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
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
\[ \frac{\partial u}{\partial t} = k\, \frac{\partial^2 u}{\partial x^2}, \qquad -\infty< x< \infty, \qquad t>0, \qquad k\in\mathbb{R} \]
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
Show Hint
Transform the heat equation in x first to get a simple ODE in t, then separately transform the initial condition e^{-a|x|}.
Updated On: Jul 21, 2026
- \[ \dfrac{2a}{a^2+w^2}\, e^{-kw^2t} \]
- \[ \dfrac{a}{a^2+w^2}\, e^{-kw^2t} \]
- \[ \dfrac{2a}{a^2+w^2}\, e^{-ka^2t} \]
- \[ \dfrac{1}{\sqrt{4\pi kt}}\, e^{\frac{-w^2}{4kt}} \]
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question.
We are given the heat equation and an initial condition, and need the Fourier transform of the solution \( u(x,t) \) with respect to \( x \), using the convention \( \hat u(w,t)=\int u(x,t)e^{iwx}\,dx \).
Step 2: Key Formula or Approach.
With this convention, taking the Fourier transform turns \( \partial_x \) into multiplication by \( -iw \), so \( \partial_x^2 \) becomes multiplication by \( (-iw)^2=-w^2 \). Taking the transform of both sides of the heat equation in \( x \) turns the PDE into an ordinary differential equation in \( t \) for each fixed \( w \). We also need the Fourier transform of the initial data \( e^{-a|x|} \).
Step 3: Detailed Explanation.
Applying the Fourier transform in \( x \) to both sides of \( u_t = k u_{xx} \) gives
\[ \frac{\partial \hat u}{\partial t} = k(-w^2)\hat u = -kw^2 \hat u \]
This is a simple first order linear ODE in \( t \) (with \( w \) fixed as a parameter), whose solution is
\[ \hat u(w,t) = \hat u(w,0)\, e^{-kw^2 t} \]
Now find \( \hat u(w,0) \), the transform of \( u(x,0)=e^{-a|x|} \). Split the integral at \( x=0 \):
\[ \hat u(w,0) = \int_0^{\infty} e^{-ax}e^{iwx}\,dx + \int_{-\infty}^0 e^{ax}e^{iwx}\,dx = \frac{1}{a-iw}+\frac{1}{a+iw} = \frac{2a}{a^2+w^2} \]
Combining the two results,
\[ \hat u(w,t) = \frac{2a}{a^2+w^2}\, e^{-kw^2t} \]
which is option (A). Option (B) is missing the factor of \( 2 \) that comes from adding the two halves of the integral for the initial data. Option (C) puts \( a^2 \) in the time-decay exponent instead of \( w^2 \), but the time decay must depend on \( w \) (the Fourier variable dual to \( x \)), not on \( a \) (which only measures how fast the initial data itself decays). Option (D) is the transform of the heat kernel alone, the solution for a delta function initial condition; it has no dependence on \( a \) at all, so it cannot be correct once the actual initial data \( e^{-a|x|} \) is used.
Step 4: Final Answer.
The Fourier transform of the solution is the transform of the initial data multiplied by the time-decay factor from solving the transformed heat equation.
\[ \boxed{\hat u(w,t) = \dfrac{2a}{a^2+w^2}\, e^{-kw^2t}} \]
We are given the heat equation and an initial condition, and need the Fourier transform of the solution \( u(x,t) \) with respect to \( x \), using the convention \( \hat u(w,t)=\int u(x,t)e^{iwx}\,dx \).
Step 2: Key Formula or Approach.
With this convention, taking the Fourier transform turns \( \partial_x \) into multiplication by \( -iw \), so \( \partial_x^2 \) becomes multiplication by \( (-iw)^2=-w^2 \). Taking the transform of both sides of the heat equation in \( x \) turns the PDE into an ordinary differential equation in \( t \) for each fixed \( w \). We also need the Fourier transform of the initial data \( e^{-a|x|} \).
Step 3: Detailed Explanation.
Applying the Fourier transform in \( x \) to both sides of \( u_t = k u_{xx} \) gives
\[ \frac{\partial \hat u}{\partial t} = k(-w^2)\hat u = -kw^2 \hat u \]
This is a simple first order linear ODE in \( t \) (with \( w \) fixed as a parameter), whose solution is
\[ \hat u(w,t) = \hat u(w,0)\, e^{-kw^2 t} \]
Now find \( \hat u(w,0) \), the transform of \( u(x,0)=e^{-a|x|} \). Split the integral at \( x=0 \):
\[ \hat u(w,0) = \int_0^{\infty} e^{-ax}e^{iwx}\,dx + \int_{-\infty}^0 e^{ax}e^{iwx}\,dx = \frac{1}{a-iw}+\frac{1}{a+iw} = \frac{2a}{a^2+w^2} \]
Combining the two results,
\[ \hat u(w,t) = \frac{2a}{a^2+w^2}\, e^{-kw^2t} \]
which is option (A). Option (B) is missing the factor of \( 2 \) that comes from adding the two halves of the integral for the initial data. Option (C) puts \( a^2 \) in the time-decay exponent instead of \( w^2 \), but the time decay must depend on \( w \) (the Fourier variable dual to \( x \)), not on \( a \) (which only measures how fast the initial data itself decays). Option (D) is the transform of the heat kernel alone, the solution for a delta function initial condition; it has no dependence on \( a \) at all, so it cannot be correct once the actual initial data \( e^{-a|x|} \) is used.
Step 4: Final Answer.
The Fourier transform of the solution is the transform of the initial data multiplied by the time-decay factor from solving the transformed heat equation.
\[ \boxed{\hat u(w,t) = \dfrac{2a}{a^2+w^2}\, e^{-kw^2t}} \]
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