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?

  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.