We know that to solve any, and all, problems of dynamics, all we have to do is solve an elementary oscillator equation. And its generic form, right here, is quite simple, (q)ddot + q = f, with the initial conditions q(0) = q_0 and (q)dot(0) = (q)dot_0. Here the loading f is known as well as the initial conditions, and we are looking for the time evolution q(t). Can we solve this differential equation for any loading f(t)? For any initial condition? This is precisely what will be able to do at the end of the video. This does not seem so simple. But there is a way. Let us do that. There is actually a very powerful tool to build solutions of this very generic equation. This tool is called the Laplace transform. What is the Laplace transform, and more particularly when we want to use in dynamics? It is quite simple. Consider something that evolves with time, say u(t). You may define the Laplace transform of this evolution, U, as the sum over all times from zero to infinity of u(t) e^(- s t), where s is a positive quantity. Of course, U depends on the variable s, so that I write simply U(s) = L[u(t)]. Of course, this requires that the integral is not infinite. To summarize, I transform the function u of time t to a function U of the variable s. This Laplace transform has some simple properties that we shall use to solve our oscillator problems. First of all, if you recall here how the Laplace transform is defined, we find that the Laplace transform of a sum of functions u and v is the sum of their transform. I can write L(u + v) = L(u) + L(v). And of couse, L(lambda u) = lambda L(u). There is a also a very interesting property on the derivatives. Let us compute the Laplace transform of u dot, the time derivative of u. It is the sum of (u)dot e^(- s t). By integrating by parts, this is equal to s sum[ u e^(- s t) ] - u(t = 0). So, L[(u)dot] = s L(u) - u(0). What have we done here? We have shown that a time derivation is replaced by a product by s in the Laplace domain. That is certainly going to simplify our equations! There is another interesting property that we can take advantage of using the Laplace transform. Let us consider two functions of time, say u(t) and v(t). You know that their convolution product which I write u star v, reads u*v(t) = sum_zero^t[ u(tau) v(t - tau) ] d tau. What is the Laplace transform of this convolution product? A simple calculation yields L(u*v) = L(u) L(v). The convolution product in the time domain is transformed into a simple product in Laplace domain. Everything seems much simpler in Laplace domain - time derivation, convolution. We can also try and compute the Laplace transform of some simple functions, by just computing the sum over all times of the function times e^(- s t). Here are some examples. A dirac delta function becomes a constant, equal to one. A heavyside step function becomes 1/s. sin(t) becomes 1/( 1+s^2 ). cos(t) becomes s/(1 + s^2). Let us now use all these in the solution of our oscillator equation. We are go back to our equation of motion of the oscillator. We are looking for the oscillator motion, q(t). Let capital Q be the Laplace transform of q. What is Q(s)? Let us start with the oscillator equation itself. (q)ddot + q = f(t). By taking the Laplace transform of the oscillator equation we have L[(q)ddot + q] = L[f]. We know f of t, the loading on the oscillator. We can take its Laplace transform, F = L[f], so that we have to solve now L[(q)ddot + q] = F. Can we go any further? Yes, because the Laplace transform has all these very nice properties. First, we can separate the two terms inside the left hand side of the equation, and now we have L[(q)ddot]+ L (q) = F. What is L[(q)ddot]? We know how the Laplace transform operates on derivatives - it is just a multiplication by s. So, we would have L[(q)dot] = s L(q) - q(0). By using this twice we get L[(q)ddot] = s^2 L(q) - s q(0) - (q)dot(0). To summarize, the elementary oscillator equation becomes in the Laplace domain s^2 L(q) - s q(0) - (q)dot(0) + L(q) = L(f), or using, my capital notation for the Laplace transforms, which are functions of s: s^2 Q - s q(0) - (q)dot(0) + Q = F. Let us summarize. What have we done? We have just moved our oscillator equation from the time domain to the Laplace domain. It was (q)ddot + q = f, with initial conditions. It is now s^2 Q - s q(0) - (q)dot(0) + Q = F. Can we solve for Q in the Laplace domain? Yes, certainly, because this is no more a differential equation but just an algebraic equation. The solution is straightforward Q = [1 /(1 + s^2)] F + s q(0) + (q)dot(0). This is very nice, we have the solution in the Laplace domain Q(s). But we want the solution in the time domain, q(t). Can we go back to the time domain? Can we do the inverse Laplace? We have three parts in the solution. We can invert them one by one, of course. The first one is [1/(1+s^2)] F(s). This is the product of two functions of s, namely [1 /(1+s^2)] and F(s). You remember that a product in the Laplace domain is equivalent to a convolution product in the time domain. So, in the time domain we have the convolution product between, the inverse Laplace of 1/(1+s^2) and the inverse Laplace of F(s). This is easy. Who has a Laplace transform equal to 1/(1+s^2)? This is just sin(t)! And who has a Laplace transform equal to F(s)? This is of course my loading f(t)! So, the first term is just therefore the convolution product between sin(t) and f(t). now the next two terms, namely [s/(1 + s^2)]q(0) and [1/(1 + s^2)](q)dot(0). The Inverse Laplace is easy. Remember that q_0 and (q)dot_0 are just quantities. Back in time domain the two terms become respectively (q)dot_0 sin(t) and q_0 cos(t). To summarize, here was the oscillator equation to solve, with its initial conditions. Here is its transform in the Laplace domain Here is the solution in the Laplace domain. And back in the time domain here is the general solution of our oscillator equation. This is the general solution of the oscillator equation. And because the oscillator equation is the elementary equation of vibrations, this is the elementary solution to all vibration problems! The elementary brick to everything in dynamics! You have everything in it - the loading f, the initial conditions, everything. And you can do all dynamics with that. We shall use this extensively in the next videos. We have done this just by using the Laplace transform. By the way, did you know that Pierre-Simon Laplace was a professor at Ecole polytechnique, here in Paris, two centuries ago ? Thank you, Mr Laplace. You have somehow come to the center point of this course. By going from a general problem to a series problems of modal oscillators last week, we had already cracked the problem. And now that you have the solution of the elementary oscillator problem you can rebuild the general solution from bricks to bricks. But let us go by steps. Before using the brick let us see what it looks like in some particular cases, in the next two videos.