\left(p(x)y'(x)\right)'+q(x)y(x)=λr(x)y(x),$$where p(x), q(x), r(x) should be determined in terms of A(x), B(x), C(x).
Multiply by $-\frac pA$:\[-py''-\frac BApy'-\frac CApy=-λ\frac pAy\]The first two terms can now be combined into an exact derivative −(py′)′ if p satisfies$$p'(x)=\frac{B(x)}{A(x)}p(x)⇒p(x)=e^{∫\frac{B(x)}{A(x)}dx}$$\begin{aligned}q(x) & =-\frac CAp\\ r(x) & =-\frac pA\end{aligned}
(b) Show that the operator $\hatℒ$ is self-adjoint.
\begin{align*}⟨u,\hatℒv⟩&=\int_a^bu(x)\hatℒv(x)\d x\\&=\int_a^b{-\left(p(x)v'(x)\right)'u(x)}+q(x)u(x)v(x)\d x\\&=\left.-p(x)u(x)v'(x)\right|_a^b+\int_a^bp(x)u'(x)v'(x)+q(x)u(x)v(x)\d x\\⟨\hatℒu,v⟩&=\int_a^bv(x)\hatℒu(x)\d x\\ &=\int_a^b{-\left(p(x)u'(x)\right)'v(x)}+q(x)u(x)v(x)\d x\\ &=\left.-p(x)u'(x)v(x)\right|_a^b+\int_a^bp(x)u'(x)v'(x)+q(x)u(x)v(x)\d x\end{align*}
(c) If $ℒy(x)=f(x)$ for some function f, what is the equivalent Sturm-Liouville problem?
\begin{aligned}p(x)&=e^{∫\frac{B(x)}{A(x)}dx}\\q(x) &=-\frac{C(x)}{A(x)}p(x)\\ r(x) & =-\frac{p(x)}{A(x)}f(x)\end{aligned}
6. FAT and existence. Determine the parameter values (A, B) that yield existence of a solution for each of the following inhomogeneous BVPs.
(a) For 0 ≤ x ≤ 2π:
Integrating by parts\begin{aligned}⟨ℒy,w⟩&=\int_0^{2π}[y''(x)+y(x)] w(x) \d x\\&=\int_0^{2π}y(x)[w''(x)+w(x)]\d x+\left[y'(x)w(x)-y(x)w'(x)\right]_0^{2π}≡\left<y, ℒ^* w\right>\end{aligned}To ensure that the boundary terms vanish\[y'(2π)w(2π)-y(2π)w'(2π)=y'(0)w(0)-y(0)w'(0)\]Using y(0) = y(2π), y′(0) = y′(2π)\[y'(0)[w(2π)-w(0)]-y(0)[w'(2π)-w'(0)]=0\]The adjoint problem\[ℒ^*w=w''+w,w(2π)-w(0)=0,w'(2π)-w'(0)=0\]has solutions w(x) = sin x, cos x.\begin{gather*} \int_0^{2π}\left(A\sin x+B\cos x+2\sin(x+\fracπ3)+\sin^3x\right)\sin x\d x=πA+0+π+\frac{3π}4=0⇒A=-\frac74\\ \int_0^{2π}\left(A\sin x+B\cos x+2\sin(x+\fracπ3)+\sin^3x\right)\cos x\d x=0+πB+\sqrt3π+0=0⇒B=-\sqrt3 \end{gather*}
(b) For 0 ≤ x ≤ 1:
Integrating by parts\begin{aligned}⟨ℒy,w⟩&=\int_0^1[y''(x)+2y'(x)+y(x)]w(x) \d x\\&=\int_0^1y(x)[w''(x)-2w'(x)+w(x)]\d x\\& +\left[y'(x)w(x)-y(x)w'(x)+2y(x)w(x)\right]_0^1≡\left<y, ℒ^* w\right>\end{aligned}Put the boundary terms into the form $(K^*_1 w)(B_1y) + (K^*_2 w)(B_2y) + (K_1y)(B^*_1w) + (K_2y)(B^*_2w)$ \begin{align*} &[y'(1)w(1)-y(1)w'(1)+2y(1)w(1)]-[y'(0)w(0)-y(0)w'(0)+2y(0)w(0)]\\ &=-w(0)[y'(0)+y(0)]+w(1)[y'(1)+y(1)]\\ &+y(0)[w'(0)-w(0)]+y(1)[w(1)-w'(1)] \end{align*}The adjoint problem\[ℒ^*w=w''-2w'+w,w'(0)-w(0)=0,w(1)-w'(1)=0\]has a non-trivial solution w(x) = ex. So\[⟨ℒy,w⟩=\int_0^1e^x\d x=e-1\]By 2.34\[e-1=-w(0)[y'(0)+y(0)]+w(1)[y'(1)+y(1)]=-A+3e⇒A=2e+1\]
Just make sure you understand the process of dealing with Q6—we're trying to find solvability conditions, and so ultimately we have a degree of freedom to deal with any issues.The usefulness of this process is that we can deal with the homogeneous case to lead to the homogeneous adjoint problem and then deal with the inhomogeneous boundary conditions.