Although there are methods for solving some nonlinear equations, it is impossible to find useful formulas for the solutions of most. Whether we are looking for exact solutions or numerical approximations, it is useful to know conditions that imply the existence and uniqueness of solutions of initial value problems for nonlinear equations. In this section we state such a condition and illustrate it with examples. Some terminology: an open rectangle \(R\) is a set of points \((x,y)\) such that \[a<x<b\quad\text{and}\quad c<y<d\nonumber \] (Figure 2.3.1 ). We’ll denote this set by \(R: \{ a < x < b, c < y < d \}\). “Open” means that
the boundary rectangle (indicated by the dashed lines in Figure 2.3.1 ) is not included in \(R\). The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for first order nonlinear differential equations. We omit the proof, which is beyond the scope of this book.
It’s important to understand exactly what Theorem 2.3.1 says.
\[\label{eq:2.3.2}y=f(x,y),\quad y(b)=\overline{y}\] that differ on every open interval that contains \(b\). Therefore \(f\) or \(f_y\) must have a discontinuity at some point in each open rectangle that contains \((b, y)\), since if this were not so, \ref{eq:2.3.2} would have a unique solution on some open interval that contains \(b\). We leave it to you to give a similar analysis of the case where \(a > −∞\). Example 2.3.1Consider the initial value problem \[\label{eq:2.3.3} y'={x^2-y^2 \over 1+x^2+y^2}, \quad y(x_0)=y_0.\] Since \[f(x,y) = {x^2-y^2 \over 1+x^2+y^2} \quad \text{and} \quad f_y(x,y) = -{2y(1+2x^2)\over (1+x^2+y^2)^2} \nonumber\] are continuous for all \((x,y)\), Theorem 2.3.1 implies that if \((x_0,y_0)\) is arbitrary, then Equation \ref{eq:2.3.3} has a unique solution on some open interval that contains \(x_0\). Example 2.3.2Consider the initial value problem \[\label{eq:2.3.4} y'={x^2-y^2 \over x^2+y^2}, \quad y(x_0)=y_0.\] Here \[f(x,y) = {x^2-y^2 \over x^2+y^2}\quad \text{and} \quad f_y(x,y) = -{4x^2y \over (x^2+y^2)^2} \nonumber\] are continuous everywhere except at \((0,0)\). If \((x_0,y_0) \ne(0,0)\), there’s an open rectangle \(R\) that contains \((x_0,y_0)\) that does not contain \((0,0)\). Since \(f\) and \(f_y\) are continuous on \(R\), Theorem 2.3.1 implies that if \((x_0,y_0)\ne(0,0)\) then Equation \ref{eq:2.3.4} has a unique solution on some open interval that contains \(x_0\). Example 2.3.3Consider the initial value problem \[\label{eq:2.3.5} y'={x+y\over x-y},\quad y(x_0)=y_0.\] Here \[f(x,y) = {x+y\over x-y}\quad \text{and} \quad f_y(x,y) = {2x\over (x-y)^2} \nonumber\] are continuous everywhere except on the line \(y=x\). If \(y_0\ne x_0\), there’s an open rectangle \(R\) that contains \((x_0,y_0)\) that does not intersect the line \(y=x\). Since \(f\) and \(f_y\) are continuous on \(R\), Theorem 2.3.1 implies that if \(y_0\ne x_0\), Equation \ref{eq:2.3.5} has a unique solution on some open interval that contains \(x_0\). Example 2.3.4In Example 2.2.4, we saw that the solutions of \[\label{eq:2.3.6} y'=2xy^2\] are \[y\equiv0\quad \text{and} \quad y=-{1 \over x^2+c}, \nonumber\] where \(c\) is an arbitrary constant. In particular, this implies that no solution of Equation \ref{eq:2.3.6} other than \(y\equiv0\) can equal zero for any value of \(x\). Show that Theorem \(\PageIndex{1b}\) implies this. We’ll obtain a contradiction by assuming that Equation \ref{eq:2.3.6} has a solution \(y_1\) that equals zero for some value of \(x\), but is not identically zero. If \(y_1\) has this property, there’s a point \(x_0\) such that \(y_1(x_0)=0\), but \(y_1(x)\ne0\) for some value of \(x\) in every open interval that contains \(x_0\). This means that the initial value problem \[\label{eq:2.3.7} y'=2xy^2,\quad y(x_0)=0\] has two solutions \(y\equiv0\) and \(y=y_1\) that differ for some value of \(x\) on every open interval that contains \(x_0\). This contradicts Theorem 2.3.1 (b), since in Equation \ref{eq:2.3.6} the functions \[f(x,y)=2xy^2 \quad \text{and} \quad f_y(x,y)= 4xy. \nonumber\] are both continuous for all \((x,y)\), which implies that Equation \ref{eq:2.3.7} has a unique solution on some open interval that contains \(x_0\). Example 2.3.5Consider the initial value problem \[\label{eq:2.3.8} y' = {10\over 3}xy^{2/5}, \quad y(x_0) = y_0.\]
Solution a Since \[f(x,y) = {10\over 3}xy^{2/5} \nonumber\] is continuous for all \((x,y)\), Theorem 2.3.1 implies that Equation \ref{eq:2.3.8} has a solution for every \((x_0,y_0)\). Solution b Here \[f_y(x,y) = {4 \over 3}xy^{-3/5} \nonumber\] is continuous for all \((x,y)\) with \(y\ne 0\). Therefore, if \(y_0\ne0\) there’s an open rectangle on which both \(f\) and \(f_y\) are continuous, and Theorem 2.3.1 implies that Equation \ref{eq:2.3.8} has a unique solution on some open interval that contains \(x_0\). If \(y=0\) then \(f_y(x,y)\) is undefined, and therefore discontinuous; hence, Theorem 2.3.1 does not apply to Equation \ref{eq:2.3.8} if \(y_0=0\). Example 2.3.6Example 2.3.5 leaves open the possibility that the initial value problem \[\label{eq:2.3.9} y'={10 \over 3}xy^{2/5}, \quad y(0)=0\] has more than one solution on every open interval that contains \(x_0=0\). Show that this is true. Solution By inspection, \(y\equiv0\) is a solution of the differential equation \[\label{eq:2.3.10} y'={10 \over 3} xy ^{2/5}.\] Since \(y\equiv0\) satisfies the initial condition \(y(0)=0\), it is a solution of Equation \ref{eq:2.3.9}. Now suppose \(y\) is a solution of Equation \ref{eq:2.3.10} that is not identically zero. Separating variables in Equation \ref{eq:2.3.10} yields \[y^{-2/5}y'={10 \over 3}x \nonumber\] on any open interval where \(y\) has no zeros. Integrating this and rewriting the arbitrary constant as \(5c/3\) yields \[{5\over 3}y^{3/5} = {5\over 3}(x^2+c) . \nonumber\] Therefore \[\label{eq:2.3.11} y = (x^2+c)^{5/3}. \] Since we divided by \(y\) to separate variables in Equation \ref{eq:2.3.10}, our derivation of Equation \ref{eq:2.3.11} is legitimate only on open intervals where \(y\) has no zeros. However, Equation \ref{eq:2.3.11} actually defines \(y\) for all \(x\), and differentiating Equation \ref{eq:2.3.11} shows that \[\begin{aligned}y'={10 \over 3}x(x^2+c)^{2/3}={10 \over 3}xy^{2/5},\,-\infty < x < \infty\end{aligned} \] Therefore Equation \ref{eq:2.3.11} satisfies Equation \ref{eq:2.3.10} on \((-\infty,\infty)\) even if \(c\le 0\), so that \(y(\sqrt{|c|})=y(-\sqrt{|c|})=0\). In particular, taking \(c=0\) in Equation \ref{eq:2.3.11} yields \[y=x^{10/3} \nonumber\] as a second solution of Equation \ref{eq:2.3.9}. Both solutions are defined on \((-\infty,\infty)\), and they differ on every open interval that contains \(x_0=0\) (Figure 2.3.2 ). In fact, there are four distinct solutions of Equation \ref{eq:2.3.9} defined on \((-\infty,\infty)\) that differ from each other on every open interval that contains \(x_0=0\). Can you identify the other two? Example 2.3.7From Example 2.3.5 , the initial value problem \[\label{eq:2.3.12} y'={10 \over 3}xy^{2/5}, \quad y(0)=-1\] has a unique solution on some open interval that contains \(x_0=0\). Find a solution and determine the largest open interval \((a,b)\) on which it is unique. Solution Let \(y\) be any solution of Equation \ref{eq:2.3.12}. Because of the initial condition \(y(0)=-1\) and the continuity of \(y\), there’s an open interval \(I\) that contains \(x_0=0\) on which \(y\) has no zeros, and is consequently of the form Equation \ref{eq:2.3.11}. Setting \(x=0\) and \(y=-1\) in Equation \ref{eq:2.3.11} yields \(c=-1\), so \[\label{eq:2.3.13} y=(x^2-1)^{5/3}\] for \(x\) in \(I\). Therefore every solution of Equation \ref{eq:2.3.12} differs from zero and is given by Equation \ref{eq:2.3.13} on \((-1,1)\); that is, Equation \ref{eq:2.3.13} is the unique solution of Equation \ref{eq:2.3.12} on \((-1,1)\). This is the largest open interval on which Equation \ref{eq:2.3.12} has a unique solution. To see this, note that Equation \ref{eq:2.3.13} is a solution of Equation \ref{eq:2.3.12} on \((-\infty,\infty)\). From Exercise 2.2.15, there are infinitely many other solutions of Equation \ref{eq:2.3.12} that differ from Equation \ref{eq:2.3.13} on every open interval larger than \((-1,1)\). One such solution is \[y = \left\{ \begin{array}{cl} (x^2-1)^{5/3}, & -1 \le x \le 1, \\[6pt] 0, & |x|>1. \end{array} \right. \nonumber\] Example 2.3.8From Example 2.3.5 ), the initial value problem \[\label{eq:2.3.14} y'={10 \over 3}xy^{2/5}, \quad y(0)=1\] has a unique solution on some open interval that contains \(x_0=0\). Find the solution and determine the largest open interval on which it is unique. Solution Let \(y\) be any solution of Equation \ref{eq:2.3.14}. Because of the initial condition \(y(0)=1\) and the continuity of \(y\), there’s an open interval \(I\) that contains \(x_0=0\) on which \(y\) has no zeros, and is consequently of the form Equation \ref{eq:2.3.11}. Setting \(x=0\) and \(y=1\) in Equation \ref{eq:2.3.11} yields \(c=1\), so \[\label{eq:2.3.15} y=(x^2+1)^{5/3}\] for \(x\) in \(I\). Therefore every solution of Equation \ref{eq:2.3.14} differs from zero and is given by Equation \ref{eq:2.3.15} on \((-\infty,\infty)\); that is, Equation \ref{eq:2.3.15} is the unique solution of Equation \ref{eq:2.3.14} on \((-\infty,\infty)\). Figure 2.3.4 ) shows the graph of this solution. What is existence and uniqueness theorem in differential equations?1. Existence and Uniqueness Theorem. The system Ax = b has a solution if and only if rank (A) = rank(A, b). The solution is unique if and only if A is invertible.
How do you prove existence and uniqueness?To prove uniqueness and existence, we also need to show that ∃x ∈ S such that P(x) is true. Example: Suppose x ∈ R − Z and m ∈ Z such that x<m<x + 1. Show that m is unique.
What is uniqueness of differential equation?a solution to the equation must have at each point. By plotting the directions of the various points in the plane, the direction field for the equation is constructed. A direction field shows uniqueness when there is only one choice of path to follow through the direction field for a given initial point.
What is existence solution?We've been acting as though just by specifying an initial condition, there must be a solution, and it must be unique (that is, the only one corresponding to that initial condition). And, in fact, this is typically true for any ``nice'' differential equation.
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