Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2005 / Differential equations

On $[-a,a]$ let $F=\Vert f{\Vert }_{\infty }$ and bound $|K|$ by $M$ on the compact integration region. Write $Th(x)={\int }_0^{x}K(x,s)h(s)ds$. The iteration gives $y_n={\sum }_{j=0}^n{T}^{j}f$. Induction, integrating absolute values also when $x<0$, givesThe exponential series is summable uniformly for $|x|\le a$. The Weierstrass M-test therefore proves uniform convergence to a continuous function $y$. Also $\Vert T(y_n-y){\Vert }_{\infty }\le Ma\Vert y_n-y{\Vert }_{\infty }\to 0$, so passing to the limit yields $y=f+Ty$.

For the indicated kernel, twice differentiating its integral givesConversely integrating the differential equation twice, with these initial conditions, gives the integral equation. This is valid also for negative $x$, either by oriented integration or by differentiating the double-integral identity.

For every finite $a$ the continuous kernel $(x-s)s$ meets the preceding hypotheses. Solutions on overlapping intervals agree: if $d=Td$ is their difference, iteration gives $\Vert d{\Vert }_{\infty }\le \Vert d{\Vert }_{\infty }(Ma{)}^j/j!$ for every $j$, forcing $d=0$. The solutions therefore patch to one solution defined on all of $\mathbb{R}$.

PutLeibniz differentiation gives $z^{\prime }=-{\int }_0^x\cos (x-s)f(s)ds$ and $z^{\prime \prime }=-f(x)+{\int }_0^x\sin (x-s)f(s)ds=-f-z$. Thus $z$ satisfies both zero initial conditions and the inhomogeneous equation. Every solution with $y(0)=0$ is $y=A\sin x+z(x)$.

On $[0,1]$, the second endpoint determines $A=-z(1)/\sin 1$ uniquely, since $\sin 1\ne 0$. Collecting the integrals, and using the sine addition formula for $s<x$, givesThis continuous kernel vanishes at the endpoints, satisfies $G_{xx}+G=0$ off $x=s$, and has derivative jump $-1$, as required for $-y^{\prime \prime }-y=f$. The preceding construction proves existence, and a homogeneous solution with both endpoints zero must vanish, proving uniqueness.

On $[0,\pi ]$, instead, the second boundary value becomesThus solvability is equivalent to ${\int }_0^{\pi }f(s)\sin sds=0$. When it holds, all solutions areThere is one free homogeneous mode because the endpoint length is resonant.

For a nonzero principal part the discriminant $\Delta =b^2-ac$ classifies the equation: it is hyperbolic for $\Delta >0$, parabolic for $\Delta =0$, and elliptic for $\Delta <0$. These are the three definitions requested by the inline roman labels.

In the hyperbolic case the characteristic equation $a(dy{)}^2-2bdxdy+c(dx{)}^2=0$ gives two distinct real tangent directions. Locally, with suitably smooth coefficients, solve the corresponding characteristic ordinary differential equations to obtain two independent families of curves. Label their level sets $\varphi =\text{constant}$ and $\psi =\text{constant}$. Their gradients satisfyand the coordinate Jacobian is nonzero. The chain rule therefore eliminates $u_{\varphi \varphi }$ and $u_{\psi \psi }$. The remaining mixed coefficient $2[a{\varphi }_x{\psi }_x+b({\varphi }_x{\psi }_y+{\varphi }_y{\psi }_x)+c{\varphi }_y{\psi }_y]$ is nonzero: otherwise the invertible coordinate change would send the nondegenerate principal quadratic form to zero. Divide by this coefficient and absorb all first-order terms into $g$, giving $u_{\varphi \psi }=g(\varphi ,\psi ,u,u_{\varphi },u_{\psi })$. This canonical reduction is local; a global chart is not automatic.

For the specific equation $a=y,b=0,c=-1$, so it is hyperbolic exactly on $y>0$, elliptic for $y<0$, and degenerate parabolic on $y=0$. For $\varphi =3x-2y^{3/2}$ and $\psi =3x+2y^{3/2}$, the Jacobian is $18\sqrt{y}\ne 0$. Direct differentiation givesSince $\psi -\varphi =4y^{3/2}$, the canonical form isThe characteristic families are $3x-2y^{3/2}=c$ and $3x+2y^{3/2}=c$, with slopes $dx/dy=\pm \sqrt{y}$. They are distinct for $y>0$ and meet the degeneracy line with coincident vertical tangent directions.