Given by theorem 2 for existence and uniqueness: given \(\frac{dy}{dx}=f\left ( x,y\right ) \), with initial value \(y\left ( x_{0}\right ) =y_{0}\). Let \(f\) and \(\frac{\partial f}{\partial x}\)be continuous in the rectangle \(R:t\leq t\leq t_{0}+a,\left \vert y-y_{0}\right \vert \leq b\). Compute \(M=\max _{\left ( x,y\right ) }\left \vert f\left ( x,y\right ) \right \vert \) and set \(\alpha =\min \left ( a,\frac{b}{M}\right ) \) then ODE has at least one solution in interval \(t\leq t\leq t_{0}+\alpha \) and this solution is unique. (I do not know why book split this into theorem 2 and 2’).
Notice in the above, if \(f\) or \(\frac{\partial f}{\partial x}\) not continuous in the range (the range must include the initial point) then not unique solution exist. For example \(y^{\prime }=\sin \left ( 2t\right ) y^{\frac{1}{3}}\) with \(y\left ( 0\right ) =0\). Here \(f^{\prime }\) is not continuous at \(y=0\).
How to use the above The first step is to find \(M\). This is done by finding maximum in \(R\). This is normally done by inspection from looking at \(f\left ( x,y\right ) \). Next, let \(g\left ( y\right ) =\frac{b}{M}\). Find where this one is maximum. set its value in \(\alpha =\min \left ( a,\frac{b}{M}\right ) \) and this finds \(\alpha \). Done. Example. Show \(y\left ( t\right ) \) solution to \(\frac{dy}{dt}=t^{2}+e^{-y^{2}},y\left ( 0\right ) =0\) exists in \(0\leq t\leq \frac{1}{2}\) and \(\left \vert y\left ( t\right ) \right \vert \leq 1\). Here it is clear \(M=\frac{5}{4}\) and hence \(\alpha =\min \left ( \frac{1}{2},\frac{b}{M}\right ) \) but \(b=1\), hence \(\alpha =\min \left ( \frac{1}{2},\frac{1}{\frac{5}{4}}\right ) =\alpha =\min \left ( \frac{1}{2},\frac{4}{5}\right ) =\frac{1}{2}\). Therefore solution exist for \(0\leq t\leq 0+\alpha \) or \(0\leq t\leq \frac{1}{2}\).
Show that the solution \(y\left ( t\right ) \) exists on \(y\left ( 0\right ) =0;0\leq t\leq \frac{1}{3}\)\[ y^{\prime }=1+y+y^{2}\cos t \] solution
Here \(a=\frac{1}{3}\).\begin{align*} M & =\max \left ( f\left ( t,y\right ) \right ) \\ & =1+b+b^{2} \end{align*}
Hence \begin{align*} \alpha & =\min \left ( \frac{1}{3},\frac{b}{M}\right ) \\ & =\min \left ( \frac{1}{3},\frac{b}{1+b+b^{2}}\right ) \end{align*}
Let \(g\left ( b\right ) =\frac{b}{1+b+b^{2}}\) then \(\frac{dg}{dp}=\frac{\left ( 1+b+b^{2}\right ) -b\left ( 1+2b\right ) }{\left ( 1+b+b^{2}\right ) ^{2}}\). Setting this to zero and solving for \(b\)\begin{align*} \left ( 1+b+b^{2}\right ) -b\left ( 1+2b\right ) & =0\\ 1+b+b^{2}-b-2b^{2} & =0\\ 1-b^{2} & =0 \end{align*}
Hence \(b=1\). At \(b=1\), then \(g\left ( b\right ) =\frac{1}{1+1+1}=\frac{1}{3}\). Therefore \begin{align*} \alpha & =\min \left ( \frac{1}{3},\frac{1}{3}\right ) \\ & =\frac{1}{3} \end{align*}
Therefore \(y\left ( t\right ) \,\) solution exist for \(0\leq t\leq 0+\alpha \) or \(0\leq t\leq \frac{1}{3}\).
Consider \(y^{\prime }=t^{2}+y^{2},y\left ( 0\right ) =0\) and let \(R\) be rectangle \(0\leq t\leq a,-b\leq y\leq b\). (a) Show the solution exist for \(0\leq t\leq \min \left ( a,\frac{b}{a^{2}+b^{2}}\right ) \) (b) Show the maximum value of \(\frac{b}{a^{2}+b^{2}}\,\), for \(a\) fixed is \(\frac{1}{2a}\). (c) Show that \(\alpha =\min \left ( a,\frac{1}{2}a\right ) \) is largest when \(a=\frac{1}{\sqrt{2}}\)\[ y^{\prime }=1+y+y^{2}\cos t \] solution
(a)
\begin{align*} M & =\max \left ( f\left ( t,y\right ) \right ) \\ & =a^{2}+b^{2} \end{align*}
Hence \begin{align*} \alpha & =\min \left ( a,\frac{b}{M}\right ) \\ & =\min \left ( \frac{1}{3},\frac{b}{a^{2}+b^{2}}\right ) \end{align*}
Hence solution exist for \(0\leq t\leq \min \left ( a,\frac{b}{a^{2}+b^{2}}\right ) \).
(b) Let \(g\left ( b\right ) =\frac{b}{a^{2}+b^{2}}\) then \(\frac{dg}{dp}=\frac{\left ( a^{2}+b^{2}\right ) -b\left ( 2b\right ) }{\left ( 1+b+b^{2}\right ) ^{2}}\). Setting this to zero and solving for \(b\)\begin{align*} \left ( a^{2}+b^{2}\right ) -b\left ( 2b\right ) & =0\\ a^{2}+b^{2}-2b^{2} & =0\\ a^{2} & =b^{2} \end{align*}
Hence \(b=\pm a\). At \(b=a\), then \(g\left ( b\right ) =\frac{a}{a^{2}+a^{2}}=\frac{a}{2a^{2}}=\frac{1}{2a}\).
(c)
\begin{align*} \alpha & =\min \left ( a,g_{\max }\left ( b\right ) \right ) \\ & =\min \left ( a,\frac{1}{2a}\right ) \end{align*}
Solving \(a=\frac{1}{2a}\) or \(a^{2}=\frac{1}{2}\). Hence \(a=\frac{1}{\sqrt{2}}\) gives largest value.
Prove that \(y\left ( t\right ) =-1\) is only solution for \(y^{\prime }=t\left ( 1+y\right ) ,y\left ( 0\right ) =-1\)
solution
Since \(f=t\left ( 1+y\right ) \) is continuous for all \(t,y\) and \(f_{y}=t\) is continuous for all \(y\), then if we find a solution, it will be unique solution by theorem 2’. But \(y\left ( t\right ) =-1\) is a solution since we can show easily it satisfies the ODE. Hence it is the only solution over all \(t\) by theorem 2’
Find solution of \(y^{\prime }=t\sqrt{1-y^{2}},y\left ( 0\right ) =1\) other than \(y\left ( t\right ) =1\). Does this violate theorem 2’?
solution
\begin{align*} \frac{dy}{dt} & =t\sqrt{1-y^{2}}\\ \int \frac{dy}{\sqrt{1-y^{2}}} & =\int tdt\\ \arcsin \left ( y\right ) & =\frac{t^{2}}{2}+C \end{align*}
At \(t=0\)\[ \arcsin \left ( 1\right ) =C \] Hence solution is \(\arcsin \left ( y\right ) =\frac{t^{2}}{2}+\arcsin \left ( 1\right ) \) or \begin{align*} y\left ( t\right ) & =\sin \left ( \frac{t^{2}}{2}+\arcsin \left ( 1\right ) \right ) \\ & =\sin \left ( \frac{t^{2}}{2}+\frac{\pi }{2}\right ) \\ & =\sin \left ( \frac{1}{2}\left ( t^{2}+\pi \right ) \right ) \end{align*}
This does not violate theorem 2’ because \(f\left ( t,y\right ) =t\sqrt{1-y^{2}}\), hence \(f_{y}=\frac{-ty}{\sqrt{1-y^{2}}}\) which is not continuous at \(y=\pm 1\). But \(y=-1\) is the initial conditions. Hence theorem 2’ do not apply. Theorem 2’ applies in the region where both \(f,f_{y}\) are continuous.