HigherLin4

We then make the change of variable t t τ in the

This preview shows pages 22–24. Sign up to view the full content.

We then make the change of variable t = t τ in the inner definite integral to obtain integraldisplay T 0 e st ( f g )( t ) d t = integraldisplay T 0 e g ( τ ) integraldisplay T τ 0 e st f ( t ) d t d τ . Upon formally letting T → ∞ above, definition (8.2) of the Laplace transform shows that the inner integral converges to F ( s ), which is independent of τ . The double integral thereby converges to G ( s ) F ( s ), yielding (8.16). square Remark. Because the upper endpoint of the inner integral depends on the variable of integration τ of the outer integral, properly passing to the limit above requires greater care than we took here. The techniques one needs are taught in Advanced Calculus courses. The argument given above suits our purposes because it illuminates why (8.16) holds. The convolution theorem can be used to help evaluate inverse Laplace transforms. For example, suppose that you know for a given F ( s ) and G ( s ) that f ( t ) = L 1 [ F ]( t ) and g ( t ) = L 1 [ G ]( t ). Then (8.16) implies that L 1 bracketleftbig F ( s ) G ( s ) bracketrightbig ( t ) = ( f g )( t ) . (8.17) You can use this fact to express inverse Laplace transforms as convolutions. You may still have to evaluate the convolution integral, but some of you might find that easier than using partial fraction identities to express F ( s ) G ( s ) in basic forms.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
23 Example. Compute y ( t ) = L 1 [ Y ]( t ) for Y ( s ) = 2 s 2 ( s 2 + 4) . Because you know from table (8.13) that L 1 bracketleftbigg 1 s 2 bracketrightbigg = t , L 1 bracketleftbigg 2 s 2 + 2 2 bracketrightbigg = sin(2 t ) , it follows from (8.17) and an integration by parts that y ( t ) = L 1 bracketleftbigg 2 s 2 ( s 2 + 4) bracketrightbigg = L 1 bracketleftbigg 1 s 2 2 s 2 + 2 2 bracketrightbigg = integraldisplay t 0 ( t τ ) sin(2 τ ) d τ = ( τ t ) cos(2 τ ) 2 vextendsingle vextendsingle vextendsingle vextendsingle t 0 integraldisplay t 0 cos(2 τ ) 2 d τ = t 2 sin(2 t ) 4 . This is the same result we got on page 18 using a partial fraction identity. 8.9: Natural Fundamental Sets. The convolution theorem also gives us a new way to understand Green functions. We have used the Green function to construct a particular solution of the nonhomogeneous equation L y = p (D) = f ( t ) by the formula y P ( t ) = integraldisplay t 0 g ( t τ ) f ( τ ) d τ . Notice that the right-hand side above is exactly ( g f )( t ). Taking the Laplace transform of this formula, the Convolution Theorem then yields L [ y P ]( s ) = L [ g f ]( s ) = G ( s ) F ( s ) = F ( s ) p ( s ) , where F ( s ) = L [ f ]( s ) . But this agrees with formula (8.10). Indeed, because y P ( t ) given by the above formula satisfies the initial conditions y P (0) = 0 , y P (0) = 0 , · · · y ( n 2) P (0) = 0 , y ( n 1) P (0) = 0 , it follows that the polynomial q ( s ) appearing in (8.10) vanishes. Formula (8.10) can generally be recast as y ( t ) = y H ( t ) + y P ( t ) , where y H ( t ) = L 1 bracketleftbigg q ( s ) p ( s ) bracketrightbigg , y P ( t ) = L 1 bracketleftbigg F ( s ) p ( s ) bracketrightbigg .
24 This is the decomposition of y ( t ) into the solution y H ( t ) of the associated homogeneous equation whose initial data agree with y ( t ) and the particular solution y P ( t ) whose initial data is zero. Because q ( s ) = ( s n 1 + a 1 s n 2 +
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern