Existence and Uniqueness Theorem Overview

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From our Elementary standards we are taught to solve ODE(stands for Ordinary Differential Equation) by some set of rules. But neither they told you nor you asked them about why these ODEs possesses even a solution and if it possesses then why solution should be unique meaning why the technique taught to us are the only way to find solution. I know upto this stage you are quite confused. But lets look at the below examples to be shocked as well.

Example 1 [Non-Existance of Solution of ODE]

Consider the following ODE

$$\begin{equation} \frac{dy}{dx}=\frac{1}{y}, \quad y(0)=0 \end{equation}$$

You might be so happy and jumped into solving this noob equation in a sec by your so called elementary methods. Lets see what we get:

By seperation of variable method (😃 your eyes are already glittering right?)
$$ydy=dx$$
$$\int ydy=\int dx +C$$
$$\frac{y^2}{2}=x +C$$

Now using the initial condition given $y(0)=0$ we have to find the C.
$$\frac{0^2}{2}=0 +C$$
$$C=0$$
So the solution we get is:
$$\frac{y^2}{2}=x$$
$$\implies y=\pm\sqrt{2x}$$

Now, what will be the solution among these two?
Both seems correct right? But here is the catch. Your so got solution $y(x)=\pm \sqrt{2x}$ is not differentiable at $x=0$. What we know ? If your y(x) is satisfying the ODE (1) then $y$ must be differentiable at all the points where the ODE (1) is defined.

Conclusion: We finally can conclude that (1) is not defined at $y(0)=0$ and this is the heart of Existence Theorem. The initial condition is the culprit. So, is our method of solving ODE that we learned so far is wrong?

Short answer is no. It is a general method where the ODE with initial condition is well-defined. In other words, suppose we forget about the initial condition and find the solution in general $y(x)=\pm \sqrt{2x+C}$, then we just will set the initial conditon such that the so got solution remains differentiable in a neighbourhood of that point. This condition is called local existence of ODE. So, you see from our elementary schools we just solve ODEs forgetting about the initial condition and then in the end after finding the solution then check for initial condition. It is something like define a function first $f(x)=x+1 \forall x\in \mathbb{R}$ then say $f(0)=0$, so if you forget about the initial condition $f(0)=0$ your definition of the function is ok otherwise you have to omit the point $x=0$ from your domain of definition of the function and then redefine the functon at that point like:

$$f(x) = \begin{cases} x+1, & x \neq 0 \\ 0, & x=0 \\ \end{cases}$$

You now might be tempted to redefine your solution of ODE (1). So, yes we can try but what is the main theme of the long discussion above is to make you aware of the fact that just blindly starting solving any differential equation not right approach as the initial condition is the integral part of the ODE, so we also have to respect the initial condition before starting solving it by our general method.

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Now, the imprtant question comes that-

Ok, you say problem occurs for initial condition. But what about our method to be unique or not? Is there anyother way to find solutions of our ODE?

Unfortunately, the answer is YES. Sorry guys!
If you carefully observe - our process of solving ODE is just a kind of back calculation using kind of inverse method of derivative - that is integration.

Redundant: I said it kind of inverse as if you have a function that is bounded in [0,1] and Reimann Integrable but doesn't satisfy intermediate value property then from such function after integration then differentiation you will not get back your original function that you started with due to Darboux theorem that says derivative of a function must satisfies IVP(Intermediate Value Property)

for example consider the function:
$$f(x)= \begin{cases} 0 && \forall x \in [0,1] \\ 1 && \forall x \in (1,2] \end{cases}$$

This is Riemann integrable. If we try to find its anti-derivative we get

$$F(x)= \int_0^x f(t)dt = \begin{cases} 0 && \forall x \in [0,1] \\ x && \forall x \in (1,2] \end{cases}$$

But $F(x)$ is not differentiable at $x=1$, hence $F'(x)=f(x), \forall x\neq 1$. So, $F(x)$ is not anti-derivative of $f(x)$.

If we now consider the ODE:
$$\frac{dy}{dx}=f(x) \quad y(1)=0$$
now, solving as above we should get $y(x)=F(x)$ but is $F$ differentiable at $x=1$? No, infact $F(x)$ is not continous at $x=1$. So, our process of doing blidly antiderivative is not always possible. Infact there is no solution of this ODE, according to Darboux Theorem. But we know by Fundamental Theorem of Integral Calculus if $f(x)$ were continous then $F(x)$ would be differentiable and solution would exists using our method as well.

Lets see an example of this non-uniqueness:

Example2 [Non-uniqueness of Solution of ODE]

Consider the following ODE
$$\begin{equation} \frac{dy}{dx}=\sqrt{y} \quad y(0)=0 \end{equation}$$

By seperation of variable method (😃 your eyes are already glittering right?) Also, now you are careful about initial condition that might be culprit in the end. Lets not to fear and try boldly solving then finally checking if the initial condition gives us any wierdness or not?

$$\frac{dy}{\sqrt{y}}=dx$$
$$\int \frac{dy}{\sqrt{y}}=\int dx +C$$
$$2\sqrt{y}=x +C$$
Now using the initial condition given $y(0)=0$ we have to find the C.
$$2\sqrt{0}=0 +C$$
$$C=0$$
So the solution we get is:
$$2\sqrt{y}=x$$
$$\implies y=\frac{x^2}{4}$$
Now, lets see the following function:

$$y(x)= \begin{cases} 0 && x\in [0,C] \\ \frac{(x-C)^2}{4} && x\in (C,\infty) \end{cases}$$

See below different values of $C=1,2,3,4,5,6,7$

Each $y(x)$ is differentiable and satisfies $y(0)=0$. But none of these solutions can be made from our back calculation method [integration] unless for the case $C=0$. There is another solution of (2) which is $y(x)=0$. It's trivial solution. So, our technique is not unique.

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Existence and Uniqueness Theorem

Finally we want to investigate what are those conditions that will guarantee the existence and uniqueness of the solution of an ODE so, that only by our Elementary Method if we find a solution that will be valid and unique so that no need to worry about other possible solution.(😃)

Theorem (Existence and Uniqueness Theorem): Let $f(x,y)$ and $\frac{\partial f}{\partial y}$ be continuous in some rectangle containing the point $(x_0,y_0)$ then there exists an interval $I$ around $x_0$ in which the initial value problem

$$\frac{dy}{dx}=f(x,y), \quad y(x_0)=y_0$$

has a unique solution.

Above theorem can be devided into two parts -

1. Existence Theorem: If $f(x,y)$ is continous in a rectangle around $(x_0,y_0)$, then existence of the solution of ODE is confirmed.
   Note: The converse is not true.
2. Uniqueness Theorem: If both $f(x,y)$ and $\frac{\partial f}{\partial y}$ are continous in a rectangle around $(x_0,y_0)$, then the solution of ODE is unique.
   Note: Converse is not true.

Remark: This guarantees a solution to exists in a small neighbourhoob around $x_0$. That is existance of the solution is confirmed locally, not globally. For global existence we will require something more.