Basic Laplace transforms table

Additional Laplace transforms table

More exercise problems

1. Solve the following differential equation:
   
   

   ⋅⋅ y   (t) + y(t) = sin t,     y(0)  and ⋅ y   (0)   given.

   (3)

   (Solution) Laplace transforms of the both sides yield:
   
   

   s2 - y   (s) − s y(0) − ⋅ y   (0) + - y   (s) = 1 1+s2 ,

   (4)

   which can be solved for ―y(s) as
   

   - y   (s) = 1 (1+s2)2 + s 1+s2 y(0) + 1 1+s2 ⋅ y   (0).

   (5)

   Hence, the inverse of the above is
   

   y(t) = sin t*sin t + y(0) cos t + ⋅ y   (0) sin t = ⌠ ⌡ t 0  sin (t − τ) sin τ dτ2 + y(0) cos t + ⋅ y   (0) sin t = 1 2 (−t cos t +sin t) + y(0) cos t + ⋅ y   (0) sin t. (6)

   
   
2. Solve the following integral equation.
   

   y(t) = t + ⌠ ⌡ t 0  sin (t − τ) y(τ) dτ.

   (7)

   (Solution) Laplace transform the both sides of the equation to get
   

   - y   = 1 s2 + 1 s2 + 1 - y   ,

   (8)

   from which it follows
   

   - y   = s2 + 1 s4 = 1 s2 + 1 s4 ,

   (9)

   which can be inverted to yield
   

   y(t) = t + t3 6 .

   (10)
3. Prove
   

   lim s → ∞  s - f   (s) = f(0).

   (Solution) Use the definition,
   

   L(f  ′(t)) = ⌠ ⌡ ∞ 0  f′(t) e−s t d t = s   - f   (s) − f(0), (11)

   and let s → ∞.
4. Prove
   

   lim s → ∞  - f   (s) = 0.

   (Solution)
   Use the definition,
   

   L(f(t)) = ⌠ ⌡ ∞ 0  f(t) e−s t d t = - f   (s), (12)

   and let s → ∞.
5. Find the inverse Laplace transform for
   

   1 (s + 3)4 .

   (Solution)
   From
   

   L(t3) = 3! s4 ,     L(ea t f(t)) = - f   (s − a),

   
   

   L−1 ⎛ ⎝ 1 (s + 3)4 ⎞ ⎠ = e−3 t t3 3! .
6. Find the inverse Laplace transform for
   

   s + 2 s3 (s − 1) .

   (Solution) Use partial fraction to get
   

   s + 2 s3 (s − 1) = 3 s − 1 − 2 + 3 s + 3 s2 s3 = 3 s − 1 − − 3 s − 3 s2 − 2 s3 , (13)

   so
   
   

   L−1 ⎛ ⎝ s + 2 s3 (s − 1) ⎞ ⎠ = 3 et −3 − 3 t − t2.
7. Express the Laplace transform for
   

   t f′(t).

   (Solution)
   Note that L(t f(t)) = −[(d)/(ds)] ―f(s). Therefore,
   

   L(t f′(t)) = − d ds L(f′(t)) = − d ds ⎛ ⎝ s - f   (s)−f(0) ⎞ ⎠ = − ⎛ ⎜ ⎝ - f   (s)+ s d - f   (s) ds ⎞ ⎟ ⎠ .
8. Solve the diffusion equation,
   

   ∂u ∂t = ∂2 u ∂x2 ,

   with
   

   u = 0   at   t = 0,     u = 0   at   x = 0,     u = 100   at   x = 1.

   by the Laplace transform.
   (Solution) Laplace transform the original equation with respect to t to get
   

   s - u   = d2 - u   d x2 ,

   which can be solved for ―u as
   

   - u   = A e√s x + B e− √s x.

   The boundary conditions are
   

   - u   |x = 0 = A + B = 0,

   
   

   - u   |x = 1 = A e√s + B e−√s = 100 s ,

   which can be solved for A and B as
   

   A = 100 s (e√s − e−√s) ,    B = − 100 s (e√s − e−√s) .

   Therefore,
   

   - u   = 100 s e√sx − e−√s x e√s − e−√s

   The inverse of ―u needs to wait for the spring semester (ME5332).
9. Problem (Variable-coefficient equation) Consider the problem
   
   

   t x" + x′+ t x = 0     (0 ≤ t < ∞)

   (14)

   
   

   x(0)=1,     x′(0)=0

   where our special interest lies in seeing whether or not we can solve (14) by the Laplace transform method even though differential equation has nonconstant coefficients.
   (a) Take the Laplace transform of the differential equation. Note that the transforms of t x"(t) and t x(t),
   
   

   L{tx"(t)}= ⌠ ⌡ ∞ 0  tx"e−stdt

   
   

   L{tx(t)}= ⌠ ⌡ ∞ 0  txe−stdt

   present a difficulty in that we cannot express them in terms of X(s) the way we can express L{x′(t)}=sX(s)−x(0) L{x"(t)}=s²X(s)−sx(0)−x′(0). Nevertheless, terms can be handled as follows. Observe that
   
   

   L{tx"(t)} = ⌠ ⌡ ∞ 0  tx"e−stdt = − ⌠ ⌡ ∞ 0  d ds (x"e−st)dt = − d ds ⌠ ⌡ ∞ 0  x"e−stdt = − d ds [s2X(s)−sx(0)−x′(0)] = − d ds [s2X(s)−s],

   
   

   (15)

   if we assume that the unknown x(t) is sufficiently well behaved for the third equality (where we have interchanged the order of two limit processes, the s differentiation and the t integration) to be justified. Handling the L{tx(t)} term in the same way, show that the application of the Laplace transform to (14) leads to the equation
   
   

   (s2 + 1) dX ds + s X = 0

   (16)

   on X(x). Note that whereas the Laplace transform method reduces constant-coefficient differential equations to linear algebraic equations on X(s), here the nonconstant coefficients result in the equation on X(s) being itself a linear differential equation! However, it is a simple one. Solving (16), show that
   
   

   X(s)= C √ s2+1

   (17)

   (b) From Appendix C, we find the inverse as x(t)=C J₀(t), where J₀ is the Bessel function of the first kind, of order zero. Applying the initial condition once again gives x(0) = 1 = C J₀(0) = C, so C=1, and the desired solution of (14) is x(t) = J₀(t). Here,however, we ask you to proceed as though you do not know about Bessel functions. Specifically, re-express (17) as
   
   

   X(s)= C s √ 1+(1/s2) = C s ⎛ ⎝ 1− 1 2 1 s2 + … ⎞ ⎠

   (18)

   where the last equality amounts to the Taylor expansion of √{1+r} in the quantity r, about r=0, where r=1/s². Carry that expansion further; invert the resulting series term by term (assuming that that step is valid), and thus show that
   
   

   x(t)=C ⎡ ⎣ 1− t2 22 + 1 (2!)2 t4 24 − 1 (3!)2 t6 26 +... ⎤ ⎦

   Setting x(0)=1 gives C=1, and the result is that we have obtained the solution in power series form. Of course, that power series is the Taylor series of the Bessel function J₀(t). NOTE: Observe that rather than pulling an s out of the square root in (18), and then expanding 1/√{1+(1/s²)} in powers of 1/s², we could have expanded (17) directly in powers of s as X(s)=C (1−[1/2]s²+...). However, positive powers of s are not invertible, so this form is of no use. [We will see, in Theorem 5.7.6, that to be invertible a transform must tend to zero as s→∞. Positive powers of s do not satisfy this condition, but negative powers do.] Also, observe that the degree to which nonconstant-coefficient differential equations are harder than constant-coefficient ones can be glimpsed from the fact that coefficients proportional to t cause the equation on X(s) to be a first-order differential equation; coefficients proportional to t² will cause the equation on X(s) to be a second-order differential equation,and so on.
   Solution
   (a) Just follow the hint.
   

   (s2+1) dX ds + sX=0

   is equivalent to
   

   dX X + s s2+1 ds = 0

   which can be integrated to yield
   

   logX + 1 2 log(s2+1) = C1

   or
   

   log ⎛ ⎝ X √   1+s2   ⎞ ⎠ = C1

   or
   

   X = C √ 1+s2 .

   (b) Follow the hint until you feel comfortable.

Z-transform

Z-transform

(1) Linearity


Z(αfn + βgn) = αZ(fn) + βZ(gn).


(2) Shift (equivalent to derivative)


Z(fn+1) = z - f   (z) − z f0.

(Proof)


Z(fn+1) = ∞ ∑ n=0  fn+1 zn = z ∞ ∑ n=0  fn+1 zn+1 = z ∞ ∑ n=1  fn zn = z ⎛ ⎝ ∞ ∑ n=0  fn zn − f0 ⎞ ⎠ = z - f   (z) − z f0. (26)

Note that


Z(fn+2) = z Z(fn+1) − z f1 = z ⎛ ⎝ z - f   (z)−z f0 ⎞ ⎠ −z f1 = z2 - f   (z) − z2 f0 − z f1. (27)


(3) Convolution theorem



Z(fn * gn) = - f   (z) - g   (z),

where


fn*gn ≡ n ∑ m=0  fn−m gm.

Footnotes: