Existence and Uniqueness Theorem: If \(f(x,y)\) is cont. on the open rectangle
\(R=\{(x,y)|a<x<b,c<y<d\}\) containing \((x_{0},y_{0})\) then the IVP \(y'=f(x,y)\), \(y(x_{0})=y_{0}\) has a solution on some open interval in \((a,b)\) containing \(x_{0}\). If, in addition, the partial derivative \(f_{y}\) is also continuous on \(R\) then the IVP has a unique solution on some open interval containing \(x_0\) contained in an open interval \((a,b)\) on which it has a solution.

(We skip the proof.)

Consider the IVP \(y'=2x\sqrt [3]{y}\), \(y(x_{0})=y_{0}\). For what points \((x_{0},y_{0})\) does this IVP have a solution (on some open interval containing \(x_{0}\))? For what points \((x_{0},y_{0})\) does this IVP have a unique solution (on some open interval containing \(x_{0}\))?

Since \(f(x,y)=2x\sqrt [3]{y}\) is continuous for all points in the \(xy\)-plane, there exists a solution to the IVP (on some open interval containing \(x_{0}\)) for any point \((x_{0},y_{0})\).

Notice that \(f_{y}(x,y)=\dfrac {2x}{3\sqrt [3]{y^{2}}}\), which is continuous everywhere except where \(y=0\). So for any \((x_{0},y_{0})\) such that \(y_{0}\neq 0\) there is an open rectangle \(R\) on which both \(f\) and \(f_{y}\) are continuous, and hence there is a unique solution to the IVP on some open interval containing \(x_{0}\).

Consider the IVP \(y'=2x\sqrt [3]{y}\), \(y(0)=0\). One solution is \(y=0\), as it makes both sides equal to \(0\). We can check that yet another solution \(\displaystyle y=\sqrt {\frac {8}{27}}\;x^3\):

\(\displaystyle y=\sqrt {\frac {8}{27}} \; x^3 \rightarrow \) \(\displaystyle y'=\sqrt {\frac {8}{3}} x^2\rightarrow \) \(\displaystyle y'=2x\left (\sqrt {\frac {2}{3}}\;x\right )\rightarrow \) \(\displaystyle y'=2x\sqrt [3]{y}\).

Hence there are (at least) two solutions to this IVP (both work on \((-\infty ,\infty )\)):
\(y=0\) and \(\displaystyle y=\sqrt {\dfrac {8}{27}}x^{3}.\)

Why didn’t the Existence and Uniqueness Theorem guarantee a unique solution on some interval containing \(0\)? Because for
\(f(x,y)=2x\sqrt [3]{y}\), the partial \(f_{y}=\dfrac {2x}{3\sqrt [3]{y^{2}}}\) isn’t continuous at \(y=0\).

Consider the IVP \(\displaystyle y'=\frac {y}{x+y+1}\), \(y(0)=1\). Does this IVP have a solution on some open interval? If so, is such a solution unique?

Since \(\displaystyle f(x,y)=\frac {y}{x+y+1}\) is continuous near \((0,1)\) as the point doesn’t lie on the line \(y=-x-1\), we know this IVP has a solution.

Using the Quotient Rule, \(\displaystyle f_y=\frac {(x+y+1)-y}{(x+y+1)^2}=\frac {x+1}{(x+y+1)^2}\). Since \(f_y\) is continuous near \((0,1)\) (same reason as above) this IVP has a unique solution.

Now consider the slightly modified IVP \(\displaystyle y'=\frac {y-1}{x+y+1}\), \(y(0)=1\).

By the Existence and Uniqueness Theorem, this IVP has a unique solution solution on some open interval containing \(0\) (as both \(f(x,y)=\frac {y-1}{x+y+1}\) and its \(y\)-partial \(f_y\) are continuous near \((0,1)\) as the point doesn’t lie on \(y=-x-1\)).

By looking at this de we see that \(y=1\) is one solution to it. Well, it must be the ONLY solution.

Notes:

-This theorem doesn’t provide for a method for finding solutions.

-If \(f_y\) is not continuous at \((x_0,y_0)\) then we get no conclusion regarding uniqueness, that is, the corresponding IVP may or may not have a unique solution. It might, but its not guaranteed by this Theorem.

This is optional supplementary material exploring the uniqueness part of the theorem:

Consider the IVP \(y'=f(x,y)\), \(y(x_0)=y_0\).

Assume \(f\) is defined on \(R=\{(x,y)|x_0-a\leq x\leq x_0+a,y_0-b\leq y\leq y_0+b\}\) for some positive \(a,b\).

Assume that both \(f\) and \(f_y\) are continuous on \(R\).

We make use of the fact (from real analysis) that a continuous function on a closed and bounded domain is bounded. So there exists positive constant \(N\) such that \(|f_y(x,y)|\leq N\) on all of \(R\).

Lemma: Let \(y\) be a function defined on \(I_h=[x_0-h,x_0+h]\) where \(0<h\leq a\) with a continuous derivative \(y'\) and values in \([y_0-b,y_0+b]\). Then \(y\) satisfies the IVP \(y'=f(x,y)\), \(y(x_0)=y_0\) if and only if \(y\) satisfies the so-called integral equation \(\displaystyle y=y_0+\int _{x_0}^x f(t,y)\,dt\).

Proof: Let \(y\) be a function defined on \(I_h=[x_0-h,x_0+h]\) with a continuous derivative \(y'\) and values in \([y_0-b,y_0+b]\). First assume \(y\) satisfies the IVP \(y'=f(x,y)\), \(y(x_0)=y_0\). Then for \(|x-x_0|\leq h\) we have that
\(\displaystyle \int _{x_0}^x y'(t) \,dt=\int _{x_0}^x f(t,y)\,dt\). By the Fundamental Theorem of Calculus (FTC) this is equivalent to \(\displaystyle y(x)-y(x_0)=\int _{x_0}^x f(t,y)\,dt\) so \(\displaystyle y=y_0+\int _{x_0}^x f(t,y)\,dt\). Next assume \(y\) satisfies \(\displaystyle y=y_0+\int _{x_0}^x f(t,y)\,dt\). Then \(y(x_0)=y_0\). By taking the derivative of both sides, again using the FTC, \(y'=f(x,y)\) on \(I_h\). That completes the proof of this lemma\(\;_{\square }\)

Let’s prove that our IVP \(y'=f(x,y)\), \(y(x_0)=y_0\) has at most one solution on some open interval \(I\) containing \(x_0\).

Proof: Let \(\alpha ,\beta \) be two functions and \(I\) be an open interval within \([x_0-a,x_0+a]\) and containing \(x_0\) such that the values of \(\alpha ,\beta \) on \(I\) are in \([y_0-b,y_0+b]\). Assume \(\alpha ,\beta \) both satisfy \(\displaystyle y=y_0+\int _{x_0}^x f(t,y)\,dt\) on \(I\). So

Let’s show \(\alpha (x)=\beta (x)\) for \(x\geq x_0\) on \(I\) and leave the argument for \(x\leq x_0\) on \(I\) to the reader (that’s you!)

Suppose \(x\geq x_0\) on \(I\). From the equations above, properties of definite integrals, and the Mean Value Theorem, we get

since by the Mean Value Theorem there exists \(\gamma \) between \(\alpha (t)\) and \(\beta (t)\) such that
\(f(t,\alpha )-f(t,\beta )=f_y (t,\gamma )\, (\alpha (t)-\beta (t))\) and as \(|f_y (t,\gamma )|\leq N\), we get that
\(|f(t,\alpha )-f(t,\beta )|\leq N|\alpha (t)-\beta (t)|\), where all this holds for all \(t\geq x_0\) on \(I\).

Now we focus on the function \(\displaystyle \psi (x) =\int _{x_0}^x |\alpha (t)-\beta (t)|\,dt\) for \(x>x_0\) on \(I\), which is non-negative. By the FTC again, for \(x\geq x_0\) on \(I\), we have \(\psi '(x)=|\alpha (x)-\beta (x)|\leq N\psi (x)\), which is equiv. to \(\psi '(x)-N\psi (x)\leq 0\), which is also equivalent to \([\psi '(x)-N\psi (x)]e^{-N(x-x_0)}\leq 0\) since \(e^{-N(x-x_0)}>0\) for all \(x\geq x_0\) on \(I\). This can be rewritten as \(\displaystyle \frac {d}{dx}\left [\psi (x)e^{-N(x-x_0)}\right ]\leq 0\).

Then \(\displaystyle \psi (x)e^{-N(x-x_0)}-\psi (x_0)e^{-N(x_0-x_0)}=\int _{x_0}^x \frac {d}{dt}\left [\psi (t)e^{-N(t-x_0)}\right ]\,dt \leq 0\). Since \(\psi (x_0)=0\) this latest inequality becomes \(\psi (x)e^{-N(x-x_0)}\leq 0\) for \(x\geq x_0\) on \(I\), but this implies that \(\psi (x)\leq 0\) for \(x\geq x_0\) on \(I\). But \(\displaystyle \psi (x)=\int _{x_0}^x |\alpha (t)-\beta (t)|\,dt\geq 0\) for \(x\geq x_0\) on \(I\). Thus \(\psi (x)=0\), and since the integrand is continuous, it follows that \(|\alpha (x)-\beta (x)|=0\) for all \(x\geq x_0\) on \(I\). Hence, \(\alpha (x)=\beta (x)\) for \(x\geq x_0\) on \(I\). A similar argument left to the reader shows \(\alpha (x)=\beta (x)\) for \(x\leq x_0\) on \(I\).

So there is at most one solution on \(I\).