[math]v_t(x, t)-v_{xx} (x, t)=0 , x\in \mathbb{R}, t>0, v (x, 0) = sin^2(3x), x\in \mathbb{R}[/math] can this be solved using separation of variables?

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MathNugget

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Trying to solve this Cauchy problem:

[math]v_t (x, t) - v_{xx} (x, t) = 0 , x \in \mathbb{R}, t>0[/math][math]v (x, 0) = sin^2 (3x), x \in \mathbb{R}[/math].
I have seen a somewhat similar problem, solved by separating the variables. Meaning [imath]v(x, t) = A(x) B(t)[/imath].
Then I get [imath]v (x, 0) = A(x) B(0) = sin^2(3x)[/imath] so [imath]A(x) = \frac{sin^2(3x)}{B(0)}[/imath].
so then I calculate [imath]v_t, v_{xx}[/imath], plug them in first equation (and there probably is a problem here when I differentiate, or the method cannot be applied here):
[math]v_x(x, t)= \frac{6sin(3x)cos(3x)}{B(0)} = \frac{3sin(6x)}{B(0)}[/math][math]v_{xx} (x, t)= \frac{18cos(6x)}{B(0)}[/math]
And [imath]\frac{sin^2(3x)}{B(0)}B'(t) - \frac{18cos(6x)}{B(0)} B(t) = 0[/imath]
I suppose it's time to switch [imath]sin^2(3x)[/imath] or [imath]18cos(6x)[/imath] so that they're in the same thing. Let's take the first...
[math]sin^2(3x) = \frac{1-cos(6x)}{2}.[/math][math](1-cos(6x))B'(t) - 36cos(6x)B(t)= 0[/math]
From what I understood, I have to get rid of x for this to make sense. But I don't think I am getting there...
 
You define [imath]v(x,t) = A(x)B(t)[/imath] but later seem to assume that [imath]v_x(x,t) = A(x)[/imath],
 
blamocur said:
You define [imath]v(x,t) = A(x)B(t)[/imath] but later seem to assume that [imath]v_x(x,t) = A(x)[/imath],
Click to expand...
You're right, I think I just skipped the part that doesn't get affected by differentiation through x (I forgot it).
But in the end I switched them correctly, I believe. It's just on the [imath]v_x[/imath] and [imath]v_{xx}[/imath] lines that I forgot a B(t)
 
MathNugget said:
From what I understood, I have to get rid of x for this to make sense. But I don't think I am getting there...
Click to expand...
Why to get rid of 'x' ? You've found A(x), now you need to find B(t) to get the solution [imath]v=Ab[/imath]. Do you see how to find B(t) ?
 
blamocur said:
Why to get rid of 'x' ? You've found A(x), now you need to find B(t) to get the solution [imath]v=Ab[/imath]. Do you see how to find B(t) ?
Click to expand...
No, I don't see it... That's what I mean when I said "get rid of x". In [imath](1−cos(6x))B' (t)−36cos(6x)B(t)=0[/imath] I have to set a common factor, such that the ecuation becomes sort of C(x) times (something without x) = 0, and then I solve the parenthesis. (I've seen an example of a similar exercise, but sin(3x) had power 1, so the whole thing was far easier).
 
Maybe I can do some magic again.
[math](1−cos(6x))B'(t)=36cos(6x)B(t)[/math][math]\frac{d}{dt}((1−cos(6x))B(t))=\frac{d}{dx}(6sin(6x)B(t))[/math]And integrate?
[math]\int ((1−cos(6x))B(t))dx=\int (6sin(6x)B(t))dt[/math]
[math](x−\frac{sin(6x)}{6})B(t)=(6sin6x)\int B(t)dt[/math]It looks to me like B(t) is constantly 0... which doesn't work.
 
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Hi Nugget.

Take this advice from someone who has solved [imath]1000[/imath] PDEs.

When the domain is [imath]a < x < b[/imath], separation of variables can work.

When the domain is [imath]0 < x < \infty[/imath], separation of variables may work.

When the domain is [imath]-\infty < x < \infty[/imath], use Fourier or Laplace transform.

[imath]\displaystyle x \in \mathbb{R}[/imath] means [imath]-\infty < x < \infty[/imath].

Get the Fourier transform of [imath]\sin^2(3x)[/imath] and the original PDE. Solve the problem peacefully. Then get the inverse Fourier transform of that. The latter may pay you a high price, but this is the standard way of solving PDEs.

Note: Someone like Dan (topsquark) or blamocur can guess the solution from the first glance without using any of these techniques. It's experience.

😎
 
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I am pretty rusty on this topic myself, but it seems that the standard approach (e.g. here and here) is to first express [imath]v_t[/imath] and [imath]v_{xx}[/imath] through [imath]A[/imath] and [imath]B[/imath], transform the original equation into two ordinary equations, integrate both and only then look at the border conditions.
 
mario99 said:
Hi Nugget.

Take this advice from someone who has solved [imath]1000[/imath] PDEs.

When the domain is [imath]a < x < b[/imath], separation of variables can work.

When the domain is [imath]0 < x < \infty[/imath], separation of variables may work.

When the domain is [imath]-\infty < x < \infty[/imath], use Fourier or Laplace transform.

[imath]\displaystyle x \in \mathbb{R}[/imath] means [imath]-\infty < x < \infty[/imath].

Get the Fourier transform of [imath]\sin^2(3x)[/imath] and the original PDE. Solve the problem peacefully. Then get the inverse Fourier transform of that. The latter may pay you a high price, but this is the standard way of solving PDEs.

Note: Someone like Dan (topsquark) or blamocur can guess the solution from the first glance without using any of these techniques. It's experience.

😎
Click to expand...
Good to know, thanks. Looks like the easy approach doesn't work here.
What are the rules in this forum? Should I delete the thread, or let it hanging, if there's nothing going on anymore? (I'm giving up on this pde for now)
 
MathNugget said:
Good to know, thanks. Looks like the easy approach doesn't work here.
What are the rules in this forum? Should I delete the thread, or let it hanging, if there's nothing going on anymore? (I'm giving up on this pde for now)
Click to expand...
There is no need to delete the thread. There is a chance that someone gets interested and posts a meaningful answer after one week from the original post.
 
I solved the problem today. I will post my solution on Saturday.

😎

Hint: The solution is a combination of trigonometric and exponential functions.

 
mario99 said:
Hint: The solution is a combination of trigonometric and exponential functions.
Click to expand...
That's what one usually gets, often combined with polynomials, from (P)DEs with constant coefficients.

Congrats!

P.S I believe that being the OP you don't need to wait a week for posting the solution.
 
blamocur said:
There is no need to delete the thread. There is a chance that someone gets interested and posts a meaningful answer after one week from the original post.
Click to expand...
Well, had the test today. I tried this method (I had exactly this [imath]sin^2(3x)[/imath]). And...
I decided to commit atrocities...
since A'(x) was a lot nicer than A(x), I integrated A'(x) to get a (nicer) A(x).
so now instead of:
A(x)=[imath]sin^2(3x)[/imath] => A(x) = [imath]-\frac{3cos(6x)}{6}[/imath] .
A'(x)=[imath]3sin(6x)[/imath]
A''(x)=[imath]18cos(6x)[/imath]
And now when I plug them in, I can get rid of x... I didn't even add the constant though when I integrated, so it might be (a bit) wrong, although I feel it's only incomplete...
in my little experience with math, there's no way of "pulling" from under sin or cos... (aka to say sin(x) = sin(y) + c, with [imath]y \neq x[/imath] and [imath]c \neq 0[/imath] of course.
I did get stuck anyway after that (I have to solve an ordinary differential equation).
 
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MathNugget said:
Well, had the test today. I tried this method (I had exactly this [imath]sin^2(3x)[/imath]). And...
I decided to commit atrocities...
since A'(x) was a lot nicer than A(x), I integrated A'(x) to get a (nicer) A(x).
so now instead of:
A(x)=[imath]sin^2(3x)[/imath] => A(x) = [imath]-\frac{3cos(6x)}{6}[/imath] .
A'(x)=[imath]3sin(6x)[/imath]
A''(x)=[imath]18cos(6x)[/imath]
And now when I plug them in, I can get rid of x... I didn't even add the constant though when I integrated, so it might be (a bit) wrong, although I feel it's only incomplete...
in my little experience with math, there's no way of "pulling" from under sin or cos... (aka to say sin(x) = sin(y) + c, with [imath]y \neq x[/imath] and [imath]c \neq 0[/imath] of course.
I did get stuck anyway after that (I have to solve an ordinary differential equation).
Click to expand...
It took me two hours to solve this problem while I am so experienced. Therefore, I doubt that this problem was in your test unless you were given a big hint that would help you to solve it in 10 minutes.

It is impossible to solve this problem by separation of variables because separation of variables depends on at least one finite boundary condition.

If the problem was written like this:

[math]v_t (x, t) - v_{xx} (x, t) = 0 , \ \ \ 0 < x < \infty, \ \ \ t>0[/math]
[math]v (0, t) = e^{-3t}[/math].
[math]v (x, 0) = \sin^2 (3x)[/math].

You would be allowed to say that this problem may be solvable by separation of variables. And may be means there is no guarantee that you will get a solution by this method, but at least you can try it first because it is the easiest approach.

You are not listening to what I am saying. It is OK to ignore my suggestions if you were not taught to solve PDEs by other methods. My method is based on Fourier transform. Once you know how to solve this problem by this method, you will be able to solve all problems of this type and many problems of other types.

I am not sure if I am allowed to post my solution now. Therefore, it will be available on Saturday. You would be surprised how simple it is. The steps to get it is a little long, but straightforward.

I am so glad to see someone else other than Mario who is also interested in partial differential equations.

🤗
 
mario99 said:
It took me two hours to solve this problem while I am so experienced. Therefore, I doubt that this problem was in your test unless you were given a big hint that would help you to solve it in 10 minutes.

It is impossible to solve this problem by separation of variables because separation of variables depends on at least one finite boundary condition.

If the problem was written like this:

[math]v_t (x, t) - v_{xx} (x, t) = 0 , \ \ \ 0 < x < \infty, \ \ \ t>0[/math]
[math]v (0, t) = e^{-3t}[/math].
[math]v (x, 0) = \sin^2 (3x)[/math].

You would be allowed to say that this problem may be solvable by separation of variables. And may be means there is no guarantee that you will get a solution by this method, but at least you can try it first because it is the easiest approach.

You are not listening to what I am saying. It is OK to ignore my suggestions if you were not taught to solve PDEs by other methods. My method is based on Fourier transform. Once you know how to solve this problem by this method, you will be able to solve all problems of this type and many problems of other types.

I am not sure if I am allowed to post my solution now. Therefore, it will be available on Saturday. You would be surprised how simple it is. The steps to get it is a little long, but straightforward.

I am so glad to see someone else other than Mario who is also interested in partial differential equations.

🤗
Click to expand...
nope , no hint given. Although I am pretty certain we were taught other methods too (but I didn't understand anything from them). It was supposed to be solved in 1 or 2 pages at most (knowing the teacher).
 
MathNugget said:
Trying to solve this Cauchy problem:

[math]v_t (x, t) - v_{xx} (x, t) = 0 , x \in \mathbb{R}, t>0[/math][math]v (x, 0) = sin^2 (3x), x \in \mathbb{R}[/math].
I have seen a somewhat similar problem, solved by separating the variables. Meaning [imath]v(x, t) = A(x) B(t)[/imath].
Then I get [imath]v (x, 0) = A(x) B(0) = sin^2(3x)[/imath] so [imath]A(x) = \frac{sin^2(3x)}{B(0)}[/imath].
so then I calculate [imath]v_t, v_{xx}[/imath], plug them in first equation (and there probably is a problem here when I differentiate, or the method cannot be applied here):
[math]v_x(x, t)= \frac{6sin(3x)cos(3x)}{B(0)} = \frac{3sin(6x)}{B(0)}[/math][math]v_{xx} (x, t)= \frac{18cos(6x)}{B(0)}[/math]
And [imath]\frac{sin^2(3x)}{B(0)}B'(t) - \frac{18cos(6x)}{B(0)} B(t) = 0[/imath]
I suppose it's time to switch [imath]sin^2(3x)[/imath] or [imath]18cos(6x)[/imath] so that they're in the same thing. Let's take the first...
[math]sin^2(3x) = \frac{1-cos(6x)}{2}.[/math][math](1-cos(6x))B'(t) - 36cos(6x)B(t)= 0[/math]
From what I understood, I have to get rid of x for this to make sense. But I don't think I am getting there...
Click to expand...
How about separation of variables by addition - i.e. assume

v(x,t) = u(x) + w(t) → uxx = wt = Constant
 
MathNugget said:
nope , no hint given. Although I am pretty certain we were taught other methods too (but I didn't understand anything from them). It was supposed to be solved in 1 or 2 pages at most (knowing the teacher).
Click to expand...
That's unfortunate to understand nothing. 🥺
Tomorrow, I will post my solution.


khansaheb said:
How about separation of variables by addition - i.e. assume

v(x,t) = u(x) + w(t) → uxx = wt = Constant
Click to expand...
Hi professor Khan.

This assumption is used when one or more of the boundary conditions are nonhomogeneous or if the original PDE is nonhomogeneous. I don't see any of these there!
 
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