Upper and Lower Solutions

Theorem: Assume $f(t,x,x')$ is continuous and bounded on $[a,b] \times \mathbb{R}^2$ . Then the BVP

$$x'' = f(t,x,x')$$

$$x(a) = A, x(b) = B$$

has a solution.
Proof: We will only do the proof in the case where we assume $f_x(t,x,x')$ and $f_{x'}(t,x,x')$ are continuous on $[a,b] \times \mathbb{R}^2$ . In this case, solutions of IVPs are unique and they depend continuously on the initial point and the initial conditions.
Define

$$f(t,x,x') = \begin{cases}f(b,x,x'), &t \geq b, x,x' \in \mathbb{R... ...b, x,x' \in \mathbb{R}\ f(a,x,x'), & t \leq a, x,x' \in \mathbb{R}\end{cases}$$

This new $f$ is continuous on $\mathbb{R}\times \mathbb{R}^2$ .
Let $\phi(t,v)$ be the solution of the IVP

$$x'' = f(t,x,x'), x(a) = A, x'(a) = v.$$

We will use the ``shooting method" of proof. Choose $P > 0$ so that $\vert f(t,x,x')\vert \leq P$ for all $(t,x,x') \in [a,b] \times \mathbb{R}^2$ . Consider for $t \in [a,b] \cap [a, \omega_v]$ ,

$$\vert \phi'(t,v) - \phi'(a,v) \vert = \vert \phi''(\xi,v)(t-a) \vert \leq P (b - a)$$

Then,

$$\vert \phi'(t,v) \vert \leq \vert \phi'(t,v) - v\vert + \vert v\vert \leq P(b-a) + \vert v \vert$$

$$\phi(t,v) - \phi(a,v) = \int_a^t \phi'(s,v) ds$$

$$\phi(t,v) = A + \int_a^t \phi'(s,v) ds$$

$$\vert \phi(t,v) \vert \leq \vert A \vert + \int_a^t \vert \phi'(s,v) \vert ds$$

So,

$$\vert \phi(t,v) \vert \leq \vert A \vert + P (b-a)^2 + \vert v \vert (b - a)$$

for $t \in [a,b] \cap [a,\omega_v)$ .
This implies $\phi(t,v)$ exists on $[a,b]$ . Consider,

$$\phi'(a,v) - \phi'(t,v) = \phi''(\xi)(a-t) \leq P(b-a)$$

$$\phi'(t,v) \geq v - P(b-a)$$

Integrate from $a$ to $t$ ,

$$\phi(t,v) - \phi(a,v) \geq v(t-a) - P(b-a)(t-a)$$

Let $t = b$ :

$$\phi(b,v) - A \geq v(b-a) - P(b-a)^2$$

$$\phi(b,v) \geq A + v(b-a) - P(b-a)^2$$

Pick $v_1$ sufficiently large so that

$$\phi(b,v_1) > B$$

Similarly, there is a $v_2$ sufficiently small so that

$$\phi(b,v_2) < B$$

By Kneser's Theorem, the BVP has a solution.

Definition: We say that $\alpha(t)$ is a lower solution of $x'' = f(t,x,x')$ on an interval $I$ provided

$$\alpha''(t) geq f(t,\alpha(t), \alpha'(t))$$

$\forall t \in I$ . Similarly, we define an upper solution $\beta(t)$ of $x'' = f(t,x,x')$ on $I$ provided

$$\beta''(t) \leq f(t,\beta(t),\beta'(t))$$

$\forall t \in I$ .

Motivating Example: $f(t,x,x') = 0$ . ($x'' = 0$ ).
Claim: $\alpha(t) = t^2$ and $\beta(t) = -t^2 + 3$ are lower and upper solutions of $x'' = 0$ :

$$\alpha''(t) = 2 \geq 0 = f(t,\alpha(t),\alpha'(t)), I = [-1,1].$$

$$\beta''(t) = -2 \leq 0 = f(t,\beta(t),\beta'(t)), I = [-1,1].$$

Notice that the graph of $\alpha(t)$ is concave upward and that the graph of $\beta(t)$ is concave downward.

Theorem: Assume that $f(t,x)$ is continuous on $[a,b] \times \mathbb{R}$ . Assume $\alpha(t)$ and $\beta(t)$ are lower and upper solutions, respectively of $x'' = f(t,x)$ on an interval $[a,b]$ with $\alpha(t) \leq \beta(t)$ on $[a,b]$ . Then, if $\alpha(a) \leq A \leq \beta(a)$ and $\alpha(b) \leq B \leq \beta(b)$ , it follows that the BVP

$$x'' = f(t,x), x(a) = A, x(b) = B$$

has a solution $x(t)$ and $\alpha(t) \leq x(t) \leq \beta(t), \forall t \in [a,b]$ .

Proof: Let $\alpha(t), \beta(t)$ be lower and upper solutions of $x'' = f(t,x)$ on $[a,b]$ with $\alpha(t) \leq \beta(t)$ on $[a,b]$ .
Define the modification $F(t,x)$ of $f(t,x)$ with respect to $\alpha(t), \beta(t)$ as follows:

$$F(t,x) = \begin{cases}f(t,\beta(t)) + \frac{x - \beta(t)}{1 + \ve... ... \alpha(t)}{1 + \vert x\vert}, & x \leq \alpha(t), a \leq t \leq b \end{cases}$$

Note $F(t,x)$ is continuous and bounded on $[a,b] \times \mathbb{R}$ , and $F(t,x) = f(t,x)$ if $\alpha(t) \leq x \leq \beta(t), a \leq t \leq b$ .
By the previous theorem, the BVP

$$x'' = F(t,x), x(a) = A, x(b) = B$$

has a solution $x(t)$ .
Claim: $\alpha(t) \leq x(t) \leq \beta(t)$ on $[a,b]$ .
We will only show that $x(t) \leq \beta(t)$ on $[a,b]$ . Assume not.
Then, there are points in $(a,b)$ such that $x(t) . \beta(t)$ on $[a,b]$ . Let

$$w(t) := x(t) - \beta(t)$$

Then,

$$w(a) = x(a) - \beta(a) = A - \beta(a) \leq 0$$

$$w(b) = x(b) - \beta(b) = B - \beta(b) \leq 0$$

and $w(t) > 0$ at some points in $(a,b)$ . From calculus, we know that $w(t)$ must have a positive maximum somewhere $t_0 \in (a,b)$ . Therefore,

$$w(t_0) > 0, w'(t_0) = 0, w''(t_0) \leq 0.$$

$$x(t_0) \geq \beta(t_0), x'(t_0) = \beta'(t_0), x''(t_0) \leq \beta''(t_0)$$

But,

$$x''(t_0) - \beta''(t_0) = F(t_0,x(t_0)) - \beta''(t_0) \geq f(t_0,\beta(t_0)) + \frac{x(t_0) - \beta(t_0)}{1 + \vert x(t_0)} - f(t_0,\beta(t_0))$$

$$= \frac{x(t_0) - \beta(t_0)}{1 + \vert x(t_0)\vert} > 0$$

which is a contradiction, completing the proof. $\square$

Example: Show that the following BVP

$$x'' = - \cos x$$

$$x(0) = 0, x(1) = 0$$

has a solution and find bounds on a solution of this BVP.
The BVP has a solution since $f(t,x,x') = - \cos x$ is continuous and bounded on $[0,1] \times \mathbb{R}^2$ . Let $\alpha(t) = 0$ . Then,

$$\alpha''(t) = 0 \geq -1 = - \cos \alpha(t) = - \cos 0 = f(t,\alpha(t))$$

$\alpha(t) = 0$ is a lower solution. Let

$$\beta(t) = \frac{t(1-t)}{2}$$

$$\beta''(t) = -1 \leq -\cos(\beta(t)) = f(t,\beta(t))$$

By the above theorem our given BVP has a solution $x(t)$ and

$$\alpha(t) \leq x(t) \leq \beta(t), t \in [a,b]$$

So,

$$0 \leq x(t) \leq \frac{t(1-t)}{2}, t \in [0,1]$$

Monday, 11-15-2004 Homework: 7.19 - 7.22.

Example Let $p$ be the width of a patch of plankton and let $x(t)$ be the density of the plakon $t$ units from one end of the patch of plankton. This leads to the BVP

$$x'' = -rx(1 - \frac{x}{K}), r>0, K > 0$$

$$x(0) = 0, x(p)=0$$

If th BVP has a nontrivial solution, then we say that the patch of plankton is viable. We will show for $r$ sufficiently large, this BVP has a nontrivial solution. Let

$$\beta(t) = K, t \in [0,p].$$

$$\beta''(t) = 0 = -r \beta(t) (1 - \frac{\beta(t)}{K}) = -r K (1 - \frac{K}{K})$$

Thus, $\beta$ is an upper solution to this BVP. We are looking for a lower solution of the form

$$\alpha(t) = \sin (\frac{\pi t}{p} ), 0 < a < K$$

$$\alpha'(t) = a \frac{\pi}{p} \cos(\frac{\pi}{p} t )$$

$$\alpha''(t) = a \frac{\pi^2}{p^2} \sin( \frac{\pi}{p} t )$$

$$\alpha + r \alpha( 1 - \frac{\alpha(t)}{K}) = - a \frac{\pi^2}{p^... ... \sin (\frac{\pi t}{p} ) \left( 1 - \frac{a \sin(\frac{\pi}{p} t )}{K} \right)$$

$$= a \sin(\frac{\pi}{p} t ) \left( -\frac{\pi^2}{p^2} + r(1 - \frac{a}{K} \sin(\frac{\pi}{p} t)) \right)$$

Hence for $0 < a < K$ and $a$ sufficiently small,

$$\alpha''(t) + r \alpha(t) \left( 1 - \frac{\alpha(t)}{K} \right) \geq 0$$

for $t \in [0,p]$ . We conclude that $\alpha(t)$ is a lower solution of $x'' = -rx(1 - \frac{x}{K})$ on $[0,p]$ .
By the previous theorem, if $r > \frac{\pi^2}{p^2}$ , then the given BVP has a nontrivial solution $x(t)$ and

$$\alpha(t) = a \sin(\frac{\pi}{p} t) \leq x(t) \leq \beta(t) = K$$

on $[0,p]$ .

Theorem (Uniqueness Theorem) Assume $f(t,x,x')$ is continuous on $[a,b] \times \mathbb{R}^2$ and for each fixed $(t,x') \in [a,b] \times \mathbb{R}$ , $f(t,x,x')$ is strictly increasing with respect to $x$ . Then the BVP

$$x'' = f(t,x,x')$$

$$x(a) = A, x(b) = B$$

has at most one solution.
Proof: Assume not; then the given BVP has at least 2 distinct solutions $x(t)$ and $y(t)$ . WLOG,

$$x(t) > y(t)$$

at some points in $(a,b)$ . Let

$$w(t) := x(t) - y(t)$$

Then,

$$w(t) > 0$$

at some points in $(a,b)$ and

$$w(a) = x(a) - y(a) = A - A = 0$$

$$w(b) = x(b) - y(b) = B - B = 0$$

$w(t$ has a positive maximum at some point, say $d$ , in $(a,b)$ . Then,

$$w(d) > 0, w'(d) = 0, w''(d) \leq 0$$

Using the definition of $d$ ,

$$x(d) > y(d), x'(d) = y'(d)$$

But,

$$w''(d) = x''(d) - y''(d) = f(d,x(d), x'(d)) - f(d,y(d), y'(d))$$

$$= f(d,x(d), x'(d)) - f(d,y(d), x'(d)) > 0$$

as $f$ is strictly increasing with respect to the second variable.

Example This show that one can't replace ``strictly increasing" in the last theorem by simply ``increasing".

$$x'' = \vert x'\vert^p$$

where $0 < p < 1$ . By exercise 7.21, this differential equation has a solution of the form $x(t) = \kappa \vert t^\alpha$ where $\kappa > 0$ and $\alpha > 2$ . Constant functions are also solutions with this differential equation. Hence, for appropriately selected BVPs, we can have more than one solution. Here,

$$f(t,x,x') = \vert x' \vert ^p$$

Hence, for each fixed $(t,x')$ , $f(t,x,x')$ is constant and is nondecreasing in $x$ . Notice that $f$ does not satisfy a Lipschitz condition.

One can prove the following uniqueness theorem:
Theorem: Assume $f(t,x,x')$ is continuous on $[a,b] \times \mathbb{R}^2$ , for each fixed $(t,x') \in [a,b] \times \mathbb{R}$ , $f(t,x,x')$ is nondecreasing, and $f9t,x,x')$ satisfies a Lipschitz condition with respect to $x'$ on each compact subset of $[a,b] \times \mathbb{R}^2$ . Then the BVP

$$x'' = f(t,x,x'), x(a) = A, x(b) = B$$

has at most one solution.

Theorem: Assume $f(t,x)$ is continuous on $[a,b] \times \mathbb{R}^2$ and for each fixed $t$ , $f(t,x)$ is nondecreasing with respect to $x$ , then the BVP

$$x'' = f(t,x), x(a) = A, x(b) = B$$

has a unique solution.
Proof: Note that $f(t,x)$ satisfies a Lipschitz condition with respect to $x$ so by the previous theorem, the BVP has at most one sol.
It remains to show that the given BVP actually has a solution.

Let $L$ be the set of points on the line segment from $(a,A)$ to $(b,B)$ . Let

$$M := \max \{ f(t,x) : (t,x) \in L \}$$

$$m := \max \{ f(t,x) : (t,x) \in L \}$$

If $(t,x)$ is a point which is above $L$ , then

$$f(t,x) \geq m$$

If $(t,x)$ is a point which is below $L$ , then

$$f(t,x) \leq M$$

Pick $p$ sufficiently large and $q$ sufficiently small so that the graphs of

$$\beta(t) := \frac{m}{2} t^2 + p$$

$$\alpha(t) := \frac{M}{2} t^2 + q$$

are above and below, respectively, the line $L$ .

$$\alpha(t) \leq \beta(t)$$

on $[a,b]$ , so

$$\alpha(a) \leq A \leq \beta(a), \alpha(b) \leq B \leq \beta(b).$$

$$\beta''(t) = m \leq f(t,\beta(t)), t \in [a,b]$$

so $\beta(t)$ is an upper solution on $[a,b]$ . Similarly, $\alpha(t)$ is a lower solution of $x'' = f(t,x)$ on $[a,b]$ .
By a previous theorem, there exist a solution to this problem.

Example: Consider the BVP

$$x'' = c(t) x + d(t) x^3 + e(t)$$

$$x(a) = A, x(b) = B$$

where $c(t), d(t), e(t)$ are continuous on $[a,b]$ and

$$c(t) \geq 0, d(t) \geq 0$$

on $[a,b]$ .
Claim: given BVP has a unique solution:
Here,

$$f(t,x) = c(t) x + d(t) x^3 + e(t)$$

is continuous on $[a,b] \times \mathbb{R}$ . Also,

$$f_x(t,x) = c(t) + 3 d(t) x^2 \geq 0$$

therefore, for each $t \in [a,b]$ , $f(t,x)$ is nondecreasing with respect to $x$ . So, the given BVP has a unique solution.

Example: A BVP that arises in combustion thoery is

$$\epsilon x'' = x^2 - t^2$$

$$x(-1) = x(1) = 1$$

where $\epsilon > 0$ is a small parameter related to the speed of the reaction, $x(t)$ is related to mass at distance $t$ to the flame.

Here,

$$f(t,x) = \frac1\epsilon x^2 - \frac1\epsilon t^2$$

is continuous on the slab $[-1,1] \times \mathbb{R}$ ;

$$f_x(t,x) = \frac1\epsilon 2 x$$

which may be negative if $x$ is negative. The previous theorem does not apply.
Note $\alpha(t) = 0$ on $[-1,1]$ is a lower sol: Proof: $\alpha''(t0 = 0$ ; $f(t,\alpha(t)) = f(t,0) = -\frac1\epsilon t^2$ and $0 \geq \frac1\epsilon t^2$ , so $\alpha$ is indeed a lower solution.

On the half slab

$$\{ (t,x) : -1 \leq t \leq 1, x \geq \}$$

$f(t,x)$ for each fixed $t \in [-1,1]$ is nondecreasing with respect to $x$ . By the proof, we can find an upper solution $\beta(t)$ such that the graph of $\beta(t)$ is above $L$ . Thus, there is a solution $x(t)$ of the BVP with $0 \leq x(t) \leq \beta(t)$ on $[-1,1]$ .
Note that the previous theorem does not apply over the whole slab, so the given BVP may have more than one solution, but there is a unique non-negative solution.

Definition: Assume $f(t,x,x')$ is continuous on $[a,b] \times \mathbb{R}^2$ . Let $\alpha, \beta \in \mathcal{C}[a,b]$ with $\alpha(t) \leq \beta(t), t \in [a,b]$ , and assume $c > 0$ is a constant. Then we say $F(t,x,x')$ is the modification of $f(t,x,x')$ with respect to $(\alpha,\beta,c)$ if

$$F(t,x,x') = \begin{cases}g(t,\beta(t),x' + \frac{x - \beta(t)}{1 ... ...'(t)) + \frac{x - \alpha(t)}{1 + \vert x\vert}, & x \leq \alpha(t) \end{cases}$$

where

$$g(t,x,x') = \begin{cases}f(t,x,c), & x' \geq c \ f(t,x,x'), & \vert x'\vert < c \ f(t,x,-c), & x' \leq -c \end{cases}$$

Friday, 11-19-2004 7.23 - 7.25

Theorem: Assume $f(t,x,x')$ is continuous on $[a,b] \times \mathbb{R}^2$ and solutions of IVPs in $[a,b] \times \mathbb{R}^2$ are unique. Further assume $\alpha(t) \leq \beta(t)$ on $[a,b]$ are lower and upper solutions, respectively, of $x'' = f(t,x,x')$ on $[a,b]$ and assume there is a $t_0 \in [a,b]$ such that $\alpha(t_0) = \beta(t_0)$ , $\alpha'(t_0) = \beta'(t_0)$ , then

$$\alpha(t) = \beta(t) \forall t \in [a,b]$$

Proof: Let $\alpha(t), \beta(t)$ , and $t_0$ be as in the statement of the theorem.
Claim: $\alpha(t) = \beta(t)$ on $[t_0,b]$ ($t_0 < b$ ).
Assume not, then $\beta(t) > \alpha(t)$ for at least one point in $(t_0,b]$ . There exists a $t_0 \leq t_1 < t_2 \leq b$ such that $\alpha(t_1) = \beta(t_1), \alpha'(t_1) = \beta'(t_1)$ and $\alpha(t) < \beta(t)$ on $(t_1, t_2]$ .
Pick $x_2 \in (\alpha(t_2), \beta(t_2))$ and let $x(t)$ be a solution of the BVP

$$x'' = F(t,x,x')$$

$$x(t_1) = \alpha(t_1), x(t_1) = x_2$$

where $F(t,x,x')$ is the modification of $f(t,x,x')$ with respect to the triple $\alpha(t), \beta(t), c$ on $[t_1,t_2]$ , where

$$c > \max \{ \max \vert \alpha'(t) \vert, \max \vert \beta'(t) \vert \}$$

Claim: $\alpha(t) \leq x(t) \leq \beta(t)$ on $[t_1,t_2]$ ; we will only show that $x(t) \leq \beta(t)$ . The other inequality follows similarly.
Assume not; then, there is a $t_3 \in (t_1,t_2]$ . so that

$$x(t_3) > \beta(t_3).$$

Let $w(t) := x(t) - \beta(t)$ . Then, $w(t)$ has a positive max at some point, say $d \in (t_1, t_2)$ .

$$w(d) > 0, w'(d) = 0, w''(d) \leq 0$$

But,

$$w''(d) = x''(d) - \beta''(d) \geq F(d,x(d),x'(d)) - f(d,\beta(d),\beta'(d))$$

$$= f(d,x(d,x'(d)) + \frac{ x(d) - \beta(d) }{1 + \vert \beta(d) \vert } - f(d, \beta(d), \beta'(d))$$

$$\frac{x(d) - \beta(d)}{1 + \vert \beta(d) \vert } > 0$$

which is a contradiction.
Hence, $x(t) \leq \beta(t)$ on $[t_1,t_2]$ . Similarly, $\alpha(t) \leq x(t)$ on $[t_1,t_2]$ . Thus,

$$\alpha(t) \leq x(t) \leq \beta(t), t \in [t_1, t_2]$$

Pick $t_4 \in [t_1, t_2)$ such that

$$\alpha(t_4) = x(t_4), \alpha'(t_4) = x'(t_4)$$

and

$$\alpha(t) < x(t)$$

on $(t_4, t_2]$ . Note,

$$\vert x'(t_4) \vert = \vert \alpha'(t_4) \vert < c$$

Pick $t_5 \in (t_4, t_2]$ so that

$$\vert x'(t) \vert < c$$

for $t \in [t_4, t_5]$ . Note that $x(t)$ is a solution of $x'' = f(t,x,x')$ on $[t_4,t_5]$ . $x(t)$ is a solution of $x'' = f(t,x,x')$ on $[t_4,t_5]$ implies $x(t)$ is an upper solution of $x'' = f(t,x,x')$ on $[t_4,t_5]$ .
Pick $x_5 \in (\alpha(t_5),x(t_5))$ . Then let $y(t)$ be a solution of the BVP

$$y'' = F_1(t,y,y'), y(t_4) = \alpha(t_4), y(t_5) = x_5$$

where $F_1(t,x,x')$ is the modification of $f(t,x,x')$ with respect to $(\alpha(t), x(t), c)$ . Just like earlier, we can show

$$\alpha(t) \leq y(t) \leq x(t),$$

on $[t_4,t_5]$ . Now, we can pick $t_6 \in (t_4, t_5]$ so that $x$ and $y$ differ at some points in $(t_4, t_6]$ and

$$y'(t) < c,$$

for $t \in [t_3, t_5]$ . Then $y$ is a solution of $x'' = f(t,x,x')$ on $[t_4, t_6]$ . But now we have that $x,y$ are two distinct solutions of the same IVP (same ICs at $t_4$ ), which contradicts the uniqueness of the solutions of IVPs.

Monday, 11-22-2004 Homework: 7.26.

In Exercise 7.23, you saw that

$$x(t) = 4 - \sqrt{4 - t}$$

is a solution of the IVP

$$x'' = 2 \vert x' \vert^3, x(0) = 2$$

with right maximal interval of existence $[0,4)$ . The solution $x(t)$ is bounded on $[0,4)$ but $x'(t)$ is unbounded on $[0,4)$ .
Here,

$$f(t,x,x') = 2 \vert x' \vert^3$$

This grows ``too fast" with respect to $x'$ .

Definition: We say that $f(t,x,x')$ satisfies a Nagumo condition with respect to the pair $\alpha(t)$ , $\beta(t)$ on $[a,b]$ provided $\alpha, \beta \in \mathcal{C}[a,b]$ , $\alpha(t) \leq \beta(t)$ , and there is a function $h: [0, \infty] \rightarrow (0, \infty)$ such that

$$\vert f(t,x,x')\vert \leq h(\vert x'\vert), t \in [a,b], \alpha(t) \leq x \leq \beta(t), x' \in \mathbb{R}$$

where

$$\int_\lambda^\infty \frac{s ds}{h(s)} > \max_{t \in [a,b]} \beta(t) - \min_{t \in [a,b]} \alpha(t)$$

where

$$\lambda := \max \{ \frac{\vert \beta(b) - \beta(a)\vert}{b - a}, \frac{\vert\alpha(b) - \beta(a)\vert}{b-a} \}$$

Theorem: Assume $f: [a,b] \times \mathbb{R}^2 \rightarrow \mathbb{R}$ is continuous and satisfies a Nagumo condition with respect to the pair $\alpha(t), \beta(t)$ on $[a,b]$ . Then, there is a constant $N > 0$ such that for any solution of $x'' = f(t,x,x')$ with

$$\alpha(t) \leq x(t) \leq \beta(t)$$

it follows that

$$\vert x'(t)\vert \leq N$$

for all $t \in [a,b]$ .
Note: We call $N$ an a priori bound on the derivative.
Proof:
Let $h$ and $\lambda$ be as in the definition of $f(t,x,x')$ satisfies a Nagumo condition with respect to the pair $\alpha(t), \beta(t)$ on $[a,b]$ .
We know that

$$\int_\lambda^\infty \frac{s ds}{h(s)} > \max_{t \in [a,b]} \beta(t) - \min_{t \in [a,b]} \alpha(t)$$

So pick $N > \lambda$ so that

$$\int_\lambda^N \frac{s ds}{h(s)} > \max_{t \in [a,b]} \beta(t) - \min_{t \in [a,b]} \alpha(t)$$

Let $x(t)$ be a solution of $x'' = f(t,x,x')$ satisfying

$$\alpha(t) \leq x(t) \leq \beta(t).$$

Claim: $\vert x'(t)\vert \leq N$ on $[a,b]$ :
Assume not; then there are values of $t$ in $[a,b]$ such that

$$\vert x'(t)\vert > N$$

We will only consider the case where $x'(t) < -N$ . By the Mean Value Theorem, there is a point $t_0 \in (a,b)$ such that

$$\left\vert \frac{x(b) - x(a)}{b-a} \right\vert = \vert x'(t_0)\vert \leq \max\{ \vert m_1\vert, \vert m_2\vert\} = \lambda$$

We limit our scope again to where there is a $t_1 \in [a,t_0)$ such that $x'(t_1) < -N$ . This implies that we can pick $t_1 < t_2 < t_3 \leq t_0$ such that

$$x'(t_2) = -N, x'(t_3) = -\lambda,$$

and $-N \leq x'(t) \leq -\lambda$ for $t \in (t_2, t_3)$ .

$$x''(t) \leq \vert x''(t)\vert = \vert f(t,x(t),x'(t)) \vert \leq h(\vert x'(t)\vert$$

for $t_2 \leq t \leq t_3$ . Then,

$$-x'(t) x''(t) \leq -x'(t) h(\vert x'(t)\vert)$$

$$\frac{ - x'(t) x''(t) }{h(-x'(t))} \leq - x'(t), t \in [t_2, t_3]$$

$$\int_{t_2}^{t_3} \frac{-x''(t) x'(t) }{h(-x'(t))} dt \leq - \int_{t_2}^{t_3} x'(t)dt$$

Let $s = -x'(t)$ ; then, $ds = -x''(t) dt$ .

$$\int_{-x'(t_2)}^{-x'(t_3)} \frac{ -s ds }{h(s) } \leq - [ x(t_3) - x(t_2) ]$$

$$\int_N^\lambda \frac{ -s ds}{ h(s) } \leq x(t_2) - x(t_3)$$

$$\int_\lambda^N \frac{ -s ds }{ h(s) } \leq x(t_2) - x(t_3)$$

So,

$$\int_\lambda^N \frac{s ds}{ h(s) } \leq \max_{t \in [a,b]} \beta(t) - \min_{t \in [a,b]} \alpha(t)$$

which is a contradiction of one of our earlier statements.

Definition: A linear D.E.

$$x'' + p(t) x' + q(t) x = 0$$

where $p \in \mathcal{C}[a,b]$ and $q \in \mathcal{C}[a,b]$ is said to be disconjugate provided no nontrivial solution of the D.E. has two or more zeros in $[a,b]$ .

Theorem: If the above D.E. is disconjugate on $[a,b]$ , then every boundary value problem

$$x'' + p(t) x' + q(t) x = 0$$

$$x(a) = A, x(b) = B$$

has a unique solution.

Theorem: If the linear D.E has a positive upper solution $\beta(t)$ on $[a,b]$ , then it is disconjugate on $[a,b]$ .

Monday, 11-29-2004
7.27, 7.28

Theorem: Assume $f(t,x,x')$ is continuous on $[a,b] \times \mathbb{R}^2$ and $\alpha(t), \beta(t)$ are lower and upper solutions of $x'' = f(t,x,x')$ respectively on $[a,b]$ with $\alpha(t) \leq \beta(t)$ on $[a,b]$ . Further assume $f(t,x,x')$ satisfies a Nagumo condition with respect to $\alpha(t), \beta(t$ . Then if

$$\alpha(a) \leq A \leq \beta(a), \alpha(a) \leq B \leq \beta(b)$$

then the BVP

$$x'' = f(t,x,x'), x(a) = A, x(b) = B$$

has a solution satisfying

$$\alpha(t) \leq x(t) \leq \beta(t)$$

on $[a,b]$ .
Proof: Pick $N_1 > 0$ so that

$$\vert\alpha'(t)\vert < N_1, \vert\beta'(t)\vert < N_1$$

on $[a,b]$ . By our Nagumo condition, there exists a function $h:[0, \infty) \rightarrow (0, \infty)$ such that

$$\vert f(t,x,x') \vert \leq h(\vert x'\vert)$$

for $t \in [a,b]$ , $\alpha(t) \leq x \leq \beta(t)$ , $x' \in \mathbb{R}$ , where

$$\int_\lambda^\infty \frac{s ds}{h(s)} > \max_{t \in [a,b]} \beta(t) - \min _{t \in [a,b]} \alpha(t),$$

where

$$\lambda = \max \{ \frac{\vert\beta(b) - \alpha(a)\vert}{b - a}, \frac{\vert\alpha(b) - \beta(a)\vert}{b - a} \}$$

Pick $N_2 > \lambda$ sufficiently large so that

$$\int_\lambda^{N_2} \frac{s ds}{h(s)} > \max_{t \in [a,b]} \beta(t) - \min_{t \in [a,b]} \alpha(t)$$

Let $N := \max \{ N_1, N_2 \}$ . Let $F(t,x,x')$ be the modification of $f(t,x,x')$ with respect to the triple $(\alpha(t), \beta(t), N)$ . Recall that $F$ is bounded on $[a,b] \times \mathbb{R}^2$ . Also,

$$F(t,x,x') = f(t,x,x')$$

for $t \in [a,b]$ , $\alpha(t) \leq x \leq \beta(t)$ , and $\vert x'\vert \leq N$ .
Let $x(t)$ be a solution of the BVP

$$x'' = F(t,x,x')$$

$$x(a) = A, x(b) = B$$

Claim: $x(t) \leq \beta(t)$ on $[a,b]$ :
Assoume not. Then, there are points in $(a,b)$ such that $x(t) > \beta(t)$ . So, defining $w(t) := x(t) - \beta(t)$ , we get that

$$w(a) \leq 0, w(b) \leq 0, w(t) > 0$$    at some point in $$(a,b)$$

Therefore, $w(t)$ has a positive maximum at some point $\xi \in (a,b)$ , and

$$w(\xi) \geq 0, w'(\xi) = 0, w''(\xi) \leq 0$$

Therefore,

$$x(\xi) > \beta(\xi), x'(\xi) = \beta'(\xi), x''(\xi) \leq \beta''(\xi)$$

But

$$w''(\xi) = x''(\xi) - \beta''(\xi)$$

$$\geq f(\xi, \beta(\xi), \beta'(\xi)) + \frac{x(\xi) - \beta(\xi)}{1 + \vert x(\xi) \vert } - f(\xi, \beta(\xi), \beta'(\xi)) > 0$$

This is a contradiction, so $x(t) \leq \beta(t)$ on $[a,b]$ . Similarly, $x(t) \geq \alpha(t)$ on $[a,b]$ .
It remains to show that $\vert x'(t)\vert \leq N$ on $[a,b]$ . By the MVT, there is a $t_0 \in (a,b)$ such that

$$\left\vert \frac{x(b) - x(a) }{b - a} \right\vert = \vert x'(t_0)\vert \leq \lambda \leq N$$

Pick a maximal subinterval $[t_3, t_4]$ where

$$a \leq t_3 \leq t_0 \leq t_4 \leq b$$

such that

$$-N \leq x'(t) \leq N,$$    on $$[t_3,t_4]$$

If $t_3 = a$ and $t_4 = b$ , we're done.
Claim: $t_4 = b$ ; assume not. Then, $t_4 < b$ and $\vert x'(t_4)\vert = N$ . We will only consider the case $x'(t_4) = N$ . Pick $t_0 \leq t_5 \leq t_6 \leq t_4$ such that

$$x'(t_5) = \lambda, x'(t_6) = N$$

and

$$\lambda < x'(t) , N$$

for $t_5 \leq t \leq t_6$ .
Notice

$$x''(t) \leq \vert x''(t) \vert = f(t,x(t), x'(t))$$

=

$$\leq h(\vert x'(t)\vert = h(x'(t))$$

$$\frac{x''(t)}{h(x'(t))} \leq 1$$

So,

$$\frac{x(t) x''(t)\vert}{h(x'(t))} \leq x'(t)$$

$$\int_{t_5}^{t_6} \frac{x'(t) x''(t)}{h(x'(t))} dt \leq x(t_6) - x(t_5)$$

Taking $s = x'(t)$ , so $ds = x''(t) dt$ ,

$$\int_{x'(t_5)}^{x'(t_6)} \frac{s ds}{h(s)} \leq \max_{t \in [a,b]} \beta(t) - \min_{t \in [a,b]} \alpha(t)$$

$$\int_\lambda^N \frac{s ds}{ h(s) } \leq \max_{t \in [a,b]} \beta(t) - \min_{t \in [a,b]} \alpha(t)$$

This is a contradiction. The other cases follow similarly. $\square$

Example Consider the BVP

$$x'' = x^2 + (x')^4$$

$$x(0) = A, x(1) = B$$

where $A > 0, B > 0$ .
Let $c > \max\{ A, B \}$ . Let $\alpha(t) = 0$ and $\beta(t) = c$ on $[0,1]$ . Note

$$\alpha''(t) = 0 = f(t,\alpha(t), \alpha'(t))$$

$$\beta''(t) = 0 \leq f(t,\beta(t),\beta'(t)) = c^2$$

Hence, $\alpha(t), \beta(t)$ are lower and upper solutions of $x'' = x^2 + (x')^4$ respectively on $[0,1]$ . Also,

$$0 = \alpha(t) \leq \beta(t) = c$$

on $[0,1]$ . Also,

$$\vert f(t,x,x')\vert = \vert x^2 + (x')^4\vert = x^2 = (x')^4$$

$$\leq c^2 + (x')^4 \leq h(\vert x'\vert)$$

if we define

$$h(s) = c^2 + s^4$$

Notice that $\lambda = c$ in the Nagumo condition. We want

$$\int_\lambda^\infty \frac{s ds}{h(s)} > \max_{t \in [0,1]} \beta(t) - \min_{t \in [0,1]} \alpha(t) = c$$

$$\int_c^\infty \frac{s}{c^2}{c^2 + (s^2)^2} ds > c$$

Using the substitution $u = s^2, du = 2s ds$ ,

$$\frac12 \frac1c \arctan \frac{s^2}{c} \vert _{s = c}^{s = \infty}$$

$$\frac1{2c} \frac\pi2 - \frac1{2c} \arctan c > c$$

$$\frac\pi2 - \arctan c > 2c^2$$

So, if

$$0 < c < c_0 = .69424$$

then the inequality holds. Then, $0 < x(t) < c_0$ for $0 \leq t \leq 1$ .

Wednesday, 12-1-2004 7.29

Let $\mathbb{X}$ be a Banach space.

Definition: We say $T: \mathbb{X}\rightarrow \mathbb{X}$ is a compact mapping provided whenever $\{ x_n \} \subset \mathbb{X}$ is bounded, then it follows that $\{ Tx_n \}$ has a convergent subsequence.

Example: Assume that $T : \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuous; then $T$ is a compact mapping.

Let $\{ x_n \}_{n = 1}^\infty \subset \mathbb{R}^n$ be a bounded sequence, then the closure of the sequence is compact. So, $\{ Tx_n \}$ is compact; hence, we know it has a convergent subsequence. Therefore, $T : \mathbb{R}^n \rightarrow \mathbb{R}^n$ is compact.

Schaefer's Fixed Point Theorem: Let $\mathbb{X}$ be a Banach space and assume $T: \mathbb{X}\rightarrow \mathbb{X}$ is continuous and compact. If the set

$$S := \{ x \in \mathbb{X}: x = \lambda Tx$$    for some $$\lambda \in [0,1) \}$$

is bounded, then $T$ has a fixed point $\bar x \in X$ . Furthermore, if

$$M > \sup\{ \Vert x \Vert : x \in S \},$$

then

$$\Vert \bar x \Vert < M.$$

Example: Need $S$ s bounded in Schaefers Theorem. Let $\mathbb{X}= \mathbb{R}$ . Define $T: \mathbb{X}\rightarrow \mathbb{X}$ by

$$T(x) = \sqrt{ 1 + x^2 }, x \in \mathbb{X}= \mathbb{R}$$

$T: \mathbb{X}\rightarrow \mathbb{X}$ is continuous, so $T: \mathbb{X}\rightarrow \mathbb{X}$ is a compact mapping.
For $0 \leq \lambda < 1$ , consider

$$x = \lambda T x$$

$$x = \lambda \sqrt{1 + x^2}$$

$$x_\lambda = \frac{\lambda}{\sqrt{1 - \lambda^2}}$$

So, $S$ is unbounded. We claim that $T$ does not have a fixed point.

$$x = Tx = \sqrt{1 + x^2}$$

$$x^2 = 1 + x^2$$

$$1 = 0$$

So, there are no fixed points.

Demonstration of Theorem
Let $\{ \lambda_n\} \subset [0,1)$ with $\lim_{n \rightarrow \infty} \lambda_n = 1$ . Assume that for each $n \geq 1$ the equation

$$x = \lambda_n T x$$

has a solution, $x_n$ .
Define,

$$x_n = \lambda_n T x_n$$

Then, $\{ x_n \}^\infty_{n = 1} \subset S$ . Since $S$ is bounded, we get that the sequence $\{ x_n \}$ is bounded. Because $T$ is compact, $\{ Tx_n \}$ has a convergent subsequence. Let $\{ T x_{n_k} \}$ be a convergent subsequence. Let

$$\bar x = \lim_{k \rightarrow \infty} T x_{n_k}$$

$$x_{n_k} = \lambda_{n_k} T x_{n_k}$$

$$\lim_{k \rightarrow \infty} x_{n_k} = \bar x$$

Hence,

$$\bar x = T \bar x$$

and $T$ has a fixed point.

We will consider the second order vector BVP,

$$x'' = f(t,x), x(a) = A, x(b) = B$$

Here $x$ is an $n$ -vector, $f : [a,b] \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuous, and $A, B$ are constant $n$ vectors. So, to study this BVP, we are interested in the so-called Green's function $G(t,s)$ for the BVP

$$x'' = 0, x(a) = 0, x(b) = 0$$

This Green's function is given by

$$G(t,s) = \begin{cases}-\frac{(t-a)(b-s)}{b-a}, & a \leq t \leq s \leq b \ -\frac{(b-t)(s-a)}{b-a}, & a \leq s \leq t \leq b. \end{cases}$$

Assume $h : [a,b] \rightarrow \mathbb{R}^n$ is continuous. Then the solution of the vector BVP

$$x'' = h(t), x(a) = A, x(b) = B$$

is given by

$$x(t) = \phi(t) + \int_a^b G(t,s) h(s) ds$$

where $\phi(t)$ solve the vector BVP

$$\phi '' = 0, \phi(a) = A, \phi(b) = B$$

So, in order to solve our original problem, we examine

$$x'' = f(t,x(t)), x(a) = A, x(b) = B$$

This has a solution if and only if $x$ is a solution of

$$x(t) = \phi(t) + \int_a^b G(t,s) f(t,x(t)) ds$$

Define $T : C[a,b] \rightarrow C[a,b]$ by

$$(Tx)(t) := \phi(t) + \int_a^b G(t,s) f(s,x(s)) ds, t \in [a,b]$$

Then we have a solution if and only if $T$ has a fixed point. Note

$$\phi(t) = A + \frac{B - A}{b-a}(t - a)$$

$$\phi'(t) = \frac{B - A}{b-a}$$

$$\phi''(t) = 0$$

$$\phi(a) = A, \phi(b) = B$$

So, the homogeneous BVP

$$\phi'' = 0, \phi(a) = A, \phi(b) = B$$

has a unique solution. Note we can rewrite this into

$$\phi(t) = \frac{A(b-a) + (B-A) (t-a)}{ b - a} = \frac{A(b-t) + B(t-a) }{b-a}$$

Let

$$\beta = \max \{ \Vert A \Vert, \Vert B \Vert \}.$$

Then,

$$\Vert \phi(t) \Vert \leq \frac{1}{b - a} [ (b-t) \Vert A \Vert + (t - a) \Vert B \Vert ]$$

$$\Vert \phi(t) \Vert \leq \frac{\beta}{b-a} [ (b - t) + (t-a) ] = \beta$$

So,

$$\Vert \phi(t) \Vert \leq \beta,$$    for $$t \in [a,b]$$

In Schaefer's Theorem, we are concerned with the equation

$$x = \lambda Tx, \lambda \in [0,1).$$

$$x(t) = \lambda(Tx)(t), t \in [a,b]$$

$$x(t) = \lambda[ \phi(t) + \int_a^b G(t,s) f(s,x(s)) ds ]$$

$$x(t) = \lambda \phi(t) + \int_a^b G(t,s) \lambda f(s,x(s)) ds$$

for $t \in [a,b]$ .
$x$ is a solution of $x = \lambda Tx$ if and only if $x$ is a solution of the BVP

$$x'' = \lambda f(t,x)$$

$$x(a) = \lambda A, x(b) = \lambda B$$

When we apply Schaefer's Theorem to get that the BVP has a solution, we will show that all solutions of the above BVP are bounded.

Friday, 12-3-2004
Define the norm

$$\Vert x \Vert = \sqrt{\sum_{i=1}^n x_i^2 }$$

$$\langle x,y \rangle = \sum_{i=1}^n x_i y_o$$

Lemma Assume $f : [a,b] \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuous and there are constants $\alpha \geq 0, K \geq 0$ such that

$$\Vert f(t,x)\Vert \leq 2 \alpha \langle x, f(t,x) \rangle + K$$

$\forall (t,x) \in [a,b] \mathbb{R}^n$ . Then if $x$ is a solution of the BVP

$$x'' = \lambda f(t,x)$$

$$x(a) = \lambda A, x(b) = \lambda B$$

for some $\lambda \in [0,1]$ , then

$$\Vert x(t) \Vert \leq M,$$    on $$[a,b],$$

where

$$M := \beta + \alpha \beta^2 + K \max_{t \in [a,b]} \int_a^b \vert G(t,s)\vert ds$$

where $G(t,s)$ is the Green's function for the BVP

$$x'' = 0, x(a) = 0, x(b) = 0$$

and

$$\beta := \max \{ \Vert A\Vert, \Vert B\Vert \}$$

Proof: Assume $x(t)$ is a solution of the BVP for some $\lambda \in [0,1]$ . Then $x(t)$ is a solution of

$$x = \lambda T x,$$

where

$$(Tx)(t) = \phi(t) + \int_a^b G(t,s) f(s,x(s)) ds, t \in [a,b]$$

where $\phi(t)$ is the solution of the vector BVP

$$\phi'' = 0, \phi(a) = A, \phi(b) = B$$

Recall last time that we proved that

$$\Vert \phi(t) \Vert \leq \beta, t \in [a,b]$$

Then,

$$x(t) = \lambda Tx(t) = x(t) + \lambda\phi(t) + \lambda \int_a^b G(t,s) f(s,x(s)) ds$$

for $t \in [a,b]$ . Taking norms,

$$\Vert x \Vert = \lambda \Vert \phi(t) \Vert + \lambda \int_a^b \Vert G(t,s) \Vert \Vert f(s,x(s))\Vert ds$$

$$\leq \beta + \lambda \int_a^b \Vert G(t,s) \Vert [ 2 \alpha \langle x(s), f(s,x(s)) \rangle + K ] ds$$

Then,

$$\Vert x(t) \Vert \leq \beta + \int_a^b \Vert G(t,s)\Vert [ 2\alpha \langle x(s), \lambda f(s,x(s) \rangle + K ] ds$$

Let $r(t) := \Vert x(t) \Vert^2$ for $t \in [a,b]$ . Then,

$$r(t) = \langle x(t), x(t) \rangle , t \in [a,b]$$

$$r'(t) = \langle x(t), x'(t) \rangle + \langle x'(t), x(t) \rangle = 2 \langle x(t), x'(t) \rangle$$

$$r''(t) = 2 \langle x'(t), x'(t) \rangle + 2 \langle x(t), x''(t) \rangle = 2 \Vert x'(t) \Vert^2 + 2 \langle x(t), \lambda f(t,x(t)) \rangle$$

$$r''(t) \geq 2 \langle x(t), \lambda f(t,x(t)) \rangle$$

Returning back to the first inequality we were working on,

$$\Vert x(t) \Vert \leq \beta + \int_a^b \Vert G(t,s) \Vert [ \alpha r''(s) + K ] ds$$

Let

$$I(t) = \int_a^b \Vert G(t,s) \Vert r''(s) ds = \int_a^t \vert G(t,s) \vert r''(s) ds + \int_t^b \Vert G(t,s) \Vert r''(s) ds$$

$$=: I_1(t) + I_2(t)$$

$$I_1(t) := \int_a^t \Vert G(t,s) \Vert r''(s) ds = \int_a^t \frac{(b - t)(s - a)}{b - a} r''(s) ds$$

$$= \frac{b-t}{b-a} \int_a^t (s - a) r''(s) ds$$

We now integrate by parts. Let $u = s - a$ and $dr = r''(s) ds$ . Then we get,

$$I_1(t) = \frac{b-t}{b-a} \left\{ \left[ (s-a) r'(s) \right]_a^t - \int_a^t r(s) ds \right\}$$

$$= \frac{b-t}{b-a} \left\{ (t-a)r'(t) - r(t) - r(a)\right\}$$

$$= \frac{(b-t)(t-a)}{b-a} r'(t) - \frac{b-t}{b-a}r(t) + \frac{b-t}{b-a} r(a)$$

Now, consider $I_2$ :

$$I_2(t) = \int_t^b \Vert G(t,s) \Vert r''(s) ds = \frac{t-a}{b-a} \int_t^b (b-s) r''(s) ds$$

Doing the same integration by parts procedure, we get

$$= - \frac{(t-a)(b-t)}{b-a} r'(t) + \frac{t-a}{b-a} r(b) - \frac{t-a}{b-a} r(t)$$

Hence,

$$I(t) = I_1(t) + I_2(t) = -r(t) + \frac{b-t}{b-a}r(a) + \frac{t-a}{b-a}r(b)$$

$$I(t) \leq \frac{b-t}{b-a}\Vert\beta^2 + \frac{t-a}{b-a} \beta^2 \leq \beta^2$$

Returning to our original inequality,

$$\Vert x(t) \Vert \leq \beta + \alpha\beta^2 + K \max_{t \in [a,b]} \int_a^b \Vert G(t,s) \Vert ds = M$$

Theorem: Assume $f : [a,b] \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuous and there are constants $K \geq 0, \alpha \geq 0$ such that

$$\Vert f(t,x)\Vert \leq 2 \alpha \langle x, f(t,x) \rangle + K$$

$\forall (t,x) \in [a,b] \times \mathbb{R}^n$ . Then the BVP

$$x'' = f(t,x), x(a) = A, x(b) = B$$

has a solution, $x(t)$ . Furthermore,

$$\Vert x(t) \Vert \leq M,$$    on $$[a,b],$$

where

$$M = \beta + \alpha \beta^2 + K \max_{t \in [a,b]} \int_a^b \Vert G(t,s) \Vert ds$$

Proof: (Uses Schaefer's Theorem)
Let $\mathbb{X}= C([a,b] \times \mathbb{R}^n)$ with norm $\Vert \cdot \Vert _\mathbb{X}$ defined by

$$\Vert x\Vert _\mathbb{X}= \max\{\Vert x(t) \Vert : a \leq t \leq b \}.$$

Define $T: \mathbb{X}\rightarrow \mathbb{X}$ by

$$(Tx)(t) = \phi(t) + \int_a^b G(t,s) f(s,x(s)) ds$$

for $t \in [a,b]$ . It can be shown that $T$ is continuous and compact.
Let

$$S := \{ x \in \mathbb{X}: x = \lambda Tx$$    for some $$\lambda \in [0,1) \}$$

So $x \in S$ if and only if $x = \lambda Tx$ for some $\lambda \in [0,1)$ . This is true if and only if $x$ is a solution of the BVP

$$x'' = \lambda f(t,x), x(a) = \lambda A, x(b) = \lambda B$$

for $\lambda \in [0,1)$ . But, by the last Lemma, $\Vert x(t) \Vert \leq M$ for all $t \in [a,b]$ , so $\Vert x \Vert _\mathbb{X}\leq M$ for all $x \in S$ .
Thus, the set $S$ is bounded in the Banach space. By Schaefer's Theorem, $T$ has a fixed point in $\mathbb{X}$ , so the given BVP

$$x'' = f(t,x)$$

$$x(a) = A, x(b) = B$$

has a solution. By the Lemma, the solution is bounded in the norm by $M$ ; $\Vert x(t) \Vert \leq M$ for all $t \in [a,b]$ .

Monday, 12-6-2004
7.28 - 7.30

Scalar Case:

$$\mathbb{X}= C[a,b]$$

$$\Vert x \Vert = \max\{ \vert x(t)\vert : a \leq t \leq b \}.$$

$$T : \mathbb{X}\rightarrow \mathbb{X}$$

$$(Tx)(t) = \phi(t) + \int_a^b G(t,s) f(s,x(s)) ds, t \in [a,b]$$

From the above theorem, we have yet to prove $T$ is continuous and compact.

Claim: $T: \mathbb{X}\rightarrow \mathbb{X}$ is continuous:
Assume $x_n \rightarrow x$ in $\mathbb{X}$ . Then,

$$\Vert x_n - x \Vert \rightarrow 0$$    as $$n \rightarrow \infty$$

$$\lim_{n \rightarrow \infty} x_n(t) = x(t)$$

uniformly on $[a,b]$ . Let

$$K = \{ (t,x) : a \leq t \leq b, x(t) -1 \leq x \leq x(t) + 1$$

$f(t,x)$ is continuous on the compact set $K$ . Therefore, $f(t,x)$ is uniformly continuous on $K$ . Let $\epsilon > 0$ be given. Then, there is a $\delta > 0$ such that

$$f(t,x) - f(\tau,y) < \epsilon$$

whenever $(t,x), (\tau,y) \in K, \vert t - \tau\vert < \delta, \vert x - y\vert < \delta$ .

$$\lim_{n \rightarrow \infty} x_n(t) = x(t)$$    uniformly on $$[a,b]$$

So, there is a positive integer $N$ sufficiently large so that

$$(t,x_n(t)) \in K, \forall t \in [a,b], n \geq N$$

Consider,

$$\vert(Tx_n)(t) - (Tx)(t)\vert = \vert \int_a^b G(t,s) [ f(s,x_n(s)) - f(s,x(s)) ] ds \vert$$

$$\leq \int_a^b \vert G(t,s) \vert \vert f(s,x_n(s)) - f(s,x(s)) \vert ds$$

Since $$\lim_{n \rightarrow \infty} x_n(t) = x(t)$$ uniformly on $[a,b]$ , there exists $N_2 \geq N$ such that

$$\vert x_n(t) - x(t) \vert < \delta$$

$\forall t \in [a,b]$ , $\forall n \geq N_2$ . Then,

$$\vert (Tx_n)(t) - (Tx)(t) \vert \leq \left( \int_a^b \vert G(t,s)\vert ds \right) \epsilon \leq \frac{(b-a)^2}{8} \epsilon$$

So,

$$\lim_{n \rightarrow \infty} T x_n = T x$$

and thus $T$ is continuous in $\mathbb{X}$ .

Claim: $T$ is compact. Let $\{ x_n \}$ be a bounded sequence in $\mathbb{X}$ . We want to show that $\{ Tx_n \}$ has a convergent subsequence.
Suppose $\{ x_n \} \subset \mathbb{X}$ is bounded. Then, there exists a constant $M_1 > 0$ such that

$$\Vert x_n \Vert \leq M_1, \forall n \geq 1.$$

$$\vert x_n(t)\vert \leq M_1, \forall n \geq 1, \forall t \in [a,b]$$

So, $\{ x_n(t) \}$ is uniformly bounded on $[a,b]$ .

$$(T x_n)(t) = \phi(t) + \int_a^b G(t,s) f(s,x_n(s)) ds$$

$$(T x_n)(t) \leq \vert\phi(t)\vert + \int_a^b \vert G(t,s) \vert \vert f(s,x_n(s)) \vert ds \leq M_2$$

for some constant $M_2 > 0$ , $\forall t \in [a,b]$ .
So, $\{ (T x_n)(t) \}_{n=1}^\infty$ is uniformly bounded on $[a,b]$ .

We now need to show equicontinuity of the sequence $\{ Tx_n \}$ . To this end,

$$(T x_n)'(t) = \phi'(t) + \int_a^b G_t(t,s) f(s,x(s)) ds$$

$$\vert Tx_n)'(t) \vert = \vert \phi'(t) \vert + \int_a^b \vert G_t(t,s) \vert f(s,x_n(s) \vert ds \leq M_3$$

for some constant $M_3$ .
Consider,

$$\vert (T x_n)(t) - (T x_n)(\tau) \vert = \vert (T x_n)'(\xi) \vert \vert t - \tau\vert$$

$$\leq M_3 \vert t - \tau\vert, t, \tau \in [a,b], n \geq 1.$$

This implies $\{(T x_n)(\tau) \}$ is equicontinuous on $[a,b]$ . By the Ascoli-Arzela Theorem, $\{ Tx_n \}$ has a subsequence $\{ T x_{n_k} \}$ such that $\{ T x_{n_k} \}$ converges uniformly on $[a,b]$ .
This implies that $\{ T x_{n_k} \}$ converges in $\mathbb{X}$ .

Example: The derived equation for the spring with external force problem is:

$$m x'' + kx = f(t)$$

From this, we say that

$$( p(t) x')' + q(t) x = f(t)$$

is the forced second-order (formally) self-adjoint DE. We want to find conditions on $p(t)$ , $q(t)$ , and $f(t)$ such that the second-order self-adjoint DE has a solution $x(t)$ with

$$\lim_{t \rightarrow \infty} x(t) = 0$$

Motivation for how we pick our operator $T$ (informal):
Assume there is a solution $x(t)$ with $\lim_{t \rightarrow \infty} x(t) = 0$ .

$$(p(t) x'(t))' = -q(t) x(t) + f(t), t \in [a,b]$$

$$= \left[ - \int_b^t q(s) x(s) ds - \int_t^\infty f(s) ds \right]'$$

Note we need the integral of $f$ to infinity converges. Integrate from $t$ to $\infty$ .

$$p(\infty) x'(\infty) - p(t)x'(t) = \left[- \int_b^t q(s) x(s) ds - \int_t^\infty f(s) ds\right]_t^\infty$$

we assume $p(\infty) x'(\infty)$ goes to 0. Then,

$$- p(t) x'(t) = - \int_b^\infty q(s) x(s) ds + \int_b^t q(\tau) x(\tau) d\tau + \int_t^\infty f(\tau) d\tau$$

We define $F(t) = \int_t^\infty f(\tau) d\tau$ . Then,

$$x'(t) = \frac{1}{p(t)} \int_b^\infty q(s) x(s) ds - \frac{1}{p(t)} \int_b^t q(\tau) x(\tau) d\tau - \frac{F(t)}{p(t)}$$

Intgrating from $t$ to $\infty$ again,

$$x(\tau) - x(t) = \int_t^\infty \frac{1}{p(s)} \int_b^\infty q(\ta... ...}{p(s)} \int_b^s q(\tau) x(\tau) d\tau ds - \int_t^\infty \frac{F(s)}{p(s)} ds$$

We define

$$K(t) = - \int_t^\infty \frac{F(s)}{p(s)} ds$$

Solving for $x(t)$ ,

$$x(t) = - \int_t^\infty \frac{1}{p(s)} \int_b^\infty q(\tau) x(\ta... ...tau ds + \int_t^\infty \frac{1}{p(s)} \int_b^s q(\tau) x(\tau) d\tau ds + K(t)$$

Integrate by parts.

Monday, 12-8-2004 7.7

Consider the forced 2nd order linear DE

$$( p(t) x')' + q(t) x = f(t)$$

Assume $p, q, f \in C[a,\infty)$ and $p(t) > 0$ .

Theorem 7.6: Assume

• $p(t) > 0$ , $q(t) \geq 0$ on $[a,\infty)$
  
• $\int_a^\infty \frac{1}{p(t)} dt < \infty$
  
• $\int_a^\infty q(t) P(t) dt < \infty$ , where $P(t) := \int_t^\infty \frac{1}{p(s)}ds$
• $\int_a^\infty f(t) dt$ converges.

hold, then the forced 2nd-order linear DE has a solution $x(t)$ satisfying

$$\lim_{t \rightarrow \infty} x(t) = 0$$

Proof: Application of the CMT.
Since $\int_a^\infty q(t) P(t) dt < \infty$ , there is a $b \geq a$ such that

$$\alpha := \int_b^\infty q(t) P(t) dt < 1$$

Let

$$\mathbb{X}:= \{ x \in C[b,\infty) : \lim_{t \rightarrow \infty} x(t) = 0 \}$$

and define a norm on $\mathbb{X}$ by

$$\Vert x \Vert = \max\{ \vert x(t)\vert : t \in [b,\infty)$$

Define an operater $T$ on $\mathbb{X}$ by

$$(Tx)(t) := K(t) + P(t) \int_t^b q(s) x(s) ds + \int_t^\infty P(s) x(s) q(s) ds$$

for $t \in [b,\infty)$ where

$$K(t) = \int_t^\infty \frac{F(s)}{p(s)} ds,$$

$$F(t) := \int_t^\infty f(s) ds,$$

$$P(t) := \int_t^\infty \frac{1}{p(s)}ds$$

for $t \geq b$ . Note the existence of all the integrals in the proof.

Let $x \in \mathbb{X}$ . Then,

$$Tx \in C[b, \infty)$$

Claim: $$\lim_{t \rightarrow \infty} (Tx)(t) = 0$$ :
The only hard part here is to show that

$$\lim_{t \rightarrow \infty} y(t) = 0$$

where

$$y(t) := P(t) \int_b^t q(s) x(s) ds, t \geq b$$

Let $\epsilon > 0$ be given. Since $$\lim_{t \rightarrow \infty} x(t) = 0$$ , there is a $c \geq b$ sufficiently large so that $\vert x(t)\vert < \epsilon$ for all $t \geq c$ . Since $$\lim_{t \rightarrow \infty} P(t) = 0$$ , there is a $d \geq c$ sufficiently large so that

$$P(t) \int_b^c q(s) \vert x(s)\vert ds < \epsilon$$

for all $t \geq d$ . Notice,

$$\vert y(t) \vert \leq P(t) \int_b^t q(s) \vert x(s)\vert ds = P(t... ...[ \int_b^c q(s) \vert x(s) \vert ds + \int_c^t q(s) \vert x(s)\vert ds \right]$$

$$< \epsilon + P(t) \int_c^t q(s) \vert x(s)\vert ds \leq \epsilon + \epsilon P(t) \int_c^t q(s) ds$$

Using the fact that $P(t)$ is a decreasing positive function, we get

$$\leq \epsilon + \epsilon \int_c^t P(s) q(s) ds$$

From our selection of $b$ , we that

$$\leq \epsilon + \alpha \epsilon < 2 \epsilon$$

So,

$$\lim_{t \rightarrow \infty} (Tx)(t) = 0.$$

Hence, $T: \mathbb{X}\rightarrow \mathbb{X}$ .

Claim: $T$ is a contraction mapping on the Banach space $\mathbb{X}$ .
Proof: Let $x,y \in \mathbb{X}$ and consider for $t \geq b$ ,

$$\vert (Tx)(t) - (Ty)(t) \vert \leq P(t) \int_b^t q(s) \vert x(s) - y(s) \vert ds + \int_t^\infty P(s) \vert x(s) - y(s) \vert q(s) ds$$

$$\leq \left[P(t) \int_b^t q(s) ds + \int_t^\infty P(s) q(s) ds \right] \Vert x - y \Vert$$

$$\leq \int_b^\infty P(s) q(s) ds \Vert x - y \Vert = \alpha \Vert x - y \Vert$$

where $\alpha$ , as defined above, is less than $1$ . As this holds for all $t$ , we get

$$\Vert Tx - Ty \Vert \leq \alpha \Vert x - y \Vert$$

so $T$ is a contraction mapping. By the CMT, $T$ has a unique fixed point $x \in \mathbb{X}$ .

We know

$$x = Tx$$

Therefore, $x(t) = (Tx)(t)$ for all $t \geq b$ . Then,

$$x(t) = K(t) P(t) \int_b^t q(s) x(s) ds + \int_t^\infty P(s) q(s) x(s) ds$$

$$x'(t) = - \frac{F(t)}{p(t)} + P(t) q(t) x(t) + - \frac{1}{p(t)} \int_b^t q(s) x(s) ds - P(t) q(t) x(t)$$

$$p(t) x'(t) = - F(t) - \int_b^t q(s) x(s) ds = -\int_t^\infty f(s) ds - \int_b^t q(s) x(s) ds$$

$$(p(t) x'(t))' + q(t) x(t) = f(t), t \geq b$$

Friday, 12-10-2004
Final Exam, 8:00 - 10:00, Avery 343

Notation: If $A(t)$ is a matrix function then we write $A(t) \leq 0$ on an interval $I$ provided that

$$a_{i,j}(t) \leq 0, t \in I$$

for all $i, j$ .

Theorem: Assume $A(t)$ is a continuous $n \times n$ matrix function of $[a,b)$ and $A(t) \leq 0$ on $[a,b)$ , where $a \leq b \leq \infty$ . Then the vector differential equation $x' = A(t)x$ has a nontrivial solution $x(t)$ satisfying

$$x(t) \geq 0, x'(t) \leq 0$$

for all $t \in [a,b)$ .
Proof: Let $a \leq \tau \leq b$ .
Let $x(t,\tau)$ be the solution of the IVP

$$x' = A(t) x, x(\tau) = x_0$$

where $x_0 > 0$ .
Claim: $x(t) > 0$ on $[a,\tau)$ .
Proof: Assume not; then there exists a $t_0 \in [a,\tau)$ such that

$$x(t) > 0$$    on $$(t_0, \tau]$$

and at least one component of $x(t,\tau)$ , say $x_i(t,\tau)$ satisfies

$$x_i(t_0,\tau) = 0$$

But

$$x'(t,\tau) = A(t) x(t,\tau) \leq 0$$

on $[t_0, \tau)$ since $x(t,\tau) \geq 0$ and $A(t) \leq 0$ . So, each component is nonincreasing. But,

$$x(t_0,\tau) \geq x(\tau,\tau) = x_0 > 0$$

Hence, $x(t,\tau) > 0$ on $[a,\tau)$ .
Let $\{\tau_n\} \subset (a,b)$ with

$$\lim_{n \rightarrow \infty} \tau_n = b$$

Let

$$y(t,\tau_n) = \frac{x(t,\tau_n)}{\Vert x(a,\tau_n) \Vert}$$

So, for each $n$ , $y(t,\tau_n)$ is a solution of $x' = A(t)x$ and

$$\Vert y(a,\tau_n) \Vert = 1$$

Also,

$$y(t,\tau_n) > 0$$    on $$[a,\tau_n)$$

$$\{ y(a,\tau_n) \}_{n=1}^\infty \subset S^1$$

and $S^1$ is compact. This implies that there exists a subsequence $\{y(a, \tau_{n_k})\}_{k=1}^\infty$ such that

$$\lim_{k \rightarrow \infty} y(a, \tau_{n_k}) = y_0$$

where

$$\Vert y_0 \Vert = 1$$

so $y_0 \neq 0$ . Let $x(t)$ be the solution of the IVP

$$x' = A(t)x, x(t_0) = y_0$$

Because $y_0 \neq 0$ , $x(t)$ is a nontrivial solution of $x' = A(t)x$ . Also,

$$\lim_{k \rightarrow \infty} y(t,\tau_{n_k}) = x(t)$$

uniformly on compact subsets of $[a,b)$ .
Thus,

$$x(t) \geq 0$$    on $$[a,b)$$

So,

$$x'(t) = A(t) x(t) \leq 0$$

because $A(t) \leq 0$ and $x(t) \geq 0$ .

Application: Assume $p(t) \geq 0$ and $q(t) \leq 0$ on $[a,b)$ where $a < b \leq \infty$ , then the scalar DE

$$u'' + p(t) u' + q(t) u = 0$$

has a solution satisfying

$$u(t) > 0, u'(t) \leq 0, u''(t) \geq 0$$

Proof: Let $u(t)$ be a solution of the scalar DE $u'' + p(t) u' + q(t) u = 0$ and set

$$x(t) = \begin{bmatrix}u(t) \ -u'(t) \end{bmatrix}$$

Then,

$$x'(t) = \begin{bmatrix}u'(t) \ u''(t) \end{bmatrix} = \begin{bmatrix}u'(t) \ p(t) u'(t) + q(t) u(t) \end{bmatrix}$$

$$= \begin{bmatrix}0 & -1 \ q(t) & -p(t) \end{bmatrix} \begin{bmatrix}u(t) \ -u'(t) \end{bmatrix}$$

$$= A(t) x(t)$$

Note that $A(t) \leq 0$ on $[a,b)$ . By our previous theorem, there exists a nontrivial solution $x(t)$ with $x(t) \geq 0, x'(t) \leq 0$ on $[a,b)$ .
So, there exists a solution $u(t)$ of the scalar DE such that

$$\begin{bmatrix}u(t) \ - u(t) \end{bmatrix} \geq 0, \begin{bmatrix}u'(t) \ -u''(t) \end{bmatrix} \leq 0,$$    on $$[a,b)$$

So, $u(t) \geq 0$ , $u'(t) \leq 0$ , and $u''(t) \geq 0$ on $[a,b)$ . As we know that $u$ is not the trivial solution, we know that $u(t) > 0$ (we get strict inequality).