Question

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Answer 1

First let us try to prove the formula. As given in the hint we can replace
(r) cos COS
by
elir)
because

$\int_{-\infty}^{\infty} e^{-\alpha x^2} \cos{(x)} dx = \mathbb{Re}\left[ \int_{-\infty}^{\infty} e^{-\alpha x^2}\cdot e^{ix}dx\right ]$

where $\mathbb{R}$ denotes real part of a complex number.

Hence we evaluate the integral $\int_{-\infty}^{\infty} e^{-\alpha x^2}\cdot e^{ix}$

We have

$\int_{-\infty}^{\infty} e^{-\alpha x^2}\cdot e^{ix}dx = \int_{-\infty}^{\infty} e^{-\alpha x^2+ix}dx$

we use the fact that

$x^2 -\frac{ix}{\alpha} = \left(x-\frac{i}{\alpha}\right)^2 + \frac{1}{4\alpha^2}$

since $i^2=-1$

thus we have

$\int_{-\infty}^{\infty} e^{\left( x^2 -\frac{ix}{\alpha}\right ) } dx= e^{-1/4\alpha}\int_{-\infty}^{\infty}e^{-\alpha \left( x-\frac{i}{\alpha}\right )^2} dx$

Let

$I = \left[\int_{-\infty}^{\infty}e^{-\alpha\left(x-\frac{i}{\alpha} \right )^2} \right ]dx$

take

y = (z-a)

Then

$I =\int_{-\infty -\frac{i}{\alpha}}^{\infty-\frac{i}{\alpha}}e^{-\alpha y^2} dy$

The above integral is calculated in complex z- plane with along the line $y = - \alpha$ , where z = (x,y) is a complex no in the plane. In order to calculate this integral we use the property integral of an analytic function
g(e)
along a line curve in the z-plane such that there are no singularities for
g(e)
in the interior domain of i.e. $\int_{C} g(z)dz = 0$ .

Let $g(z) = e^{-\alpha z^2}$

Now we must choose judiciously.

Let C be a rectangular loop with four parts $C_1$,$C_2$,$C_3$,$C_4$

$C_1 \rightarrow z=-R\ to \ z=R$

$C_2 \rightarrow z=R\ to \ z=R-\frac{i}{\alpha}$

$C_3 \rightarrow z=R-\frac{i}{\alpha}\ to \ z=-R-\frac{i}{\alpha}$

$C_4 \rightarrow z=-R-\frac{i}{\alpha}\ to \ z=-R$

Here R is a purely real number. We are interested in $C_3$.

We have

$\int_{C_1 +C_2 + C_3 + C_4} g(z) dz = 0$

$\implies \int_{C_3 } g(z) dz = -\int_{C_1 +C_2 + C_4} g(z) dz$

Moreover, the integral along $C_2$ & $C_4$ is merely same except for a change of sign

Hence,

$\int_{C_2 + C_4} g(z) dz = 0$

Thus

$\implies \int_{C_3 } g(z) dz = -\int_{C_1} g(z) dz$

Note that if we took $R \rightarrow \infty$ , we will recover, our integral .

Doin that we have

$\int_{C_3}g(z)dz=I = \int_{-\infty-\frac{i}{\alpha}}^{\infty-\frac{i}{\alpha}}e^{-\alpha z^2} dz = -\int_{C_1}g(z)dz =- \int_{\infty}^{-\infty}e^{-\alpha z^2} dz = \int_{-\infty}^{\infty}e^{-\alpha z^2} dz$

Now replace the extreme rhs integral is merely the value of

$\frac{\Gamma (\frac{1}{2})}{\sqrt{\alpha}} = \sqrt{\frac{\pi}{\alpha}}$

Since this is a real number

we have

$\int_{-\infty}^{\infty} e^{-\alpha x^2} \cos{(x)} dx = \mathbb{Re}\left[ \int_{-\infty}^{\infty} e^{-\alpha x^2}\cdot e^{ix}dx\right ]$

$\implies \mathbb{R}\left[ \int_{-\infty}^{\infty} e^{\alpha x^2-ix} dx \right ] = \sqrt{\frac{\pi}{\alpha}}e^{-1/4\alpha}$

Now lets go back to our heat equation :

$u_t = \Delta u + e^t$

Assume solution in variable separated form as

$u = \cos{x_1}\sin{x_2}f(t)$

Where f(t) is the function depended on variable t only

Hence ,

$\Delta u = -f(t)\left[ \cos{x_1}\sin{x_2} + \cos{x_1}\sin{x_2} \right ] = -2f(t) \cos{x_1}\sin{x_2}}$

$u_t = f'(t)\cos{x_1}\sin{x_2}$

Puttting these values back into the heat equation we have

$u_t - \Delta u = e^t$

$\implies f'(t) +2f(t) = \frac{e^t}{\cos{x_1}\sin{x_2}}$

Multiply $e^{2t}$ on both sides

$\implies e^{2t} \left( f'(t) +2f(t) \right ) = \frac{e^{3t}}{\cos{x_1}\sin{x_2}}$$\implies \frac{d \left( e^{2t} f(t)) \right )}{dt}= \frac{e^{3t}}{\cos{x_1}\sin{x_2}}$

Hence the general solution is

$e^{2t} f(t)= \frac{e^{3t}}{3\cos{x_1}\sin{x_2}} + K$

where K is the arbitrary constant dependent on initial condition

For initial condition we have

$u (x_1,x_2,0) = \cos{x_1}\sin{x_2} \implies f(0) = 1$

This gives K as

$K = 1 -\frac{1}{3\cos{x_1}\sin{x_2} }$

Final solution

$u(x_1,x_2,t) = \cos{x_1}\sin{x_2}\left( \frac{e^{-t}}{3\cos{x_1}\sin{x_2}} + e^{-2t} -\frac{e^{-2t}}{3\cos{x_1}\sin{x_2} } \right )$