# Definition of time global solution for PDE heat

Thank you for your kindness!! In detail, i consider the following problem $$\begin{cases} \partial_{t}u(x,t)=\Delta u(x,t),\ (x,t)\in[0,1]\times (0,\infty),\\ u(0,t)=u(1,t)=0,\ [0,\infty),\\ u(x,0)=u_{0}(x),\ x\in[0,1], \end{cases}$$ this is a one-dimension mixed problem of Heat equation. Now, Let $X:=\{u_{0}\in C^{2}[0,1]\ ;\ u_{0}(0)=u_{0}(1)=0\}$be a initial data space. Then, since Fourier expansion theorem(for above initial data $u_{0}(x)$), we obtained the following fact: $$u(x,t):= \begin{cases} \sum_{n=0}^{\infty} 2e^{-(n\pi )^{2}t}(u_{0}(\cdot),\sin (n\pi(\cdot)))_{L^2}\sin (n\pi x),\ t>0,\\ u_{0}(x),\ t=0. \end{cases}$$ where $(f,g)_{L^2[0,1]}:=\int_{0}^{1}f(x)g(x)dx$.Then u(x,t) is satisfy the above conditions.(and it have a uniqueness ). My question: if for any T>0, defined the solution spaces as $C^{0}([0,T];X):=\{u(x,t)\ ;u(t) is\ conti.\ on [0,T]\ and\ u(t)\in X\},\ where\ \|u\|:=\underset{t\in[0,T]}{\max}\max_{x\in[0,1]}|u(x,t)|.$ want to know the definition of $u(x,t)$ is global time solutions, in this situation.

First notice that for the space $X$ we have the Poincare inequality. Multiplying our equation by $u$ and integrating by parts yields $$\frac{d}{dt}\left(\frac{1}{2}\|u(t)\|_{L^2(0,1)}^2\right)+\|u_x(t)\|_{L^2(0,1)}^2 = 0,$$ poincare inequality gives us $$\frac{d}{dt}\left(\frac{1}{2}\|u(t)\|_{L^2(0,1)}^2\right)+c\frac{1}{2}\|u(t)\|_{L^2(0,1)}^2 \le 0,$$ then Gronwall's lemma yields $$\|u(t)\|_{L^2(0,1)}^2\le e^{-ct}\|u(0)\|_{L^2(0,1)}^2,$$ for all $t\in(0,\infty)$. Note that the fourier expansion is smooth, so we can differentiate it twice with respect to $x$, let $v^i=\frac{\partial^i}{\partial x^i}(u)$ for $i=1,2$, then $v^i$ solves $$v^i_t-\Delta v^i=0.$$ and $v^2(0,t)=v^2(1,t)=0,$ with $v^2(x,0)=\frac{d^2}{dx^2}u_0(x).$ We can repeat the above set up to deduce that $$\|v^2(t)\|_{L^2(0,1)}^2\le Ce^{-ct}\|v^2(0)\|_{L^2(0,1)}^2.$$ Now notice that \begin{align} \|v^1(t)\|_{L^2(0,1)}^2 &=\|u_x\|_{L^2(0,1)}^2=-\frac{1}{2}\int_0^1uu_t \\ &\le\frac{1}{4}\left(\|u(t)\|_{L^2(0,1)}^2+\|u_t(t)\|_{L^2(0,1)}^2\right) \quad \text{(by AM-GM inequality)} \\ &=\frac{1}{4}\left(\|u(t)\|_{L^2(0,1)}^2+\|u_{xx}(t)\|_{L^2(0,1)}^2\right) \\ &=\frac{1}{4}\left(\|u(t)\|_{L^2(0,1)}^2+\|v^2(t)\|_{L^2(0,1)}^2\right). \end{align} Now we have that \begin{align} \|u(t)\|_{H^2(0,1)}^2&=\|u(t)\|_{L^2(0,1)}^2+\|v^1\|_{L^2(0,1)}^2+\|v^2(t)\|_{L^2(0,1)}^2 \\ &\le C(\|u(t)\|_{L^2(0,1)}^2+\|v^2(t)\|_{L^2(0,1)}^2) \\ &\le Ce^{-ct}\|u_0\|_{H^2(0,1)}^2 \\ &\le Ce^{-ct}\|u_0\|_{C^2[0,1]}^2. \end{align} Sobolev embedding gives us $$\|u(t)\|_{C[0,1]}^2\le C\|u(t)\|_{H^2(0,1)}^2\le Ce^{-ct}\|u_0\|_{C^2[0,1]}^2,$$ which holds for all $t\in(0,\infty)$, and so we obtain $$\|u\|_{C(0,1;X)}\le C\|u_0\|_{C^2[0,1]}.$$ To see that the solution is unique, let $u_1$ and $u_2$ solve the equation, and set $W=u_1-u_2$, using the first energy estimate on $W$, we find that $W=0$.
• Could you please show how you obtain $-\int_0^1uu_t\le\|u(t)\|_{L^2(0,1)}^2+\|u_t(t)\|_{L^2(0,1)}^2$? I get the first term on the right, but not the second term. Also, for the first term, should you not have a constant scalar $c$ in the front coming from the Poincare's Inequality? – Hans Dec 27 '15 at 21:46
• it follows directly from the inequality $ab\le\frac{1}{2}(a^2+b^2)$, I just accidentally forgot to include the extra factor of $\frac{1}{2}$ but I'll edit that now – Ellya Dec 28 '15 at 18:24