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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

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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).

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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 $y\to x+\sqrt{4kt}y$ 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!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Mar 24 '16 at 15:06

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