Integral Equations, Functional Equations and Inequalities

Transcript

0:00Introduction

Hello everyone! Here is a strange equation for you. f of x equals the integral from zero to x of f of t, d t, plus e to the power x. The unknown function is hiding inside its own integral! So how on earth do we solve this? It takes just one simple move. We turn three new kinds of problems into differential equations: integral equations, functional equations, and inequalities. Questions like these do come up in JEE, so let's begin.

0:34Integral equations: differentiate, then put x=a

First, look at the integral from a to x of f of t, d t. The tall S is the integral sign, the lower limit is a, the upper limit is x, and t inside is just a dummy variable. The d by d x in front means differentiate with respect to x. So what do we get? Yes, exactly, we just get f of x! Differentiating undoes the integral, and t becomes x. Okay? So, step one: when an equation has such an integral, with x as the upper limit, differentiate both sides. The integral disappears, and a differential equation is left. Right? But wait, solving that differential equation brings in an arbitrary constant. To fix it we need an initial condition, and the question gives none. So where is it hiding? It is inside the question itself! Step two: put x equal to a, the lower limit, in the original equation. Careful, the original one, because the differentiated equation has lost that information. Got it? This works because the integral from a to a, with both limits the same, is zero. So only the value of f at the lower limit is left, and that is our initial condition. Clear?

1:57Functional equations: differentiate, then put y=0

Now the second kind, a functional equation. Here, f of x plus y equals f of x times f of y, for all x and y. And f must be differentiable, or we could not differentiate it. Okay? Differentiate both sides with respect to y, keeping x constant. The little dash, the prime, means the derivative. The left side gives f prime of x plus y. On the right, f of x is a constant, so we get f of x times f prime of y. With me? Now, there are still two variables here. Which value of y should we put in, to make it simple? Yes, y equals zero, exactly! Then x plus y is just x, and f prime of y becomes f prime of zero. So f prime of x equals f prime of zero, times f of x. Right? And f prime of zero is just a constant, so let us call it k. Then f prime equals k times f, a differential equation with variables separable. Clear?

3:05Inequalities: e to the minus kx times f is increasing

Alright, the third kind, an inequality. Suppose f prime minus k times f is greater than zero, where k is a constant. The symbol before the zero means greater than. Okay? Now, you have seen f prime minus k f before, in linear equations. Which factor turns it into one single derivative? Yes, the integrating factor, e to the power minus k x! By the product rule, the derivative of e to the power minus k x, times f, is e to the power minus k x, times the bracket, f prime minus k f. Check it in one line, okay? Now, e to the power minus k x is always positive, and the bracket is positive too; that was given. So this derivative is greater than zero. With me? And a positive derivative means an increasing function. So e to the power minus k x, times f, is increasing. That is the whole trick: an inequality on f prime becomes one increasing product. Clear?

4:11Example 1: f(x) = integral 0 to x of f(t) dt + e to the x

Now, Example one, and it is our puzzle from the start. Find f, if f of x equals the integral from zero to x of f of t, d t, plus e to the power x. Pause the video and try it first. Done? Okay, let's check. Step one: differentiate both sides. The integral gives f of x, and e to the power x stays as it is. So f prime of x equals f of x, plus e to the power x. Okay? Now the initial condition: put x equal to zero in the original equation. The integral from zero to zero is zero, so f of zero is zero, right? No, no, careful, that is the trap! The integral is zero, but e to the power x is still there. So f of zero equals zero, plus e to the power zero, and that makes it one. Clear? Now, f prime minus f equals e to the power x is a linear equation, with integrating factor e to the power minus x. Multiply through, and the left side becomes the derivative of f times e to the power minus x. On the right, the two exponentials cancel and give one. Right? Integrate both sides, and f times e to the power minus x equals x, plus capital C, the constant. Now use f of zero equals one. At x equal to zero, this reads one equals capital C, so capital C is one. Got it? That gives f of x equals the bracket, x plus one, times e to the power x. And there it is: the function hiding inside its own integral is found! Well done if you got this one.

6:00Example 2: f' at most f, f(0) = 0, show f(x) at most 0

Now a harder one, Example two. f is differentiable, f of zero is zero, and f prime of x is less than or equal to f of x, for x greater than or equal to zero. The symbols with a line under them mean less than or equal to, and greater than or equal to. Show that f of x is less than or equal to zero for these x. Pause the video and try it first. Done? Okay, let's check. Bring f to the left: f prime minus f is less than or equal to zero. This is our pattern with k equal to one, so multiply by e to the power minus x. Okay? But wait, with inequalities we must be careful. When we multiply by e to the power minus x, does the inequality sign flip? No, because e to the power minus x is always positive; only a negative number flips the sign. So the derivative of e to the power minus x times f, which is e to the power minus x times the bracket, f prime minus f, is less than or equal to zero. Right? And a derivative less than or equal to zero means the function never increases. So e to the power minus x, times f, never increases from x equal to zero onwards. Clear? So its value is at most its value at zero. At zero, e to the power zero is one, and f of zero is zero, so that value is zero. So e to the power minus x times f of x is less than or equal to zero. Got it? Finally, since e to the power minus x is positive, f of x has the same sign as this product. The double arrow means implies, so f of x is less than or equal to zero. And that is exactly what we had to show!

7:57Quick revision: integral, functional equations, inequalities

So, quick revision. For an integral equation, differentiate, then put x equal to a in the original equation. For a functional equation, differentiate in y, then put y equal to zero. And for an inequality like f prime minus k f greater than zero, multiply by e to the power minus k x, and that product is increasing. Now go and try the practice problems on this page. All the best!