An introduction to partial differential equations. PDE playlist: ruclips.net/user/view_play_list... Part 10 topics: -- derivation of d'Alembert's formula
I have my masters admissions interview tomorrow, and I watch these to prepare! Really helped me to learn PDE from 0 to a manageable level just in two days! Thank you!
I JUST did this today in class!!! Please make more PDE videos! You are saving my math career here. I have to wake up every morning at 6:30am and fall asleep in class at 8:00am to 12:00am (summer class, so it's long).
I have a hard time understanding how we are able to change arguments of x into arguments of x+-ct at 7:37. Weren't our initial position and initial velocity equations formed on the condition that they are equations of the SAME argument? how can we derive new equations and suddenly involve time in the arguments for p and q after deriving our equations at t=0?
we are just replacing x with x+/-ct in the function argument. Say I have f(x) = 2*x + 1. Now I am saying f(x+/-ct) = 2*(x+/-ct)+1. It is a function, which is like an operator, does not matter what you feed to it, an ant or an elephant, form stays the same.
I'm sure after 8 years you've figured this out, but incase anyone needs to know its simply whatever the highest order derivative is. Eg. If the equation has an d²x/dt², a dy/dx and a d³y/dt³ then it is a third order DE
Good video, there's just one thing that I don't understand. At 6:52, when you divided the bottom eqn by C and added them together, you will have P(x)-P(x) which =0 and also if you divided it by C, then how are you still getting C in the equation?
You already did this in previous videos but I just realized: In order for (x+ct) to make sense (and since x is an element of R^n as I understand) c needs to be in R^n as well, right? Because t is also just a real number, not a vector. So what do you mean by 1/c? Sorry if I'm missing something obvious...
glad to know someone has the same question every book just jumps to it and performs the same algebra using f' but the derivatives are with respect to different variables....
thee video is very helpful but i have a ques: in most of the case i found that this is the general wave equation u(x,t) = f(c t - x) + g(c t + x) but here (in the video) its written u(x,t) = f(x-ct) + g(c t + x) help please...
+Yonatan Younessy i doesn't matter at all. because f is an arbitrary function. only difference would be the signs of this function is the opposite in the two equations you wrote. but as it is arbitrary, sign of f doesn't matter so both is supposed to be true.
As stated before, the arbitrary functions, f & g, allow either order by absorbing or emitting a (-1). However, the form u(x, t) = f(x - ct) + g(x + ct) gives a physical insight that there are two wave pulses moving as speed "c". Function f has a wave moving to the left (negative) & g is moving to the right.
When you find u_t you differentiate p and q with respect to t but you used prime which is usually reserved for x derivatives and then you undo that by integrating with respect to x which makes no sense
Think about it as ψ(x, t) = G(η) + F(ξ) where η: (x, t) ⟶ (x - vt) and ξ: (x, t) ⟶ (x + vt) Now ∂ψ/∂t = ∂G(η)/ ∂t + ∂F(ξ)/∂t = (∂G/∂η)(∂η/∂t) + (∂F/∂ξ)(∂ξ/∂t) by chain rule But ∂η/∂t ≔ -v and ∂ξ/∂t ≔ v Therefore ∂ψ/∂t = -v (∂G/∂η) + v (∂F/∂ξ) However we want ∂ψ/∂t(x,0) Now we take an approach that’s non-standard in nature: ∂ψ/∂t = -v (∂G/∂η) + v (∂F/∂ξ) = -v (∂G/∂(x-vt)) + v (∂F/∂(x+vt)) ⇒ ∂ψ/∂t(x,0) = -v (∂G/∂(x)) + v (∂F/∂(x)) Now we can obviously use the prime notation ∂ψ/∂t(x,0) = -v G’ + v F’
Glad to know someone else has the same question. I'm not following; every book just jumps to it and performs the same algebra using f' but the derivatives are with respect to different variables....
@@frankchen4229 Think about it as ψ(x, t) = G(η) + F(ξ) where η: (x, t) ⟶ (x - vt) and ξ: (x, t) ⟶ (x + vt) Now ∂ψ/∂t = ∂G(η)/ ∂t + ∂F(ξ)/∂t = (∂G/∂η)(∂η/∂t) + (∂F/∂ξ)(∂ξ/∂t) by chain rule But ∂η/∂t ≔ -v and ∂ξ/∂t ≔ v Therefore ∂ψ/∂t = -v (∂G/∂η) + v (∂F/∂ξ) However we want ∂ψ/∂t(x,0) Now we take an approach that’s non-standard in nature: ∂ψ/∂t = -v (∂G/∂η) + v (∂F/∂ξ) = -v (∂G/∂(x-vt)) + v (∂F/∂(x+vt)) ⇒ ∂ψ/∂t(x,0) = -v (∂G/∂(x)) + v (∂F/∂(x)) Now we can obviously use the prime notation ∂ψ/∂t(x,0) = -v G’ + v F’
I have my masters admissions interview tomorrow, and I watch these to prepare! Really helped me to learn PDE from 0 to a manageable level just in two days! Thank you!
I'm a student teacher of undergraduate degree and I've copied this to copy it in the exam if it's tested in the pde paper 😩😩🙏... guys pray for me.
I JUST did this today in class!!! Please make more PDE videos! You are saving my math career here. I have to wake up every morning at 6:30am and fall asleep in class at 8:00am to 12:00am (summer class, so it's long).
I was having hard time in understanding this proof from apostol calculus volume 2, u made it simple dude, thanks a ton😍
While introducing G(x) you integrated it to g(x)=c(p(x)+q(x)) without considering integration constant. How is it justified?
I'm watching all of your videos. They're really excellent. Thank you
thanks bro for the lecture i didn't attend my college's lec. so it really helped me .
Crystal clear. Thank you so much!
which software do you use to write this?
Where did 1/2 and 1/2c come from?
where do we get our general sollution of EM wave? is it from the Helmholtz then find the solution of E (electric field)?
I don't understand how the choice of G(x)=-cp(x)+c q(x) is justified.(How it was thought to be the suitable G(x))
Hello, what is the software that you are using?
Thank You so much!, it was really helpful
I have a hard time understanding how we are able to change arguments of x into arguments of x+-ct at 7:37. Weren't our initial position and initial velocity equations formed on the condition that they are equations of the SAME argument? how can we derive new equations and suddenly involve time in the arguments for p and q after deriving our equations at t=0?
we are just replacing x with x+/-ct in the function argument. Say I have f(x) = 2*x + 1. Now I am saying f(x+/-ct) = 2*(x+/-ct)+1. It is a function, which is like an operator, does not matter what you feed to it, an ant or an elephant, form stays the same.
What about Heat equation : d elembert solution
So excellent!!
what is "s" now?
If uT=x
We write the same or change
Thanks for your video. I got confused how to classify the PDEs. is there any relations between first order PDEs and second order PDE?
I'm sure after 8 years you've figured this out, but incase anyone needs to know its simply whatever the highest order derivative is.
Eg. If the equation has an d²x/dt², a dy/dx and a d³y/dt³ then it is a third order DE
thank you this was helpful.
excellent explanation
Good video, there's just one thing that I don't understand. At 6:52, when you divided the bottom eqn by C and added them together, you will have P(x)-P(x) which =0 and also if you divided it by C, then how are you still getting C in the equation?
He adds G(x)/c to f(x) to get q(x). Then subtract G(x)/c from f(x) to get p(x).
thank you. very, very helpful.
So useful!!
Really helpful!!! :) Thanks n.n
Much appreciated.
thank you very much
Excellent !
You already did this in previous videos but I just realized: In order for (x+ct) to make sense (and since x is an element of R^n as I understand) c needs to be in R^n as well, right? Because t is also just a real number, not a vector. So what do you mean by 1/c? Sorry if I'm missing something obvious...
This is the 1-dimensional wave equation, so _x_ is simply a real number, and thus _c_ is also.
Thank you for this
You're computing the derivative with respect to t and then saying that the antiderivative is respecto to x??
glad to know someone has the same question
every book just jumps to it and performs the same algebra using f' but the derivatives are with respect to different variables....
thee video is very helpful
but i have a ques:
in most of the case i found that this is the general wave equation u(x,t) = f(c t - x) + g(c t + x)
but here (in the video) its written u(x,t) = f(x-ct) + g(c t + x)
help please...
+Yonatan Younessy i doesn't matter at all. because f is an arbitrary function. only difference would be the signs of this function is the opposite in the two equations you wrote. but as it is arbitrary, sign of f doesn't matter so both is supposed to be true.
thank u :)
As stated before, the arbitrary functions, f & g, allow either order by absorbing or emitting a (-1). However, the form u(x, t) = f(x - ct) + g(x + ct) gives a physical insight that there are two wave pulses moving as speed "c". Function f has a wave moving to the left (negative) & g is moving to the right.
Gracias
When you find u_t you differentiate p and q with respect to t but you used prime which is usually reserved for x derivatives and then you undo that by integrating with respect to x which makes no sense
Think about it as
ψ(x, t) = G(η) + F(ξ)
where η: (x, t) ⟶ (x - vt) and ξ: (x, t) ⟶ (x + vt)
Now ∂ψ/∂t = ∂G(η)/ ∂t + ∂F(ξ)/∂t = (∂G/∂η)(∂η/∂t) + (∂F/∂ξ)(∂ξ/∂t) by chain rule
But ∂η/∂t ≔ -v and ∂ξ/∂t ≔ v
Therefore ∂ψ/∂t = -v (∂G/∂η) + v (∂F/∂ξ)
However we want ∂ψ/∂t(x,0)
Now we take an approach that’s non-standard in nature:
∂ψ/∂t = -v (∂G/∂η) + v (∂F/∂ξ) = -v (∂G/∂(x-vt)) + v (∂F/∂(x+vt))
⇒ ∂ψ/∂t(x,0) = -v (∂G/∂(x)) + v (∂F/∂(x))
Now we can obviously use the prime notation
∂ψ/∂t(x,0) = -v G’ + v F’
Thank you
good
lol I'm PDE10
5:41 but g(x) is from derivative of p and q w.r.t. t, and you integrate p’ and q’ w.r.t. x, and you say p’ and q’ will go back to p and q??????
?????
Glad to know someone else has the same question. I'm not following; every book just jumps to it and performs the same algebra using f' but the derivatives are with respect to different variables....
@@frankchen4229 Think about it as
ψ(x, t) = G(η) + F(ξ)
where η: (x, t) ⟶ (x - vt) and ξ: (x, t) ⟶ (x + vt)
Now ∂ψ/∂t = ∂G(η)/ ∂t + ∂F(ξ)/∂t = (∂G/∂η)(∂η/∂t) + (∂F/∂ξ)(∂ξ/∂t) by chain rule
But ∂η/∂t ≔ -v and ∂ξ/∂t ≔ v
Therefore ∂ψ/∂t = -v (∂G/∂η) + v (∂F/∂ξ)
However we want ∂ψ/∂t(x,0)
Now we take an approach that’s non-standard in nature:
∂ψ/∂t = -v (∂G/∂η) + v (∂F/∂ξ) = -v (∂G/∂(x-vt)) + v (∂F/∂(x+vt))
⇒ ∂ψ/∂t(x,0) = -v (∂G/∂(x)) + v (∂F/∂(x))
Now we can obviously use the prime notation
∂ψ/∂t(x,0) = -v G’ + v F’
mega based
I think im the only one who doesnt understand any of this at all ffs xD
thanks bro for the lecture i didn't attend my college's lec. so it really helped me .