Exercise 5.28

Question

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Exercise 28: Formulate and prove an analogous uniqueness theorem for systems of differential equations of the form $$y_j'\phi_j(x,y_1,\ldots,y_k),\quad y_j(a)=c_j\quad
    (j=1,\ldots,k).$$ Note that this can be rewritten in the form $$\mathbf y'=\boldsymbol\phi(x,\mathbf y),\quad\mathbf y(a)=\mathbf c$$ where $\mathbf y=(y_1,\ldots,y_k)$ ranges
    over a $k$-cell, $\boldsymbol\phi$ 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 $(c_1,\ldots,c_k)$.

analambanomenos · Accepted Answer

If there is a constant $A$ such that $$\big|\boldsymbol\phi(x,\mathbf y_2)-\boldsymbol\phi(x,\mathbf y_1)\big|\le A|\mathbf y_2-\mathbf y_1|,$$ then $\mathbf y'=
    \boldsymbol\phi(x,\mathbf y)$, $\mathbf y(a)=\mathbf c$ has at most one solution. For suppose $\mathbf f_1$ and $\mathbf f_2$ are two such solutions, and let $\mathbf f=
    \mathbf f_1-\mathbf f_2$. Then $\mathbf f$ is differentiable on $[a,b]$, $\mathbf f(a)=\mathbf 0$, and $$\big|\mathbf f'(x)\big|= \big|\boldsymbol\phi(x,\mathbf f_1(x))-
    \boldsymbol\phi(x,\mathbf f_2(x))\big|\le A|\mathbf f_1-\mathbf f_2|=A|\mathbf f|.$$ Hence by the vector-valued version of Exercise 26 shown above, $\mathbf f(x)=\mathbf 0$
    for all $x\in[a,b]$.

Exercise 5.28

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Exercise 5.28

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