# Heat Equation (Partial Differential Equation) - Fourier Transform Solution Form.

Heat Equation is

$$Au_{xx} = u_{t},$$ for $-\infty<x<\infty$ and $t>0$. Also, $u(x,0) \rightarrow 0$ as $x \rightarrow \pm \infty$.

The initial condition:

$$u(x,0) = \exp{(-|x|)},$$

for $-\infty<x<\infty$.

Want the solution form: (Using the Fourier Transform)

$$u(x,t) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{cos(wx)}{1+w^2} \exp{(-Aw^2t) dw}.$$

The fundamental FT equations are:

$$F(s) = \mathcal{F[f(x)]} = \int_{-\infty}^{\infty} f(x) \exp{(-iwx)}dx$$ $$f(x) = \mathcal{F^{-1}[f(x)]} = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(s) \exp{(iwx)}dw$$

Attempt:

Taking fourier transform of both sides of the PDE,

$$-Aw^2U(w,t) = \frac{\partial U}{\partial t} (w,t)$$ Then by solving this as an ODE and using the integrating factor, $$U(w,t) = U(w,0)\exp{(-Aw^2t)}$$

Taking inverse fourier transform of both sides,

$$u(x,t) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{1+w^2} \exp{(iwx)} \exp{(-Aw^2t)} dw.$$

But clearly this is not the required answer. Euler's formula can be used to get

$$u(x,t) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{cos(wx)}{1+w^2}\exp{(-Aw^2t)} dw + \frac{i}{\pi}\int_{-\infty}^{\infty} \frac{sin(wx)}{1+w^2} \exp{(-Aw^2t)} dw.$$

However, that means that the integral on the right must be equal to zero (which computationally it indeed is). But lack of knowledge about that type of integral makes this impossible. Feel as if there's something easy that is being missed. Any help would be appreciated.

• The thing you have called the boundary conditions $u(x,0) = \exp(- \lvert x \rvert)$ is actually the initial condition. And there should be no $e^{i w x}$ in the integrals after 'Euler's formula can be used to get ...'. Finally, notice that $\sin(\cdot)$ is odd, where as $1/(1 + w^{2})$ and $\exp(-Aw^{2}t)$ are even, the product of those function is odd, and the integral of an odd function over a symmetric interval is $\dots$ – mattos Mar 30 '18 at 5:09
• Yep sorry, all the things you mentioned were just typos. Ah okay so the second integral is indeed equal to zero. – MATHSUSER Mar 30 '18 at 6:13