Solve u_t=a^2u_(xx) on x∈(-∞,∞) , t>0 , with u(x,0)=varphi(x)
Solve $u_t=a^2u_{xx}$ on $x\in(-\infty,\infty)$, $t>0$, with $u(x,0)=\varphi(x)$.
Explanation of Method
Apply the Fourier transform in $x$ to reduce the PDE to a first-order linear ODE in $t$ for each frequency, solve it, compute the inverse transform of the resulting Gaussian multiplier (the heat kernel), and write the solution as a convolution.
Step 1 · Transform the PDE
With $\hat u(\lambda,t)=\int u\,e^{-i\lambda x}dx$: \[ \hat u_t = -a^2\lambda^2\hat u \implies \hat u(\lambda,t)=\hat\varphi(\lambda)e^{-a^2\lambda^2t}. \]
Step 2 · Invert the Gaussian multiplier (heat kernel)
Completing the square in $-a^2\lambda^2t+i\lambda x$ and using the Gaussian integral: \[ K(x,t) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-a^2\lambda^2t+i\lambda x}\,d\lambda = \frac{1}{2a\sqrt{\pi t}}\,e^{-x^2/(4a^2t)}. \]
Step 3 · Convolve
Since $\hat u=\hat\varphi\cdot\hat K$, by the convolution theorem \[ u(x,t) = \int_{-\infty}^\infty \varphi(\xi)\,K(x-\xi,t)\,d\xi. \]
Verification
$K$ satisfies $K_t=a^2K_{xx}$ by direct differentiation, and $\int_{-\infty}^\infty K(x,t)\,dx=1$ for all $t>0$ (Gaussian normalisation), with $K(x,t)\to\delta(x)$ as $t\to0^+$, so $u(x,0^+)=\varphi(x)$. \checkmark
Special case: point source
For $\varphi=\delta$, $u=K$, a Gaussian of standard deviation $\sigma=a\sqrt{2t}$ spreading with time.
Result
The Poisson integral solution is: