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I would like to define the notion of weak solution to initial value problems. Consider the PDE $$u_t - \Delta_p u = f \quad \text{on} \ \Omega \times [0,\infty)$$ with the condition $u(x,0) = 0$ and $u|_{\partial \Omega}(\cdot,t) = 0$.

In the book by DiBenedetto (page 21), he defines the weak solution as $$\int_{\Omega} u\phi(x,t) dx + \int_0^t\int_{\Omega}-u \phi_t + |\nabla u|^{p-2} \nabla u\cdot\nabla \phi \ dx \ dt = \int_{\Omega} u(x,0) \phi(x,0) dx$$ for all $\phi \in W^{1,2}(0,T; L^2(\Omega)) \cap L^p(0,T; W_0^{1,p}(\Omega))$. Note that $\phi(x,0)$ is not assumed to be $0$.

On the other hand, in the paper (http://users.jyu.fi/~miparvia/Julkaisuja/final_singularhigherint.pdf) in equation 2.2, he takes $\phi \in C_c^{\infty}(\Omega \times (0,T))$.

My question is what is the right test function? Do I need $\phi(\cdot,0)=0$ or not?

Also what is the right definition of weak solution's to the above problem under the use of Steklov averages? Where does the test function exactly lie in?

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  • $\begingroup$ Something is bizarre about your description. The definition you attribute to DiBenedetto does not involve the inhomogeneity $f$, so something is wrong. Furthermore, as you specified that $u(x,0) = 0$, you have that the right hand side $\int_{\Omega} u(x,0) \phi(x,0) dx \equiv 0$ irregardless of what $\phi$ is at $t = 0$. Can you double check and make sure you are asking the question you want to ask? $\endgroup$ – Willie Wong Aug 15 '17 at 14:17
  • $\begingroup$ DiBemedetto has the inhomogeneous term $f$ within the nonlinear structure $A(x,u,\nabla u)$ and $B(x,u,\nabla u)$. My concern was about the requirement $\phi(x,0)=0$ and that's not needed. $\endgroup$ – Adi Aug 16 '17 at 14:39
  • $\begingroup$ Since i am very new to parabolic equations, i was trying to understand the notion of weak and distributional solutions. The book by Ladyzenskaya has the necessary clarification that i was looking for. $\endgroup$ – Adi Aug 16 '17 at 14:42

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