Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2004 / Differential equations

The first iterate isOn the prescribed box, $F(x,y)=1+x^2+y^2$ is continuous, $|F|\le 102$, and $|F(x,y)-F(x,v)|\le 20|y-v|$. The initial value is distance nine from the nearer horizontal edge. Thus on $|x|\le h=9/102=3/34$, the Picard map keeps functions within the box: its values differ from one by at most $102h=9$.

Successive differences satisfy, by induction,The Weierstrass M-test proves uniform convergence on this local interval. The uniform limit is continuous, and the Lipschitz estimate allows passage to the limit inside the integral, giving the displayed integral equation. Differentiation gives $y^{\prime }=1+x^2+y^2$, $y(0)=1$. The same local Lipschitz bound proves local uniqueness.

Direct differentiation gives $u^{\prime }={\sec }^2(x+\pi /4)=1+u^2$, $u(0)=1$, and $u\to +\infty$ as $x\uparrow \pi /4$. On the common positive existence interval put $w=y-u$. It satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$, so the integrating-factor formula givesThus $y\ge u$. Let $x_{\ast }>0$ be the maximal forward existence endpoint. The solution is strictly increasing because $y^{\prime }\ge 1$. If $x_{\ast }$ were finite and $y$ remained bounded, its finite limit and the smooth right side would allow local continuation, contradicting maximality. Comparison with $u$ prevents existence through $\pi /4$. ConsequentlyThere is no contradiction with Picard's theorem, which gives local existence and uniqueness. Its box estimate gives only a short guaranteed interval, and the Lipschitz constant for $y^2$ is not globally bounded in $y$.

First solve for $v=y^{\prime }$. Multiplication by $e^x$ and use of $v(1)=0$ giveIntegrating from zero to $x$, and using $y(0)=0$, yields $y(x)=-{\int }_0^x{e}^{-s}{\int }_s^1{e}^{t}f(t)dtds$. Reversing the order of integration givesThus $G(x,t)$ equals $G_1$ for $t<x$ and $G_2$ for $x<t$. It is continuous at $x=t$, vanishes at $x=0$, and has $G_x(1,t)=0$. Its derivative jumps from $-1$ to zero across $x=t$, so the jump is $+1$, matching the coefficient of $y^{\prime \prime }$. Both pieces satisfy $G_{xx}+G_x=0$ away from the jump. These facts also verify the Green representation directly. The homogeneous solutions are $A+B{e}^{-x}$, and the two boundary conditions force $A=B=0$, proving uniqueness.

For $f(t)=e^{-t}$, the formula givesHence $y(x)=-x{e}^{-x}$. Indeed $y^{\prime }=(x-1)e^{-x}$ and $y^{\prime \prime }=(2-x)e^{-x}$, which verify the equation and both boundary conditions.

This is the heat equation, with $y$ as time. The stated zero solution is unique. Define $E(y)=\frac{1}{2}{\int }_0^{1}u(x,y{)}^{2}dx$. For $0<y<1$, integration by parts and $u_y=u_{xx}$ giveThe boundary term is zero because $u(0,y)=0$ and $u_x(1,y)=0$. Also $E(0)=0$. Nonnegativity forces $E(y)=0$, and continuity gives $u=0$ throughout the square. The unknown final-time boundary values are determined by evolution and need not be prescribed. Under the stated regularity the integrations can first be made on an interior interval and then passed to the endpoints using continuity of $u,u_x,u_y$.

This is Laplace's equation. Zero is not unique: with $k=\pi /2$,is nonzero and harmonic. It vanishes on $y=0$ and $x=0$, while $u_x(1,y)=k\cos (k)\sinh (ky)=0$.

A sufficient additional condition is $u(x,1)=0$ on the top edge. Green's first identity then givesOn the bottom, left and top edges $u=0$; on the right edge the normal derivative is zero. Thus the gradient vanishes and $u$ is constant; the bottom value makes that constant zero. For $C^2(D)\cap C^1(\overline{D})$, apply the identity on inset rectangles and pass to the boundary using continuity of the function and its gradient.

This is the wave equation, again taking $y$ as time. Zero is not unique: for $k=\pi /2$,satisfies $u_{xx}=u_{yy}$ and all the stated boundary conditions. The initial displacement is zero but the initial velocity is not prescribed.

A sufficient extra condition is $u_y(x,0)=0$. For the resulting homogeneous data use wave energyThe equation and integration by parts give $E^{\prime }(y)=[u_x{u}_y{]}_0^1=0$. At $x=1$ the factor $u_x$ vanishes, and at $x=0$ the factor $u_y$ vanishes because $u(0,y)=0$. The initial displacement implies $u_x(x,0)=0$, so the extra initial velocity condition gives $E(0)=0$. Consequently $u_x=u_y=0$ and $u=0$. With the stated boundary regularity, integrate the local energy conservation law over inset rectangles and let their edges approach the boundary; the energy and flux terms converge by continuity of the first derivatives.