If I am not mistaken, you forgot to divide by Pi when you are finding An using the Fourier cosine series. Also, I think that it should be -8 instead of 8.
Hi William. Loved the video. I have a question. I am dealing with a 3-D Laplace equation with homogeneous Neumannn condition on all the X and Y faces. The two Z faces have Robin conditions on them. Can a unique solution exist to this configuration ? In general will the presence of these 4 Neumann conditions give me a trivial solution?
Hi, this is just a little use of the Fourier Series. In that expression of 18:05 there is some kind of thing multiplying the cos. This is sth similar to the typical Fourier series sumatory of some "a priori" unknown function (then we'll see he forces that function to be sin(2x)). If you are not familiarized with Fourier series treatment, let me tell you this is a way of expressing a periodic function (of any shape in a certain interval repeated all the time), as an infinite addition of sinus and/or cosinus times a specific coefficient. Those coefficients A_n for sinus and B_n for cosinus (some may write them the other way around) are calculated by integrating the function you want to reexpress times some sin(nx) or cos(nx). And that's what he does, but here the function to reexpress is that of sin(2x), and he just calculates the expected A_n and divides by the things that he already had there written. In this case the variable A_n is already taken so let me say that the coefficients I said before A_n and B_n, now I'll call them C_n and D_n. So what we find is D_n=2 integral(cos(nx)*sin(2x)) = A_n n sinh(nπ). Now he just takes out the A_n, getting that expression A_n= 2/[n sinh(nπ)] * integral(cos(nx)*sin(2x)). And that's it ;)
You should write the constatnt as negative lambda. If you do it this way you solved the eigen probelms kind o wrong. p''=lambda*p has an exponential solution. Now p''=-lambda*p has the sin/cos solution.
If I am not mistaken, you forgot to divide by Pi when you are finding An using the Fourier cosine series. Also, I think that it should be -8 instead of 8.
Thanks. Was looking for whether Laplace's was unique with just Neumann and this answered that.
Hi William. Loved the video. I have a question. I am dealing with a 3-D Laplace equation with homogeneous Neumannn condition on all the X and Y faces. The two Z faces have Robin conditions on them. Can a unique solution exist to this configuration ? In general will the presence of these 4 Neumann conditions give me a trivial solution?
I just could not follow the passage at 18'05" to 18'40". Could someone point me to a clarification of this passage, please?
Hi, this is just a little use of the Fourier Series. In that expression of 18:05 there is some kind of thing multiplying the cos. This is sth similar to the typical Fourier series sumatory of some "a priori" unknown function (then we'll see he forces that function to be sin(2x)).
If you are not familiarized with Fourier series treatment, let me tell you this is a way of expressing a periodic function (of any shape in a certain interval repeated all the time), as an infinite addition of sinus and/or cosinus times a specific coefficient. Those coefficients A_n for sinus and B_n for cosinus (some may write them the other way around) are calculated by integrating the function you want to reexpress times some sin(nx) or cos(nx).
And that's what he does, but here the function to reexpress is that of sin(2x), and he just calculates the expected A_n and divides by the things that he already had there written. In this case the variable A_n is already taken so let me say that the coefficients I said before A_n and B_n, now I'll call them C_n and D_n. So what we find is D_n=2 integral(cos(nx)*sin(2x)) = A_n n sinh(nπ). Now he just takes out the A_n, getting that expression A_n= 2/[n sinh(nπ)] * integral(cos(nx)*sin(2x)).
And that's it ;)
You should write the constatnt as negative lambda. If you do it this way you solved the eigen probelms kind o wrong. p''=lambda*p has an exponential solution. Now p''=-lambda*p has the sin/cos solution.
Thanks a lot for this.
Liked the video, thank you.
Great