This was a great solution, but you were fortunate that the final expression was easy to inverse Laplace transform at the end. A different initial condition would have made this much more challenging.
@@maths_505 When I taught engineering math I used to solve the same problem by Laplace transforms, but with a constant initial condition. The solution, which can also be found by combination of variables, involves the complementary error function. But solving it with Laplace transforms and then inverting the final answer without use of a table of transforms made it a challenging and interesting problem.
Great to see Laplace transform used to solve a PDE. This said, if use the method of separation of variables with an added constant: u(x,t)=f(x)g(t)+c, we can very quickly arrive at the same solution. Things works out nicely because of the initial conditions. So I wonder if given the initial conditions are defined in such a way the laplace method gives a simple solution is equivalent to using the separation of variable method ?
You could also skip using the variation of parameters method if you introduced the Fourier transform of u/U, and do it totally algebraically. That would have brought you to about 10:54 much faster. I agree, however, that variation of parameters method is an important one to know.
The PDE I think is fun (and insightful) to some via Laplace transforms is u_xx = (1/c^2) u_tt + δ(x - v t) , u(0,t) = 0, u(x,0) = 0, u_t(x,0), x >= 0. Take LT wrt t ( think). Then the cases v c supersonic are different. I hope I remembered everything right. And yes δ is the dirac delta function
Interesting that the boundary conditions create a situation where the homogenous solution doesn’t matter anymore and we only look the particular solution when need to invert it back.
Heaviside did this operationally, getting (as you did) e^Sqrt[s] and e^-Sqrt(s) , but then interpreted s = d/dt, so we need to compute e^(d^1/2/dt^1/2), the exponential of the 1/2 derivative. See Electromagnetic Theory By O. Heaviside.
This was a great solution, but you were fortunate that the final expression was easy to inverse Laplace transform at the end. A different initial condition would have made this much more challenging.
That's pretty much how I designed it. Challenging cases make good full length videos in their own right.
@@maths_505 When I taught engineering math I used to solve the same problem by Laplace transforms, but with a constant initial condition. The solution, which can also be found by combination of variables, involves the complementary error function. But solving it with Laplace transforms and then inverting the final answer without use of a table of transforms made it a challenging and interesting problem.
I strongly believe the initial conditions were not randomly chosen, let's just say. :D
Yeah you have to do stuff with Contour Integration if the Laplace Function isn’t so nice. Some Bromwich contours
@@emanuellandeholm5657 how you doin homie? I hope everything's good.
I'd appreciate some more videos on differential equations, you don't see many math channels touching them beyond a surface level.
Great to see Laplace transform used to solve a PDE.
This said, if use the method of separation of variables with an added constant: u(x,t)=f(x)g(t)+c, we can very quickly arrive at the same solution. Things works out nicely because of the initial conditions.
So I wonder if given the initial conditions are defined in such a way the laplace method gives a simple solution is equivalent to using the separation of variable method ?
You could also skip using the variation of parameters method if you introduced the Fourier transform of u/U, and do it totally algebraically. That would have brought you to about 10:54 much faster.
I agree, however, that variation of parameters method is an important one to know.
I haven't used that method in a long time so I really wanted to employ it here😂
The PDE I think is fun (and insightful) to some via Laplace transforms is u_xx = (1/c^2) u_tt + δ(x - v t) , u(0,t) = 0, u(x,0) = 0, u_t(x,0), x >= 0. Take LT wrt t ( think). Then the cases v c supersonic are different. I hope I remembered everything right. And yes δ is the dirac delta function
Corrected. LT WRT t., u(0,t) = 0 , u (x,t0 = u_t(x,0) = 0 Please note
Interesting that the boundary conditions create a situation where the homogenous solution doesn’t matter anymore and we only look the particular solution when need to invert it back.
That's how I designed them. I wanted to convey the solution development comprehensively while still getting a nice function as the solution.
Masterpiece!!!!
May I ask what is the "blackboard" software you use in this video? I also need to teach the Engineering Math class next semester. Thank you!
❗️❗️🗣️🗣️WE GETTING OUT OF THE ODES WITH THIS ONE🗣️🗣️❗️❗️
PDEs are actually my second favourite objects.
@@maths_505 So at some point, we'll be getting Rodrigues polynomials and spherical harmonics? (We don't want to leave out Bohr...I mean...).
Am here to learn big good things
Can you plz explain how do you find the value of v1 and v2 in particular solution.
Heaviside did this operationally, getting (as you did) e^Sqrt[s] and e^-Sqrt(s) , but then interpreted s = d/dt, so we need to compute e^(d^1/2/dt^1/2), the exponential of the 1/2 derivative. See Electromagnetic Theory By O. Heaviside.
That sounds extremely cool and I'll definitely check that out.
Smart solution. Thank you.
What software do you use for drawing?
Samsung notes
How would you do it with ft tho? I know how for R(1,1) but not for R(1,3), since my integrals always seem to diverge...
Can you do a playlist about the methods to solve the various types of PDE?
Sure. This is my first PDE video and there are lots more to come
i alwats wondered gow it was solved