Solving a partial differential equation using laplace transforms

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Күн бұрын

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Пікірлер: 31

@michaelihill3745 9 ай бұрын

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 9 ай бұрын

That's pretty much how I designed it. Challenging cases make good full length videos in their own right.

@michaelihill3745 9 ай бұрын

@@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.

@emanuellandeholm5657 9 ай бұрын

I strongly believe the initial conditions were not randomly chosen, let's just say. :D

@SuperSilver316 9 ай бұрын

Yeah you have to do stuff with Contour Integration if the Laplace Function isn’t so nice. Some Bromwich contours

@maths_505 9 ай бұрын

@@emanuellandeholm5657 how you doin homie? I hope everything's good.

@mcalkis5771 9 ай бұрын

I'd appreciate some more videos on differential equations, you don't see many math channels touching them beyond a surface level.

@riadsouissi 9 ай бұрын

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 ?

@Calcprof 9 ай бұрын

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

@Calcprof 9 ай бұрын

Corrected. LT WRT t., u(0,t) = 0 , u (x,t0 = u_t(x,0) = 0 Please note

@nadavslotky 9 ай бұрын

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.

@maths_505 9 ай бұрын

I haven't used that method in a long time so I really wanted to employ it here😂

@SuperSilver316 9 ай бұрын

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.

@maths_505 9 ай бұрын

That's how I designed them. I wanted to convey the solution development comprehensively while still getting a nice function as the solution.

@danielrosado3213 9 ай бұрын

❗️❗️🗣️🗣️WE GETTING OUT OF THE ODES WITH THIS ONE🗣️🗣️❗️❗️

@maths_505 9 ай бұрын

PDEs are actually my second favourite objects.

@AB-nu5we 9 ай бұрын

@@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...).

@AttiaNaz-pw6do 2 ай бұрын

Can you plz explain how do you find the value of v1 and v2 in particular solution.

@Calcprof 9 ай бұрын

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.

@maths_505 9 ай бұрын

That sounds extremely cool and I'll definitely check that out.