Vrftsjtryk

using the Fourier transform, solve
$$dfrac{partial u}{partial t}=frac{partial^{2}u}{partial x^{2}},hspace{3mm}-infty<x<infty,hspace{2mm}t>0 \u(x,0)=left{begin{array}{lll}u_{0},&|x|<a\0,&|x|geq aend{array}right.hspace{4mm}|u(x,t)|<M$$

I know this question is old now but perhaps somebody else may use it later too. I may rewrite the problem as
begin{cases}
displaystyle frac{partial u}{partial t} = frac{partial^2 u}{partial x^2} \
u(x,0) = u_0Big(h(x+a)-h(x-a)Big)
end{cases}
where $h(cdot)$ is the unit step and it is given that $|u|<M$ where $Minmathbb{R}$ is some bound on $u$.

Applying the transform as $mathcal{F}[u(x,t)]=U(lambda,t)$,

$$mathcal{F}big[u_t-u_{xx}big]=frac{dU}{dt}-(-ilambda)^2 U=0$$
and
$$mathcal{F}big[u(x,0)big] = u_0 int_{-infty}^infty big(h(x+a) - h(x-a)big)e^{ilambda x}dx = u_0 int_{-a}^a e^{ilambda x}dx.$$

Evaluating this transform directly as you did gives

$$
begin{align}
mathcal{F}big[u(x,0)big] &= u_0 frac{e^{ilambda a} - e^{-ilambda a}}{ilambda} \
&= u_0frac{coslambda a + isinlambda a - coslambda a + isinlambda a}{ilambda} \
&= frac{2u_0}{lambda}sin lambda a.
end{align}
$$

Alternatively, we can say that because the limits are symmetric and $sinlambda x$ is odd in $x$ and $cos lambda x$ is even in $x$,
$$mathcal{F}big[u(x,0)big] = 2u_0 int_0^a coslambda x dx = frac{2u_0}{lambda}sinlambda a.$$

So the problem becomes

begin{cases}
U'=-lambda^2 U \
displaystyle U(lambda,0) = frac{2u_0}{lambda}sinlambda a
end{cases}

Which has the solution
$$U(lambda, t) = frac{2u_0}{lambda}sinlambda a e^{-lambda^2 t}.$$

Now the solution in the time domain is simply the inverse transform of this expression.

$$u(x,t) = frac{1}{2pi} int_{-infty}^infty frac{2 u_0}{lambda}sinlambda a , e^{-lambda^2 t} e^{-i lambda x} dlambda$$

Multiplying by $a/a$ helps us see that this integral definitely exists.

$$u(x,t) = frac{u_0 a}{pi} int_{-infty}^infty frac{sin a lambda}{a lambda} e^{-lambda^2 t} e^{-i lambda x} dlambda$$

Under the same reasoning, $frac{sin a lambda}{a lambda} e^{-lambda^2 t}$ is even in $lambda$, so the $sin -lambda x$ term contributes nothing to the integral, and still $cos lambda x$ is even.

$$u(x,t) = frac{2u_0 a}{pi} int_0^infty frac{sin a lambda}{a lambda} cos x lambda , e^{-tlambda^2} , dlambda$$

Alternatively, the solution may be expressed as a convolution.

$$u(x,t)=2u_0a , frac{1}{2pi}int_{-infty}^inftyfrac{sin alambda}{alambda}e^{-i x lambda} , dlambda , * , frac{1}{2pi}int_{-infty}^infty e^{-t lambda^2} e^{-i x lambda} , dlambda $$

Evaluating the convolution involves evaluating each involved inverse Fourier transform independently.

To begin, let

$$
begin{align}
f(x) &= frac{1}{2pi}int_{-infty}^inftyfrac{sin alambda}{alambda}e^{-i x lambda} , dlambda \
&= frac{1}{pi}int_0^infty frac{sin alambda}{alambda}cos xlambda , dlambda \
&= frac{1}{2pi a}int_0^infty frac{sinlambda(a + x) + sinlambda(a - x)}{lambda} , dlambda \
&= frac{1}{2pi a}left((a + x)int_0^infty frac{sinlambda(a + x)}{lambda(a + x)} , dlambda + (a - x)int_0^infty frac{sinlambda(a - x)}{lambda(a - x)} , dlambdaright) \
&= frac{1}{2pi a}left((a + x)int_0^infty operatorname{sinc}lambda(a + x) , dlambda + (a - x)int_0^infty operatorname{sinc}lambda(a - x) , dlambdaright)
end{align}
$$

Knowing that $operatorname{sinc}y = operatorname{sinc}-y$ and

$$int_{0}^infty operatorname{sinc}by , dy = frac{pi}{2|b|},$$

we know the value of these integrals

$$
begin{align}
int_0^infty operatorname{sinc}lambda(a + x) , dlambda &=
begin{cases}
displaystyle -frac{pi}{2(a + x)}, &a + x < 0\
infty, &a + x = 0\
displaystyle frac{pi}{2(a + x)}, &a + x > 0\
end{cases} \
&= frac{pi}{2(a + x)}operatorname{sgn}(a + x) \
\
int_0^infty operatorname{sinc}lambda(a - x) , dlambda &= frac{pi}{2(a - x)}operatorname{sgn}(a - x)
end{align}
$$

So

$$f(x) = frac{operatorname{sgn}(a + x) + operatorname{sgn}(a - x)}{4a}.$$

Now let

$$g(x,t) = frac{1}{2pi}int_{-infty}^infty e^{-t lambda^2} e^{-i x lambda} dlambda = frac{1}{pi}int_0^infty e^{-t lambda^2} cos xlambda , dlambda.$$

Note that

$$
begin{align}
frac{partial g}{partial x}(x,t) &= -frac{1}{pi}int_0^infty lambda e^{-tlambda^2}sin xlambda , dlambda \
&= frac{1}{2pi t}int_0^infty sin xlambda , de^{-tlambda^2} \
&= frac{1}{2pi t}left(sin xlambda , e^{-tlambda^2}Bigvert_0^infty - int_0^infty xcos xlambda , e^{-tlambda^2} , dlambdaright) \
&= -frac{x}{2pi t}int_0^infty e^{-tlambda^2} cos xlambda , dlambda \
&= -frac{x}{2t}g(x,t).
end{align}
$$

Knowing that $e^{-t lambda^2} = e^{-t (-lambda)^2}$ and

$$int_{-infty}^infty e^{-by^2} , dy = sqrt{frac{pi}{b}},$$

we know $g(x,t)$ if we can solve

$$
begin{cases}
displaystyle frac{partial g}{partial x}(x,t) = -frac{x}{2t}g(x,t) \
displaystyle g(0,t) = frac{1}{2sqrt{pi t}}.
end{cases}
$$

The solution is readily obtainable as

$$g(x,t) = frac{e^{-frac{x^2}{4t}}}{2sqrt{pi t}}.$$

Then we have the solution

$$u(x,t) = 2 u_0 aint_{-infty}^infty f(s)g(x-s,t) , ds = frac{u_0}{4sqrt{pi t}}int_{-infty}^infty big(operatorname{sgn}(a + s) + operatorname{sgn}(a - s)big)e^{-frac{(x-s)^2}{4t}} , ds.$$

If $|s| > a$, then the integrand is exactly $0$. Otherwise each $operatorname{sgn}$ function evaluates to $1$ except at $|s| = a$ where one is $1$ and the other is $0$.
$$
begin{align}
u(x,t) &= frac{u_0}{4sqrt{pi t}}int_{-a}^a big(operatorname{sgn}(a + s) + operatorname{sgn}(a - s)big)e^{-frac{(x-s)^2}{4t}} , ds \
&= frac{u_0}{2sqrt{pi t}} int_{-a}^a e^{-frac{(x-s)^2}{4t}} , ds \
&= frac{u_0}{2}left(operatorname{erf}left(frac{a + x}{2sqrt{t}}right) + operatorname{erf}left(frac{a - x}{2sqrt{t}}right) right)
end{align}
$$

As $operatorname{erf}y$ is bounded, the solution $u(x,t)$ is bounded.

$hskip 1 in$