# PDE Fourier transform(heat equation)

I solved the heat equation using the fourier transform method and got my solution to be $$u(x,t)= \int_{-\infty}^{\infty}\frac{1}{\sqrt{4\pi kt}}\exp(\frac{-(x-y)^2}{4kt})\tag{1}$$ however I have intial condition $$u(x,0) =\phi(x)$$ . I can't seem to show how the intial condition is satisfied

Here, we present a way forward that does not rely on knowledge of the theory or generalized functions (i.e., distributions). Rather, we apply standard tools from real analysis.

Starting from the integral expression for $$u(x,t)$$, we have

$$u(x,t)=\frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^\infty \phi(y) e^{-\frac{(x-y)^2}{4kt}}\,dy$$

Enforcing the substitution $$y\mapsto x+\sqrt{4kt}y$$ yields

\begin{align} u(x,t)&=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty \phi(x+\sqrt{4kt}y) e^{-y^2}\,dy\\\\ &=\phi(x)+\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty \left(\phi(x+\sqrt{4kt}y) -\phi(x)\right)\,e^{-y^2}\,dy\\\\ \end{align}

If $$\phi(x)$$ is continuous and bounded, then the Dominated Convergence Theorem guarantees that

\begin{align} \lim_{t\to 0^+}\int_{-\infty}^\infty \left(\phi(x+\sqrt{4kt}y) -\phi(x)\right)\,e^{-y^2}\,dy&=\int_{-\infty}^\infty \lim_{t\to 0^+}\left(\phi(x+\sqrt{4kt}y) -\phi(x)\right)\,e^{-y^2}\,dy\\\\ &=0 \end{align}

Therefore, we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{t\to 0^+}u(x,t)=\phi(x)}$$

as was to be shown!

• Hi, I know its too late, but I dont understand why does it follow from the continuity and boundedness of the function $\phi$ that you can use Dominated Convergence Theorem. For that Theorem you need some integrable function $g$ such that $\phi_n \leq g$, right? May 17, 2023 at 21:22
• Since $\phi$ is bounded in absolute value, say by $M>0$, then surely the integrand is less than $g(y) = 2Me^{-y^2}\in L^1$. And the DCT applies. May 17, 2023 at 22:52
• Oh, thank you, I understand now. :) May 18, 2023 at 5:39

You should have gotten $$u(x,t) = \int^\infty_{-\infty} \frac{1}{\sqrt{4\pi k t}} e^{\frac{-(x-y)^2}{4kt}} \phi(y) dy;$$ i.e., the initial condition should be included in there. Then $u(x,0) = \phi(x)$ since $\frac{1}{\sqrt{4\pi k t}} e^{\frac{-(x-y)^2}{4kt}} \to \delta(x)$ as $t \to 0^+$ in the distributional sense, where $\delta(x)$ is the Dirac mass centered at $x$ (any decent early graduate level PDE book should have a proof that the heat kernel tends to the Dirac mass).