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Let $\varphi(x,t)$ be a sufficiently smooth solution of the problem $$\left\{\begin{array}{rcll} \frac{\partial\varphi}{\partial t}(x,t) - \frac{\partial^2\varphi}{\partial x^2}(x,t) & = & f(x,t) & 0 < x< 1,\ 0 < t,\\ \varphi(x,0) & = & u_0(x), & 0\leq x\leq 1,\\ \varphi(t,0)\ =\ \varphi(t,1) & = & 0, & t\geq 0, \end{array}\right.$$ where $0 \leq t \leq T$, $T > 0$ fixed. Prove that there exist two constants $C_1(T)$ and $C_2(T)$, independent of the functions $\varphi$, $u_0$ and $f$, such that \begin{eqnarray*} \sup_{0\leq t\leq T}\left(\int_0^1|\varphi(x,t)|^2dx\right)^{1/2} & \leq & C_1(T)\left(\int_0^1|u_0(x)|^2dx\right)^{1/2}\\ & & +\ C_2(T)\sup_{0\leq t\leq T}\left(\int_0^1|f(x,t)|^2dx\right)^{1/2}. \end{eqnarray*}


Can anybody please help me to solve this problem? The only thing I have could do is prove that, for each $t\in [0,T]$, that $$\int_0^1|\varphi(x,t)|^2dx\ =\ \int_0^1|\varphi(x,0)|^2dx + \int_0^t\left\{\frac{d}{d\lambda}\int_0^1|\varphi(x,\lambda)|^2dx\right\}d\lambda.$$ Thanks so much in advance.

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Form $$\int_0^1|\varphi(x,t)|^2dx\ =\ \int_0^1|\varphi(x,0)|^2dx + \int_0^t\left\{\frac{d}{d\lambda}\int_0^1|\varphi(x,\lambda)|^2dx\right\}d\lambda,$$ you have that \begin{eqnarray*} \int_0^1|\varphi(x,t)|^2dx & = & \int_0^1|u_0(x)|^2dx + \int_0^t\int_0^1\frac{d}{d\lambda}\left\{[\varphi(x,\lambda)]^2\right\}dxd\lambda,\\ & = & \int_0^1|u_0(x)|^2dx + 2\int_0^t\int_0^1\varphi(x,\lambda)\left\{\frac{d}{d\lambda}\varphi(x,\lambda)\right\}dxd\lambda,\\ & = & \int_0^1|u_0(x)|^2dx + 2\int_0^t\int_0^1\varphi(x,\lambda)\left\{\frac{\partial^2\varphi}{\partial x^2}(x,\lambda) + f(x,\lambda)\right\}dxd\lambda,\\ & = & \int_0^1|u_0(x)|^2dx + 2\int_0^t\int_0^1\varphi\frac{\partial^2\varphi}{\partial x^2}dxd\lambda + 2\int_0^t\int_0^1\varphi f\ dxd\lambda. \end{eqnarray*} Now, integrating by parts the second term of the right-side and applying Cauchy-Schwarz inequality to the third term of the right-side, we have \begin{eqnarray*} \int_0^1|\varphi(x,t)|^2dx & \leq & \int_0^1|u_0(x)|^2dx - 2\underbrace{\int_0^t\int_0^1\left\{\frac{\partial\varphi}{\partial x}\right\}^2dxd\lambda}_{\geq 0} + 2\int_0^t\int_0^1\varphi f\ dxd\lambda,\\ & \leq & \int_0^1|u_0(x)|^2dx + 2\int_0^t\left\{\int_0^1\varphi f\ dx\right\}d\lambda,\\ & \leq & \int_0^1|u_0(x)|^2dx + 2\int_0^t\left\{\left(\int_0^1|\varphi(x,\lambda)|^2dx\right)^{1/2}\left(\int_0^1|f(x,\lambda)|^2dx\right)^{1/2}\right\}d\lambda. \end{eqnarray*} Next, we know that $$\left(\int_0^1|\varphi(x,\lambda)|^2dx\right)^{1/2}\ \leq\ \sup_{0\leq t\leq T}\left(\int_0^1|\varphi(x,t)|^2dx\right)^{1/2}\ =\ M, \mbox{ and}$$ $$\left(\int_0^1|f(x,\lambda)|^2dx\right)^{1/2}\ \leq\ \sup_{0\leq t\leq T}\left(\int_0^1|f(x,t)|^2dx\right)^{1/2}\ =\ F, \mbox{ then}$$ \begin{eqnarray*} \int_0^1|\varphi(x,t)|^2dx & \leq & \int_0^1|u_0(x)|^2dx + 2\int_0^tM\cdot F\ d\lambda,\\ & = & \int_0^1|u_0(x)|^2dx + (M\cdot F)2\underbrace{\int_0^td\lambda}_{=\ t\ \leq\ T},\\ & = & M\left(\frac{\int_0^1|u_0(x)|^2dx}{M} + (2T)F\right). \end{eqnarray*} So, taking sup in the left-side, we have \begin{eqnarray*} M^2 & \leq & M\left(\frac{\int_0^1|u_0(x)|^2dx}{M} + (2T)F\right),\\ M & \leq & \frac{\int_0^1|u_0(x)|^2dx}{M} + (2T)F,\\ \sup_{0\leq t\leq T}\left(\int_0^1|\varphi(x,t)|^2dx\right)^{1/2} & \leq & \frac{\int_0^1|u_0(x)|^2dx}{M} + (2T)\sup_{0\leq t\leq T}\left(\int_0^1|f(x,t)|^2dx\right)^{1/2}. \end{eqnarray*} Finally, note that $$\frac{\int_0^1|u_0(x)|^2dx}{M} \leq \frac{\int_0^1|u_0(x)|^2dx}{\left(\int_0^1|\varphi(x,0)|^2dx\right)^{1/2}}\ =\ \left(\int_0^1|u_0(x)|^2dx\right)^{1/2},$$ therefore take $C_1(T) = 1$, and $C_2(T) = 2T$.

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I'm not sure that you really need absolute value inside of squares; is your solution complex-valued? Even then, you can separate into real and imaginary parts. I'll assume it's real.

Let $M=\sup_{0\leq t\leq T} \int_0^1|\varphi(x,t)|^2dx$. Fix $t$ that attains $M$, so that
$$M=\int_0^1 \varphi(x,t) ^2dx = \int_0^1 \varphi(x,0) ^2dx + \int_0^t\left\{\frac{d}{d\lambda}\int_0^1 \varphi(x,\lambda) ^2dx\right\}d\lambda$$ The first integral on the right looks good: this is the integral of $u_0^2$ that you were expected to have. What to do with the second? Use the PDE, of course: $$ \frac{d}{d\lambda} \varphi(x,\lambda) ^2 = 2\varphi(x,\lambda) \frac{d}{d\lambda} \varphi(x,\lambda) = 2\varphi(x,\lambda)\, \varphi_{xx}(x,\lambda) + 2\varphi(x,\lambda)\, f(x,\lambda) $$ Once you integrate $\int \varphi \varphi_{xx}\,dx$ by parts, it becomes $-\int \varphi_x^2\,dx$, so that's good (negative).

For the last term on the right use $2\varphi f\le \epsilon \varphi^2+\epsilon^{-1}f^2$, the souped-up version of $2ab\le a^2+b^2$. Let's see where we are now: $$M \le \int_0^1 u_0(x)^2\,dx + \int_0^t \epsilon M+\epsilon^{-1}\int f(x,\lambda)^2\,d\lambda $$ The only thing that looks out of place is $\int_0^t \epsilon M$ on the right. But this is at most $t\epsilon M\le \epsilon M$. Absorb it in the left side: $$(1-\epsilon)M \le \int_0^1 u_0(x)^2\,dx + \epsilon^{-1}\int f(x,\lambda)^2\,d\lambda $$ and that's about it. Choose some convenient value for $\epsilon$ to tidy things up.

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What about the exponent 1/2? –  user70195 May 7 '13 at 12:18
    
Thanks for your answer! –  FASCH May 7 '13 at 16:35
1  
@ShanaKugimiya Exponent does not really matter: $x\le C_1y+C_2z$ and $\sqrt{x}\le C_1\sqrt{y}+C_2\sqrt{z}$ are interchangeable at the cost of constants $C_1,C_2$. –  75064 May 7 '13 at 17:18

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