Question

Let \( u(x,t) \) be the solution of the initial value problem for the heat equation on the real line:
\[ \frac{\partial u}{\partial t} = k\, \frac{\partial^2 u}{\partial x^2}, \qquad -\infty< x< \infty, \qquad t>0, \qquad k\in\mathbb{R} \]
with the initial condition
\[ u(x,0) = e^{-a|x|}, \qquad a>0. \]
If \( \hat u(w,t) = \displaystyle\int_{-\infty}^{\infty} u(x,t)\, e^{iwx}\, dx \) is the Fourier transform of \( u(x,t) \) with respect to \( x \), then \( \hat u(w,t) \) is equal to

• A. \[ \dfrac{2a}{a^2+w^2}\, e^{-kw^2t} \]
• B. \[ \dfrac{a}{a^2+w^2}\, e^{-kw^2t} \]
• C. \[ \dfrac{2a}{a^2+w^2}\, e^{-ka^2t} \]
• D. \[ \dfrac{1}{\sqrt{4\pi kt}}\, e^{\frac{-w^2}{4kt}} \]

Hint:

Transform the heat equation in x first to get a simple ODE in t, then separately transform the initial condition e^{-a|x|}.

Answer 1

The Correct Option is A

Here is a different way to reach the same transform, by first writing the solution in real space as a convolution and then using the convolution theorem, instead of transforming the PDE directly.

The general solution of the heat equation $u_t=ku_{xx}$ on the whole real line, with initial data $u(x,0)=g(x)$, is the convolution of $g$ with the heat kernel:

\[ u(x,t) = \int_{-\infty}^{\infty} g(y)\, K(x-y,t)\, dy, \qquad K(x,t)=\dfrac{1}{\sqrt{4\pi kt}}\,e^{-x^2/(4kt)} \]

Taking the Fourier transform of a convolution turns it into a product of the two individual transforms, so

\[ \hat u(w,t) = \hat g(w)\cdot \hat K(w,t) \]

The Fourier transform of the heat kernel $K(x,t)$, with the convention $\hat f(w)=\int f(x)e^{iwx}dx$, works out to $\hat K(w,t)=e^{-kw^2t}$; this is the standard fact that a Gaussian in $x$ transforms into a decaying exponential in $w$.

Now find $\hat g(w)$, the transform of the initial data $g(x)=e^{-a|x|}$. Split the integral at $x=0$:

\[ \hat g(w) = \int_0^\infty e^{-ax}e^{iwx}dx + \int_{-\infty}^0 e^{ax}e^{iwx}dx = \dfrac{1}{a-iw}+\dfrac{1}{a+iw} = \dfrac{2a}{a^2+w^2} \]

Multiplying the two pieces together:

\[ \hat u(w,t) = \dfrac{2a}{a^2+w^2}\cdot e^{-kw^2t} \]

Let's check this against the other options. Option B is missing the factor of $2$ that comes from adding the two halves of the integral for $g$. Option C puts $a^2$ in the exponent instead of $w^2$, but the exponential decay in time must depend on $w$ (the Fourier variable dual to $x$), not on $a$ (which only measures how fast the initial condition itself decays). Option D is just $\hat K(w,t)$ alone, the transform for a point source (delta function) initial condition; it has no $a$ in it at all, so it cannot be the right answer once the actual initial data $e^{-a|x|}$ is folded in.

Let's summarize:

• $u(x,t)$ is the convolution of the initial data with the heat kernel, so its transform is the product of the two transforms.
• $\hat g(w) = \dfrac{2a}{a^2+w^2}$ and $\hat K(w,t)=e^{-kw^2t}$, giving $\hat u(w,t)=\dfrac{2a}{a^2+w^2}e^{-kw^2t}$.

So the correct expression is $\hat u(w,t) = \dfrac{2a}{a^2+w^2}e^{-kw^2t}$.

\[ \boxed{\hat u(w,t)=\dfrac{2a}{a^2+w^2}\,e^{-kw^2t}} \]