On the real line, u t=ku xx, t > 0, k ∈ ℝ, with u(x,0)=e-a x , a > 0. For Fourier convention u(w,t)= -u(x,t)eiwxdx, find the transform.

$\frac1{\sqrt{4\pi kt}}e^{-w^2/(4kt)}$

GATE 2026 MA – Question 15

Partial Differential Equations · Heat and Wave Equations · 1 mark · Multiple choice

On the real line, $u_t=ku_{xx}$, t > 0, k ∈ ℝ, with $u(x,0)=e^{-a|x|}$, a > 0. For Fourier convention $\hat u(w,t)=\int_{-\infty}^{\infty}u(x,t)e^{iwx}dx$, find the transform.

$\frac{2a}{a^2+w^2}e^{-kw^2t}$
$\frac{a}{a^2+w^2}e^{-kw^2t}$
$\frac{2a}{a^2+w^2}e^{-ka^2t}$
$\frac1{\sqrt{4\pi kt}}e^{-w^2/(4kt)}$

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Correct answer: (A) $\frac{2a}{a^2+w^2}e^{-kw^2t}$

Explanation

Fourier transformation yields $\partial_t\hat u=-kw^2\hat u$. Initial transform is $\int e^{-a|x|}e^{iwx}dx=1/(a-iw)+1/(a+iw)=2a/(a^2+w^2)$. Multiplying by the time factor e^(−kw²t) gives **A**. This is the forward heat solution for k ≥ 0; for negative k the displayed transformed evolution is formal unless a suitable solution/transform exists.

GATE 2026 MA – Question 15

Explanation

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