Here is Prob. 28, Chap. 5, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Formulate and prove an analogous uniqueness theorem for systems of differential equations of the form $$ y_j^\prime = \phi_j \left( x, y_1, \ldots, y_k \right), \qquad y_j (a) = c_j \qquad (j = 1, \ldots, k).$$ Note that this can be rewritten in the form $$ \mathbf{y}^\prime = \mathbf{\phi} (x, \mathbf{y} ), \qquad \mathbf{y}(a) = \mathbf{c}$$ where $\mathbf{y} = \left( y_1, \ldots, y_k \right)$ ranges over a $k$-cell, $\mathbf{\phi}$ is the mapping of a ($k+1$)-cell into the Euclidean $k$-space whose components are the functions $\phi_1, \ldots, \phi_k$, and $\mathbf{c}$ is the vector $\left( c_1, \ldots, c_k \right)$. Use Exercise 26, for vector-valued functions.

Here is Prob. 27, Chap. 5, in Baby Rudin, 3rd edition:

Let $\phi$ be a real function defined on a rectangle $R$ in the plane, given by $a \leq x \leq b$, $\alpha \leq y \leq \beta$. A solution of the initial-value problem $$ y^\prime = \phi(x, y), \qquad y(a) = c \qquad (\alpha \leq c \leq \beta) $$ is, by definition, a differentiable function $f$ on $[a, b]$ such that $f(a) = c$, $\alpha \leq f(x) \leq \beta$, and $$ f^\prime(x) = \phi\left( x, f(x) \right) \qquad (a \leq x \leq b). $$ Prove that such a problem has at most one solution if there is a constant $A$ such that $$ \left| \phi \left(x, y_2 \right) - \phi \left(x, y_1 \right) \right| \leq A \left| y_2 - y_1 \right| $$ whenever $\left( x, y_1 \right) \in R$ and $\left( x, y_2 \right) \in R$.

Hint: Apply Exercise 26 to the difference of two solutions. Note that this uniqueness theorem does not hold for the initial-value problem $$ y^\prime = y^{1/2}, \qquad y(0) = 0, $$ which has two solutions: $f(x) = 0$ and $f(x) = x^2/4$. Find all other solutions.

Here is the link to my recent post on Math SE on this very problem.

Prob. 27, Chap. 5, in Baby Rudin: The initial-value problem $y^\prime = \phi(x,y)$, $y(a)=c$

And, here is Prob. 26, Chap. 5:

Suppose $f$ is differentiable on $[a, b]$, $f(a) = 0$, and there is a real number $A$ such that $\left| f^\prime(x) \right| \leq A \left| f(x) \right|$ on $[a, b]$. Prove that $f(x) = 0$ for all $x \in [a, b]$. Hint: Fix $x_0 \in [a, b]$, let $$ M_0 = \sup \left| f(x) \right|, \qquad M_1 = \sup \left| f^\prime(x) \right|$$ for $a \leq x \leq x_0$. For any such $x$, $$ \left| f(x) \right| \leq M_1 \left( x_0 - a \right) \leq A \left( x_0 - a \right) M_0.$$ Hence $M_0 = 0$ if $A \left( x_0 - a \right) < 1$. That is, $f = 0$ on $\left[ a, x_0 \right]$. Proceed.

Here is the link to my Math SE post on this problem.

Prob. 26, Chap. 5 in Baby Rudin: If $\left| f^\prime(x) \right| \leq A \left| f(x) \right|$ on $[a, b]$, then $f = 0$

First, I'll try to formulate and prove the result in Prob. 26 for vector-valued functions.

Suppose $\mathbf{f}$ is a differentiable mapping of the interval $[a, b]$ into $\mathbb{R}^k$ such that $\mathbf{f}(a) = 0$ and there is a real number $A$ such that $$\left| \mathbf{f}^\prime(x) \right| \leq A \left| \mathbf{f}(x) \right| \tag{0}$$ on $[a, b]$. Then $\mathbf{f}(x) = \mathbf{0}$ for all $x \in [a, b]$.

Is this statement correct and what Rudin has meant?


Let $n$ be a natural number such that $n > A(b-a)$, and let $$ a = x_0 < x_1 < \cdots < x_{n-1} < x_n = b$$ be a partition of $[a, b]$ such that $$\Delta x_j \colon= x_{i} - x_{i-1} = \frac{b-a}{n}$$ for $i = 1, \ldots, n$.

Let $x$ be any point of $\left(a , x_1 \right]$. Let $$ M_0 \colon= \sup \left\{ \ \left| \mathbf{f} (t) \right| \ \colon \ a \leq t \leq x \ \right\}, \qquad M_1 \colon= \sup \left\{ \ \left| \mathbf{f}^\prime (t) \right| \ \colon \ a \leq t \leq x \ \right\}. $$ Then from (0) we can conclude that, for all $t \in [a, x]$, $$ \left| \mathbf{f}^\prime (t) \right| \leq A \left| \mathbf{f} (t) \right| \leq A M_0,$$ which implies that $$ M_1 \leq A M_0. \tag{1}$$

Now as $\mathbf{f}$ is continuous on $\left[a, x \right]$ and differentiable on $\left( a, x \right)$, so there is a point $c \in \left( a, x \right)$ such that $$ \left| \mathbf{f} (x) \right| = \left| \mathbf{f} (x) - \mathbf{f} (a) \right| \leq \left| \mathbf{f}^\prime (c) \right| ( x- a ) \leq M_1 \left( x_1 - a \right) = M_1 \frac{b-a}{n}, $$ which implies that $$ M_0 \leq M_1 \frac{b-a}{n}. \tag{2}$$ And, from (1), we can conclude that, if $M_0 > 0$, then we have $$ M_1 \frac{b-a}{n} \leq A M_0 \frac{b-a}{n} < M_0 \tag{3}$$ because $n$ has been chosen such that $A (b-a)/n < 1$.

Now if $M_0 > 0$, then from (2) and (3), we have $$ M_0 \leq M_1 \frac{b-a}{n} < M_0, $$ which gives rise to a contradiction. So $M_0$ must be $0$, and thus $\mathbf{f}(t) = \mathbf{0}$ for all $t \in [a, x]$.

Therefore by taking $x = x_1$, we can conclude that $\mathbf{f}(t) = \mathbf{0}$ for all $t \in \left[a, x_1\right]$.

Now as $\mathbf{f}\left( x_1 \right) = \mathbf{0}$, so we can apply the above reasoning to $\mathbf{f}$ on $\left[ x_1, x_2 \right]$ instead of $\left[ a, x_1 \right]$ and obtain the conclusion that $\mathbf{f}(t) = \mathbf{0}$ for all $t \in \left[ x_1, x_2 \right]$ also.

And, continuing the above procedure $n$ times, we conclude that $\mathbf{f} = \mathbf{0}$ for all $x \in [a, b]$, as required.

Is this proof correct and rigorous enough for Rudin?

If so, then we can formulate our required result as follows:

Let $\mathbf{\phi} = \left(\phi_1, \ldots, \phi_k \right)$ be a mapping of a ($k+1$)-cell $$R \colon= [a, b] \times \left[ \alpha_1, \beta_1 \right] \times \cdots \times \left[ \alpha_k, \beta_k \right] \subset \mathbb{R}^{k+1}$$ into $\mathbf{R}^k$, where $\phi_j$ is a real function defined on $R$, for each $j = 1, \ldots, k$.

If there exists a real number $A$ such that $$ \left| \mathbf{\phi} \left(x, y_1, \ldots, y_k \right) - \mathbf{\phi} \left(x, z_1, \ldots, z_k \right) \right| \leq A \left| \mathbf{y} - \mathbf{z} \right|,$$ where $\mathbf{y} = \left( y_1, \ldots, y_k \right)$ and $\mathbf{z} = \left( z_1, \ldots, z_k \right)$, whenever $\left(x, y_1, \ldots, y_k \right) \in R$ and $\left(x, z_1, \ldots, z_k \right)\in R$, then the given IVP has at most one solution.

Is this formulation correct and as required by Rudin?


if $\mathbf{f}, \mathbf{g} \colon [a, b] \to \left[ \alpha_1, \beta_1 \right] \times \cdots \times \left[ \alpha_k, \beta_k \right]$, where $\mathbf{f} = \left( f_1, \ldots, f_k \right)$ and $\mathbf{g} = \left( g_1, \ldots, g_k \right)$, are any two solutions of the given system, then we have $$ \mathbf{f}^\prime(x) = \mathbf{\phi} \left( x, f_1(x), \ldots, f_k(x) \right), \tag{4}$$ and $$ \mathbf{g}^\prime(x) = \mathbf{\phi} \left( x, g_1(x), \ldots, g_k(x) \right) \tag{5}$$ for all $x \in [a, b]$. Moreover, $$ \mathbf{f}(a) = \mathbf{g}(a) = \mathbf{c}. \tag{6} $$

Let $\mathbf{h} \colon= \mathbf{f} - \mathbf{g}$. Then from (4) and (5) we find that $$ \begin{align} \left| \ \mathbf{h}^\prime(x) \ \right| &= \left|\ \mathbf{f}^\prime(x) - \mathbf{g}^\prime(x) \ \right| \\ &= \left| \ \mathbf{\phi} \left( \ x, f_1(x), \ldots, f_k(x) \ \right) - \mathbf{\phi} \left( \ x, g_1(x), \ldots, g_k(x) \ \right) \ \right| \\ &\leq A \left| \ \left( \ f_1(x), \ldots, f_k(x) \ \right) - \left( \ g_1(x), \ldots, g_k(x) \ \right) \ \right| \\ &= A \left| \ \mathbf{f} (x) - \mathbf{g}(x) \ \right| \\ &= A \left| \ \mathbf{h}(x) \ \right|. \end{align}$$ for all $x \in [a, b]$. And, from (6) we find that $$ \mathbf{h}(a) = \mathbf{0}.$$

So, by the version of the result in Prob. 26 for vector-valued functions, we can conclude that $\mathbf{h}(x) = \mathbf{0}$ for all $x \in [a, b]$, which implies that $\mathbf{f} = \mathbf{g}$.

Is this proof correct and what is intended by Rudin? If so, then is my presentation accurate and lucid enough? If not, then where is (are) the problem(s)?



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