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Given $g : [0,\infty) \to \mathbb{R}$, with $g(0)=0$, derive the formula $$u(x,t)=\frac{x}{\sqrt{4\pi}}\int_0^t \frac 1{(t-s)^{3/2}}e^{-\frac{x^2}{4(t-s)}}g(s)\,ds$$ for a solution of the initial/boundary-value problem \begin{cases}u_t-u_{xx}=0 & \text{in }\mathbb{R}_+ \times (0,\infty) \\ \qquad \quad \, u=0 & \text{on } \mathbb{R}_+ \times \{t=0\} \\ \qquad \quad \, u= g & \text{on }\{x=0\} \times [0,\infty). \end{cases} (Hint: Let $v(x,t):=u(x,t)-g(t)$ and extend $v$ to $\{x<0\}$ by odd reflection.)

This is Chapter 2, Exercise 15 of PDE Evans, 2nd edition (pages 87-88).

Following the hint and taking odd reflection, we have $$v(t)=\begin{cases}u(x,t)-g(t) & \text{if }x > 0 \\ -[u(-x,t)-g(t)] & \text{if }x<0 \end{cases}$$ and so we obtain $$v_t=\begin{cases}u_t(x,t)-g'(t) & \text{if }x > 0 \\ -[u_t(-x,t)-g(t)] & \text{if }x<0 \end{cases} \quad \text{ and } \quad v(t)=\begin{cases} u_{xx}(x,t) & \text{if }x > 0 \\ -u_{xx}(-x,t) & \text{if }x<0. \end{cases}$$ Thus, $\require{cancel} v(x,0)=\cancelto{0}{u(x,0)}-\cancelto{0}{g(0)}=0$ and $\require{cancel} v(0,t)=\cancelto{g'(t)}{u(0,t)}-g'(t)=0$. Our PDE is now $$\begin{cases}v_t - v_{xx} = \begin{cases} -g'(t) & \text{if } x > 0 \\ g'(t) & \text{if }x < 0 \end{cases} & \text{in } \mathbb{R}_+ \times (0,\infty) \\ \, \, \, v(x,0)=0 & \text{on } \mathbb{R}_+ \times \{t=0\} \\ \quad v(0,t) = 0& \text{on } \{x=0\} \times [0,\infty)\end{cases}$$


Now, page 49 of PDE Evans, 2nd edition asserts that, for $x \in \mathbb{R}^n$, $t > 0$, $$u(x,t)=\int_0^t \frac 1{4\pi(t-s))^{n/2}} \int_{\mathbb{R}^n} e^{-\frac{|x-y|^2}{4(t-s)}} f(y,s) \, dy \, ds \tag{13}$$ is a solution to the nonhomogeneous initial-value problem $$\begin{cases}u_t - \Delta u = f & \text{in } \mathbb{R}^n \times (0,\infty) \\ \qquad \quad u = 0 & \text{on } \mathbb{R}^n \times \{t=0\}. \tag{12} \end{cases}$$ (The numbers $\text{(13)}$ and $\text{(12)}$ are from page 49 of the textbook. Also, these $u$'s for $\text{(13)}$ and $\text{(12)}$ are different from the $u$'s in the exercise problem.)


Applying $\text{(13)}$ to our given initial/boundary value problem with $n=1$, we obtain \begin{align} v(x,t)&= \int_0^t \frac 1{(4\pi(t-s))^{1/2}} \left[\int_{-\infty}^0 e^{-\frac{|x-y|^2}{4(t-s)}}g'(y,s)\,dy - \int_0^{\infty} e^{-\frac{|x-y|^2}{4(t-s)}}g'(y,s)\,dy \right]ds \end{align}

By the lemma on page 46, $\int_{-\infty}^{\infty} \frac 1{(4\pi(t-s))^{1/2}} e^{-\frac{|x-y|^2}{4(t-s)}} \, dy=1$. So we have \begin{align} g(t)&=\int_0^t g(s) \, ds \int_{-\infty}^{\infty} \frac 1{(4\pi(t-s))^{1/2}} e^{-\frac{|x-y|^2}{4(t-s)}} \, dy \\ &= \int_0^t \frac 1{(4\pi(t-s))^{1/2}} \left[ \int_{-\infty}^0 e^{-\frac{|x-y|^2}{4(t-s)}} \, dy \, ds + \int_0^{\infty} e^{-\frac{|x-y|^2}{4(t-s)}} \, dy \, ds \right]. \end{align}

Thus, \begin{align} u(x,t)&= v(x,t)+g(t)\\ &= 2 \int_0^t \frac 1{(4\pi(t-s))^{1/2}} \int_{-\infty}^0 e^{-\frac{|x-y|^2}{4(t-s)}} \, dy \, g'(s) \, ds \end{align}

But I cannot seem to derive at this point the desired formula of $$u(x,t)=\frac{x}{\sqrt{4\pi}}\int_0^t \frac 1{(t-s)^{3/2}}e^{-\frac{x^2}{4(t-s)}}g(s)\,ds.$$ I think integration by parts would need to be used at this point. Am I on the right track though?

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I will post the answer from math24.files.wordpress.com/2013/02/evans_solutions-ch2.pdf. All credit to Joe Benson, Denis Bashkirov, Minsu Kim, Helen Li and Alex Csar.

We define\begin{align*} v(x,t)\equiv\begin{cases} u(x,t)-g(t)&x>0\\ -u(-x,t)+g(t)&x\leq 0. \end{cases} \end{align*} Thus \begin{align*} v_t(x,t)= \begin{cases} u_t(x,t)-g'(t)&x>0\\ -u_t(-x,t)+g'(t)&x\leq 0 \end{cases} \end{align*} and \begin{align*} v_{xx}(x,t)= \begin{cases} u_{xx}(x,t)&x>0\\ -u_{xx}(-x,t)&x\leq 0. \end{cases} \end{align*} Hence,\begin{align*} \begin{cases} v_t(x,t)-v_{xx}(x,t)=\begin{cases}-g'(t)&x>0\\g'(t)&x\leq 0 \end{cases}\\ v(x,0)=0\\ v(0,t)=0. \end{cases} \end{align*} Using the formula of the solution to the nonhomogeneous problem, we get$$ v(x,t)=\int_0^t\frac{1}{\sqrt{4\pi (t-s)}} \left(\int_{-\infty}^0e^{\frac{-(y-x)^2}{4(t-s)}}g'(s)\,dy-\int_0^\infty e^{\frac{-(y-x)^2}{4(t-s)}}g'(s)\,dy \right) \,ds. $$ The integral of the fundamental solution satisfies$$ \int_{-\infty}^\infty \frac{1}{\sqrt{4\pi(t-s)}}e^\frac{-(y-x)^2}{4(t-s)}\,dy=1, $$ and so, for $x>0$, letting $y-x\equiv-z$ leads to\begin{align*} u(x,t)&=v(x,t)+g(t)\\ &=v(x,t)+\int_0^tg'(s)\,ds\int_{-\infty}^\infty \frac{1}{\sqrt{4\pi(t-s)}}e^\frac{-(y-x)^2}{4(t-s)}\,dy\\ &=2\int_0^t \frac{1}{\sqrt{4\pi}}(t-s)^{-\frac12}\int_{-\infty}^0e^\frac{-(y-x)^2}{4(t-s)}\,dy\, g'(s)\,ds\\ &=\int_0^t \frac{1}{\sqrt{\pi}}(t-s)^{-\frac12}\int_x^\infty e^\frac{-z^2}{4(t-s)}\,dzdg(s). \end{align*} Integrating by parts, we get\begin{align*} u(x,t)&=\frac{1}{\sqrt{\pi}}(t-s)^{-\frac12}\int_x^\infty e^\frac{-z^2}{4(t-s)}\,dz\,g(s)|_{s=0}^{s=t}\\ &\hspace{10mm}-\int_0^tg(s)\frac{1}{2\sqrt{\pi}}(t-s)^{-\frac32}\,ds\int_X^\infty e^\frac{-z^2}{4(t-s)}\,dz\\ &\hspace{20mm}-\int_0^tg(s)\frac{1}{\pi}(t-s)^{-\frac12}\,ds\int_x^\infty e^\frac{-z^2}{4(t-s)}\frac{-z^2}{4(t-s)^2}\,dz\\ &\equiv I_1-\int_0^tg(s)\frac{1}{2\sqrt{\pi}}(t-s)^{-\frac32}\,ds\int_X^\infty e^\frac{-z^2}{4(t-s)}\,dz\\ &\hspace{10mm}-\int_0^tg(s)\frac{1}{\pi}(t-s)^{-\frac12}\,ds\int_x^\infty \frac{-z}{2(t-s)}\,de^\frac{-z^2}{4(t-s)}\\ &=I_1-\int_0^tg(s)\frac{1}{2\sqrt{\pi}}(t-s)^{-\frac32}\,ds\int_X^\infty e^\frac{-z^2}{4(t-s)}\,dz\\ &\hspace{10mm}+\int_0^tg(s)\frac{1}{\sqrt{4\pi}}(t-s)^{-\frac32}\,ds(-z)e^\frac{-z^2}{4(t-s)}|_{z=x}^{z=\infty}\\ &\hspace{20mm}+\int_0^tg(s)\frac{1}{2\sqrt{\pi}}(t-s)^{-\frac32}\,ds\int_x^\infty e^\frac{-z^2}{4(t-s)}\,dz\\ &=I_1+\frac{x}{\sqrt{4\pi}}\int_0^t\frac{1}{(t-s)^\frac32}e^\frac{-x^2}{4(t-s)}g(s)\,ds. \end{align*} Now we focus on $I_1$ and set $w^2\equiv\frac{z^2}{4\epsilon}$,\begin{align*} I_1&=\lim_{\epsilon\to 0^+}\frac{1}{\sqrt{\pi}}\epsilon^{-\frac12}\int_x^\infty e^\frac{-z^2}{4\epsilon}\,dz g(t-\epsilon)\\ &=g(t)\lim_{\epsilon\to 0^+}\frac{1}{\sqrt{\pi}}\int_{\frac{x^2}{4\epsilon}}^\infty 2e^{-w^2}\,dw=0. \end{align*} Thus we proved$$ u(x,t)=\frac{x}{\sqrt{4\pi}}\int_0^t\frac{1}{(t-s)^\frac32}e^\frac{-x^2}{4(t-s)}g(s)\,ds $$ for $x>0$. Next, we need to show that\begin{align*} \lim_{x\to 0^+}u(x,t)&=\lim_{x\to 0^+}\frac{x}{\sqrt{4\pi}}\int_{t-\delta}^t\frac{1}{(t-s)^\frac32}e^\frac{-x^2}{4(t-s)}g(s)\,ds\\ &\hspace{10mm}+\lim_{x\to 0^+}\frac{x}{\sqrt{4\pi}}\int_0^{t-\delta}\frac{1}{(t-s)^\frac32}e^\frac{-x^2}{4(t-s)}g(s)\,ds\\ &=g(t)\lim_{x\to 0^+}\frac{x}{\sqrt{4\pi}}\int_{t-\delta}^t\frac{1}{(t-s)^\frac32}e^\frac{-x^2}{4(t-s)}\,ds\\ &=g(t)\lim_{x\to 0^+}\frac{x}{\sqrt{4\pi}}\int_0^\delta \frac{1}{s^\frac32}e^\frac{-x^2}{4s\,ds}. \end{align*} For fixed $x$, let $s\equiv\frac{x^2}{w^2}$ and get\begin{align*} \lim_{x\to 0^+} u(x,t)&=g(t)\lim_{x\to 0^+}\frac{x}{2\sqrt{\pi}}\int_{-\infty}^{x^2/\delta}\frac{w^2}{x^3}e^\frac{-w^2}{4}\frac{-2x^2}{w^3}\,dw\\ &=g(t)\lim_{x\to 0^+}\frac{1}{\sqrt{\pi}}\int_{x^2/\delta}^\infty e^\frac{-w^2}{4}\,dw\\ &=g(t)\frac{1}{\sqrt{\pi}}\int_0^\infty e^\frac{-w^2}{4}\,dw\\ &=g(t) \end{align*}and we are done.

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  • $\begingroup$ Yes. I typed this answer for some personal notes a long time ago and I forgot the source. I apologize for this. Since I'm unable to delete my answer and share the link as a comment, I edited it regarding the comment above. Thank you for your observation. $\endgroup$ – sam wolfe Oct 30 '17 at 13:08
  • $\begingroup$ Everything is fine now, thanks for the update. $\endgroup$ – Jack D'Aurizio Oct 30 '17 at 13:14

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