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Let the pde $$\dfrac{\partial^2 u}{\partial t^2} - \dfrac{\partial^2 u}{\partial x^2}=f(x)$$ The question is:

Find the limit condition such that this pde admit a unique solution in $[a,b] \times [0,T].$

For this, I suppose the existence of to sulutions $u_1$ and $u_2$ and we put $v = u_1 - u_2.$ Then, $$\dfrac{\partial^2 v}{\partial t^2} - \dfrac{\partial^2 v}{\partial x^2} = 0$$ myltiply this last equation by $\dfrac{\partial v}{\partial t}$ and we integrate over $[a,b]$ and we obtained $$\displaystyle\int_a^b \dfrac{\partial^2 v}{\partial t^2} . \dfrac{\partial v}{\partial t} dx - \int_a^b\dfrac{\partial^2 v} {\partial t^2}. \dfrac{\partial^2 v}{\partial x^2} dx = 0$$

I tried integration by parts, but didn't find the limit conditions to obtain the unicity of this pde. Thanks for any help.

chridam
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jijiii
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1 Answers1

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For the wave equation there is a conserved energy $$ E(t)=\int_{-\infty}^\infty (u_t)^2+(u_x)^2 dy. $$ This is obtained multiplying by $u_t$ and integrating by parts (as you did). For $v$ this quantity is zero initially, thus it vanishes for all times. Right?

guacho
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  • no, sorry I did not understand. how we continue the calculations of the integration by parts and how we deduce the limit conditions? – jijiii May 27 '13 at 12:40
  • You have $$ \int u_{tt}u_t-\int u_tu_{xx}=\frac{1}{2}\int (u_{t})t^2+\int u{xt}u_{x} $$ $$ =\frac{1}{2}\int (u_{t})t^2+\frac{1}{2}\int (u{x})^2_t=\frac{d}{dt}E(t)=0. $$ – guacho May 27 '13 at 13:02
  • As $E(0)=0$ you obtain that $v$ is a constant (and thus it vanishes) – guacho May 27 '13 at 13:08
  • but who are the limit conditions? please. – jijiii May 27 '13 at 13:45
  • What do you mean by "limit condition"? I think that you refer to the conditions under the functional makes sense, right? In that case you need $H^1$, isn't it? – guacho May 27 '13 at 16:47
  • the limit conditions are y(a) , y(b) or y'(a) , y'(b) for exampl... Dirichlet, or Neuman?? – jijiii May 27 '13 at 17:47
  • Ok, you mean boundary conditions :-P. Well, in that case we have to check the boundary terms in the integration by parts that I made above. We get $$ \int_a^b u_tu_{xx}=u_tu_x|{a}^b-\int_a^b u{tx}u_x. $$ Thus, you need to impose Neumann BC, right? – guacho May 27 '13 at 18:01
  • Slowly please, we have $$\int_a^b \dfrac{\partial^2 v}{\partial t^2} . \dfrac{\partial v}{\partial t} dx - \int_a^b \dfrac{\partial^2 v}{\partial x^2} . \dfrac{\partial v}{\partial t} dx = 0$$ then $$\int_a^b \dfrac{\partial^2 v}{\partial t^2} . \dfrac{\partial v}{\partial t} dx - [\dfrac{\partial v}{\partial t} . \dfrac{\partial v}{\partial x}]_a^b + \int_a^b \dfrac{\partial}{\partial t} [\dfrac{\partial v}{\partial x}] . \dfrac{\partial v}{\partial x} dx = 0$$ and what are the other steps? please – jijiii May 27 '13 at 20:11
  • why $$\dfrac{d}{dt} E(t)=0$$ – jijiii May 28 '13 at 10:42
  • Well, in the second line you have $$ \frac{1}{2}\frac{d}{dt}\int_a^b (u_t)^2+\frac{1}{2}\frac{d}{dt}\int_a^b (u_x)^2-u_tu_x|_{a}^b=0. $$ Now you impose Neumann BC ($u_x=0$ at the boundary) and you use the definition of $E$: $$ \frac{d}{dt}\int_a^b (u_t)^2+(u_x)^2=\frac{d}{dt}E(t)=0. $$ – guacho May 28 '13 at 10:47
  • if we don't know definition of $E(t),$ How we can prouve that $$\dfrac{1}{2} \dfrac{d}{dt} \int_a^b [\dfrac{\partial u}{\partial x}]^2 dx + \dfrac{1}{2} \dfrac{d}{dt} \int_a^b [\dfrac{\partial u}{\partial t}]^2 dx$$ if – jijiii May 28 '13 at 10:56
  • It's the first line in my answer: $$ E(t)=\int_a^b (u_t)^2+(u_x)^2 $$ – guacho May 28 '13 at 10:56
  • yes, but why if $E(0)=0$ then $E(t)=0$ for all $t?$ i don't understand this point please, then why $E$ depend only on $t?$ why she not depend on $x?$ – jijiii May 28 '13 at 11:00
  • Look, assume that you have two different solutions with the same BC and initial conditions. Define $v=u_1-u_2$. You have $$ E(t)=\int_a^b (v_t)^2+(v_x)^2=E(0) $$ but initially you have $v(0,x)=0$ and $v_t(0,x)=0$ ($u_1, u_2$ has the same BC and initial conditions), thus $E(0)=0$, ok? – guacho May 28 '13 at 11:02
  • $E(t)$ depends only on time because you are integrating in $x$ (and this kill this dependence), and its derivative is zero due to the equation and the computation above. – guacho May 28 '13 at 11:03
  • Sorry i dont't understand the first point: $$E(t) = E(0)$$ – jijiii May 28 '13 at 11:08
  • $$ \frac{d}{dt}E(t)=0 $$ implies $E(t)$ is a constant. This constant is $E(0)$. – guacho May 28 '13 at 11:10
  • So, we have $E(t)=0$ so $\dfrac{d}{dt} E(t)=0$ so $E(t)=c$ for all $t,$ whith $c$ constante, then $E(t)=0$ it's okay? – jijiii May 28 '13 at 11:25
  • Mmm you have $\frac{d}{dt}E(t)=0$ so $E(t)=E(0)=0$ – guacho May 28 '13 at 12:13