# Prob. 29, Chap. 5, in Baby Rudin: Uniqueness of solution to a linear ODE with initial conditions

Here is Prob. 29, Chap. 5, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Specialize Exercise 28 by considering the system \begin{align} y_j^\prime &= y_{j+1} \qquad (j = 1, \ldots, k-1), \\ y_k^\prime &= f(x) - \sum_{j=1}^k g_j(x) y_j, \end{align} \tag{1} where $f, g_1, \ldots, g_k$ are continuous real functions on $[a, b]$, and derive a uniqueness theorem for solutions of the equation $$y^{(k)} + g_k(x) y^{(k-1)} + \cdots + g_2(x) y^\prime + g_1 (x) y = f(x), \tag{2}$$ subject to initial conditions $$y(a) = c_1, \qquad y^\prime(a) = c_2, \qquad \ldots, \qquad y^{(k-1)}(a) = c_k. \tag{3}$$

Here is Prob. 28, Chap. 5, in Baby 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.

And, here is the link to my most recent Math SE post on this problem.

Prob. 28, Chap. 5, in Baby Rudin: How to formulate the result on the uniqueness of the solution to a system of differential equations?

My Attempt:

Let us put $$y_1 \colon= y, \qquad y_2 \colon= y^\prime, \qquad \ldots, \qquad y_k \colon= y^{(k-1)}. \tag{4}$$

Then the system (1) along with the initial conditions (3) has a (unique) solution if and only if the linear ODE (2) along with the initial conditions (3) has a (unique) solution.

Am I right?

So in order to derive a uniqueness theorem for the IVP (2) and (3), we refer to Prob. 28 for the relevant uniqueness theorem for the system (1) and (3).

Let $$R \colon= [a, b] \times \left[ \alpha_1, \beta_1 \right] \times \cdots \times \left[ \alpha_k, \beta_k \right] \subset \mathbb{R}^{k+1},$$ Let $\mathbf{\phi}$ be the mapping of the ($k+1$)-cell $R$ into $\mathbb{R}^k$ given by $$\mathbf{\phi} \left( x, z_1, \ldots, z_k \right) \colon= \left( z_2, \ldots, z_k, f(x) - \sum_{j=1}^k g_j(x) z_j \right) \tag{*}$$ for all $\left( x, z_1, \ldots, z_k \right) \in R$.

The system (1) and (3) becomes $$\mathbf{y}^\prime = \mathbf{\phi} \left(x, y_1, \ldots, y_k \right), \qquad \mathbf{y}(a) = \mathbf{c}, \tag{5}$$ where $\mathbf{y} \colon= \left( y_1, \ldots, y_k \right)$ and $\mathbf{c} \colon= \left( c_1, \ldots, c_k \right) \in \mathbb{R}^k$.

So by the conclusion in Prob. 28, the IVP (5) has at most one solution if there is a real constant $A$ such that $$\left\vert \ \mathbf{\phi} \left( x, z_1, \ldots, z_k \right) - \mathbf{\phi} \left( x, w_1, \ldots, w_k \right) \ \right\vert \leq A \left\vert \ \mathbf{z} - \mathbf{w} \ \right\vert \tag{6}$$ whenever $\left( x, z_1, \ldots, z_k \right) \in R$ and $\left( x, w_1, \ldots, w_k \right) \in R$. Here $$\mathbf{z} \colon= \left( z_1, \ldots, z_k \right), \qquad \mathbf{w} \colon= \left( w_1, \ldots, w_k \right).$$

Now, using the definition of $\mathbf{\phi}$ in (*), the condition in (6) says that the system (5) has at most one solution if there is a real constant $A$ such that $$\left\vert \ \left( z_2, \ldots, z_k, f(x) - \sum_{j=1}^k g_j(x) z_j \right) - \left( w_2, \ldots, w_k, f(x) - \sum_{j=1}^k g_j(x) w_j \right) \ \right\vert = \left\vert \ \left( \ z_2 - w_2, \ldots, z_k - w_k, \sum_{j=1}^k g_j(x) \left( z_j - w_j \right) \ \right) \ \right\vert \leq A \left\vert \ \mathbf{z} - \mathbf{w} \ \right\vert$$ whenever $\left( x, z_1, \ldots, z_k \right) \in R$ and $\left( x, w_1, \ldots, w_k \right) \in R$.

Thus our linear ODE (2), along with the initial conditions (3), has at most one solution if there is a real number $A$ such that $$\left\vert \ \left( \ z_2 - w_2, \ldots, z_k - w_k, \sum_{j=1}^k g_j(x) \left( z_j - w_j \right) \ \right) \ \right\vert \leq A \left\vert \ \mathbf{z} - \mathbf{w} \ \right\vert \tag{7}$$ whenever $\left( x, z_1, \ldots, z_k \right) \in R$ and $\left( x, w_1, \ldots, w_k \right) \in R$. Here $$\mathbf{z} \colon= \left( z_1, \ldots, z_k \right), \qquad \mathbf{w} \colon= \left( w_1, \ldots, w_k \right).$$

Is this solution correct and as required by Rudin? If so, then can my condition (7) be simplified any further?

If my solution is not the correct one, then what is lacking in it? Where have I missed or erred?

It is correct but you should explain in your case what is the constant $A$ and how you use the hypotheses on the functions $g_i$. The conclusion would be false if the functions $g_i$ were unbounded. Every time you prove something you need to explain how you use the hypotheses.
EDIT Here are the full details. \begin{align}\left\vert\left(z_2-w_2,\ldots,z_k-w_k,\sum_jg_j(x)(z_j-w_j)\right)\right\vert &=\sqrt{\sum_{j=2}^k(z_j-w_j)^2+\left(\sum_jg_j(x)(z_j-w_j)\right)^2} \\&\le \sqrt{\vert \mathbf{z}-\mathbf{w}\vert^2+\vert \mathbf{g}(x)\vert^2\vert \mathbf{z}-\mathbf{w}\vert^2} \\&\le \sqrt{\vert \mathbf{z}-\mathbf{w}\vert^2+\Vert \mathbf{g}\Vert_\infty^2\vert \mathbf{z}-\mathbf{w}\vert^2} \\&=(1+\Vert \mathbf{g}\Vert_\infty^2)^{1/2}\vert \mathbf{z}-\mathbf{w}\vert,\end{align} where $\mathbf{g}=(g_1,\ldots,g_k)$.
• you see, the functions $g_j$ are continuous on the closed interval $[a, b]$ and are therefore necessarily bounded. What else do we need to do? After all, from Prob. 28, we have only the sufficient condition for the uniqueness of the solution. Jun 11, 2017 at 8:12
• I know. In your solution you just have to say that $A=c+c\sum_j\Vert g_j\Vert_\infty$ where $c=c(k)$ is a constant. Jun 11, 2017 at 8:14
• can we then conclude that the given linear ODE, along with the given initial conditions, always has at most one solution? For, what you have shown actually finds that real constant $A$. Jun 11, 2017 at 10:16