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

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

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 $\epsilon _{0}$ such that for all $\epsilon \in \left[0,\epsilon _{0}\right]$ such that given two initial value problems

${\begin{cases}{\frac {\mathrm {d} y_{1}}{\mathrm {d} t}}=f(t,y_{1})\\y_{1}(a)=\alpha \end{cases}}$

${\begin{cases}{\frac {\mathrm {d} y_{2}}{\mathrm {d} t}}=f(t,y_{2})+\delta (t)\\y_{2}(a)=\alpha +\epsilon _{0}\end{cases}}$

$\max _{t\in \left[a,b\right]}\left|y_{1}(t)-y_{2}(t)\right|\leq k\epsilon$. This is true if $\max _{t}\left|\delta (t)\right|+\epsilon _{0}\leq \epsilon$.

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

$\left|f(t,y_{1})-f(t,y_{2})\right|\leq L\left|y_{1}-y_{2}\right|$ 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 $D=\left\{(t,y)\mid t\in \left[a,b\right],y\in \mathbb {R} \right\}$ 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 ${\begin{cases}{\frac {\mathrm {d} y(t)}{\mathrm {d} t}}=y(t)-t^{2}+1\\y(0)={\frac {1}{2}}\end{cases}}$ well-posed for $t\in [0,2]$?

1. $f$ is continuous with respect to $t$ since $-t^{2}+1$ is continuous
2. $\left|f(t,y_{1})-f(t,y_{2})\right|=\left|y_{1}-t^{2}+1-y_{2}+t^{2}-1\right|=\left|y_{1}-y_{2}\right|$, hence $L=1$.

Example 2

is ${\begin{cases}{\frac {\mathrm {d} y(t)}{\mathrm {d} t}}=2{\sqrt {y(t)}}\\y(0)=0\end{cases}}$ 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:

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

Chapter 5.2: Euler Method

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

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

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

$t_{i}=a+i\,h$ for $i\in 0,1,\ldots ,N$.

We set $t_{0}=a$, so $y_{0}=\alpha$. Now we use the fundamental theorem of calculus to construct solutions for $t_{i}$:

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

We approximate the value of the integral as follows:

$\int _{t_{i}}^{t_{i+1}}f(\tau ,y(\tau ))\,\mathrm {d} \tau \approx \int _{t_{i}}^{t_{i+1}}f(t_{i},y(t_{i}))\,\mathrm {d} \tau$

Hence $y_{i+1}=y_{i}+h\,f(t_{i},y_{i})$ for $i>0$.

In summary

${\begin{aligned}y_{0}&=\alpha &i=0y_{i+1}&=y_{i}+h\,f(t_{i},y_{i})&i>0\end{aligned}}!$

Example

Construct an approximate solution for ${\begin{cases}{\frac {\mathrm {d} y}{\mathrm {d} t}}=y-t^{2}+1\\y(0)={\frac {1}{2}}\end{cases}}$ for $t\in \left[0,2\right]$ 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 $\max _{t\in [a,b]}\left|y''(t)\right|\leq M$

$\left|y_{i}-y(t_{i})\right|\leq {\frac {h\,M}{2L}}\,\left(\mathrm {e} ^{L\,(t_{i}-a)}-1\right)$

Proof.

Theorem. [Discrete Gronwall]. Assume we have a sequence $a_{i}(s,r)$ that satisfies $a_{i+1}\leq (1+s)\,a_{i}+r$ for all $i$. Then $a_{i+1}\leq \mathrm {e} ^{(i+1)\,s}\,\left(a_{0}+{\frac {r}{s}}\right)-{\frac {r}{s}}$

Proof. [omitted].

quod erat demonstrandum

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

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:

${\tilde {y}}(t_{i+1})={\tilde {y}}(t_{i})+h\,f(t_{i},{\tilde {y}}(t_{i}))$

Hence the error is

$E_{i}=\left|y(t_{i})-{\tilde {y}}(t_{i})\right|$

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,

$\left|E_{i}\right|\leq \left(\mathrm {e} ^{i\,h\,L}-1\right)\,{\frac {M\,h^{\cancel {2}}}{2{\cancel {h}}\,L}}=\left(\mathrm {e} ^{(t_{i}-a)\,L}-1\right)\,{\frac {M\,h}{2L}}$

quod erat demonstrandum