Thayer Watkins
Silicon Valley
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The Solution to
the Differential Equation
u"(x) − u(x) = f(x)

The differential equation

uxx - u(x) = f(x)

is a linear inhomogeneous ordinary differential equation. Its solution can be obtained most easily using Fourier or Laplace Transforms.

Taking the Laplace Transform of the above equation gives:

(s²−1)U = F

where U is the Laplace transform of u(x) and F is the Laplace transform of f(x) and where it is assumed that u(0)=0 and u'(0)=0.

This means that

U = (s²−1)-1F = −½[1/(S+1) - 1/(S-1)]F


The Laplace transform of sinh(x) is equal to −½[1/(S+1) - 1/(S-1)] so

U = Lsinh(x)F.

By the convolution theorem then

u(x) = ∫0xsinh(x−z)f(z)dz

To verify that this a solution, first

u'(x) = sinh(0)f(x) + ∫0xcosh(x−z)f(z)dz
and since sinh(0)=0 this reduces to
u'(x) = ∫0xcosh(x−z)f(z)dz

The second derivative is given by

u"(x) = cosh(0)f(x) + ∫0xsinh(x−z)f(z)dz
and since cosh(0)=1
u"(x) = f(x) + ∫0xsinh(x−z)f(z)dz
and since
and therefore
u"(x)−u(x) = f(x).

The lower limit of 0 for the integral is arbitrary, stemming from the way the Laplace transform has to be defined. It is appropriate to take the lower limit as −∞ and to take u(−∞)=0 and u'(−∞)=0. Thus the solution is

u(x) = ∫−∞xsinh(x−z)f(z)dz

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