I love the efficiency of these videos! So much understanding in such a condensed form. I literally use them as an online reference guide whenever I get stuck while doing physics or chemistry. You're doing the world a great service Kahn.
I've seen before the limits of integrations of the erf defined from 0 to y. Is there a reason for writing the limits from y to infinity in this case? Are both integrals equivalent? Greetings and thank you for uploading these excellent videos.
3:05 I know how laplace transforms work, they translate your diffeq into a simpler alebraic equation that can be solved and inverted to find a solution. But, what do you mean by "integrating out"? As i understand it, integration creates independent variables whereas derivation eliminates them eg. f(x) = 4x, then f'(x) = 4, and f''(x) = 0.
When you solve a differential equation (DE), you basically 'integrate' it. That may not be technically true (i.e. you usually don't actually integrate the equation), but functionally, that's what happens. That's why you see arbitrary constants in the solution to your DE before applying boundary/initial conditions; those arbitrary constants are integration constants. What I'm trying to say in the video is that the Laplace transform is an *integral* transform, and you can intuitively think of it as *integrating* out one of the variables. For instance, when I apply the Laplace transform to du/dt = d^2 u/dx^2, it's as though I'm *integrating* out the t (i.e. solving the DE so that there's no more derivatives in t), resulting in a simple ODE in x. Does that clarify things?
Faculty of Khan Yes, i realize my understanding is limit since we are talking about pde's instead of ode's. So how exactly does a laplace transform change a time domain singal to a frequency? I hear this in every explanation of an LT, yet i have little intuition about it.
It's because when you look at the formula for a Laplace Transform: L[f(t)] = integral f(t)e^(-st) dt, we can see the 's' next to the 't' in the exponential. Since the argument of the exponential must be dimensionless, 's' must have units of 1/time because it's multiplying the time t. That's why when we take the Laplace Transform, we go from an independent variable in time (i.e. t) to an independent variable in 1/time, or frequency (i.e. s). Hope that helps!
How did you know U(x,s) has to be bounded in x->infinity? I understand everything except how you would jump to knowing this, which leads to C1 being equal to zero.
sorry its not a good answer, u can use the rule directly for the 2nd order diff equations with a positive discriminant, or use laplace transform on a similar looking equation you will get an exponential like answer
I have a similar problem with linear flow diffusivity equation but the inner boundary condition is (u_0) is unknown at zero? it goes zero but it's not defined at x=0
thank mr cursor
I'm glad someone can explain this in 12 minutes while my teacher couldn't do it in a month.
I love the efficiency of these videos! So much understanding in such a condensed form. I literally use them as an online reference guide whenever I get stuck while doing physics or chemistry. You're doing the world a great service Kahn.
Wow man, you really shine with the aid of Mr Cursor! Thank you both for the very lucid and informative lesson!
Our pleasure!
best applied maths channel on YT !!
Thanks for this , I have gained more in 12 minutes than in 3 weeks of lecturing
Really...well explained...very helpful
Very useful method, cheers!
Cool video. Doing some videos on priori/energy estimates would be cool. Maybe harmonic analysis?
Sure, I'll add it to my to-do list.
Can you use Laplace transform in finite domian problems?
Best explanation found , thanks !
Thank you, very much. Now, I will be able to understand the mathematical calculus in unsteady diffusion and decay of a pulse.
Thank you so much! It helps a lot!!
Glad to hear that!
simply explain!d a difficult content. Thank you sir.
What about the functions with 3 variables
Great sir....
Is the table with Laplace Transform solutions and inverses available anywhere?
What if u(x,0) is not zero, leading to a non homogenous equation? Yp and Yc?
What's the name of the program he is writing equations in?
Smoothdraw; you can also just use MS Paint.
I've seen before the limits of integrations of the erf defined from 0 to y. Is there a reason for writing the limits from y to infinity in this case? Are both integrals equivalent?
Greetings and thank you for uploading these excellent videos.
Because the error function is not the same as the complementary error function. The difference is precisely the integration limits.
Thank you sir
Can u provide this table? Laplace transform nd its inverse...
Could you kindly share the table? Can’t seem to find on web
3:05 I know how laplace transforms work, they translate your diffeq into a simpler alebraic equation that can be solved and inverted to find a solution. But, what do you mean by "integrating out"? As i understand it, integration creates independent variables whereas derivation eliminates them eg. f(x) = 4x, then f'(x) = 4, and f''(x) = 0.
When you solve a differential equation (DE), you basically 'integrate' it. That may not be technically true (i.e. you usually don't actually integrate the equation), but functionally, that's what happens. That's why you see arbitrary constants in the solution to your DE before applying boundary/initial conditions; those arbitrary constants are integration constants.
What I'm trying to say in the video is that the Laplace transform is an *integral* transform, and you can intuitively think of it as *integrating* out one of the variables. For instance, when I apply the Laplace transform to du/dt = d^2 u/dx^2, it's as though I'm *integrating* out the t (i.e. solving the DE so that there's no more derivatives in t), resulting in a simple ODE in x. Does that clarify things?
Faculty of Khan Yes, i realize my understanding is limit since we are talking about pde's instead of ode's. So how exactly does a laplace transform change a time domain singal to a frequency? I hear this in every explanation of an LT, yet i have little intuition about it.
It's because when you look at the formula for a Laplace Transform:
L[f(t)] = integral f(t)e^(-st) dt,
we can see the 's' next to the 't' in the exponential. Since the argument of the exponential must be dimensionless, 's' must have units of 1/time because it's multiplying the time t. That's why when we take the Laplace Transform, we go from an independent variable in time (i.e. t) to an independent variable in 1/time, or frequency (i.e. s). Hope that helps!
Thank you
Where did you get your table?
thank mr cursor (Y)
On what type of PDEs we can use Laplace transform. Kindly explain
Assalamoalaikum Sir, kindly upload video regarding integral transform solution of partial differential equations (Fourier). please.
I'll add it to my to-do list!
you are khan?
How did you know U(x,s) has to be bounded in x->infinity? I understand everything except how you would jump to knowing this, which leads to C1 being equal to zero.
hi how so you solve this ODE how do you get the exponential
i think he used laplace boi to get a 2nd order ODE in the s variable world, solved for it thats why it has exponentials
sorry its not a good answer, u can use the rule directly for the 2nd order diff equations with a positive discriminant, or use laplace transform on a similar looking equation you will get an exponential like answer
I have a similar problem with linear flow diffusivity equation but the inner boundary condition is (u_0) is unknown at zero? it goes zero but it's not defined at x=0
Sir can we apply laplace on product of two function u(x,t).v(x,t)
PDE用LT如果沒表的話太難了。
Thank god no Indian accent.
lol
thank mr cursor