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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 ole.gif x ole1.gif 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



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



3. Laplace transform of ∂2U/∂t2. The Laplace transform of ∂U2/∂t2 is given by





4. Laplace transform of ∂2U/∂x2. The Laplace transform of ∂U2/∂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 partial differential equation into an ordinary differential equation.

To solve 1) multiply both sides by the integrating factor


This gives


which can be written


Integration gives




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 c1 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 c1 =0, c2 = 0. Thus 2) becomes


Inversion gives 


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


  Murray R. Spiegel. Laplace Transforms. (Schaum)

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