PDE10

Transcription

The time has come to discuss the famous d'Alembert's formula for the Wave equation. Now what d'Alembert's formula does for us is it allows us to write-down a solution to a general initial value problem (IVP) for the Wave equation. Now this is a special kind of thing to be able to do because other PDEs don't have formulas like this for their solutions. So let's get right to it. The 'initial value problem' for the Wave equation that d'Alembert's formula allows us solve is (well it's) the Wave equation which is: 'u_tt' = 'c^2.u_xx' plus: initial values. Our initial values will take the form of an initial position: 'u(x,0).' So known position at time zero: and it will be some given function: 'f(x)' And an initial velocity: 'u_t(x,0)' some known velocity at time zero and that's going to be some function: g(x) Now the way to think about these physically (well it's very simple): remember that the Wave equation really is just a formulation of 'F=m.a' for a string. (as you would learn through the video of the derivation of the Wave equation) So really we're solving an 'F = m.a' problem and you know when you solve something like 'F = m.a': if you know the initial position and you know the initial velocity, then you know the trajectory of the particle. So it would make perfect sense that we would know the trajectory of the string if we know the initial position and the initial velocity. So this is: " initial position ", and this is: " initial velocity ". The tool we're going to use to derive d'Alembert's formula is the general solution to the Wave equation that we derived in the last video. So let's recall that real quick: (I'll box it off up here) So: "Recall "... the " general solution " to the " W(ave) E(quation) " is " p(x-ct) " plus " q(x+ct) " where " p, q ".. are " arbitrary ".. (sufficiently nice) .." functions". What we will do now is something which is similar to what we do in ordinary differential equations (ODE) when we have a general solution and we want to satisfy an initial value problem. We take the general solution and essentially plug in the initial values to figure out which general solution we have. So we will find 'p' and 'q' by using the fact that we must satisfy these initial conditions. So let's write down our general solution as we did before: it's: 'u(x,t)' = 'p(x-ct) + q(x+ct)'. So this is the solution to the Wave equation now we have to figure out what 'p' and 'q' have to be to satisfy these two initial conditions. So along the way we'll need 'u_t(x,t)'. (let's compute that) We'll have to use the chain-rule: "u_t(x,t) = -c.p'(x-ct) ".. plus " c.q'(x+ct) ". Okay, so let's use our initial conditions: we have " u(x,0) = f(x) ".. (but 'u(x,0) is simply plugging 't=0' into this formula) so this is the same thing as: " = p(x) + q(x) ". (alright we're getting somewhere) Similarly, plugging in 't=0' into the 'u_t(x,t)' formula that must be: " u_t(x,0) = g(x) ".. that's also the same thing as: " -c.p'(x) + c.q'(x) ". Alright so let's now box off these formulas we've got here: remember our goal is to find 'p' and 'q' in terms of 'f' and 'g' so we're part way there. The next step will be to get rid of the primes ( ' ) on p and q: and the way to do that is.. ..essentially just integrate this second equation. (I'll be a little 'vague' about it right now, but I'll ...be much more clear in a moment) For now let's just observe that we can choose a capital 'G' such that 'G prime' is equal to 'g': G'(x) => g(x) Now, there's lots of choice for 'G': many 'anti-derivatives', and if I choose correctly, choose carefully I can choose one such that I have: " G(X) = -c.p(x) + c.q'(x) " ( I have taken away the 'prime' ) and I've integrated both sides of the equation in just the right way that this anti-derivative matches up with the anti-derivative of the right-hand side. So what we'll do essentially is just.. get rid of this equation right here (let's cross it out) and ..replace it with this guy. This is 'excellent' because we can now solve for 'p' and 'q' in terms of (separate equations) 'f 'and 'G'. Let's do that: (We will work up 'here' to give more 'space') If we use a little bit of arithmetic here.. divide the bottom equation by 'c' and (then 'add' or 'subtract') we'll find: " p(x) = (1/2).f(x) - (1/2c).G(x) " and.. " q(x) = (1/2).f(x) + (1/2c).G(x) " So now we have 'p' and 'q' in terms of 'f' and 'G'. But remember if I know 'p' and 'q' then I know 'u' ... by the formula for the general solution. So if we use (the general) formula we essentially have solved our problem, (we still have this matter figuring out exactly what 'G' is) but we've essentially solved our problem - our solution is: " u(x,t) ".. ... and that's equal to 'p(x-ct)' which is: " = (1/2).f(x-ct) - (1/2c).G(x-ct) ".. plus 'q(x+ct)' so: " + (1/2).f(x+ct) + (1/2c).G(x+ct) ". So let's figure out of what's going on with this function 'G' ..and then we'll have the full d'Alembert's formula. (So let's do that on the next slide) Let's start with the last formula from the previous slide: We're concerned with this 'G': We made a 'choice' of anti-derivative, and I was pretty vague about that 'choice'. Now the 'secret' here is that in fact the choice doesn't matter. So let's choose a 'particular' anti-derivative: let's use the Fundamental theorem of Calculus and let's take 'G' to be 'an integral', from some starting point, (here's where the 'choice' is..I have to start at some point, 'x_0') up to 'x'.. (of) "g(s)ds ". Now 's' is just a 'dummy' variable for the integration. Now let's examine this 'difference' of two anti-derivatives using this 'integral' definition of the anti-derivative I've chosen. So we take: " G(x+ct) - G(x-ct) ". Now using this integral that is the integral from 'x_0' to 'x+ct' of " g(s)ds " minus the integral from 'x_0' to 'x-ct' of " g(s)ds ". Now remember that if I have a negative of an integral I can change that 'minus' to a 'plus' so long as integrate in the 'opposite direction'. So I have to switch the bounds of integration: I'll put the 'x_0' on the top and I'll put the 'x-ct' on the bottom. Also remember that if I integrate from a starting point, let's call it 'a', up to an ending point 'b' and I then I integrate from 'b' up to another endpoint, 'c': that's the same as integrating from the original starting point up to the final ending point. So I can combine these two integrals into a single integral from 'x-ct' up to 'x+ct' of " g(s)ds. Now you see that the choice of 'x' is irrelevant. It doesn't matter. It was some point in this interval that I used to define my anti-deriviative. We've got everything in the form we wanted and we're basically done. So let's to summarize what it is that we've done: we've a " Solution to IVP ".. ..for the Wave equation which is the Wave equation: " u_tt = (c^2).u_xx ".. plus initial values: " u(x,0) = " some function: " f(x) ".. and " u_t(x,0) = " some function " g(x) ": So that's our initial value problem. The solution to this initial value problem is given by: " u(x,t) ".. equals one-half..well the first part right here " (1/2).[f(x+ct) + f(x-ct)] ".. plus the second part but we'll write it in the 'integral' form that is more convenient to use: integral (from 'x-ct' up to 'x+ct'): of "g(s)ds ". Now let's box this off because it's so important: this is d'Alembert's formula for the solution to the Wave equation: Just notice one thing before we finish here. Notice that the formula can be computed just by using 'f' and 'g' So if I give you the functions 'f' and 'g' you can plug them into this formula and you get the solution 'u(x,t)'. This will allow us to produce 'interesting' examples and that will be our next 'order of buisiness'