Exercise 5.29

Question

\def\where{\,|\,}
Exercise 29: Specialize Exercise 28 by considering the system
    \begin{align*}
        y_j' &=y_{j+1}\quad(j=1,\ldots,k-1),\
        y_k' &=f(x)-\sum_{j=1}^k g_j(x)y_j,
    \end{align*}
    where $f,g_1,\ldots,g_k$ are continuous real functions on $[a,b]$, and derive a uniqueness theorem for solutions of the equation $$y^{(k)}+g_k(x)y^{(k-1)}+\cdots+g_2(x)y'+
    g_1(x)y=f(x),$$ subject to the initial conditions $$y(a)=c_1,\quad y'(a)=c_2,\quad\ldots,\quad y^{(k-1)}(a)=c_k.$$

analambanomenos · Accepted Answer

\def\where{\,|\,}

Using the notation of Exercise 28, $\boldsymbol\phi(x,\mathbf y)=\bigl(\phi_1(x,\mathbf y),\ldots,\phi_k(x,\mathbf y)\bigr)$ where
    $$\mathbf \phi_j(x,\mathbf y)=
    \begin{cases}
        y_{j+1} & j=1,\ldots,k-1 \
        f(x)+\sum_{i=1}^k g_i(x)y_i = f(x)+\mathbf g(x)\cdot\mathbf y & j=k
    \end{cases}$$
    where $\mathbf g(x)=\bigl(g_1(x),\ldots,g_k(x)\bigr)$. Then
    \begin{align*}
        \big|\boldsymbol\phi(x,\mathbf y_2)-\boldsymbol\phi(x,\mathbf y_2)\big|^2 &= \sum_{j=1}^k\big|\phi_j(x,\mathbf y_2)-\phi_j(x,\mathbf y_1)\big|^2 \
        &= \sum_{j=1}^{k-1}|y_{2,j+1}-y_{1,j+1}|^2 + \big|\mathbf g(x)\cdot(\mathbf y_2-\mathbf y_1)\big|^2 \
        &\le |\mathbf y_2-\mathbf y_1|^2+|\mathbf g(x)|^2|\mathbf y_2-\mathbf y_1|^2 \
        &=\bigl(1+|\mathbf g(x)|^2\bigr)|\mathbf y_2-\mathbf y_1|^2 \ \
        \big|\boldsymbol\phi(x,\mathbf y_2)-\boldsymbol\phi(x,\mathbf y_2)\big| &\le \sqrt{\bigl(1+|\mathbf g(x)|^2\bigr)}\;|\mathbf y_2-\mathbf y_1|
    \end{align*}
    Since the $g_i(x)$ are continuous on $[a,b]$, they are bounded, so that there is a constant $A$ such that $\sqrt{\bigl(1+|\mathbf g(x)|^2\bigr)}\le A$ for all $x\in[a,b]$.
    Hence by Exercise 28, the equation has at most one solution.

Exercise 5.29

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Exercise 5.29

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