Question

Textbook 'Applied Partial Differential Equations Revised Edition' by Jhon Ockenden, Sam Howison, Andrew Lacey, and Alexander

Show that Lespl= 14 + i:(8 – c)]dx = Vesp [.(4.70*] where i = V-1. Hint: (a) Use the Euler's formula exp[it] = cost + i sint; then apply the property of even and odd functions to simplify the integral. (b) Define = and 2 = rewrite the above integral in terms of u and 2 (times some function in t). (c) Define the integral in (b) (without the function in t) as 1(2); use integration by parts to obtain an ODE for I(2). (a) Compute (0) and solve the initial value problem for 1(2).

Answer 1

a.)

Consider the given integrable function as

exp(t+i-)) ete-) = et (cos(A(-0)+i sin(A(r-) -12t

The integration can be separated as

Atcos(A(r-) dA+i -12t e exp(-t+iX(-) dA = sin(A(r OS dA

Since is an even function, the first integral is

tcos(A(x-) e

t cos(A-) dA At cos(A( )) dA 2 COs e

Since is an odd function, the second integral is

sin((-) e

t sin((r - ) dA 0 e

So the the integral we need to find is

e-t cos(A(r )) dX 2

b.)

Define .

vdp and A

By substituting the above transformations into the integral, we get

.

2 e-At cos(A(r ) d e cos(uz) du 2 COS

c.)

Let

e cos(uz) du

Let us find the derivative of I with respect to z, i.e.,

$\frac{dI(z)}{dz}=\int_{0}^{\infty}e^{-\mu^2}\frac{d}{dz}(\cos(\mu z))\;d\mu=-\int_{0}^{\infty}\mu e^{-\mu^2}\sin(\mu z)\;d\mu$

Use the integration by parts to I, then we get

eHcos(z) du sin(uz) I (z) sin(uz) ) du e 2 2

()2dI z dz I 0 dz

Solution of this first order ODE is

$I(z)=I(0)e^{-\int \frac{z}{2}\;dz}=I(0)e^{-\frac{z^2}{4}}$

d.)

$I(0)=\int_{0}^{\infty}e^{-\mu^2}\cos(\mu (0))\;d\mu=\int_{0}^{\infty}e^{-\mu^2}\;d\mu$

Define $w=\mu^2\implies dw=2\mu\;d\mu\implies 2w^{1/2}\;d\mu$ . Substitute these into the integral to get

$I(0)=\frac{1}{2}\int_{0}^{\infty}e^{-w}w^{-1/2}\;dw=\frac{\Gamma(1/2)}{2}=\frac{\sqrt{\pi}}{2}$

The integral value is

$\frac{2}{\sqrt{t}}\frac{\sqrt{\pi}}{2}e^{-\frac{z^2}{4}}=\sqrt{\frac{\pi}{t}}exp(-\frac{(x-\xi)^2}{4t})$