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.
