Oxford Calculus: Separable Solutions to PDEs

[@TomRocksMaths]

Test yourself with some exercises on separable solutions with this FREE worksheet in Maple Learn: learn.maplesoft.com/d/COKQBLEMDIGTPODOJIPGMQHKIPBGAMIQOOIGJSKTDFIOELMPEPERDKJHLRIIEHGHMHKQGLLTNLCLOFGTOGONMMDJINDTGSALDFPR

[@weonlygoupfromhere7369]

I love how you teach complex topics simpler than a good chunk of professors. They just explain it but don't give you an in-depth and step by step explanation

[@Spectator007]

If I took my Fourier analysis class a quarter earlier this video wouldn't have been posted. Now I get MGK as my math tutor. You make me really enjoy my uni class even though you are sadly not my professor.

[@TomRocksMaths]

Glad I could help!

[@wez7279]

Was literally going to recap this topic today after finishing my first year, perfect timing!

[@jcleary3134]

I was reading the textbook for an hour+, and this video just saved me. Thank you sir.

[@sunandinighosh6037]

Yesterday I was trying to understand the solution of Schrodinger's equation for my physics exam and couldn't understand the method...thankfully today you uploaded this video. What a coincidence!!

[@cristianorlandoelpro416]

Mate, my textbook did a horrible job explaining this topic. I'm glad I came across your vid

[@bos567564]

Hi Tom. I'm from South Africa. I just wanted to thank you for your channel. I wasn't very good at maths at school. In fact, I really struggled with it, especially with geometry .Since leaving school, I have started learning maths again. It has become a sort of hobby of mine. Maths is in my opinion the most interesting subject I have ever learnt after philosophy (that will always be my first joy, because I believe it is even more fundamental than maths, although you might disagree with me haha) 😀. Your channel gives me hope that I can learn and will improve. So thanks a lot, and keep making videos for us your maths fans!

[@BlackEyedGhost0]

Had to go back and review linear differential equations before I could remember how to do this. Thanks for the practice problems. Apparently I needed the practice.

[@HuyNguyen-wj1ho]

What a great and clear lecture. Thank you very much Dr. Tom. Waiting for your next lecture.😊

[@peterhall6656]

Separability ultimately has its physical justification in the assumed independence of the functional relations. This is quite believable in all the major PDEs which arose form looking at physical phenomena. Just look at how Maxwell derived his velocity distribution law to appreciate the independence angle.

[@M.athematech]

Hi Tom, like John I am also from South Africa, but I was very good at maths at school and went on to complete my PhD at 24. I wasn't very good at begging people to give me money to do maths research though and started an IT company instead. But anyway, D = 1 doesn't follow from u(1,1) = e. The most one can say is that D and C are related by D = e^(1-3C/2).

[@TomRocksMaths]

I suppose I’m really appealing to uniqueness of the solution (my bad for failing to state this explicitly).

[@giosanchez2714]

Excited for this one!

[@nicholasifeajika1827]

Great explanation. Easy to understand

[@priscillaflores99]

Great explanation

[@aniketeuler6443]

Pretty excited sir!

[@Au-fx4pv]

Thanks for helping me!!!

[@jakobandrews2096]

When you say u(1,1) = e, wouldn't D=1/e and C=4/3 also work? I feel there is an infinite number of constants that work here

[@jperez7893]

i didn't get how you arrived at D=1

[@arjunsinha4015]

Nice video sir

[@AyodeleKayode-w1g]

Thank you sir

[@ronanmccluskey900]

Why does D=1 when u(1,1)=e??

[@TomRocksMaths]

I’m appealing to the unique solution property (without proving it - my bad).

[@felipesernabarbosa2796]

D = exp(1 - 3c/2), I believe.@@TomRocksMaths

[@two697]

How do you know the only solution is in this form though? How do you know it isn't a linear combination of them? For example f(x)+g(y)

[@TomRocksMaths]

We rely on being able to show the PDE has a unique solution, which can be done for most of the examples seen here.

[@MrFtriana]

You must check that this solution satisfies the boundary conditions. If two different solutions of a given PDE satisfies the same boundary conditions, it can be assumed that they are the same.

[@travischism]

if we label f(x) == ln(F(x)) and g(x) == ln(G(x)) then u(x,y) == f(x) + g(x) == ln(f(x)) +ln(G(x)) == ln(F(x)•G(x)) and now the solution U(x,y) == exp(u(x,y)) == F(x)•G(x)

[@camachojankowilderbeismar6167]

Hello

[@MisterTutor2010]

Logan Paul does PDEs? :)

[@raneena5079]

I feel really weird just assuming that it's separable with no justification :/

[@TomRocksMaths]

It’s a standard technique to try and if it happens to work, then we can appeal to the uniqueness of solutions to claim it is the only solution (and in some sense we made a lucky/good first guess).