Existence & uniqueness of solutions

Suppose we have a first order IVP:

What information do we want to know?

• Does a solution exist?
• If it does, is the solution unique?
• If a solution does exist, how do we find it?

The objectives of this course are mainly concerned with the last question. However, we do not want to waste time trying to find a solution that does not exist. Also, if we are looking for a solution to a physical situation, we want to know that there is a unique solution corresponding to the physical process.

In this section, we informally state the theorems that guarantee that a solution exists and is unique and show how to apply them to IVPs. The theorem proofs are rather involved and beyond the scope of this course; If interested, check any advanced differential equations textbook for a more rigorous definition and proof of the existence and uniqueness theorem for initial-value problems.

Theorems

In the following statements of the existence and uniqueness theorems, we refer back to the IVP given by Eq. (12.91).

Existence: If $f(x, y)$ is continuous 'near' a point $(a, b)$, then at least one solution to (12.91) exists. Uniqueness: Given that a solution exists 'near' $(a, b)$ then, if the partial derivative of $f(x, y)$ wrt $y$, i.e. $\partial f / \partial y$, is continuous 'near' a point $(a, b)$, then the solution is unique.

Figure 12.3 shows what we mean by 'near' $(a, b)$ in the boxed definitions above; the blue blob represents a region which contains the initial condition, $(a, b)$. This represents the largest interval within which we can check whether $f(x, y)$ and $\partial f / \partial y$ are continuous. An integral curve passing through $(a, b)$ is also shown within the blob, in red: it is guaranteed by the theorems that if $f$ and $\partial f / \partial y$ are continuous, then the solution through $(a, b)$ exists and is unique even if it is for a very small interval. Note that outside of the region (where the hypotheses of the theorems are not satisfied), we cannot guarantee whether solutions exist/are unique.

The definitions in the boxes above give sufficient conditions for existence and uniqueness. This means that if $f(x, y)$ is $\underline{n o t}$ continuous in the region, then a solution may or may not exist. Similarly, if $\partial f / \partial y$ is not continuous in the region, then a solution may or may not be unique.

Figure 12.3: Plot showing the region defined as 'near' the point $(a, b)$ in the existence and uniqueness theorems. If $f$ and $\partial f / \partial y$ are continuous there, then a solution exists and is unique.

Applying the theorems to IVPs

Consider the IVP,

Before we apply the theorems, we need to put the ODE in the form given in (12.91). So, by rearranging Eq. (12.92), the ODE is expressed as,

Clearly this is very easily solved for $y(x)$ but our objective here is to apply the theorems to state whether a solution exists and whether it is unique before attempting to solve the IVP.

Existence

In Eq. (12.93),

It is easy to see that $y / x$ is continuous near any $(a, b)$ where $a \neq 0$. This means that, the region within which $f(x, y)$ is continuous and hence the existence theorem may be applied, can be as big as we want as long as it does not contain the points $a=0$ (these lie on the vertical axis). Therefore, by the existence theorem, a solution exists near any $(a, b)$ as long as $a \neq 0$.

Note that, at $a=0, y / x$ is discontinuous and the existence theorem hypothesis is not satisfied. A solution may or may not exist but we cannot conclude anything from the theorem itself.

Uniqueness

To apply the uniqueness theorem, we first need to find $\partial f / \partial y$. Differentiating $f(x, y)$ partially wrt $y$ gives

Again, $1 / x$ is continuous near any $(a, b)$ where $a \neq 0$. Therefore, by the uniqueness theorem, the solution is unique near any $(a, b)$ as long as $a \neq 0$.

To reinforce the ideas of existence and uniqueness, let us proceed to find the general solution to the ODE. The ODE is separable so by separating variables and integrating, we obtain the following general solution

where $k$ is a constant. Figure 12.4 shows plots of Eq. (12.94) for several values of $k$, including the one that passes through $(a, b)$ given by $y=(b / a) x$.

We note the following:

1. As long as $a \neq 0$, we can solve the IVP with $y(a)=b$. Then the solution is unique: in Fig. 12.4 the orange line annotated with the equation $y=(b / a) x$ satisfies the ODE and passes through the point $(a, b)$.
2. The $y$-axis represents all the points where $a=0$. If we take a point $(0, b)$ where $b \neq 0$, then,

(a) the hypotheses of the existence and uniqueness theorems are not satisfied so we cannot conclude whether a unique (or any) solution exists just by using the theorems;

(b) having solved the ODE, we know that there exists no $k$ value to satisfy Eq. $(12.94)$ at $x=0$. The solution does not exist and therefore existence fails at $(0, b)$ where $b \neq 0$.

Figure 12.4: Plots of the general solution (12.94) for several values of $k$. Existence fails for the IVP with $y(0)=b$ where $b \neq 0$ and uniqueness fails for the IVP with $y(0)=0$ as the initial condition.

3. Let us now take the point $(0,0)$ as our initial condition to the IVP. We note that here,

(a) the hypotheses of the existence and uniqueness theorems are not satisfied because $f$ and $\partial f / \partial y$ are discontinuous. Using the theorems alone, we cannot therefore conclude whether a unique (or any) solution exists;

(b) having solved the ODE however, we know that $(0,0)$ satisfies the general solution (12.94) for any value of $k$. This means that all solutions pass through the origin and satisfy the IVP with $y(0)=0$, as shown in Fig. 12.4. A solution exists but it is not unique i.e. uniqueness fails. Example 12.9 Determine whether the following initial value problem

is guaranteed to have a unique solution.

Solution We rewrite Eq. (12.95) as

A solution to Eq. (12.96) is guaranteed to exist if $f(x, y)=x \ln (y)$ is continuous. The function $x \ln (y)$ is continuous everywhere where $y>0$. By the existence theorem, there exists at least one solution in $-\infty<x<\infty$. To address the uniqueness of the solution, we need to determine whether $\partial f / \partial y$ is continuous:

is continuous everywhere except where $y=0$. By the uniqueness theorem, a unique solution exists to the IVP as long as $y \neq 0$. According to the existence and uniqueness theorems therefore, we know that

(i) existence is satisfied if $y>0$;

(ii) uniqueness is satisfied if $y \neq 0$.

Since the $y$-value corresponding to the initial condition is $e>0$, then the hypotheses of the theorems guarantee a unique solution through $(1, e)$.