Proving Solutions Increase In Initial Value Problems For T > 0

Hey guys! Today, we're diving deep into the fascinating world of ordinary differential equations (ODEs) and initial value problems (IVPs). Specifically, we're going to explore a scenario where we can confidently say that the solution to an IVP is increasing for all times greater than zero. Buckle up, because this is going to be an exciting ride!

Setting the Stage: Our Initial Value Problem

Let's kick things off by introducing the star of our show – the initial value problem we'll be dissecting. We're dealing with the following system:

v'(t) = p - f(v)
v(0) = 0

Now, let's break down what each of these components means. First up, we have v'(t), which represents the derivative of the function v(t) with respect to time t. In simpler terms, it tells us how the function v(t) is changing at any given moment. Next, we have p, which is a positive constant. This means it's a fixed number greater than zero, acting as a driving force in our system. Then there's f(v), a non-negative, unbounded, and infinitely differentiable function (a C^${\infty}$ function, for those who like the fancy math lingo). Non-negative means it's always greater than or equal to zero, and unbounded means it can grow without limit. The C^${\infty}$ part just ensures it's super smooth and well-behaved.

Finally, we have the initial condition, v(0) = 0. This is like our starting point, telling us that at time t = 0, the value of our function v(t) is zero. Together, these pieces form our initial value problem, and our goal is to understand the behavior of the solution v(t), specifically whether it's increasing for t > 0.

Why This Matters: The Intuition Behind Increasing Solutions

Before we jump into the nitty-gritty details, let's take a step back and think about why we care whether a solution is increasing. In many real-world scenarios, differential equations model how systems change over time. For instance, v(t) could represent the velocity of an object, the population of a species, or the concentration of a chemical. If we can show that the solution is increasing, it means that quantity is growing over time.

In our case, the equation v'(t) = p - f(v) gives us a clue. The term p is a positive push, trying to make v'(t) positive, which would mean v(t) is increasing. However, f(v) acts as a resistance, pulling v'(t) downwards. The battle between these two forces determines whether v(t) increases. Since f(v) is non-negative and v(0) = 0, initially, f(v) will be small, and the positive p should win, making v'(t) positive and v(t) increase. The question is, will this trend continue for all t > 0?

The Proof: Showing v(t) is Increasing for t > 0

Alright, let's get down to the core of the matter – proving that the solution v(t) is indeed increasing for t > 0. To do this, we'll use a bit of mathematical reasoning and a touch of contradiction.

Step 1: Establishing v'(0) > 0

Our first move is to show that the derivative of v(t) at time t = 0 is positive. This will give us a solid starting point. We know that v'(t) = p - f(v) and v(0) = 0. So, let's plug in t = 0:

v'(0) = p - f(v(0)) = p - f(0)

Since f(x) is non-negative, f(0) must be greater than or equal to zero. Therefore, v'(0) = p - f(0) is greater than zero because p is a positive constant. This tells us that at the very beginning, v(t) is increasing.

Step 2: The Proof by Contradiction

Now comes the clever part – a proof by contradiction. We'll assume the opposite of what we want to prove and show that it leads to a logical absurdity. This will force us to conclude that our initial assumption must be false, and therefore, our desired conclusion must be true.

Let's assume, for the sake of contradiction, that there exists a time t* > 0 such that v'(t*) <= 0. This means that at some point after time zero, the rate of change of v(t) becomes either zero or negative. In other words, v(t) stops increasing or even starts decreasing.

Since v'(0) > 0 and we're assuming v'(t*) <= 0, there must be a first time, let's call it t0, where v'(t0) = 0. This is because v'(t) is a continuous function (since f(v) is infinitely differentiable), and if it goes from positive to non-positive, it must pass through zero at some point. So, we have:

v'(t0) = 0

Step 3: Unraveling the Contradiction

Now, let's see what this v'(t0) = 0 implies. Using our differential equation, we have:

v'(t0) = p - f(v(t0)) = 0

This means that f(v(t0)) = p. Since p is positive, this tells us that v(t0) must be a positive value (because f(0) <= p).

Here's the crucial step: Because t0 is the first time v'(t) becomes zero, it means that for all times t between 0 and t0, v'(t) must be strictly positive. In mathematical terms:

v'(t) > 0 for all t in (0, t0)

This is because if v'(t) were negative at any point before t0, then t0 wouldn't be the first time v'(t) becomes zero or negative. A positive derivative means that v(t) is strictly increasing on the interval (0, t0). Since v(0) = 0, this implies that v(t) is positive for all t in (0, t0). In particular, v(t0) must be strictly greater than 0.

Now, since v'(t0) = 0, we have p - f(v(t0)) = 0, which means f(v(t0)) = p. But we also know that v(t) is increasing on the interval [0, t0], so v(t0) > v(0) = 0. This means that f(v(t0)) = p for some v(t0) > 0. However, we also have v'(t) = p - f(v(t)) and because f(v(t0)) = p, we know that v'(t0) = 0. Now, if v(t) is ever greater than v(t0), f(v(t)) will be greater than p, which will lead to v'(t) being less than 0. So v(t) can never be greater than v(t0). This would imply v(t) is bounded.

But wait! Remember that f(x) is unbounded? This is where the contradiction arises. If v(t) were to stop increasing, it would eventually be bounded, and v'(t) couldn't switch sign again. This contradicts our assumption that there exists a t* where v'(t*) <= 0.

Step 4: The Grand Finale – Concluding v(t) is Increasing

We've reached the climax of our proof! Our assumption that there exists a time t* where v'(t*) <= 0 has led us to a contradiction. Therefore, our assumption must be false. This means that v'(t) must be strictly greater than zero for all t > 0. In other words, the solution v(t) is increasing for all t > 0.

Wrapping Up: The Power of Mathematical Reasoning

So there you have it, guys! We've successfully proven that the solution to our initial value problem is increasing for t > 0. This journey took us through the heart of differential equations, using a blend of logical deduction and proof by contradiction. We started by understanding the problem, setting up the necessary assumptions, and then carefully unraveling the implications. The key was to assume the opposite of what we wanted to prove and show that it leads to a logical absurdity, thus solidifying our original claim.

This exploration highlights the power of mathematical reasoning in understanding the behavior of dynamic systems. By carefully analyzing the equations and using rigorous proof techniques, we can gain deep insights into the world around us. And that, my friends, is what makes mathematics so incredibly fascinating!

Keywords

• Initial value problem
• Ordinary differential equation
• Solution verification
• Increasing solution
• Proof by contradiction
• Non-negative function
• Unbounded function
• Derivative