using the Fourier transform, solve

Question

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

Answer 1

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_0\Big(h(x+a)-h(x-a)\Big) \end{cases} where $h(\cdot)$ is the unit step and it is given that $|u|<M$ where $M\in\mathbb{R}$ is some bound on $u$.

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

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

Evaluating this transform directly as you did gives

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

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

So the problem becomes

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

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

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

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

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

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

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.

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

Alternatively, the solution may be expressed as a convolution.

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

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

To begin, let

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

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

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

So

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

Now let

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

Note that

$$ \begin{align} \frac{\partial g}{\partial x}(x,t) &= -\frac{1}{\pi}\int_0^\infty \lambda e^{-t\lambda^2}\sin x\lambda \, d\lambda \\ &= \frac{1}{2\pi t}\int_0^\infty \sin x\lambda \, de^{-t\lambda^2} \\ &= \frac{1}{2\pi t}\left(\sin x\lambda \, e^{-t\lambda^2}\Big\vert_0^\infty - \int_0^\infty x\cos x\lambda \, e^{-t\lambda^2} \, d\lambda\right) \\ &= -\frac{x}{2\pi t}\int_0^\infty e^{-t\lambda^2} \cos x\lambda \, d\lambda \\ &= -\frac{x}{2t}g(x,t). \end{align} $$

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

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

The solution is readily obtainable as

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

Then we have the solution

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

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 $0$. $$ \begin{align} u(x,t) &= \frac{u_0}{4\sqrt{\pi t}}\int_{-a}^a \big(\operatorname{sgn}(a + s) + \operatorname{sgn}(a - s)\big)e^{-\frac{(x-s)^2}{4t}} \, ds \\ &= \frac{u_0}{2\sqrt{\pi t}} \int_{-a}^a e^{-\frac{(x-s)^2}{4t}} \, ds \\ &= \frac{u_0}{2}\left(\operatorname{erf}\left(\frac{a + x}{2\sqrt{t}}\right) + \operatorname{erf}\left(\frac{a - x}{2\sqrt{t}}\right) \right) \end{align} $$

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

$\hskip 1 in$