In this case it could be whatever you want (The problem is symmetric). But when you solve wave equation it has to be negative otherwise you don't have a complete solution (it's easy to find out if you assume it's zero or negative)
I think it just makes the ode easier to work with once it is solved, either way, you will get the correct answer, one might require a little more work than the other.
why is Y(y) not a complex function? From my class notes, is k^2 is negative it's sin+cos, but if it's positive then the solution to the de is a function of sinh(ix) + cos(ix)
lambda=0 ends up being a trivial answer that not fulfill the boundary conditions, but lambda>0 has non trivial solutions (and that fulfill the boundary conditions)?
Exponents and trig functions are somewhat interchangeable, but each has properties that you can exploit to help with certain cases. Sometimes, due to symmetry found in the boundary conditions (ie C=D), you could combine exponential terms in a way you can't combine trig functions. Or if you need a term that vanishes as x goes to infinity. On the other hand, trig functions are useful when you need a periodic function. You can move between exponentials and trig functions, so when one is used instead of the other, it is likely to help make the math easier or to better represent the physical situation
One of the beautiful things about Laplace's equation is that a sum of solutions is still a solution! If you have two non-homogeneous boundary conditions, break it into the sum of two problems, each with only one non-homogeneous boundary condition. Then add the solutions! For a simple example, suppose you have a similar setup as the video, but now V= V_1 at y = 0 (instead of V = 0 there) in addition to V = V_0 at y = b. Turn this into two problems: (A) All sides have V = 0, except V = V_0 at y = b. This solution is V_A(x,y). (B) All sides have V = 0, execpt V = V_1 at y = 0. This solution is V_B(x,y). The full solution is then V = V_A + V_B. Amazing!
Oh wonderful! Wonderful! Thank you!
How do know that lambda < 0?
I'm glad to know I'm not the only one stumped on this haha I thought I was stupid and shouldn't be in this class
also, what I think we are supposed to do is just choose either + or -. That what was confusing me too
In this case it could be whatever you want (The problem is symmetric). But when you solve wave equation it has to be negative otherwise you don't have a complete solution (it's easy to find out if you assume it's zero or negative)
We check it in all three lambda ir,=0 but less then 0 always gives non trivial solution
Is the choice of lambda (negative or positive) based on, if it can get DE with positive or negative constant coefficient.?
I think it just makes the ode easier to work with once it is solved, either way, you will get the correct answer, one might require a little more work than the other.
Should there not be a B sub n at 6:53?
This makes sense thank you!
why is Y(y) not a complex function? From my class notes, is k^2 is negative it's sin+cos, but if it's positive then the solution to the de is a function of sinh(ix) + cos(ix)
Sir why you have not taken d^2x/dx^2 = -lamda × Y?
lambda=0 ends up being a trivial answer that not fulfill the boundary conditions, but lambda>0 has non trivial solutions (and that fulfill the boundary conditions)?
Thank you for this video!
The answer could have been sinusoidal in second ODE, (y(y)), why didn't you write it in Sine and cosine.?
+1
Exponents and trig functions are somewhat interchangeable, but each has properties that you can exploit to help with certain cases. Sometimes, due to symmetry found in the boundary conditions (ie C=D), you could combine exponential terms in a way you can't combine trig functions. Or if you need a term that vanishes as x goes to infinity. On the other hand, trig functions are useful when you need a periodic function. You can move between exponentials and trig functions, so when one is used instead of the other, it is likely to help make the math easier or to better represent the physical situation
Please how do I approach a problem with two non-homogeneous boundary conditions?
Thank you
One of the beautiful things about Laplace's equation is that a sum of solutions is still a solution!
If you have two non-homogeneous boundary conditions, break it into the sum of two problems, each with only one non-homogeneous boundary condition. Then add the solutions!
For a simple example, suppose you have a similar setup as the video, but now V= V_1 at y = 0 (instead of V = 0 there) in addition to V = V_0 at y = b. Turn this into two problems:
(A) All sides have V = 0, except V = V_0 at y = b. This solution is V_A(x,y).
(B) All sides have V = 0, execpt V = V_1 at y = 0. This solution is V_B(x,y).
The full solution is then V = V_A + V_B. Amazing!
Thanks Dr.
why is summation of n started from 1 and not zero?
i think because for the n=0 term, the sin ((n*pi*x)/a)=0. so whether you start from n=1 or n=0 the summations are equivalent
Perfect
Your proof for the separation of variables is inconclusive at point where F or G are 0. At those points you cant divide by FG
Hi, wheres the usyd squad! 🦧
still find it kinda funny we're being linked to these videos instead of the actual lecture