Inequality for solution of boundary-value problem 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.
 A: 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.
A: 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 $C_2(T) = 2T$.
