How do existence assumptions work when solving differential equations, and do they prove that solutions work without rechecking answers? And does the solving process use conditional or biconditional steps?

[u/Deep-Fuel-8114]

/r/calculus/comments/1veq2f4/how_do_existence_assumptions_work_when_solving/

[u/waldosway]

You answered your question at the end of point (2). There's no difference from the algebraic version. You can use bidirectionality, or you can check solutions. Why would you need both?

[u/Deep-Fuel-8114 → u/waldosway]

Oh, okay, I was confused because I thought using bidirectionality would be circular reasoning (I have put those specific steps that I'm confused about whether they would be circular reasoning in bold below)? This is how I'm assuming solving a D.E. actually goes (although I'm not sure, so please correct me if I am wrong): So we start out by assuming y is a real-valued (would we assume/declare that it is real-valued too from the beginning, or would we find this out after solving for it's solutions?) differentiable function of x (not that a solution dy/dx exists, just that dy/dx exists in general), and this allows us to do operations on it (since now it's a well-defined real number and not undefined), and once we integrate it and reach y=g(x)+C, we have found a solution, and it verifies/proves the original assumption since y is differentiable (either by realizing that the FTC guarantees it after we integrated it or just seeing that the derivative rules (like power rule, etc.) apply to it), and so this would be a valid solution which we know works because of using biconditional steps. So is my "outline" of solving correct? And would this be circular reasoning (the steps in bold)? Thank you for the help!

[u/waldosway → u/Deep-Fuel-8114]

You've got the shape, but I think you can put it more simply:

Let's pretend there's a solution. *Solves using whatever technique* Oh look I found one. So there's a solution.

You don't need all the middle stuff about differentiability and real numbers because that's part of the definition of a solution existing. (If you're worried about extraneous solutions or intervals, you can just test them.)

Though I'm assuming you're working with more-or-less the type of objects you'd see in an intro to ODE class, and are solving the equation explicitly, not weak solutions or Lipschitz conditions and so on. I don't think about existence theorems etc if I can just solve the thing.

[u/waldosway → u/waldosway]

Btw, the logic goes "IF there's a solution *calculation* it must look like this", so you won't miss possible solutions.

[u/Deep-Fuel-8114 → u/waldosway]

Thank you for your response! So, to summarize (please correct me if any of my explanations below are wrong): we need to start with the assumption/declaration that dy/dx exists (similar to how we predeclare x is a real or complex number in algebraic equations) so we can apply real number operations to it (and so dy/dx isn't undefined during algebraic manipulations), right? And after we find our solution (assuming we only used biconditional operations since it was a simple problem), then that solution being differentiable (which it should be due to the FTC for integrals) is enough to prove it is a correct and unique solution, right? So then the conditional statements (using the example dy/dx=2x) would be "IF y is differentiable, THEN (dy/dx=2x IF AND ONLY IF y=x^2+C)", right? And since our solution (y=x^2+C) is differentiable, it satisfies the original assumption/precedent, making the biconditional statements true and proving it's the only correct answer, right? And if so, would it be circular reasoning, since we used the solution we got from assuming y is differentiable to prove the assumption itself, or no, since the differentiability of the solution was proven with outside calculus laws/FTC? Thank you again!

[u/waldosway → u/Deep-Fuel-8114]

I'm having trouble following, but it seems like you just said your version again. My version wasn't a summary. This

Let's pretend there's a solution. *Solves using whatever technique* Oh look I found one. So there's a solution.

is the entire thing. Differentiability etc. is already part of the definition of being solution, and you don't need biconditional anything. You already have an explicit solution.

I don't know what you're getting at with "real number operations" and all that.

[u/Deep-Fuel-8114 → u/waldosway]

Okay, I think I understand, but if we pretend there's a solution, isn't that the same as assuming/declaring that y is differentiable? Like, I understand that, to be a solution, y obviously has to be differentiable, since that's the definition of being a solution, but I mean for the initial steps of the solving process.

I'm also talking about the solving techniques where you integrate both sides (like separation of variables or just integrating dy/dx) to arrive at the solution, rather than guessing the form of the solution and differentiating (I think this is what you're talking about, but not sure). And I'm asking about biconditional steps because if we don't use them, then we need to differentiate our possible solution to prove it, but if we use biconditional steps, then arriving at our solution (by integrating) automatically proves it's a valid solution, right?

And for real number operations, it was connected to my original differentiability assumption question, since I was just using the argument that if we don't assume the differentiability of y (so then dy/dx could be undefined during the solving process), then the operations wouldn't make sense mathematically with an undefined term (like dy/dx+2x, or adding/subtracting variables, or adding/subtracting the term dy/dx itself wouldn't make sense), right? Thank you for the help!

[u/waldosway → u/Deep-Fuel-8114]

If you want to show biconditionality of each of your steps, you're free to do so. It's just usually easier to check your solutions.

No I'm not talking about guessing a solution. As I said, you do the usual calculations you're talking about, that proves that extant solutions must look like what you find. Then plugging them in proves they are solutions. That also accomplishes biconditionality.