Past exam of the differential equations 2 course of the University of Oxford / 2015 / 3

By the definition of little o notation, the required quotient isAs $ϵ\to 0^{+}$, write this as $\log ϵ/(1/ϵ)$. L'Hôpital's rule applies to this infinite-over-infinite ratio and givesTherefore ${ϵ}^2\log ϵ=o(ϵ)$. The logarithm is negative on this approach, which is consistent with a quotient tending to zero from below; little o notation concerns the limiting ratio, not its sign.^[ref]

Take $ϵ\to 0^{+}$. A root on the smaller scale has $x=ϵu$, giving $u+2+ϵu^3=0$. Put $u=u_0+ϵu_1+O({ϵ}^2)$. Successive coefficients give $u_0=-2$ and $u_1+u_0^3=0$, hence $u_1=8$. The real root is thereforeThe other dominant balance is between the cubic and linear terms, requiring $x={ϵ}^{1/2}z$. Set $t={ϵ}^{1/2}$; the equation becomes $z^3+z+2t=0$. At $t=0$ the two nonzero roots are $z_0=\pm i$. With $z=z_0+t{z}_1+O(t^2)$, the first correction satisfies $(3z_0^2+1)z_1+2=0$, so $-2z_1+2=0$ and $z_1=1$. ThusThe limiting roots in both rescalings are simple, so the implicit function theorem justifies these regular expansions in their respective small parameters. The fractional powers in the original variable are a fractional-power root expansion. The real derivative $3x^2+ϵ$ is strictly positive; the cubic therefore has exactly one real root. Its other roots are the displayed complex conjugate pair, accounting for all three roots. Their leading real corrections sum to $-2ϵ+ϵ+ϵ=0$, as required by the absence of a quadratic coefficient.

For a positive eigenvalue write $\mu =\sqrt{\lambda }>0$. The general solution of the limiting constant-coefficient differential equation is $y=A\cos (\mu x)+B\sin (\mu x)$. The endpoint at zero gives $A=0$. A nonzero eigenfunction then requires $B\ne 0$, and the endpoint at $\pi$ gives $\sin (\mu \pi )=0$. ConsequentlyFor real eigenfunctions the constants may be any nonzero real amplitudes. Requiring unit L2 norm would set $|C_k|=\sqrt{2/\pi }$, but that normalization is not required by the question. The lowest eigenvalue is ${\lambda }_1=1$ and its eigenspace is spanned by $\sin x$. For completeness, zero yields only an affine solution annihilated by both endpoints, and negative eigenvalues yield hyperbolic functions whose two zero endpoint values also force the zero solution.

Choose the real leading eigenfunction $y_1=C\sin x$, keeping its amplitude $C\ne 0$ fixed. At first order in the regular perturbation expansion, the equation becomesThe operator $L=D^2+1$ with these homogeneous Dirichlet boundary conditions is self-adjoint, as allowed in the question, and its null space is spanned by $y_1$. The Fredholm solvability condition requires the forcing to be orthogonal to that null space:The same condition follows directly by integrating $y_{1}L{y}_{11}$ by parts: the endpoint term $[y_1{y}_{11}^{\prime }-y_1^{\prime }{y}_{11}{]}_0^{\pi }$ vanishes, and $L{y}_1=0$. Since a nonzero real $y_1$ has a positive squared integral,This is the requested expression and is an instance of the leading nonlinear eigenvalue shift in a Sturm-Liouville problem. In the amplitude convention above, ${\int }_0^{\pi }{\sin }^{2}xdx=\pi /2$ and ${\int }_0^{\pi }{\sin }^{3}xdx=4/3$, so ${\lambda }_{11}=-8C/(3\pi )$. For the particular choice $C=1$, it is $-8/(3\pi )$.

The amplitude dependence of a nonlinear eigenvalue correction follows here from the quadratic term: rescaling the eigenfunction changes the eigenvalue shift. There is no single numerical correction independent of the chosen leading amplitude. Once the solvability condition is met, $y_{11}$ is determined up to addition of a multiple of $y_1$; a normalization such as ${\int }_0^{\pi }{y}_1{y}_{11}dx=0$ can fix that remaining freedom.