Laplace Transform of Partial Derivatives

Laplace transform of partial derivatives. Applications of the Laplace transform in solving partial differential equations. Laplace transform of partial derivatives. Theorem 1. Given the function U(x, t) defined for a

x

b, t > 0. Let the Laplace transform of U(x, t) be

We then have the following:

1. Laplace transform of ∂U/∂t. The Laplace transform of ∂U/∂t is given by

Proof 

2. Laplace transform of ∂U/∂x. The Laplace transform of ∂U/∂x is given by

Proof 

3. Laplace transform of ∂2U/∂t2. The Laplace transform of ∂U 2/∂t2 is given by

where

Proof 

4. Laplace transform of ∂2U/∂x2. The Laplace transform of ∂U 2/∂x2 is given by

Extensions of the above formulas are easily made.

Example 1. Solve

which is bounded for x > 0, t > 0.

Solution. Taking the Laplace transform of both sides of the equation with respect to t, we obtain

Rearranging and substituting in the boundary condition U(x, 0) = 6e

-3x

, we get

 Note that taking the Laplace transform has transformed the par tial differential equation into an ordinar y differential equation.

To solve 1) multiply both sides by the integrating factor

This gives

which can be written

Integration gives

or

 Now because U(x, t) must be bounded as x → ∞, we must have u(x, s) also bounded as x → ∞. Thus we must choose c = 0. So

and taking the inverse, we obtain

Example 2. Solve

with the boundary conditions

U(x, 0) = 3 sin 2πx U(0, t) = 0 U(1, t) = 0

where 0 < x < 1, t > 0.

Solution. Taking the Laplace transform of both sides of the equation with respect to t, we obtain

Substituting in the value of U(x, 0) and rearranging, we get

where u = u(x, s) = L[U(x, t]. The general solution of 1) is

We now wish to determine the values of c 1 and c2. Taking the Laplace transform of those boundary conditions that involve t, we obtain 3)

L[U(0, t)] = u(0, s) = 0

4)

L[U(1, t)] = u(1, s) = 0

Using condition 3) [u(0, s) = 0] in 2) gives

5)

c1 + c2 = 0

Using condition 4) [u(1, s) = 0] in 2) gives

From 5) and 6) we find c 1 =0, c2 = 0. Thus 2) becomes

Inversion gives

For more examples see Murray R. Spiegel. Laplace Transforms. (Schaum). Chap. 3, 8.