Question

1. Why guessing and checking is alright in solving differential equations In lecture (and possibly in other courses), you have seen differential equations solved by looking at the equation, moving parts around, reasoning about it using an analogy with eigenvalue/eigenspaces, and then seeing that the solution that we proposed actually works — i.e. satisfies all the conditions of the differential equation problem. This process should have felt a bit different than how you have seen how systems of linear equations are solved (by doing Gaussian Elimination) where it was clear that every step was valid. Indeed, it is different. Although the eigenvalue/eigenspace analogy to differential equations can be made precise and rigorous, doing that carefully is beyond the scope of this course. In effect, all of that reasoning in between seeing the problem and checking the solution can be considered a kind of inspired guessing. This should lead you to a natural question — how can we be sure that we have found all of the solutions? We’ve checked to see that the solution we found solves the equations, but maybe there are more solutions that are different. How can we be sure? After all, we are using the solution of the differential equation for its predictive power — for example, we are using the fact of RC time constants to argue that this limits the speed of digital computation. Making such inferences is only proper if we have indeed found the only solution to the differential equation. In the mathematical literature, this is sometimes referred to as the problem of establishing the “uniqueness” of solutions. The concept is also very important for us in engineering contexts. You have already seen in EE16A’s touchscreen module that node voltages need not be unique, and that is why you need to specify a ground in your circuit. You also saw this concept in EE16A’s localization module where you learned how to approach inconsistent linear equations by the method of least squares: you started with no solutions, allowed some error and then got infinitely many potential solutions with error. To make the solution unique, you had to specify that you wanted to minimize the size of the hypothesized error. This problem walks you through an elementary proof of the uniqueness of solutions to a simple scalar differential equation of the form

d/dt x(t) = αx(t) (1) with initial condition x(0) = x0  (2)

Being able to do simple proofs is an important skill, not only in its own right, but also for the systematic logical thinking that it exercises. This problem has multiple parts, but the goal is simply to help you see how you could have come up with this proof entirely on your own.

(a) Please verify that the guessed solution xd(t) = x0e αt satisfies (1) and (2).

(b) To show that this solution is in fact unique, we need to consider a hypothetical y(t) that also satisfies (1) and (2)

Our goal is to show that y(t) = x(t) for all t ≥ 0. (The domain t ≥ 0 is where we have defined the conditions (1) and (2). Outside of that domain, we don’t have any constraints. ) How can we show that two things are equal? In the past, you have probably shown that two quantities or functions are equal by starting with one of them, and then manipulating the expression for it using valid substitutions and simplifications until you get the expression for the other one. However, here, we don’t have an expression for y(t) so that style of approach won’t work. In such cases, we basically have a couple of basic ways of showing that two things are the same. • Take the difference of them, and somehow argue that it is 0. • Take the ratio of them, and somehow argue that it is 1. We will follow the ratio approach in this problem. First assume that x0 6= 0. In this case, we are free to define z(t) = y(t) xd (t) since we are dividing by something other than zero. What is z(0)?

(c) Take the derivative d dtz(t) and simplify using (1) and what you know about the derivative of xd(t). (HINT: The quotient rule for differentiation might be helpful since a ratio is involved.) You should see that this derivative is always 0 and hence z(t) does not change. What does that imply for y and xd ?

(d) At this point, we have shown uniqueness in most cases. Just one special case is left: x0 = 0. Here, the division approach doesn’t seem to work because we are not permitted to divide by zero and xd(t) = 0. However, we want to show that y(t) = 0 here as well. Fundamentally, the argument we want to make is of the “it can’t possibly be otherwise” variety. Consequently, a proof by contradiction can be easier to start. In such proofs, we start by assuming the thing that we want to show is not possible. So assume that y(t) is not identically 0 everywhere for t > 0. What does this mean? This means that there is some t0 > 0 for which y(t0) = k 6= 0. (Otherwise, it would be zero everywhere.) We want to create a contradiction. It is clear that we will have no easy contradiction if we just move forward for t > t0 because we have no information given about such solutions y(t) that we can contradict. What do we know about? We have (2) which says something about y(0). This means, that we need to somehow move backward in time from t0. That way, we can hope to contradict the initial condition of 0. What do we have to work with? Well, we just did some work in the previous parts establishing uniqueness of solutions assuming nonzero initial conditions. How can we view what happens at t0 as a kind of nonzero initial condition? Apply the change of variablest = t0−τ to (1) to get a new differential equation for xe(τ) = x(t0−τ) that specifies how d dτ xe(τ) must relate to xe(τ). This should hold for −∞ < τ ≤ t0.

(e) Because the previous part resulted in a differential equation of a form for which we have already proved uniqueness for the case of nonzero initial condition, and since ye(0) = y(t0) = k 6= 0, we know what ye(τ) must be. Write the expressions for ye(τ) for τ ∈ [0,t0] and what that implies for y(t) for t ∈ [0,t0].

Answer 1

The images contains the required solution. In general well-posedness of a problem is critical to the existence of a solution. One must explore it in general to have wide perspective of differential equations.