Solving PDEs using Laplace Transforms, Chapter 15

Solving PDEs using Laplace Transforms, Chapter 15 Given a function u(x, t) defined for all t > 0 and assumed to be bounded we can apply the Laplace tr...
Author: Anthony Sparks
61 downloads 0 Views 87KB Size
Solving PDEs using Laplace Transforms, Chapter 15 Given a function u(x, t) defined for all t > 0 and assumed to be bounded we can apply the Laplace transform in t considering x as a parameter. Z ∞ e−st u(x, t) dt ≡ U (x, s) L(u(x, t)) = 0

In applications to PDEs we need the following: Z Z ∞ ∞ −st −st e ut (x, t) dt = e u(x, t) 0 + s L(ut (x, t) =



e−st u(x, t) dt = sU (x, s) − u(x, 0)

0

0

so we have L(ut (x, t) = sU (x, s) − u(x, 0) In exactly the same way we obtain L(utt (x, t) = s2 U (x, s) − su(x, 0) − ut (x, 0). We also need the corresponding transforms of the x derivatives: Z ∞ e−st ux (x, t) dt = Ux (x, s) L(ux (x, t)) = 0

Z



L(uxx (x, t)) =

e−st uxx (x, t) dt = Uxx (x, s)

0

Consider the following examples. Example 1. ∂u ∂u + = x, ∂x ∂t with boundary and initial condition u(0, t) = 0 t > 0,

x > 0, t > 0,

and u(x, 0) = 0, x > 0.

As above we use the notation U (x, s) = L(u(x, t))(s) for the Laplace transform of u. Then applying the Laplace transform to this equation we have x dU (x, s) + sU (x, s) − u(x, 0) = dx s

dU x (x, s) + sU (x, s) = . dx s



This is a constant coefficient first order ODE. We solve it by finding the integrating factor R

µ=e Thus we have

sdx

= esx

d sx x [e U (x, s)] = esx . dx s

We integrate both sides to get e−sx U (x, s) = s

Z

1



e r dr + Ce−sx . sr

We can use integration by parts to evaluate the integral: Z Z  sx 0 e sx x dx e x dx = s Z  sx  xesx e = − dx s s xesx esx − 2. s s So we have

e−sx U (x, s) = s



xesx esx − 2 s s



+ Ce−sx =

x 1 − 3 + Ce−sx . 2 s s

We can evaluate the constant C using the boundary condition 0 = U (0, s) = − so we have

1 +C s3

⇒ C=

1 s3

1 e−sx x − + . s2 s3 s3

U (x, s) = Taking the inverse Laplace transform we have u(x, t) = xt −

(t − x)2 t2 + H(t − x) 2 2

where H is the unit step function (or Heaviside function) ( 0, x < 0 H(x) = . 1, x ≥ 0 Example 2. ∂u ∂u + + u = 0, ∂x ∂t with boundary and initial condition u(0, t) = 0 t > 0,

x > 0, t > 0,

and u(x, 0) = sin(x), x > 0.

As above we use the notation U (x, s) = L(u(x, t))(s) for the Laplace transform of u. Then applying the Laplace transform to this equation we have dU (x, s) + sU (x, s) − u(x, 0) + U (x, s) = 0 ⇒ dx

dU (x, s) + (s + 1)U (x, s) = sin(x). dx

This is a constant coefficient first order linear ODE. We solve it by finding the integrating factor µ=e Thus we have

R

(s+1)dx

= e(s+1)x

 d  (s+1)x e U (x, s) = e(s+1)x sin(x). dx

We integrate both sides to get −(s+1)x

U (x, s) = e

Z e

(s+1)r

2



sin(r) dr + Ce−(s+1)x .

We can use integration by parts to evaluate the integral: Z x  (s + 1) sin(x) − cos(x) (s+1)r −(s+1)x e sin(r) dr = . e s2 + 2s + 2 0 So we have

(s + 1) sin(x) − cos(x) + Ce−(s+1)x . s2 + 2s + 2 We can evaluate the constant C using the boundary condition U (x, s) =

0 = U (0, s) =

s2

−1 +C + 2s + 2

⇒ C=

s2

1 . + 2s + 2

So we have

(s + 1) sin(x) − cos(x) + e−(s+1)x) . s2 + 2s + 2 Taking the inverse Laplace transform we have U (x, s) =

u(x, t) = e−t cos(t) sin(x) − e−t sin(t) cos(t) + e−t H(t − x) sin(t − x) This can be written as u(x, t) = e−t [sin(x − t) + H(t − x) sin(t − x)] . Example 3. ∂ 2u ∂u (x, t) = (x, t), 0 < x < 2, t > 0, ∂t ∂x2 u(0, t) = 0, u(2, t) = 0 u(x, 0) = 3 sin(2πx). Take the Laplace transform and apply the initial condition d2 U (x, s) = sU (x, s) − u(x, 0) = sU (x, s) − 3 sin(2πx). dx2 We write this equation as a non-homogeneous, second order linear constant coefficient equation for which we can apply the methods from Math 3354. d2 U (x, s) − sU (x, s) = −3 sin(2πx). dx2 The general solution can be written as U (x, s) = Uh (x, s) + Up (x, s) where Uh (x, s) is the general solution of the homogeneous problem √

Uh (x, s) = c1 e

sx

+ c2 e−



sx

and Up (x, s) is any particular solution of the non-homogeneous problem Up (x, s) = A cos(2πx) + B sin(2πx). 3

We first use the method of undetermined coefficients to find A and B. To this end we have d Up (x, s) = −2πA sin(2πx) + 2πB cos(2πx), dx d2 Up (x, s) = −(2π)2 A cos(2πx) + (2π)2 B sin(2πx). 2 dx Therefore d2 Up (x, s) − sUp (x, s) dx2 = (−(2π)2 − s)[A cos(2πx) + B sin(2πx)] = −3 sin(2πx). From this we conclude that −(s + (2π)2 )A = 0,

and

so that A = 0,

B=

− (s + (2π)2 )B = −3, 3 . s + 4π 2

Now we have the general solution √

U (x, s) = c1 e

sx



+ c2 e −

sx

+

3 sin(2πx) (s + 4π 2 )

We note the the Laplace transforms of the boundary conditions give u(0, t) = 0 ⇒ U (0, s) = 0,

and u(2, t) = 0 ⇒ U (2, s) = 0

So we have



0 = U (0, s) = c1 + c2 ,

0 = U (2, s) = c1 e

s2

+ c2 e−



s2

which gives c1 = c2 = 0 and we have U (x, s) =

3 sin(2πx). (s + 4π 2 )

To find our solution we apply the inverse Laplace transform   3 2 −1 u(x, t) = L sin(2πx) = 3e−4π t sin(2πx). 2 (s + 4π ) Just as we would have obtained using eigenfunction expansion methods. Example 4. Next we consider a similar problem for the 1D wave equation. 2 ∂ 2u 2∂ u (x, t) = c (x, t) + sin(πx), ∂t2 ∂x2 u(x, 0) = 0, ut (x, 0) = 0 u(0, t) = 0 u(1, t) = 0.

0 < x < 1, t > 0,

Taking the Laplace transform and applying the initial conditions we obtain d2 U sin(πx) sin(πx) 2 2 (x, s) = s U (x, s) − su(x, 0) − u (x, 0) − = s U (x, s) − . t dx2 s s 4

We need to solve the constant coefficient non-homogeneous ODE d2 U sin(πx) (x, s) − s2 U (x, s) = − 2 dx s Once again we know that U (x, s) = Uh (x, s) + Up (x, s) where Uh (x, s) is the general solution of the homogeneous problem Uh (x, s) = c1 esx + c2 e−sx and Up (x, s) is any particular solution of the non-homogeneous problem Up (x, s) = A cos(πx) + B sin(πx). We apply the method of undetermined coefficients to find A and B. To this end we have d Up (x, s) = −πA sin(πx) + πB cos(πx), dx d2 Up (x, s) = −π 2 A cos(πx) + π 2 B sin(πx). dx2 Therefore d2 Up (x, s) − s2 Up (x, s) 2 dx = (−π 2 − s2 )[A cos(πx) + B sin(πx)] sin(πx) . =− s From this we conclude that −(s2 + π 2 )A = 0,

and

1 − (s2 + π 2 )B = − , s

so that A = 0,

B=

So we have Up (x, s) =

s(s2

1 . + π2)

sin(πx) s(s2 + π 2 )

and U (x, s) = c1 esx + c2 e−sx +

sin(πx) . s(s2 + π 2 )

Next we apply the BCs to find c1 and c2 . 0 = U (0, s) = c1 + c2 ,

and 0 = U (1, s) = c1 es + c2 e−s

which implies c1 = 0 and c2 = 0. So we arrive at U (x, s) =

sin(πx) . s(s2 + π 2 ) 5

Finally we apply the inverse Laplace transform to obtain   1 −1 −1 sin(πx) u(x, t) = L (U (x, s)) = L s(s2 + π 2 )   1 −1 1 s = 2 L − sin(πx) π s (s2 + π 2 ) 1 = 2 (1 − cos(πt)) sin(πx). π Here we have done partial fractions 1 a bs + c 1 = + 2 = 2 2 2 2 s(s + π ) s (s + π ) π



1 s − 2 s (s + π 2 )

 .

Example 5. This example shows the real use of Laplace transforms in solving a problem we could not have solved with our earlier work. ∂ 2u ∂u (x, t) = (x, t), ∂t ∂x2 u(x, 0) = f (x) u(x, t) bounded.

−∞ < x < ∞, t > 0,

Under the assumption that u(x, t) is bounded we know that the Laplace transform exists and, indeed, we have Z ∞ Z ∞ M −st e−st dt = e |u(x, t)| dt ≤ M |u(x, t)| ≤ M ⇒ |U (x, s)| ≤ . s 0 0 Applying the Laplace transform we obtain d2 U (x, s) = sU (x, s) − u(x, 0) = sU (x, s) − f (x). dx2 We write this equation as a non-homogeneous, second order linear constant coefficient equation. d2 U (x, s) − sU (x, s) = −f (x). dx2 The general solution can be written as U (x, s) = Uh (x, s) + Up (x, s) where Uh (x, s) is the general solution of the homogeneous problem √

Uh (x, s) = c1 e

sx

+ c2 e−



sx

and Up (x, s) is any particular solution of the non-homogeneous problem. We find it using √the √ sx method of variation of parameters from Math 3354. For this method we use U1 = e , U2 = e− sx . U1 (x, s) U2 (x, s) √ = −2 s W (U1 , U2 ) = 0 0 U1 (x, s) U2 (x, s)

6

Z

x

[−U1 (x, s)U2 (ξ, s) + U2 (x, s)U1 (ξ, s])(−f (ξ)) dξ W (ξ, s) 0 Z xh √ √ √ √ i 1 sx − sξ − sx −e e = √ +e e sξ f (ξ) dξ 2 s 0 √ √ Z Z e sx x −√sξ e− sx x √sξ =− √ e e f (ξ) dξ f (ξ) dξ + √ 2 s 0 2 s 0

Up (x, s) =

So the general solution can be written as     Z x √ Z x √ √ √ 1 1 − sξ sx sξ e e f (ξ) dξ e− sx . U (x, s) = c1 − √ f (ξ) dξ e + c2 + √ 2 s 0 2 s 0 Recall our assumption that u(x, t) be bounded for all −∞ < x < ∞ implies that U (x, s) is also bounded for all −∞ < x < ∞ for any fixed s > 0. Now in order that the first term in the general solution stays bounded as x → ∞ we need   Z x √ 1 − sξ f (ξ) dξ = 0 e lim c1 − √ x→∞ 2 s 0 which implies 1 c1 = √ 2 s

Z





e−



f (ξ) dξ.

0

In exactly the same way we must have   Z x √ 1 sξ lim c2 + √ e f (ξ) dξ = 0 x→−∞ 2 s 0 which implies −∞ √

Z

1 c2 = − √ 2 s

e 0



1 f (ξ) dξ = √ 2 s

Z

Z

x

0

e

√ sξ

f (ξ) dξ.

−∞

Thus  U (x, s) =

1 √ 2 s 

+

Z



√ − sξ

e 0

1 √ 2 s

Z

0

√ − sξ

e −∞

1 f (ξ) dξ − √ 2 s

e

√ − sξ





f (ξ) dξ e

sx

0

1 f (ξ) dξ + √ 2 s

Z

x √

e







f (ξ) dξ e−

0

√   −√sx Z x  Z √ e sx ∞ −√sξ e sξ √ √ e f (ξ) dξ + e f (ξ) dξ = 2 s x 2 s −∞ Z ∞ √ 1 = √ e− s|x−ξ| f (ξ) dξ 2 s −∞



We want to find the inverse Laplace transform  −√s|x−ξ|  e −1 √ L . 2 s 7

sx

From our table we have L

−1



2

e−a /(4t) = √ πt



e−a s √ s



e−|x−ξ| /(4t) √ = ≡ K(|x − ξ|, t). 4πt



and if we set a = |x − ξ| then we have −1



L



e− s|x−ξ| √ 2 s

2

So we have −1

u(x, t) = L (U (x, s)) = L Z



=

L −∞

−1





−1

e− s|x−ξ| √ 2 s



1 √ 2 s

Z

f (ξ) dξ

Z ∞ 1 2 =√ e−|x−ξ| /(4t) f (ξ) dξ 4πt −∞ Z ∞ = K(|x − ξ|, t) f (ξ) dξ −∞

2

e−x /(4t) K(x, t) = √ 4πt is called the “Fundamental Heat Kernel”.

8

√ − s|x−ξ|

e −∞



The function



 f (ξ) dξ

Table of Laplace Transforms Z f (t) for t ≥ 0



fb = L(f ) =

e−st f (t) dt

0

1

1 s

eat

1 s−a n!

tn

sn+1

(n = 0, 1, . . .)

ta

Γ(a + 1) (a > 0) sa+1

sin bt

b s 2 + b2

cos bt sinh bt

s2

s + b2

s2

a − b2

cosh bt

s s 2 − b2

f 0 (t)

sL(f ) − f (0)

f 00 (t)

s2 L(f ) − sf (0) − f 0 (0) (−1)n

tn f (t)

dn F (s) dsn

eat f (t)     0 t≤a u(t − a) =    1 t>a

L(f )(s − a)

u(t − a)f (t − a)

e−as L(f )(s)

u(t − a)g(t)

e−as L(g(t + a))(s)

δ(t − a)

e−as

Z (f ∗ g)(t) =

e−as s

t

f (t − τ )g(τ ) dτ 0

9

L(f ∗ g) = L(f )L(g)

The “error function” denoted by erf(x) is given by Z x 2 2 erf(x) = √ e−x dx. π 0 Notice that we can use the properties of integrals to deduce that erf(−x) = − erf(x). The complementary error function erfc(x) defined by Z ∞ 2 2 erfc (x) = √ e−s ds. π x Notice that

Z

2 erf(x) + erfc(x) = √ π

x

−x2

e

Z



dx +

0

e

−s2

 ds = 1.

x

Additional Laplace Transforms √

2

e−a /(4t) √ πt

e−a s √ s

2

ae−a /(4t) √ 2 πt3



e−a es

erf (t)  erfc r 2

e

b2 t+ab

−e

a √ 2 t



2 /4



e−a s



   t −a2 /(4t) a e − a erfc √ π 2 t

b2 t+ab

s

  √ a erfc b t + √ 2 t

     √ a a erfc b t + √ + erfc √ 2 t 2 t

10

erfc(s/2) s s



e−a s √ s s √

e−a s √ √ s( s + b) √

be−a s √ s( s + b)