Learning PDEs from Scratch in 24 Hours

In summary, In order to do a Fourier series, you first need to use Fourier series and then use separation of variables to find the boundary values. The roots of the equation are the solutions to the homogenous differential equation.
  • #1
imsleepy
49
0
mVXNF.jpg


i basically don't know how to do pde's, so I'm learning it from scratch today for my test which is tomorrow (which is in 24hrs from now, for those who don't live in australia), and notes/the internet arent nearly as good as explaining things as people are.

so how would i go by starting these questions? any tips?
for the second one (Q4), i know that method of separation of variables basically involves putting all the x's on one side and the y's on the other side.
i don't understand how to find the boundaries, or what is insulated. i think it has something to do with those u(L,y)=0 etc constraints, which i'll look up.

some guidance on how to start these questions please?

as with my other threads, i'll post up my working out as i attempt these questions, to see how I'm going.

thanks.
 
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  • #2
I would start with the tips in the questions
- first use Fourier series
- 2nd use separation of variables
the boundary values are given in the question
u(x,0) = 0 shows the side with y=0 has u=0
 
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  • #3
for the first can you find the complimentary solution no worries? do you know what Fourier series is?
 
  • #4
for the first one (c) would i just first do the left hand side.

turn that into λ2 + 10λ = 0, solve it, so then λ = 0 and -10
so then
yh = Ae0x + Be-10x
= A + Be-10x

and so now f(x) = A + Be-10x and work this out as a Fourier question, but we have A and B and i don't know what to do wtih them, and i don't know what to do next...
 
  • #5
i would try substituting that complimentary solution back into the DE as a check... it doesn't look quite right to me, do you get zero?
 
  • #6
lanedance said:
i would try substituting that complimentary solution back into the DE as a check... it doesn't look quite right to me, do you get zero?

complimentary solution.. do you mean the λ2 + 10λ = 0, λ = 0 and -10 line?
or yhomogenous= A + Be-10x?
both look okay, unless there's something really stupid I am doing...
 
  • #7
homogenous & complimentary mean the same thing, they are the solutions to the homogenous DE
[tex] \frac{d^2y}{dx^2}+10y = 0[/tex]

if I take your homogenous solution and substitute it in
[tex] y_h = A + B e^{-10x}[/tex]
[tex](0+ (-10)(-10)Be^{-10x}+10(A+B e^{-10x}) \neq 0[/tex]
 
  • #8
also it might be worth including the start of the question

to use tex, just right click
use square brackets to open & close
[]
tex = open
/tex - close
 
  • #9
well this is awkward. i can't seem to do something so basic.

the whole question is:
FfowZ.jpg
 
  • #10
ahhhhhhh I am such a retard!
y turns into 1, not lambda, lol!

so lambda = +- root (-10)?
i hate imaginary values in ode's -_-
 
  • #11
ok so after consulting maths notes from 2 years ago, the roots are lambda = +- root (-10)

so y = (A*cos[itex]\sqrt{10}[/itex]*x + B*sin[itex]\sqrt{10}[/itex]*x)

yesh?

supposed to be e ^ alpha*x(Acosblah + Bcosblah) but alpha = 0, so the e at the beginning becomes 1
 
  • #12
imsleepy said:
ok so after consulting maths notes from 2 years ago, the roots are lambda = +- root (-10)

so y = (A*cos[itex]\sqrt{10}[/itex]*x + B*sin[itex]\sqrt{10}[/itex]*x)

yesh?
looks good up to here

imsleepy said:
supposed to be e ^ alpha*x(Acosblah + Bcosblah) but alpha = 0, so the e at the beginning becomes 1

not quite, I think the assumed form is [itex]e^{\lambda x}[/itex], but as you found [itex]\lambda = \pm \sqrt{-10}= \pm i \sqrt{10}[/itex] the exponential becomes complex which can be equivalently expressed in terms of sinusoids
 
  • #13
now I'm guessing you can sketch it no worries... what is the period? this will tell you the Fourier components to use. If the function were odd or even about the expansion point, it may save you time as you will only need to consider sin or cos terms

have you managed to find the Fourier expansion?
 
  • #14
oh, that second quote, i meant if the root is in the form λ = a ± bi. but a=0 in this case, so e^0x is just e^0, which is 1.
working off this:
M9oPo.jpg

no alpha in our roots, only beta.
 
  • #16
yess i was just going to say, it's in the other thread lol.
how's my solutions from that thread?

and are you talking about sketching y = (A*cos√10*x + B*sin√10*x)? because i wouldn't know how to do that..
 
  • #17
no the problem says sketch f(x)
 
  • #18
i think the period would be root 10 or something?

amplitiude has something to do with the A and B i think...

edit: oh, well I am not sure how to graph that. i can do them separately lol (2, 2/pi sinx, 2/pi sin3x) but not together..
 
  • #19
hang on sketching as in part a? I've done that.
part b i think I've done it.
 
  • #20
ok to avoid confusion..

question: http://i.imgur.com/FfowZ.jpg

3. a.
bAIUq.png


b.
imsleepy said:
ok so I've worked out a0, an, and bn again, and I am getting
a0 = 1/2,
an = 0,
bn = [itex]\frac{1 - (-1)n}{n*pi}[/itex]
erm it isn't showing up right for me..

bn = [1 - (-1)n] / (n*pi)

is this right?

and then i sub all into f(x)?

imsleepy said:
so I am getting

f(x) = 2 + [itex]\frac{2}{Pi}[/itex]*sin(x) + [itex]\frac{2}{Pi}[/itex]*sin(3x)

is this right?
c.
imsleepy said:
ok so after consulting maths notes from 2 years ago, the roots are lambda = +- root (-10)

so y = (A*cos[itex]\sqrt{10}[/itex]*x + B*sin[itex]\sqrt{10}[/itex]*x)

so now do i say

(A*cos[itex]\sqrt{10}[/itex]*x + B*sin[itex]\sqrt{10}[/itex]*x) = 2 + [itex]\frac{2}{Pi}[/itex]*sin(x) + [itex]\frac{2}{Pi}[/itex]*sin(3x)

?

edit: ugh the LHS is just y, so i'd have to make y" + 10y equal to f(x)
 
  • #21
ok so what ideas do you have to find yp the particular solution for f(x)?
 
  • #22
yeah that too!

ok particular solution, this one i'd have to guess depending on what f(x) is (which I've never learned how to do lol). method of undetermined coefficients, i think it is.

my guess, something related to a trig function because we have a couple of sine's in f(x)?

so according to this table in my notes, if i have kcosθx or ksinθx in f(x), my choice for yp would be Kcosθx + Msinθx...
right?
 
  • #23
ok so i haven't been through the Fourier part but you should end up with an infinite series. I would then express yp(x) as a Fourier expansion with similar terms but unknown coefficients and substitute into

you don't need the complimentary solution for thsi part, only right at the end when you find the general solution
 
  • #24
firstly is f(x) = 2 + (2/Pi)*sin(x) + (2/Pi)*sin(3x) correct?

then i'd use k = 2/pi, and theta would be 1 and 3?
 
  • #25
lanedance said:
ok so i haven't been through the Fourier part but you should end up with an infinite series. I would then express yp(x) as a Fourier expansion with similar terms but unknown coefficients and substitute into

you don't need the complimentary solution for thsi part, only right at the end when you find the general solution

http://i.imgur.com/0MiDc.jpg

my result as a sum of infinite series at the top, the rest is getting the first 4 non-zero terms.

and I am not following you from your second sentence onwards...
 
  • #26
imsleepy said:
firstly is f(x) = 2 + (2/Pi)*sin(x) + (2/Pi)*sin(3x) correct?

then i'd use k = 2/pi, and theta would be 1 and 3?

no, that is not f(x), it may be the first 4 terms in the expansion of f(x) but it is not f(x), be careful as they are different things

As i said I haven't check the Fourier expansion part, but assuming it is correct you have f in the form
[tex] f(x) = a_0 + \sum_{n} b_n sin(nx) [/tex]

i would assume a similar form or the particular solution
[tex]y_p(x) = c_0 + \sum_{n} d_n sin(nx) [/tex]

now sub both those into the full DE and see if you can relate the dn's to the bn's
 
  • #27
by the way, posting pictures is ok, but makes it difficult to reply if you start using tex, its a lot easier to cut and paste rather than type from scratch - up to you but you may find you get quicker replies
 
  • #28
im not used to tex but i'll keep that in mind.

lanedance said:
no, that is not f(x), it may be the first 4 terms in the expansion of f(x) but it is not f(x), be careful as they are different things

As i said I haven't check the Fourier expansion part, but assuming it is correct you have f in the form
[tex] f(x) = a_0 + \sum_{n} b_n sin(nx) [/tex]

i would assume a similar form or the particular solution
[tex]y_p(x) = c_0 + \sum_{n} d_n sin(nx) [/tex]

now sub both those into the full DE and see if you can relate the dn's to the bn's

yes my f is in that form.

so i have

[tex] a_0 = \frac{1}{2}[/tex]

[tex]b_n = \frac{1-(-1)^{n}}{n*Pi}[/tex]

would i just let [tex]c_0[/tex] and [tex]d_n[/tex] equal a_0 and b_n respectively?
 
  • #29
no if it was
[tex] y(x) = f(x)[/tex]

then that would be a good idea, but if i was you i would try and involve the original differential equation
 
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  • #30
ughh i really have no idea what to do now.
also I've wasted too much time on one question.

gonna try to do the PDE one.

you said
lanedance said:
I would start with the tips in the questions
- first use Fourier series
- 2nd use separation of variables
the boundary values are given in the question
u(x,0) = 0 shows the side with y=0 has u=0

i'll give this a shot and post up working out as i go.
 
  • #31
ehh my head's not working, i should go to bed soon.

mVXNF.jpg


question 4. no clue.
H and L are just arbitrary points on the y and x-axis respectively, i presume..
what do each of those initial constraints mean, please?
and how do i know what is insulated?
 
  • #32
ok found some stuff in my notes.
i think the du/dx (0,y) = 0 line tells you that something is insulated.

i've done some question in my book (obviously lecturer has done it and i was merely copying it, lol), and i have written down:

uxx + uyy = 0
u(0,y) = 0
u(a,y) = 0
u(x,0) = x (maintain temp-profile 'x')
du/dy (x,0) = 0 -> temp gradient perpendicular to side on bottom is zero, ie. no heat can flow across side => INSULATED


i might've missed something because my graph has a 'b', and i see no b's written in the question, but main thing is that from this i can gather that with question 4, the du/dx (0,y) = 0 line tells me that something is insulated :)

now to find out what side is insulated...
 
  • #33
du/dx (0,y) = 0
so, gradient at 0,y is equal to 0?
would this be just the y-axis (up to point H) that is insulated?
 
  • #34
FMJYk.jpg


from there onwards, i simply cannot understand what worked solutions are doing >_<
 
  • #35
in you question i think a=L, b=H as you have
0<x<L, 0<y<H

the insulated side is the one where the derivative is held at zero,
 

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