In the book on PDEs by L. Evans, a solution to the heat equation with Dirichlet boundary conditions: $$\tag{HP} \begin{cases}\displaystyle \frac{\partial u}{\partial t}=\Delta u & x\in U,\ t\in(0, T)\\ u=0 & \text{on } \partial U\\ u=g\in L^2(U) & \text{at time }t=0 \end{cases} $$ is constructed by means of the Galerkin method. This happens to be a weak solution in the sense that $$ u\in L^2(0, T; H^1_0(U)),\quad u'\in L^2(0, T; H^{-1}(U)) $$ (the time derivative is taken in distributional sense $^{[1]}$) and it satisfies the equation where the Laplacian is taken in the "weak sense" of elliptic theory $^{[2]}$: $$\tag{1} \int_U \frac{\partial u}{\partial t}(x, t)v(x)\, dx = -\int_U \nabla u (x, t)\cdot\nabla v(x)\, dx,\quad \forall v \in H^1_0(U). $$ (see §7.1 of the second edition).

Later in the same book, (HP) is treated by means of the semigroup approach. Namely, it is proved that the unbounded operator $A=\Delta$ defined on the domain $D(A)=H^2(U)\cap H^1_0(U)$ generates a contraction semigroup $S_t$ on $L^2(U)$ space. Therefore the function $$\tag{2}u(x, t)=(S_tg)(x)$$ is a strong solution to (HP) if $g\in D(A)$. (see §7.4).

Question. Formula (2) makes sense even if $g\notin D(A)$. Is it true that in this case the function $u$ defined by (2) is a weak solution in the sense of formula (1)?

In the book Vector-valued Laplace transforms and Cauchy problems by V.A., 2nd edition $^{[3]}$, this kind of solutions are called mild solutions (Definition 3.1.1 - see also Proposition 3.1.9). This explains the title.


$^{[1]}$ Explicitly, $$-\int_0^T u(t)\phi'(t)\, dt= \int_0^t u'(t)\phi(t)\, dt,\quad \forall \phi\in C^{\infty}_c(0, T).$$ This is somewhat related to this other question on vector-valued distributions.

$^{[2]}$ Also called energetic extension of the Laplacian.

$^{[3]}$ Recommended to me some time ago by the user lvb in this great answer.


The easiest way to show the coincidence of the two solutions is to verify that both are weak solutions of the class $L^2(0, T; H^1_0(U))$. For problem (HP), a function $u\colon\, (0,T)\to H^1_0(U)$ is called a weak solution of the class $u\in L^2(0, T; H^1_0(U))$ if $u$ satisfies the integral identity \begin{align*} -\!\!\int\limits_{Q_T}\! uv_t\,dxdt+\int\limits_{Q_T} \nabla u\cdot\nabla v\,dxdt= \!\int\limits_U g(x)v(x,0)\,dx\tag{$\ast$}\\ \forall\, v\in H^1(Q_T)\colon\,v|_{\partial U}=0,\;v(x,T)=0, \end{align*} where $Q_T=U\times (0,T)$. For the linear problem (HP), uniqueness of the weak solution $(\ast)$ is rather obvious. Indeed, take $g=0$ and notice that the choice of a test function $$ v(x,t)=\int\limits_t^T u(x,s)\,ds $$ implies that $$ \int\limits_{Q_T}\! |u(x,t)|^2\,dxdt+\frac{1}{2}\int\limits_{U} \Bigl|\int\limits_0^T\nabla u(x,s)\,ds\Bigr|^2dx=0, $$ i.e., $u=0$ a.e. in $Q_T\,$.

To verify that a weak solution $(1)$ will be a weak solution $(\ast)$, notice that the weak solution $u$ in the sense $(1)$ satisfies the integral identity $$ \int\limits_{Q_T} u_t v \, dxdt +\int\limits_{Q_T} \nabla u \cdot\nabla v\, dxdt=0 \tag{$\ast\ast$} $$ for all test functions $v=v(x,t)$ stepwise w.r.t. variable $t$, hence for all $v\in L^2(0, T; H^1_0(U))$, and hence for all $$ v\in H^1(Q_T)\colon\,v|_{\partial U}=0,\;v(x,T)=0. $$ Integrating by parts in $(\ast\ast)$ results in the identity $(\ast)$.

To verify that the mild solution will be a weak solution $(\ast)$ multiply the equation $$ \Delta\!\!\int\limits_0^t u(x,s)\,ds=u(x,t)-g(x) $$ by $v_t(x,t)$ with $v\in H^1(Q_T)\colon\,v|_{\partial U}=0,\;v(x,T)=0$, and then integrate over $Q_T$ to find $$ \int\limits_{Q_T}\Bigl(\Delta\!\int\limits_0^t u(x,s)\,ds\Bigr)v_t(x,t)\,dxdt= \int\limits_{Q_T}uv_tdxdt+\int\limits_{U}g(x)v(x,0)\,dx, $$ where the left hand side equals \begin{multline*} \int\limits_{U}dx\int\limits_0^T v_t(x,t)\,dt\int\limits_0^t\Delta u(x,s)\,ds= \int\limits_{U}dx\int\limits_0^T\Delta u(x,s)\,ds\int\limits_s^T v_t(x,t)\,dt\\ =-\int\limits_{Q_T}\Delta u(x,s)v(x,s)\,dxds= \int\limits_{Q_T}\nabla u(x,s)\cdot\nabla v(x,s)\,dxds \end{multline*} which results in the identity $(\ast)$.

  • $\begingroup$ +1: I find this answer to be very instructive. Is the terminology "very weak solution" used for the solution in form (*)? I recall having heard this terminology in some talk. $\endgroup$ – Giuseppe Negro Mar 19 '14 at 9:48
  • $\begingroup$ @Giuseppe Negro: For a heat-like equation, within the $L_2$-theory of weak locally integrable solutions to initial boundary value problems, there are only three types of weak solutions. Namely, 1)$\,u_t\,,\!\nabla_x u\in L^2(Q_T)$; 2)$\,u,\!\nabla_x u\in L^2(Q_T)$; 3) $\,u\in L^2(Q_T)$, with the latter type tagged as a "very weak solution". Of course, there are some additional options related to the fractional order of smoothness, to say nothing of the locally non-integrable weak solutions. $\endgroup$ – mkl314 Mar 19 '14 at 20:50

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