$\newcommand{\abs}[1]{\left|#1\right|}\newcommand{\d}{\mathrm{\ d}}$
- Consider the initial-value problems
\[
y'(x)=(1-2 x) y(x), \quad y(0)=1 .
\]
- Find $y_0, y_1, y_2, y_3$, where $\left\{y_n\right\}_{n≥0}$ is the sequence of Picard iterates.
- Use Picard's theorem to show that the problem has a unique solution for all $x$.
- Find the solution $y$ explicitly.
- Find an explicit expression for the Picard iterates $y_n$ and use this expression as well as the Weierstrass M-test to prove directly that $y_n$ converges uniformly on every interval $[-h,h]$ to the solution $y$ obtained in (c).
- \begin{align*} y_1(x)&=1+\int_0^x(1-2t)y_0(t)\d t=-x^2+x+1 \\y_2(x)&=1+\int_0^x(1-2t)y_1(t)\d t=\frac{x^4}{2}-x^3-\frac{x^2}{2}+x+1 \\y_3(x)&=1+\int_0^x(1-2t)y_2(t)\d t=-\frac{x^6}{6}+\frac{x^5}{2}-\frac{5 x^3}{6}-\frac{x^2}{2}+x+1 \end{align*}
- $f(x,y)=(1-2x)y$ is continuous, so P(i) holds. \[\abs{f(x,u)-f(x,v)}=\abs{1-2x}⋅\abs{u-v}≤(1+2h)\abs{u-v}\] so P(iii) holds for all $h$. So there is a unique solution for all $x$.
- Separating variables,$$y^{-1}\d y=(1-2x)\d x$$Integrating,$$\log y=x-x^2+C$$Using the initial value $y(0)=1$,$$\log y=x-x^2⇒y=\exp(x-x^2)$$
- Write $y_n(x)$ as a polynomial in $(x-x^2)$,
\begin{align*}
y_1(x)&=1+(x-x^2)
\\y_2(x)&=1+(x-x^2)+\frac{(x-x^2)^2}2
\\y_3(x)&=1+(x-x^2)+\frac{(x-x^2)^2}2+\frac{(x-x^2)^3}6
\end{align*}
We find$$y_n(x)=\sum_{i=0}^n\frac{(x-x^2)^i}{i!}$$Let $M$ be the maximum of the continuous function $\abs{x-x^2}$ on the closed interval $[-h,h]$, then $\frac{\big(x-x^2\big)^i}{i!}≤\frac{M^i}{i!}$ for $i=0,1,⋯$ and $x∈[-h,h]$.
By ratio test, the series $\sum_{i=0}^∞\frac{M^i}{i!}$ converges. By Weierstrass M-test, $y_n$ converges uniformly on $[-h,h]$.
- Consider the initial-value problem
\[
y'(x)=xy^{1/3}(x), \quad y(0)=b,
\]
- Use Picard's theorem to show that if $b>0$ then there is a unique solution on an interval $[-h, h]$, for a suitable $h>0$ which you should specify (you must check carefully that the assumptions of Picard's theorem are satisfied).
- Show that the function $F(x,y)=xy^{1/3}$ does not satisfy a Lipschitz condition on any rectangle $\{(x,y):\abs x≤h,\abs y≤k\}$ with $h>0$ and $k>0$.
- Show that if $b=0$ then for any $c>0$ there is a solution $y_c$ which is identically zero on $[-c, c]$ and positive when $\abs x>c$. Are there any other solutions of the initial value problem?
- Now return to the case $b>0$. Consider the set $R=\{(x, y):y≥b,\abs x≤h\}$. By working in this $R$, and adapting the proof of Picard's theorem, prove that in fact there is a unique solution of the problem on $\abs x≤h$ for any $h$ and hence that there is global existence of solutions.
- Pick $0< ϵ< b$, the function $y^{1/3}$ is Lipschitz on $(ϵ,2b-ϵ)$. Let $k=b-ϵ$, then $M=h(2b-ϵ)^{1/3}$, so$$Mh≤k⇔h^2(2b-ϵ)^{1/3}≤b-ϵ⇔h≤\left((b-ϵ)^3\over2b-ϵ\right)^{1/6}$$
RHS is a decreasing function of ϵ for $ϵ∈(0,b)$.
By Picard's theorem, there is a unique solution on $[-h,h]$.
Let $ϵ→0$, we get the largest interval $\left(-\frac{b^{1/3}}{2^{1/6}},\frac{b^{1/3}}{2^{1/6}}\right)$. - Assume $∃L>0,\abs{xy^{1/3}}=\abs{f(x, y)-f(x, 0)}≤L\abs y$, then $L≥\abs{xy^{-2/3}}$, but $\lim_{y→0^+}y^{-2/3}=∞$.
- For any $c>0$,$$y_c(x)=\begin{cases}0&\abs{x}≤c\\\left(x^2-c^2\over3\right)^{3/2}&\abs x>c\end{cases}$$There are other solutions, for $d≤c$,$$y(x)=\begin{cases}0&x∈[d,c]\\\left(x^2-c^2\over3\right)^{3/2}&x∉[d,c]\end{cases}$$the limit case $d→-∞$ is$$y(x)=\begin{cases}0&x≤c\\\left(x^2-c^2\over3\right)^{3/2}&x>c\end{cases}$$the limit case $c→∞$ is$$y(x)=\begin{cases}0&x>d\\\left(x^2-c^2\over3\right)^{3/2}&x≤d\end{cases}$$
- Adapted proof of Claim 1: Assume the graph of $y_n$ is in $R$, then $y_n(t)>0$. For $x≤0$, we have $-t≥0$, so$$\int_0^x t⋅y_n(t)^{1/3}\d t=\int_x^0(-t)⋅y_n(t)^{1/3}\d t≥0$$For $x≥0$, we have $t≥0$, so$$\int_0^x t⋅y_n(t)^{1/3}\d t≥0\quad∀x ∈[-h,h]$$Then $$y_{n+1}(x)=b+\int_0^x t⋅y_n(t)^{1/3}\d t≥b$$ the graph of $y_{n+1}$ is in $R$.
- Show that $f(x,y)=x^{1/3}\left(1+y^2\right)$ satisfies the Lipschitz-condition on $[-h,h]×[-k,k]$ for every $h,k>0$. Does $f$ also satisfy the global Lipschitz condition (P(iii))?
- Determine the maximal number $h_0>0$ so that Picard's theorem guarantees the existence of a unique solution on $\left[-h_0, h_0\right]$ of the initial value problem \[ y'(x)=x^{1/3}\left(1+y^2(x)\right), \quad y(0)=0 . \] Does this interval agree with the maximal interval on which a unique solution exists?
- $f_y=2x^{1/3}y$ is bounded on $[-h,h]×[-k,k]$, so $f$ satisfies the Lipschitz-condition on $[-h,h]×[-k,k]$.
$f$ doesn't satisfy the global Lipschitz condition: for $u>0,v=u+1$,\[\abs{f(1,u)-f(1,v)}=(2u+1)\abs{u-v}\]and $2u+1→∞$ as $u→∞$. - $$M=h^{1/3}\left(1+k^2\right)$$Condition for Picard's theorem:
$$Mh≤k⇔h^{4/3}\left(1+k^2\right)≤k⇔h≤\left(k\over1+k^2\right)^{3/4}$$
$$h_0=\max_{k>0}\left(k\over1+k^2\right)^{3/4}=2^{-3/4}$$
To solve the ODE, separating variables,$$\frac{\mathrm{d}y}{1+y^2}=x^{1/3}\d x$$Integrating,$$\arctan y=\frac34x^{4/3}$$Since $\arctan y∈(-π/2,π/2)$, the maximal interval on which a unique solution exists is $[-h,h],h=\left(\frac{2 \pi }{3}\right)^{3/4}$, which is larger.
When we continue building rectangles on the vertex as new initial condition, $M$ increases, so the permitted $h$ decreases, but the total length of $x$-interval$<\left(\frac{2 \pi }{3}\right)^{3/4}$.
- Suppose that $f:[a, b]→ℝ$ and $K:[a, b]×[a, b]→ℝ$ are continuous. Consider the integral equation
\[
y(x)=f(x)+\int_a^x K(x, t) y(t)\d t, \quad x∈[a, b] .
\]
For $x∈[a, b]$ define
\begin{aligned}
y_0(x) &=f(x) \\
y_{n+1}(x) &=f(x)+\int_a^x K(x, t) y_n(t)\d t .
\end{aligned}
-
Adapt the proof of Picard's theorem to show that $y_n$ converges uniformly to a solution of the integral equation for all $x∈[a,b]$.
You may assume that if $y:[a,b]→ℝ$ is continuous then so too is $f(x)+\int_a^x K(x,t)y(t)\d t$ for $x∈[a,b]$. - Show that the solution of (IE) is unique and prove that the solution depends continuously on $f$. (You will need to define what this means.)
- Let $e_0(x)=f(x),e_{n+1}(x)=y_{n+1}(x)-y_n(x)$ and $\abs{f(x)}≤M$ and $\abs{K(x,t)}≤N$, we prove $\abs{e_n(x)}≤\frac{N^nM}{n!}(x-a)^n$ inductively:\begin{align*} \abs{e_{n+1}(x)}&=\abs{\int_a^xK(x,t)⋅e_n(t)\d t}\\&≤\int_a^x\abs{K(x,t)}⋅\abs{e_n(t)}\d t\\&≤\int_a^xN⋅\frac{N^nM}{n!}(t-a)^n\d t\\&=\frac{N^{n+1}M}{(n+1)!}(x-a)^{n+1} \end{align*} Applying the Weierstrass M-test to $y_n=\sum_{i=0}^ne_i(x)$, the uniform convergence follows.
- We say the solution depends continuously on $f$, if for solutions $y_1$ and $y_2$, $∀ϵ>0,∃δ>0,\sup\abs{f_1(x)-f_2(x)}≤δ⇒\sup\abs{y_1(x)-y_2(x)}≤ϵ$. \[ y_1(x)-y_2(x)=f_1(x)-f_2(x)+\int_a^xK(x,t)(y_1(t)-y_2(t))\d t \] Let $L=\sup\abs{K(x,t)},E=\sup\abs{f_1(x)-f_2(x)}$. By triangle inequality, \[ \abs{y_1(x)-y_2(x)}≤E+\abs{\int_a^x L\abs{y_1(t)-y_2(t)}\d t} \] By Grönwall's inequality, \[ \abs{y_1(x)-y_2(x)}≤Ee^{L(x-a)} \] Since $a≤x≤b$, \[ ⇒\sup\abs{y_1(x)-y_2(x)}≤Ee^{L(b-a)} \]Let $E=0$ we have uniqueness. Since $Ee^{L(b-a)}→0$ as $E→0$, we have continuous dependence.
-
Adapt the proof of Picard's theorem to show that $y_n$ converges uniformly to a solution of the integral equation for all $x∈[a,b]$.