(f g)(t) = Example 4.5.1: Find f g the convolution of the functions f(t) = e t and g(t) = sin(t). Solution: The definition of convolution is,

Transcription

1 .5. Convolutions and Solutions Solutions of initial value problems for linear nonhomogeneous differential equations can be decomposed in a nice way. The part of the solution coming from the initial data can be separated from the part of the solution coming from the nonhomogeneous source function. Furthermore, the latter is a kind of product of two functions, the source function itself and the impulse response function from the differential operator. This kind of product of two functions is the subject of this section. This kind of product is what we call the convolution of two functions..5.. Definition and Properties. One can say that the convolution is a generalization of the pointwise product of two functions. In a convolution one multiplies the two functions evaluated at different points and then integrates the result. Here is a precise definition. Definition.5.. The convolution of functions f and g is a function f g given by (f g(t f(τg(t τ dτ. (.5. Remark: The convolution is defined for functions f and g such that the integral in (.5. is defined. For example for f and g piecewise continuous functions, or one of them continuous and the other a Dirac s delta generalized function. Example.5.: Find f g the convolution of the functions f(t e t and g(t sin(t. Solution: The definition of convolution is, (f g(t e τ sin(t τ dτ. This integral is not difficult to compute. Integrate by parts twice, [ ] e τ sin(t τ dτ e τ t [ ] cos(t τ e τ t sin(t τ that is, 2 e τ sin(t τ dτ We then conclude that Example.5.2: Graph the convolution of Solution: Notice that Then we have that e τ sin(t τ dτ, [ ] e τ t [ ] cos(t τ e τ t sin(t τ e t cos(t + sin(t. (f g(t 2[ e t + sin(t cos(t ]. (.5.2 f(τ u(τ u(τ, { 2 e 2τ for τ g(τ for τ <. g( τ { 2 e 2τ g(t τ g( (τ t for τ for τ >. { 2 e 2(τ t for τ t for τ > t.

2 2 In the graphs below we can see that the values of the convolution function f g measure the overlap of the functions f and g when one function slides over the other. Figure. The graphs of f, g, and f g. A few properties of the convolution operation are summarized in the Theorem below. But we save the most important property for the next subsection. Theorem.5.2 (Properties. For every piecewise continuous functions f, g, and h, hold: (i Commutativity: f g g f; (ii Associativity: f (g h (f g h; (iii Distributivity: f (g + h f g + f h; (iv Neutral element: f ; (v Identity element: f δ f. Proof of Theorem.5.2: We only prove properties (i and (v, the rest are left as an exercise and they are not so hard to obtain from the definition of convolution. The first property can be obtained by a change of the integration variable as follows, (f g(t f(τ g(t τ dτ.

3 3 Now introduce the change of variables, ˆτ t τ, which implies dˆτ dτ, then (f g(t so we conclude that t f(t ˆτ g(ˆτ( dˆτ g(ˆτ f(t ˆτ dˆτ, (f g(t (g f(t. We now move to property (v, which is essentially a property of the Dirac delta, (f δ(t This establishes the Theorem. f(τ δ(t τ dτ f(t The Laplace Transform. The Laplace transform of a convolution of two functions is the pointwise product of their corresponding Laplace transforms. This result will be a key part in the solution decomposition result we show at the end of the section. Theorem.5.3 (Laplace Transform. If both L[g] and L[g] exist, including the case where either f or g is a Dirac s delta, then L[f g] L[f] L[g]. (.5.3 Remark: It is not an accident that the convolution of two functions satisfies Eq. (.5.3. The definition of convolution is chosen so that it has this property. One can see that this is the case by looking at the proof of Theorem.5.3. One starts with the expression L[f] L[g], then changes the order of integration, and one ends up with the Laplace transform of some quantity. Because this quantity appears in that expression, is that it deserves a name. This is how the convolution operation was created. Proof of Theorem.5.3: We start writing the right hand side of Eq. (.5., the product L[f] L[g]. We write the two integrals coming from the individual Laplace transforms and we rewrite them in an appropriate way. [ ] [ L[f] L[g] e st f(t dt ] e s t g( t d t ( e s t g( t e st f(t dt d t ( g( t e s(t+ t f(t dt d t, where we only introduced the integral in t as a constant inside the integral in t. Introduce the change of variables in the inside integral τ t + t, hence dτ dt. Then, we get L[f] L[g] ( g( t t t e sτ f(τ t dτ d t (.5. e sτ g( t f(τ t dτ d t. (.5.5

4 Here is the key step. We must switch the order of integration. From Fig. 2 we see that changing the order of integration gives the following expression, L[f] L[g] τ e sτ g( t f(τ t d t dτ. Then, is straightforward to check that ( τ L[f] L[g] e sτ g( t f(τ t d t dτ e sτ (g f(τ dt L[g f] L[f] L[g] L[f g]. This establishes the Theorem. t t τ Figure 2. Domain of integration in (.5.5. τ Example.5.3: Compute the Laplace transform of the function u(t Solution: The function u above is the convolution of the functions f(t e t, g(t sin(t, that is, u f g. Therefore, Theorem.5.3 says that Since, L[u] L[f g] L[f] L[g]. e τ sin(t τ dτ. L[f] L[e t ], L[g] L[sin(t] s + s 2 +, we then conclude that L[u] L[f g] is given by L[f g] (s + (s 2 +. Example.5.: Use the Laplace transform to compute u(t e τ sin(t τ dτ. Solution: Since u f g, with f(t e t and g(t sin(t, then from Example.5.3, L[u] L[f g] (s + (s 2 +. A partial fraction decomposition of the right hand side above implies that L[u] ( ( s + 2 (s + (s 2 + ( 2 (s + + (s 2 + s (s 2 + ( L[e t ] + L[sin(t] L[cos(t]. 2 This says that u(t 2( e t + sin(t cos(t. So, we recover Eq. (.5.2 in Example.5., that is, (f g(t 2( e t + sin(t cos(t,

5 Example.5.5: Find the function g such that f(t s transform L[f] (s 2 + 6((s Solution: Since f(t sin(t g(t, we can write so we get that s (s 2 + 6((s 2 L[f] L[sin(t g(t] + 9 L[sin(t] L[g] (s L[g] s (s 2 + 6((s We now rewrite the right-hand side of the last equation, L[g] (s + (s L[g] ( that is, L[g] (L[cos(3t](s + 3 L[sin(3t](s sin(τ g(t τ dτ has the Laplace (s L[g], L[g] (s (s s (s (s , ( L[e t cos(3t] + 3 L[et sin(3t], 5 which leads us to g(t et ( cos(3t + 3 sin(3t.5.3. Solution Decomposition. The Solution Decomposition Theorem is the main result of this section. Theorem.5. shows one way to write the solution to a general initial value problem for a linear second order differential equation with constant coefficients. The solution to such problem can always be divided in two terms. The first term contains information only about the initial data. The second term contains information only about the source function. This second term is a convolution of the source function itself and the impulse response function of the differential operator. Theorem.5. (Solution Decomposition. Given constants a, a, y, y and a piecewise continuous function g, the solution y to the initial value problem can be decomposed as y + a y + a y g(t, y( y, y ( y, (.5.6 y(t y h (t + (y δ g(t, (.5.7 where y h is the solution of the homogeneous initial value problem y h + a y h + a y h, y h ( y, y h( y, (.5.8 and y δ is the impulse response solution, that is, y δ + a y δ + a y δ δ(t, y δ (, y δ(.

6 6 Remark: The solution decomposition in Eq. (.5.7 can be written in the equivalent way y(t y h (t + y δ (τg(t τ dτ. Also, recall that the impulse response function can be written in the equivalent way y δ L [ e cs ], c, and y δ L [ ], c. p(s p(s Proof of Theorem.5.: Compute the Laplace transform of the differential equation, L[y ] + a L[y ] + a L[y] L[g(t]. Recalling the relations between Laplace transforms and derivatives, L[y ] s 2 L[y] sy y, L[y ] s L[y] y. we re-write the differential equation for y as an algebraic equation for L[y], (s 2 + a s + a L[y] sy y a y L[g(t]. As usual, it is simple to solve the algebraic equation for L[y], L[y] (s + a y + y (s 2 + a s + a + (s 2 + a s + a L[g(t]. Now, the function y h is the solution of Eq. (.5.8, that is, L[y h ] (s + a y + y (s 2 + a s + a. And by the definition of the impulse response solution y δ we have that L[y δ ] (s 2 + a s + a. These last three equation imply, L[y] L[y h ] + L[y δ ] L[g(t]. This is the Laplace transform version of Eq. (.5.7. Inverting the Laplace transform above, y(t y h (t + L [ L[y δ ] L[g(t] ]. Using the result in Theorem.5.3 in the last term above we conclude that y(t y h (t + (y δ g(t. Example.5.6: Use the Solution Decomposition Theorem to express the solution of y + 2 y + 2 y g(t, y(, y (. Solution: We first find the impuse response function y δ (t L [ ], p(s s 2 + 2s + 2. p(s since p has complex roots, we complete the square, so we get s 2 + 2s + 2 s 2 + 2s (s + 2 +, y δ (t L [ ] (s y δ (t e t sin(t.

7 7 We now compute the solution to the homogeneous problem Using Laplace transforms we get y h + 2 y h + 2 y h, y h (, y h(. L[y h] + 2 L[y h] + 2 L[y h ], and recalling the relations between the Laplace transform and derivatives, ( s 2 L[y h ] s y h ( y h( + 2 ( L[y h] s L[y h ] y h ( + 2L[y h ], using our initial conditions we get (s 2 + 2s + 2 L[y h ] s + 2, so so we obtain L[y h ] (s + (s + (s 2 + 2s + 2 (s + 2 +, [ ] y h (t L e t cos(t. Therefore, the solution to the original initial value problem is y(t y h (t + (y δ g(t y(t e t cos(t + e τ sin(τ g(t τ dτ. Example.5.7: Use the Laplace transform to solve the same IVP as above. y + 2 y + 2 y g(t, y(, y (. Solution: Compute the Laplace transform of the differential equation above, L[y ] + 2 L[y ] + 2 L[y] L[g(t], and recall the relations between the Laplace transform and derivatives, L[y ] s 2 L[y] sy( y (, Introduce the initial conditions in the equation above, L[y ] s L[y] y(. L[y ] s 2 L[y] s ( (, L[y ] s L[y], and these two equation into the differential equation, Reorder terms to get (s 2 + 2s + 2 L[y] s + 2 L[g(t]. L[y] (s + (s 2 + 2s (s 2 + 2s + 2 L[g(t]. Now, the function y h is the solution of the homogeneous initial value problem with the same initial conditions as y, that is, L[y h ] (s + (s + (s 2 + 2s + 2 (s L[e t cos(t]. Now, the function y δ is the impulse response solution for the differential equation in this Example, that is, cl[y δ ] (s 2 + 2s + 2 (s L[e t sin(t]. If we put all this information together and we get L[y] L[y h ] + L[y δ ] L[g(t] y(t y h (t + (y δ g(t,

8 8 More explicitly, we get y(t e t cos(t + e τ sin(τ g(t τ dτ.