Wikipedia:Reference desk/Archives/Mathematics/2021 April 10

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April 10

Generalizing the solution to the partial differential equation

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When solving the partial differential equation of heat equation in this article, the heat equation is arranged as the following form:

${\frac {T'(t)}{\alpha T(t)}}={\frac {X''(x)}{X(x)}}$ (Form 1)

Since the right hand side depends only on x and the left hand side only on t, both sides are equal to some constant value −λ.

According to the aforementioned method, can we generalize it and say that both sides of any equation of the form

$f(x)=g(y)$

are equal to some constant value −λ? That is f(x) = −λ and g(y) = −λ? -- Justin545 (talk) 08:22, 10 April 2021 (UTC)

The assumption is that the equation holds universally; more precisely, for all values of the arguments in the domains of the function, provided these are non-empty. Then both functions are constant on their respective domains, with a common constant function value. The common constant value does not have to be a number; it could be a vector, or anything. Expressed more fully, using logical quantifiers:

Given functions $f:X\to Z,g:Y\to Z,X\neq \emptyset ,Y\neq \emptyset ,$

if      $\;\forall \,x\in X,y\in Y.f(x)=g(y),$

then  $\exists \,c\in Z.(\forall \,x\in X.f(x)=c)\land (\forall \,y\in Y.g(y)=c).$

This can be proved as follows. Assume that the condition is true. Let $y_{0}$ be any element of $Y$ and set $c=g(y_{0})$. Then, by instantiation of the condition, substituting $y_{0}$ for $y$ and then $c$ for $g(y_{0})$, we obtain:

$\;\forall \,x\in X.f(x)=c.$

The analogous statement for function $g$ follows similarly.  --Lambiam 10:45, 10 April 2021 (UTC)

Let

$\tau (t)={\frac {T'(t)}{\alpha T(t)}},\chi (x)={\frac {X''(x)}{X(x)}}$

And

$\tau :\mathbb {T} {\to }\mathbb {C} ,\chi :\mathbb {X} {\to }\mathbb {C}$

where

$\mathbb {C}$, the codomain of $\tau$ and $\chi$, is the set of all complex numbers,

$\mathbb {T}$ and $\mathbb {X}$ are the domains of $\tau$ and $\chi$, respectively.

Does the above satisfy $\forall \,t\in \mathbb {T} \;\forall \,x\in \mathbb {X} \;\tau (t)=\chi (x)$? If the answer is no, we should not assume that both sides of (Form 1) are equal to some constant value −λ unless one can proof that (Form 1) is an identity or holds universally, right? - Justin545 (talk) 15:28, 10 April 2021 (UTC)

The τ=χ part comes from another part of the analysis. The τ(t)=χ(x) trick occurs so frequently in PDE's that the details are often left out, especially in an encyclopedic treatment such as in Wikipedia, but it is valid and useful. The assumptions required are that there are solutions of the form T(t)X(x), that when the PDE is applied to this form of solution then the variables can be separated, that the PDE is linear so the sum of such solutions is also a solution, and that any solution can be written as a (possibly infinite) sum of such solutions. None of these are guaranteed in general, but they are true in a surprising number of physical applications suchs as the heat equation. --RDBury (talk) 19:17, 10 April 2021 (UTC)

Still, under the assumption of separability explicitly stated in the article, the situation fits the condition of my proof. The heat equation is a universal equality:

$\forall t\in \mathbb {R} ,x\in \mathbb {R} .~{\frac {\partial u}{\partial t}}-\alpha {\frac {\partial ^{2}u}{\partial x^{2}}}=0.$

Likewise, separability means:

$\forall t\in \mathbb {R} ,x\in \mathbb {R} .~u(x,t)=X(x)T(t).$

Combining these two universal equalities, we obtain:

$\forall t\in \mathbb {R} ,x\in \mathbb {R} .~{\frac {T'(t)}{\alpha T(t)}}={\frac {X''(x)}{X(x)}}.$

So, letting $\tau$ and $\chi$ be as above, we have:

$\forall t\in \mathbb {R} ,x\in \mathbb {R} .~\tau (t)=\chi (x).$

 --Lambiam 20:03, 10 April 2021 (UTC)

I'm sorry for not being able to find out the assumption of separability in the article. However, it's understandable that both sides of (Form 1) are equal to some constant value −λ if logical quantifiers are present as the above:

$\forall t\in \mathbb {R} ,x\in \mathbb {R} .~{\frac {\partial u}{\partial t}}-\alpha {\frac {\partial ^{2}u}{\partial x^{2}}}=0.$ (Form 2)

Now, let's consider another equation

$-6x+3y-15=0$

which can be arranged and separated as

$y=2x+5$ (Form 3)

Obviously, (Form 2) and (Form 3) are both equations which are expected to have similar properties. So the new question to me is that why are the logical quantifiers present in equation of (Form 2) but not present in equation of (Form 3)? What does make the difference between the PDE and the linear equation with respect to their universalities and logical quantifiers? -- Justin545 (talk) 05:23, 11 April 2021 (UTC)

The name of the article, Separation of variables, is already a hint, but the assumption is expressed in the sentence: "Let us attempt to find a solution which is not identically zero satisfying the boundary conditions but with the following property: u is a product in which the dependence of u on x, t is separated, ...". Your (Form 3) expresses a relation between values assumed by the variable $y$ and by the variable$x$; it does not represent a universal equality that holds for all pairs $(x,y)\in \mathbb {R} ^{2}$, but characterizes a subset – which is the point of the equation, its raison d'être. So it is not intended to mean $\forall y\in \mathbb {R} ,x\in \mathbb {R} .y=2x+5.$ In contrast, where the article introduces the equality $u(x,t)=X(x)T(t)\quad$(3), it is intended to mean $\forall t\in \mathbb {R} ,x\in \mathbb {R} .~u(x,t)=X(x)T(t).$ Usually, mathematicians leave such universal quantifications unstated because they are assumed to be understood in the context.  --Lambiam 07:46, 11 April 2021 (UTC)

How would mathematicians define terms separated and not separated? It seems that the separability is directly related to universal quantifications.

Consider the linear equation above and let

$v(x,y)=-6x+3y-15$

It seems that the assumption "the dependence of v on x, y is separated" implies

$\forall x\in \mathbb {R} ,y\in \mathbb {R} .\;v(x,y)=0$

whereas the assumption "the dependence of v on x, y is not separated" implies just

$v(x,y)=0$

(where universal quantifications must be removed).

If my speculation above is correct, I think that I can understand why the assumption "the dependence of u on x, t is separated" implies (Form 2) which has universal quantifications in it. -- Justin545 (talk) 16:59, 11 April 2021 (UTC)

I'm sorry, but mathematicians have a bad habit of using ambiguous notations and terminology, which is (most of the time) convenient for themselves but confusing for people in the business of mastering a subject. Most of the time they aren't even aware of the ambiguities. "Separation of variables" is not one specific technique, but a medley of tricks that have one thing in common: by manipulating an equation, all occurrences of one variable of interest are brought to one side, and those of another variable are all brought to the other side. For example, one can massage the equation $xy^{2}=x+y$ into the form $y/(y^{2}-1)=x.$ If we start with a universally valid equality (sometimes called an identity and written with an $\equiv$ sign, something I don't do) and make no mistakes, the result is also universal. But, conversely, if the original equation was not a universal equality to start with, then it will not turn into one by successful manipulation. The very first example in the article, where ${dy \over h(y)}=g(x)\,dx$ is obtained by rearranging, is not a universal equality, since here $y$ depends on $x$, being constrained by the equation $y=f(x).$ The example of the heat equation is different. The equality $u(x,t)=X(x)T(t)$ is universal here for no other reason than that we are supposed to understand that the author intended it to be understood that way. It does not constrain $x$ and $t$. This is an instance of a very common technique: guessing a kind of general form for what you are seeking and examining where that gets you. For some unknown function in one variable, someone might want to examine what the consequences are of the assumption that it is a linear function, and write, "Assume $f(x)=ax+b.$" This really means, suppose that $f(x)$ can be (universally) expressed in that form for some values $a$ and $b$. The reader is supposed to understand that thus far the coefficients $a$ and $b$ are still unconstrained, but constraints on function $f$ will then entail constraints on $a$ and $b.$ Likewise, "$u(x,t)=X(x)T(t)$" really means (here!): let us suppose that $u(x,t)$ can be (universally) expressed in the form of the product of two as-of-yet-undetermined functions, one of which depends solely on $x$ and which we name $X$, while the other depends solely on $t$ and shall go by the name $T.$ That it means all this is not explicit, but when one mathematician writes the snappy form, another mathematician will just assume that this is what they mean.  --Lambiam 22:26, 11 April 2021 (UTC)

Firstly, I have to say that I really appreciate your constant replies to my questions. A bad habit is trivial compared with the help you gave me.

Although separation of variables and related techniques are just a part or a few steps of the solution to the PDE in question, it seems to be more complicated than I can imagine...

The assumption of separability implies

$\forall t\in \mathbb {R} ,x\in \mathbb {R} .~u(x,t)=X(x)T(t)$ (Form 4)

but does (Form 4) imply (Form 2)? I cannot relate (Form 4) to (Form 2) directly in the first place. Now, my guess is that the assumption of separability assumes that elements in set $S_{1}=\{X(x)T(t)\mid x\in \mathbb {R} \land t\in \mathbb {R} \}$ are all solutions to PDE (which is without universal quantifications)

${\frac {\partial u}{\partial t}}-\alpha {\frac {\partial ^{2}u}{\partial x^{2}}}=0$. (Form 5)

By definition, a solution to an equation must satisfy the equation. Therefore, all the elements in $S_{1}$ must satisfy (Form 5), which can be written as (Form 2). So that's merely my guess for the relation between (Form 4) and (Form 2) and its correctness needs to be confirmed... -- Justin545 (talk) 00:01, 13 April 2021 (UTC)

Indeed. In general, solutions imply their equations, but not the other way around. For example, $x=y$ implies $x^{y}=y^{x}$, but $x^{y}=y^{x}$ does not imply $x=y$, since there are other solutions of the equation xʸ = yˣ. But if we know that an equation has a unique solution, then the implication becomes mutual: equation implies solution. Or there may be several solutions, but we have managed to find a "complete solution": a most general form, which requires throwing in some existential quantifiers to express the equivalence of equation and solution formally. But this works only if we have managed to find a solution, which, if the solution was based on a guess, requires that the initial guess was a lucky one. Obviously, (Form 4) cannot directly imply (Form 2). Since the functions $X$ and $T$ are not yet fixed, we could choose $X(x)=1,T(t)=t,$ and so $u(x,t)=t.$ Then (Form 2) simplifies to $1=0,$ which mathematicians know not to be universally valid. But we are trying to find a solution of the pde, so the question here is, can we find functions $X$ and $T$ such that (Form 4) implies (Form 2)? Yes, we can! It was a lucky guess, and given the boundary conditions we found a complete solution (a fact not really proved in the derivation), so the implication is, after all, mutual. Starting out with the guess $u(x,t)=X(x)+T(t)$ would have led to a dead end – the pde itself has solutions, but these do not respect the boundary conditions.  --Lambiam 08:50, 13 April 2021 (UTC)

Yes, I once doubted that (Form 4) is a valid assumption which is able to find out all the solutions to the PDE. What if $u(x,t)$ is in a different form other than $X(x)T(t)$? (such as $u(x,t)=X(x)+T(t)$ that you have mentioned) Luckily, your reply has alleviated my concern.

And I think that I have finally realized the key points to resolve my question. Thank you!

 Resolved -- Justin545 (talk) 19:19, 13 April 2021 (UTC)

The τ=χ part confused me so badly to the extent such that I was asking someone here to rescue me from being stuck on the problem. I think it would be better if the name of the analysis or the analysis itself can be noted in the the article rather than using the confusing sentence "Since the right hand side depends only on x and the left hand side only on t, both sides are equal to some constant value −λ." to explain. -- Justin545 (talk) 05:23, 11 April 2021 (UTC)