Past exam of the differential equations 2 course of the University of Oxford / 2023 / 1

The characteristic equation of the linear ordinary differential equation is $r^2+2r+2=0$, with roots $-1\pm i$. A real pair of homogeneous solutions is thereforeTheir Wronskian, with the convention $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, isThus the two functions are linearly independent throughout the interval and span its homogeneous solutions.

Use the unweighted inner product $⟨w,y⟩={\int }_0^a\overline{w}ydx$. Two applications of integration by parts giveThe given boundary conditions set $y=0$ at both ends, but leave $y^{\prime }(0)$ and $y^{\prime }(a)$ independently variable. The remaining boundary term vanishes for every admissible $y$ precisely when $w(0)=w(a)=0$. Consequently the formal adjoint and the boundary conditions specifying the adjoint operator are

Set $a=\pi /2$. To construct the Green's function from a Dirac delta function, require ${\mathcal{L}}_{x}g(x,\xi )=\delta (x-\xi )$ and zero endpoint values. Away from $x=\xi$, the homogeneous solutions satisfying the left and right boundary conditions are respectivelyWrite $g=A(\xi )u(x)$ for $x<\xi$ and $g=B(\xi )v(x)$ for $x>\xi$. A discontinuity in $g$ would generate a derivative of the Dirac delta function, which is absent from the desired equation. Thus $g$ is continuous. Integrating that equation through $\xi$ then gives the jump conditionThe resulting two equations are $Au(\xi )=Bv(\xi )$ and $B{v}^{\prime }(\xi )-A{u}^{\prime }(\xi )=1$. Since the Wronskian $u{v}^{\prime }-u^{\prime }v=-e^{-2\xi }$, they giveBoth endpoint values vanish. At $x=\xi$, subtracting the two first derivatives gives ${\sin }^2\xi +{\cos }^2\xi =1$, verifying the sign and normalization of the Dirac delta function.

The homogeneous solution $h(x)=e^{-x}\cos x$ already has the prescribed endpoint values. Thus $z=y-h$ has zero boundary conditions and satisfies $\mathcal{L}z=f$. The Green's function givesEquivalently, splitting the integral at $\xi =x$ gives the explicit formulaAny difference between two solutions would be $e^{-x}(C\cos x+D\sin x)$ with zero endpoint values, forcing $C=D=0$. The displayed solution is therefore unique.

At $a=\pi$, the nonzero homogeneous solution $e^{-x}\sin x$ satisfies both zero boundary conditions. Thus the boundary value problem has a nontrivial kernel and cannot have an inverse for arbitrary forcing.

More directly, the adjoint operator has the zero-endpoint homogeneous solution $w(x)=e^x\sin x$. If an ordinary Green's function existed, the integration by parts identity would implyBut $w(\xi )\ne 0$ for $0<\xi <\pi$. This contradiction rules out the Green's function for the unmodified operator. It is the Fredholm alternative obstruction.

The nonzero endpoint data contribute to the Fredholm alternative condition. With $w=e^x\sin x$, the same integration by parts identity givesFor the specified forcing the left side is $A{\int }_0^{\pi }{\sin }^{2}xdx=A\pi /2$. HenceTo verify sufficiency, set $y=e^{-x}u$. Direct differentiation gives $\mathcal{L}y=e^{-x}(u^{\prime \prime }+u)$. At this value of $A$, a particular solution of $u^{\prime \prime }+u=(2/\pi )\sin x$ is $-(x/\pi )\cos x$. The complete family obeying the two boundary conditions isThe arbitrary constant records the one-dimensional kernel; existence does not imply uniqueness here.