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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} y_0 &= \alpha & i = 0 y_{i+1} &= y_i + h \, f(t_i, y_i) & i > 0 \end{align}!}

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \in \left[ 0,2 \right]} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N = 4} .

Calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h = \frac{2-0}{4} = \frac{1}{2}}

$t_{0}=0$	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_0 = \frac{1}{2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t_1 = \frac{1}{2}}	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1 = \frac{5}{4}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t_2 = 1}	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2 = \frac{9}{4}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t_3 = \frac{3}{2}}	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_3 = \frac{27}{8}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t_4 = 2}	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_4 = \frac{71}{16}}

Error Estimate

Theorem. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f : D \to \mathbb{R}}

1. be continuous with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t}
2. be Lipschitz-continuous with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} with some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L}
3. there is a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M > 0} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \max_{t \in [a,b]} \left| y''(t) \right| \le M}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| y_i - y(t_i) \right| \le \frac{h \, M}{2L} \, \left( \mathrm{e}^{L \, (t_i - a)} - 1 \right)}

Proof.

Theorem. [Discrete Gronwall]. Assume we have a sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_i(s, r)} that satisfies Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{i+1} \le (1+s) \, a_i + r} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{i+1} \le \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1 = \frac{\mathrm{d}}{\mathrm{d}\tau} \left( \tau - t_{i+1} \right)} :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{y}(t_{i+1}) = \tilde{y}(t_i) + h \, f(t_i, \tilde{y}(t_i))}

Hence the error is

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_i = \left| y(t_i) - \tilde{y}(t_i) \right|}

Expanding this and simplifying gives

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| E_i \right| \le \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