MATH 417 Lecture 15

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Chapter 5.1: Initial Value Problems

Definition: A differential equation of the form $\frac{\mathrm{d}y(t)}{\mathrm{d}t}=f(t,y(t))$ given $y(a)=\alpha$

Do we have a solution for $y(t)$? If so, is it unique?

Example: Population Growth Model

$\frac{\mathrm{d}y(t)}{\mathrm{d}t}=ky(t)$

If $k>0$, the population grows, and if $k<0$, the population decays.

These are the types of differential equations we will focus on.

Example: Harmonic Oscillator

$\frac{\mathrm{d}\theta }{\mathrm{d}t}=v(t)$ and $\frac{\mathrm{d}v(t)}{\mathrm{d}t}=-\frac{g}{L}\theta (t)$

Well-Posedness

An initial value problem is well-posed if and only if

1. it has a unique solution (existence and uniqueness)
2. continuous dependence of the solution with respect to (the initial data) + $f(x)$.

there should exist a $k$ and an ${ϵ}_0$ such that for all $ϵ\in [0,{ϵ}_0]$ such that given two initial value problems

$\{\begin{array}{l}\frac{\mathrm{d}{y}_1}{\mathrm{d}t}=f(t,y_1)\\ y_1(a)=\alpha \end{array}$

$\{\begin{array}{l}\frac{\mathrm{d}{y}_2}{\mathrm{d}t}=f(t,y_2)+\delta (t)\\ y_2(a)=\alpha +{ϵ}_0\end{array}$

$\underset{t\in [a,b]}{\max }|y_1(t)-y_2(t)|\le kϵ$. This is true if $\underset{t}{\max }|\delta (t)|+{ϵ}_0\le ϵ$.

Lipschitz Continuity

Let $D\subset {\mathbb{ℝ}}^2$. Then $f:D\to \mathbb{ℝ}$ is Lipschitz continuous with respect to the second variable if there exists a $L$ such that

$|f(t,y_1)-f(t,y_2)|\le L|y_1-y_2|$ for all $(t,y_1),(t,y_2)\in D$.

Note: $D$ represents a collection of points on the $t-y$ plane

Cauchy-Lipschitz / Picard-Lindelof Theorem

Theorem. Let $D=\{(t,y)\mid t\in [a,b],y\in \mathbb{ℝ}\}$ be a stripe of the $t,y$ plane. Let $f:D\to \mathbb{ℝ}$.

If

1. $f$ is continuous with respect to $t$
2. $f$ is Lipschitz continuous with respect to $y$

then the initial value problem is well-posed.

Proof. [omitted].

quod erat demonstrandum

Example 1

is $\{\begin{array}{l}\frac{\mathrm{d}y(t)}{\mathrm{d}t}=y(t)-t^2+1\\ y(0)=\frac{1}{2}\end{array}$ well-posed for $t\in [0,2]$?

1. $f$ is continuous with respect to $t$ since $-t^2+1$ is continuous
2. $|f(t,y_1)-f(t,y_2)|=|y_1-t^2+1-y_2+t^2-1|=|y_1-y_2|$, hence $L=1$.

Example 2

is $\{\begin{array}{l}\frac{\mathrm{d}y(t)}{\mathrm{d}t}=2\sqrt{y(t)}\\ y(0)=0\end{array}$ well-posed for $t\in {\mathbb{ℝ}}^{+}\times {\mathbb{ℝ}}^{+}$

We can find solutions $y(t)=0$ and $y(t)=t^2$ that satisfy the equation, so the solution is not unique, and therefore the problem is not well-posed.

Observe that $f$ is continuous with respect to $t$, so there must be a discrepancy in the Lipschitz continuity:

$|f(t,y)-f(t,0)|=|2\sqrt{y}-0|\ge L\left|y\right|$ holds true for $y\le \frac{4}{L^2}$.

Chapter 5.2: Euler Method

In real life, analytical solutions to ODEs are rare. Instead, we approximate solutions.

$\{\begin{array}{l}\frac{\mathrm{d}y(t)}{\mathrm{d}t}=f(t,y(t))\\ y(a)=\alpha \end{array}$ for $t\in [a,b]$.

We define $h=\frac{b-a}{N}$ as our time-step function to discretize time into $N$ samples:

$t_i=a+ih$ for $i\in 0,1,\dots ,N$.

We set $t_0=a$, so $y_0=\alpha$. Now we use the fundamental theorem of calculus to construct solutions for $t_i$:

$\underset{t_i}{\overset{t_{i+1}}{\int }}\frac{\mathrm{d}y(\tau )}{\mathrm{d}t}\mathrm{d}\tau =y(t_{i+1})-y(t_i)=\underset{t_i}{\overset{t_{i+1}}{\int }}f(\tau ,y(\tau ))\mathrm{d}\tau$

We approximate the value of the integral as follows:

$\underset{t_i}{\overset{t_{i+1}}{\int }}f(\tau ,y(\tau ))\mathrm{d}\tau \approx \underset{t_i}{\overset{t_{i+1}}{\int }}f(t_i,y(t_i))\mathrm{d}\tau$

Hence $y_{i+1}=y_i+hf(t_i,y_i)$ for $i>0$.

In summary

$\begin{array}{rlrlr}{y}_0& =\alpha & i=0y_{i+1}& =y_i+hf(t_i,y_i)& i>0\end{array}!$

Example

Construct an approximate solution for $\{\begin{array}{l}\frac{\mathrm{d}y}{\mathrm{d}t}=y-t^2+1\\ y(0)=\frac{1}{2}\end{array}$ for $t\in [0,2]$ with $N=4$.

Calculate $h=\frac{2-0}{4}=\frac{1}{2}$

$t_0=0$	$y_0=\frac{1}{2}$
$t_1=\frac{1}{2}$	$y_1=\frac{5}{4}$
$t_2=1$	$y_2=\frac{9}{4}$
$t_3=\frac{3}{2}$	$y_3=\frac{27}{8}$
$t_4=2$	$y_4=\frac{71}{16}$

Error Estimate

Theorem. Let $f:D\to \mathbb{ℝ}$

1. be continuous with respect to $t$
2. be Lipschitz-continuous with respect to $y$ with some $L$
3. there is a $M>0$ such that $\underset{t\in [a,b]}{\max }|y^{″}(t)|\le M$

$|y_i-y(t_i)|\le \frac{hM}{2L}({\mathrm{e}}^{L(t_i-a)}-1)$

Proof.

Theorem. [Discrete Gronwall]. Assume we have a sequence $a_i(s,r)$ that satisfies $a_{i+1}\le (1+s)a_i+r$ for all $i$. Then $a_{i+1}\le {\mathrm{e}}^{(i+1)s}(a_0+\frac{r}{s})-\frac{r}{s}$

Proof. [omitted].

quod erat demonstrandum

The exact solution has form $y(t_{i+1})=y(t_i)+\underset{t_i}{\overset{t_{i+1}}{\int }}f(\tau ,y(\tau ))\mathrm{d}\tau$. Let's integrate by parts substituting $1=\frac{\mathrm{d}}{\mathrm{d}\tau }(\tau -t_{i+1})$:

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} y(t_{i+1}) = y(t_i) + \int_{t_i}^{t_{i+1}} \left( \frac{\mathrm{d}}{\mathrm{d}\tau} \left( \tau - t_{i+1} \right) \right) \, \left( f(\tau, y(\tau)) \right) \,\mathrm{d}\tau &= y(t_i) + \left( t_{i+1} - t_i \right) \, f(t_i, y(t_i)) - \int_{t_i}^{t_{i+1}} \left( \tau - t_{i+1} \right) \, \frac{\mathrm{d}}{\mathrm{d}\tau} \left( f(\tau, y(\tau) \right) \right) \,\mathrm{d}\tau \\ &= y(t_i) + h \, f(t_i, y(t_i)) + \int_{t_i}^{t_{i+1}} \left( t_{i+1} - \tau \right) \, \frac{\mathrm{d}^2 y(t)}{\mathrm{d}\tau^2} \,\mathrm{d}\tau \end{align}}

Our approximation is as follows:

$\tilde{y}(t_{i+1})=\tilde{y}(t_i)+hf(t_i,\tilde{y}(t_i))$

Hence the error is

$E_i=|y(t_i)-\tilde{y}(t_i)|$

Expanding this and simplifying gives

Failed to parse (syntax error): {\displaystyle \left| E_{i+1} \right| &\le \left| E_i \right| + h \, \left| f(t_i, y(t_i)) - f(t_i, \tilde{y}(t_i)) \right| + \int_{t_i}^{t_{i+1}} \left( t_{i+1} - \tau \right) \, \left| \frac{\mathrm{d}^2 y}{\mathrm{d}\tau^2} \right| \,\mathrm{d}\tau \\ &\le \left| E_i \right| + h\,L \, \left| y(t_i) - \tilde{y}(t_i) \right| + M \, \int_{t_i}^{t_{i+1}} \left( t_{i+1} - \tau \right) \,\mathrm{d}\tau \\ &= \left| E_i \right| + h \, L \, \left| E_i \right| + M \, \frac{h^2}{2} \\ &= \left( 1 + h \, L \right) \, \left| E_i \right| + \frac{M \, h^2}{2}}

By the lemma above,

$\left|E_i\right|\le ({\mathrm{e}}^{ihL}-1)\frac{M{h}^{\cancel{2}}}{2\cancel{h}L}=({\mathrm{e}}^{(t_i-a)L}-1)\frac{Mh}{2L}$

quod erat demonstrandum