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We have the folllwing problem: $\begin{cases} & \dfrac{\partial u}{\partial t} = k \dfrac{\partial^2 u}{\partial x^2}, 0 < x < l, t > 0\\ & u(0,t)=0,\\ & \dfrac{\partial u}{\partial x} (l,t)=0, t \geq 0\\ & u(x,0) = f(x), 0 \leq x \leq l \end{cases} $ The question is to prove the unicity of the solution of problem. For this, I suppose that they are two solutions $u_1$ and $u_2$, and we put $v=u_1- u_2$. By multiplying the equation by $v(x,t)$ and integrated, we obtained $$ \displaystyle\int_0^l \left[\displaystyle\int_0^T \dfrac{\partial v}{\partial t} (x,t) v(x,t) dt \right] dx - k \displaystyle \int_0^T \left[\displaystyle\int_0^l \dfrac{\partial^2 v}{\partial x^2} (x,t) v(x,t) dx \right] dt=0 $$ wich implies $$ \displaystyle\int_0^T \dfrac{\partial v}{\partial t}(x,t) v(x,t) = \dfrac{1}{2} \left[v^2(x,T)- v^2(x,0)\right] $$ and $$ \displaystyle\int_0^l \dfrac{\partial^2 v}{\partial x^2}. v(x,t) dx = \dfrac{\partial v}{\partial x}(l,t). v(l,t) - \dfrac{\partial v}{\partial x}(0,t).v(0,t) - \displaystyle\int_0^l \left[\dfrac{\partial v}{\partial x}\right]^2 dx. $$ Then we conclude: $$ \dfrac{1}{2} \displaystyle\int_0^l v^2(x,T) dx - \dfrac{1}{2} \displaystyle\int_0^l v^2(x,0) dx + k \displaystyle\int_0^l \left(\dfrac{\partial v}{\partial x}(x,t)\right)^2 dx=0. $$ My question is: How can we conclude that $v(x,t)=0$ for all $(x,t)$? Thanks

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2 Answers 2

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Since the initial condition is the same for $u_1$ and $u_2$ you have $v(x,0)=0$ for all $x$ and thus the second term of the last equality vanishes. You are left with to integrals that are positive, because the integrand is positive. So the only continuous function satisfying the last equality is $v\equiv0$.

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For $u$ equal to any of $u_1,u_2,u_1-u_2$, you have \begin{align} \frac{d}{dt}\int_{0}^{l}u^{2}(x,t)dx & = 2\int_{0}^{l}u_{t}(x,t)u(x,t)dx \\ & = 2\int_{0}^{l}ku_{xx}(x,t)u(x,t)dx \\ & = 2ku_{x}(x,t)u(x,t)|_{x=0}^{l} -2\int_{0}^{l}ku_{x}^{2}(x,t)dx \\ & = -2\int_{0}^{l}ku_{x}^{2}(x,t)dx \le 0. \end{align} Therefore, if $u=u_1-u_2$ the function $F(t)=\int_{0}^{1}u^{2}(x,t)dx$ satisfies $$ F(0)=0,\;\; F(t) \ge 0,\;\; F'(t) \le 0. $$ It follows that $F(t)=0$ for all $t \ge 0$.

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