Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 4

The Picard-Lindelöf theorem gives local existence and uniqueness for an initial value problem when $f$ is continuous on a rectangle around $(a,b)$ and is Lipschitz continuous in its second variable there, with a constant uniform in the first variable. A sufficient hypothesis is that $f$ and $\partial f/\partial y$ are continuous on that rectangle.

More quantitatively, if the rectangle is $|x-a|\le h_0$, $|y-b|\le k$, and $|f|\le M$, a sufficiently small $h>0$ with $h\le h_0$ and $hM<k$ yields a unique differentiable solution on $[a-h,a+h]$ remaining inside the vertical strip. The proof uses the integral equationand Picard iteration, or the Banach fixed-point theorem after shrinking $h$ to make the integral operator a contraction. The continuity and Lipschitz hypotheses must hold near the actual initial point.

Dividing by $x^2$, which is allowed precisely away from zero, givesThis function is smooth on each of the two half-planes $x>0$ and $x<0$, and is affine in the dependent variable $y$. Its derivative with respect to $y$ is $(1-2x)/x^2$.

For fixed $a\ne 0$ and $B>0$, the rectangleis compact and stays away from $x=0$. The function $f$ in part (i) is continuous there. The extreme value theorem makes $|f|$ attain a finite maximum $M_0$ on $K$; choosing $\boxed{M=M_0+1}$ gives the requested strict bound.

The absolute values make this a positive-$x$ rectangle even when $a<0$. Its boundedness is still true, but for a negative initial point the Picard-Lindelöf theorem must instead be applied to a rectangle centered at that negative point, as in part (iii).

Take any $a\ne 0$ and any $b\in \mathbb{R}$. Choose $0<h_0<|a|$, so $[a-h_0,a+h_0]$ avoids zero. Onboth $f$ and $f_y=(1-2x)/x^2$ are continuous. They are bounded by compactness, and the bound on $f_y$ gives the uniform Lipschitz continuity in $y$ required by the Picard-Lindelöf theorem.

If $M$ is a bound for $|f|$, shrink $h$ so $0<h<h_0$ and $hM<1$. Then the solution has $|y(x)-b|\le hM<1$, which supplies even strict inequalities at both edges of the requested vertical strip.

At $a=0$, neither $f$ nor the original exponential expression is defined. Thus the theorem applies exactly for $a\ne 0$, with no restriction on $b$. Failure of the hypotheses alone would not prove nonexistence; the explicit solutions below settle that issue.

On either connected half-line, the homogeneous linear differential equation isIntegration gives $\boxed{y_h=C{x}^{-2}{e}^{-1/x}}$, including the zero solution. For the inhomogeneous equation choose the positive integrating factorIt satisfies ${\mu }^{\prime }/\mu =2/x-1/x^2$. Multiplication gives the exact derivativeHence all solutions on either half-line areFor $a\ne 0$, impose $y(a)=b$ to obtain $C=a^2{e}^{1/a}b-\arctan a$. The complete initial value problem solution is thereforeIt is smooth throughout the maximal connected interval $(0,\infty )$ if $a>0$, and $(-\infty ,0)$ if $a<0$.

For every $a\ne 0$ and every real $b$, there is exactly one solution on the maximal connected interval containing $a$. The formula in part (i) proves existence on the entire appropriate half-line, and the initial condition fixes its sole constant uniquely. This agrees with the local Picard-Lindelöf theorem.

For $a=0$, there is no classical solution on an interval containing the initial point, whatever $b$ is: the printed equation is not defined there. Even if one only imposes the equation for $x\ne 0$ and asks for a finite continuous extension across zero, the negative side rules it out. For $x\to 0^{-}$:

Thus no choice of the negative-side integration constant admits a finite left limit.

There is a useful distinction with a separately extended one-sided problem. On $x>0$, every member of the solution family tends to zero with all its derivatives as $x\downarrow 0$, since the exponential dominates every inverse power of $x$. If the forcing is separately defined to be zero at the right endpoint, that modified one-sided problem has infinitely many smooth solutions with $y(0)=0$, and none with $y(0)\ne 0$. Those one-sided extensions do not give solutions on the two-sided intervals in the question.