Laplace transform is useful in equation solving. By definition, Laplace transform transforms a function defined on into another function by calculating the convolution

or symbolically,

A table of important Laplace transforms can be found on mathworld.wolfram.com. Here we steal some of the commonly used.

It can also be applied to differentials.

Laplace Transform of Differentials

The general form is

Integrals with upper limit as argument also transform nicely.

Laplace transform is useful in solving differential equations because it transforms many equations into fractions and polynomials. A simple example is the harmonic osicllators. The equation of motion is

which is transformed into

The solution to it is

We can spot sin and cos from the solution, or more generally perform an inverse Laplace transform,

This example is too simple sometimes naive. However, it shows the spirit.

Caveats

Laplace transform has a lot of counter intuitive expressions.

Laplace transform of product of two functions is

**NOT**the product of the Laplace transforms . However, if one of the functions is a constant, say , we can prove that the Laplace transform of is .The product of two Laplace transforms is the Laplace transform of a convolution

Small s corresponds to large x, due to the nature of the exponential suppression in Laplace transform. For example, for small argument x, the function becomes almost 1, meanwhile, the Laplace transform of the function becomes under large s. We can see that the two limits are consistant since the Laplace transform of 1 is .

In so many circumstance the Laplace transform doesn’t exist simple because the integral doesn’t converge. Please beware of this and use the transform only when it exists.

Inverse Laplace Transform

We do not usually use the general form of inverse Laplace transform since we can find it in the table. Nevertheless we write it down here.

By defining , we can rewrite the formula

The more interesting application is to solve matrix differential equations. For any equations

Laplace transform takes it to the form

The solution is

So the final solution for is

We could work out the Taylor expansion of solution,

The inverse Laplace transform can be done simply term by term,

Finally we obtain the formal solution of the system, which is

Only Works for Constant Coefficients

This result only works for constant coefficients. In general, if the matrix depends on the argument , the solution can be systematically calculated using the so called Magnus Expansion. However, it is as tedious as a numerical solution.

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