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Let $T>0$. Consider the following semilinear parabolic equation $$ \partial_tu(x,t)-\Delta u(x,t)=f(u(x,t)) \quad \forall (x,t) \in (0,1) \times (0,T), $$ with homogeneous Neumann boundary conditions and given initial conditions in $L^2(0,1)$.

Assume that $f:\mathbb{R}\longrightarrow\mathbb{R}$ is continuously differentiable. Prove that the solution to the stated PDE is unique.

Attempt:

Let $u_1,u_2$ be two solutions. Then, using Green formula, we obtain $$ \begin{split} \dfrac{1}{2} \dfrac{d}{dt}\int_{0}^1 \vert u_1(x,t)-u_2(x,t)\vert^2 dx&=\int_{0}^1 f(u_1(x,t))-f(u_2(x,t)) (u_1(x,t)-u_2(x,t))dx\\ &\leq C \int_{0}^1 \vert u_1(x,t)-u_2(x,t)\vert^2 dx \end{split} $$ where we have used the mean value theorem and such that $$C:=\underset{s \in [u_1(x,t),u_2(x,t)]}{\sup} f'(s)$$
which exists since $f$ is continuously differentiable.

By applying Gronwall inequality, we obtain $u_1=u_2.$

Is this proof correct? I can't spot the mistake.

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1 Answer 1

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$$ \dfrac{1}{2} \dfrac{d}{dt} \int_0^1 \vert u_1(x,t)-u_2(x,t)\vert^2 dx\leq \int_{0}^1C(x,t) \vert u_1(x,t)-u_2(x,t)\vert^2dx $$

The mistake is assuming that the constant $C$ is uniform.

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