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?


1 Answer 1


The test function $\varphi$ is $C^\infty$ and has compact support, but its domain is not an open set. Its domain is $$ \mathcal D=\mathbb R^n\times[0,\infty). $$ So, $\varphi$ is compactly supported in $\mathcal D$, and thus its support could be any compact subset of $\mathcal D$.

Hence $\varphi$ does not have to vanish for $t=0$.

  • 1
    $\begingroup$ Thanks for the clarification. I just found a nice reference for this, Linear and Quasilinear Equations of Parabolic Type by Ladyzenskaja, Solonnikov, Ural'ceva on page 25. $\endgroup$
    – Adi
    Aug 15, 2017 at 8:36

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