Eignvalues of Laplacian operator and Sobolev spaces

Let $\Omega$ an open bounded in $\mathbb{R}^n$ and $I$ un interval in $\mathbb{R}$, let $t_0 \in I$, $f\in H^1_0(\Omega)$, $g\in L^2(\Omega)$. Let $(\lambda_n)_{n\in \mathbb{N}}$ the eigen values od the Laplacian operator $\Delta$, and $(e_n)_n$ the vector associates to the eigen values. We denote $$u=\sum_{n=1}^{+\infty} [(f,e_n) \cos[(t-t_0) \sqrt{\lambda_n}] + \dfrac{(g,e_n)}{\sqrt{\lambda_n}}\sin[(t-t_0)\sqrt{\lambda_n}]]e_n$$

How we can prouve that $u \in C^0(I,H^1_0(\Omega))$ and $\dfrac{\partial u}{\partial t} \in C^0(I,L^2(\Omega))$?

II

I undestand the answer of this question. I have an seconde question please.

How we prouve that $$\Box u = 0 \quad\mbox{in} \quad \mathcal{D}'(I^{0} \times \Omega)$$ where $\Box$ is wave operator, and $I^{0}$ is the interior od $u$.

Thank's for the help beaucause i lost.

• No need to put "EDIT EDIT EDIT" in the title. You can add "EDIT" into the body of the question, indicated what was edited. Also, note that different questions should be placed in separate threads; not edited into previous questions. – Asaf Karagila May 26 '14 at 13:46
• the edited is in my first question. Why T.A.E. use it? – varphi May 26 '14 at 13:50
• (1) You can ask T.A.E. in a comment to their answer; or if that doesn't help, or the explanation seems too complicated for a comment exchange (2) you can post a new question with a link to this one, saying that in such and such answer this and that was used, and you don't understand why. On a different topic, you shouldn't accept answers if you don't understand them. – Asaf Karagila May 26 '14 at 13:54
• It is not nice to change the question after someone answered it. Since you asked a new one already, I am going to revert the edit. – Willie Wong May 26 '14 at 14:50

I'll assume $\Omega$ has a smooth boundary (or something similar.) The Laplacian has a natural domain $\mathcal{D}(\Delta)=H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$ because $$-\Delta : \mathcal{D}(\Delta) \subset L^{2}(\Omega)\rightarrow L^{2}(\Omega)$$ is selfadjoint and non-negative on its domain. The domain of the positive square root of $-\Delta$ is $H_{0}^{1}(\Omega)$. Furthermore, $-\Delta$ has a complete orthonormal set of eigenfuntions $\{ e_{n}\}_{n=0}^{\infty}$ with corresponding eigenvalues $\{ \lambda_{n} \}_{n=0}^{\infty} \subset (0,\infty)$. By the spectral theorem, $$f \in \mathcal{D}(-\Delta) \iff \sum_{n=0}^{\infty}|\lambda_{n}(f,e_{n})|^{2} < \infty,\;\;\; f \in \mathcal{D}(\sqrt{-\Delta}) \iff \sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(f,e_{n})|^{2} < \infty.$$ Using Friedrichs extensions, one finds that $\mathcal{D}(\sqrt{-\Delta})=H_{0}^{1}(\Omega)$. So the above characterization of $\mathcal{D}(\sqrt{-\Delta})$ becomes a characterization of $H_{0}^{1}$. Because $\sqrt{-\Delta}$ is selfadjoint, then $f(t)=e^{it\sqrt{-\Delta}}f_{0}$ is a continuous function from $[0,\infty)$ to $L^{2}(\Omega)$ which is given by Functional Calculus: $$f(t) = e^{it\sqrt{-\Delta}}f_{0}=\sum_{n=0}^{\infty} e^{i\sqrt{\lambda_{n}}t}(f_{0},e_{n})e_{n}.$$ If $f_{0} \in \mathcal{D}(\sqrt{-\Delta})=H_{0}^{1}(\Omega)$, then $f$ is a continuous function from $[0,\infty)$ to $H_{0}^{1}(\Omega)$, and $f'(t)=\sqrt{-\Delta}f(t)=e^{it\sqrt{-\Delta}}(\sqrt{-\Delta}f_{0})$ is a continuous function from $[0,\infty)$ to $L^{2}(\Omega)$. Your situation is a mild variation of this.

Your function $u$: (Discussion added after your questions.) So you have defined $u=v+w$ where $$v(t) = \sum_{n=0}^{\infty}(f,e_{n})\cos((t-t_{0})\sqrt{\lambda_{n}})e_{n}, \;\;\; f \in H_{0}^{1}(\Omega)\\ w(t) = \sum_{n=0}^{\infty}\frac{(g,e_{n})}{\sqrt{\lambda_{n}}}\sin((t-t_{0})\sqrt{\lambda_{n}})e_{n},\;\;\; g \in L^{2}(\Omega).$$ Both $v$ and $u$ are in $L^{2}(\Omega)$ for all $t\in\mathbb{R}$ because the Fourier coefficents with respect to the complete orthonormal set $\{ e_{n}\}_{n=0}^{\infty}$ in each series are square integrable: $$\|v(t)\|^{2}_{L^{2}}=\sum_{n=0}^{\infty}|(f,e_{n})|^{2}\cos^{2}((t-t_{0})\sqrt{\lambda_{n}}) \le \sum_{n} |(f,e_{n})|^{2} = \|f\|^{2},\\ \|w(t)\|_{L^{2}}^{2}=\sum_{n=0}^{\infty}\frac{|(g,e_{n})|^{2}}{\lambda_{n}}\sin^{2}((t-t_{0})\sqrt{\lambda_{n}}) \le \sum_{n=0}^{\infty}\frac{1}{\lambda_{1}}|(g,e_{n})|^{2} = \frac{1}{\lambda_{1}}\|g\|^{2}.$$ Because $f \in H^{1}_{0}$, then the equivalent condition $\sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(f,e_{n})|^{2} < \infty$ holds. This forces $v(t) \in H^{1}_{0}$ for all $t$ because $$\sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(v(t),e_{n})|^{2}= \sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(f,e_{n})|^{2}\sin^{2}((t-t_{0})\sqrt{\lambda_{n}})\le \sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(f,e_{n})|^{2} < \infty.$$ Similarly, $w(t) \in H^{1}_{0}$ for all $t$ because $g\in L^{2}$ and you're dividing the Fourier coefficients of $g$ by $\sqrt{\lambda_{n}}$, which puts the result in $H^{1}_{0}$. So everything is properly defined and $u(t) \in H^{1}_{0}$ for all $t$.

As I pointed out before, $h \in H^{1}_{0}$ iff $h\in\mathcal{D}(\sqrt{-\Delta})$, and, in that case, $$\|h\|^{2}_{H^{1}_{0}}=\|h\|^{2}_{L^{2}}+\|\nabla h\|^{2}_{L^{2}} = \|h\|^{2}_{L^{2}}+\|\sqrt{-\Delta}h\|^{2}_{L^{2}}=\sum_{n=0}^{\infty}(1+\lambda_{n})|(f,e_{n})|^{2}.$$ In fact, for $h \in L^{2}$, the sum on the right is finite iff $h \in H^{1}_{0}$. To shorten the expressions, let $s_{n}(t)=\sin((t-t_{0})\sqrt{\lambda_{n}})$ and $c_{n}(t)=\cos((t-t_{0})\sqrt{\lambda_{n}})$. What you've got is \begin{align} \|u(t)-u(t')\|_{H^{1}_{0}} & \le\left[\sum_{n=0}^{\infty}(1+\lambda_{n})|(f,e_{n})|^{2}|c_{n}(t)-c_{n}(t')|^{2}\right]^{1/2} \\ & +\left[\sum_{n=0}^{\infty}\frac{1+\lambda_{n}}{\lambda_{n}}|(g,e_{n})|^{2}|s_{n}(t)-s_{n}(t')|^{2}\right]^{1/2}. \end{align} The sums are absolutely and uniformly convergent with respect to $t$, $t'$ because $f \in H^{1}_{0}$ and $g\in L^{2}$. It follows that the above is continuous in $t,t'$. So $u : \mathbb{R}\rightarrow H^{1}_{0}$ is a continuous vector function. And when you take a derivative of $u$, you end up with a continuous function $u' : \mathbb{R}\rightarrow L^{2}$. The reduced regularlity is because of the added $\sqrt{\lambda_{n}}$ which appears in the derived sums. I'll let you work out the details.

Your New Modifications: The new mention of $\Box$ requires that you be able to take two time derivatives. This can be done if $f\in \mathcal{D}(-\Delta)$ and $g \in \mathcal{D}(\sqrt{-\Delta})=H_{0}^{1}(\Omega)$. If $f$, $g$ are as stated, then taking two time derivatives gives the same result as applying $-\Delta$ (both multiply the n-th Fourier coefficient by $\lambda_{n}$.) The expression $\Box$ is now interpreted in terms of a derivative of a vector function in $t$ and in terms of an operator $-\Delta$, which is more subtle than classical derivatives. So $\frac{d^{2}}{dt^{2}}u(t) = \Delta u(t)$ in this operator language, where $u : [t_{0},\infty)\rightarrow L^{2}(\Omega)$ is a vector function whose value lies in the domain of $-\Delta$ for all $t > 0$ so that $-\Delta u(t)$ makes sense, provided $f \in \mathcal{D}(-\Delta)$ and $g\in H_{0}^{1}$.

• Thank's for the answer, please, how you use Friedrichs extensions? – varphi May 22 '14 at 16:30
• Without getting too far into that, you can just look at $\|\nabla f\|^{2}=(-\Delta f,f)=\|\sqrt{-\Delta}f\|^{2}$ for $f \in H^{2}\cap H^{1}_{0}$. That's essentially all you need to see that the domain of $\sqrt{-\Delta}$ is $H_{0}^{1}$. – DisintegratingByParts May 22 '14 at 17:36
• But $g$ in in $L^2(\Omega)$ not $H^1(\Omega)$, so there is an problem for the second therm of $u$. No? – varphi May 22 '14 at 18:10
• When you divide $(g,e_{n})$ by $\sqrt{\lambda_{n}}$ and form the Fourier series, the result function is in $H^{1}_{0}$ because of the characterization of $H^{1}_{0}$ as $\mathcal{D}(\sqrt{-\Delta})$. – DisintegratingByParts May 22 '14 at 18:40
• Please, help me beaucause i don't understand this last point. Why $(g,e_n) \in H^1_0$? – varphi May 22 '14 at 18:58

I'm assuming that $f \in \mathcal{D}(-\Delta)$ and that $g \in \mathcal{D}(\sqrt{-\Delta})=H^{1}_{0}(\Omega)$. That way, $u(t)$ as you have defined is a continuously differentiable vector function of $t$ from $\mathbb{R}$ to $\mathcal{D}(-\Delta)$. And, \begin{align} (-\Delta)u(t) & =(-\Delta) \sum_{n=1}^{\infty}\left[(f,e_{n})\cos((t-t_{0})\sqrt{\lambda_{n}}) + \frac{(g,e_{n})}{\sqrt{\lambda_{n}}}\sin((t-t_{0})\sqrt{\lambda_{n}})\right]e_{n} \\ & = \sum_{n=1}^{\infty}\left[(f,e_{n})\cos((t-t_{0})\sqrt{\lambda_{n}}) + \frac{(g,e_{n})}{\sqrt{\lambda_{n}}}\sin((t-t_{0})\sqrt{\lambda_{n}})\right]\lambda_{n}e_{n} \\ & = -\frac{d^{2}}{dt^{2}}\sum_{n=1}^{\infty}\left[(f,e_{n})\cos((t-t_{0})\sqrt{\lambda_{n}}) + \frac{(g,e_{n})}{\sqrt{\lambda_{n}}}\sin((t-t_{0})\sqrt{\lambda_{n}})\right]e_{n}\;. \end{align} All of the sums involved are convergent in $L^{2}$ because of the assumptions on $f$, $g$. Suppose that $\psi \in C^{\infty}_{c}(I\times \Omega)$. Then $t\mapsto \psi(x,t)$ can be considered to be an infinitely differentiable function of $t$ with values in $L^{2}(\Omega)$. Using the inner-product $(\cdot,\cdot)$ on $L^{2}(\Omega)$, the function $t\mapsto (u(t),\psi(\cdot,t))$ is a twice continuously function of $t$ which is compactly supported in $I$. Furthermore, \begin{align} (\frac{d^{2}}{dt^{2}}u(t),\psi) &=\frac{d}{dt}(\frac{d}{dt}u(t),\psi) -(\frac{d}{dt}u(t),\frac{\partial}{\partial t}\psi) \\ (u(t),\frac{\partial^{2}}{\partial t^{2}}\psi) &=\frac{d}{dt}(u(t),\frac{\partial}{\partial t}\psi) -(\frac{d}{dt}u(t),\frac{\partial}{\partial t}\psi) \end{align} Integrating over $I$ then gives $$\int_{I}(\frac{d^{2}}{dt^{2}}u,\psi)\,dt=-\int_{I}(\frac{d}{dt}u,\frac{\partial}{\partial t}\psi)\,dt = \int_{I}(u,\frac{\partial^{2}}{\partial t^{2}}\psi)\,dt.$$ For fixed $t$, one can consider $\psi(t,\cdot)$ to be a vector in $\mathcal{D}(-\Delta)$, and, because $-\Delta$ is selfadjoint on $L^{2}(\Omega)$, $$(\frac{d^{2}}{dt^{2}}u(t),\psi)=(\Delta u(t),\psi)=(u(t),\Delta\psi).$$ Combining these last two equations gives $$\int_{I}(u,\Delta\psi)\,dt = \int_{I}(u,\frac{\partial^{2}}{\partial t^{2}}\psi)\,dt$$ In other words, $$(u,\Box \psi)_{L^{2}(I\times\Omega)}=0,\;\;\; \psi \in C_{c}^{\infty}(I\times\Omega).$$ Therefore $T_{u} \in \mathcal{D}'(I^{0}\times\Omega)$ defined by $T_{u}(\psi)=(u,\psi)_{L^{2}(I\times\Omega)}$ satisfies the weak equation $\Box T_{u}=0$.

You've added more to your question. Assuming $F=F(t)$ is in $C^{0}(I,L^{2}(\Omega))$, you must now solve for $u(t) \in L^{2}(\Omega)$ such that $$\Box u = F,\;\;\; u(0)=u'(0)=0.$$ This $u$ can be found by assuming the separation of variables solution $u(t)=\sum_{n}a_{n}(t)e_{n}$. Formally, the equation reduces to $$\sum_{n}(a_{n}''+\lambda_{n}a_{n})e_{n} = \sum_{n}(F,e_{n})e_{n},\\ a_{n}(0)=a_{n}'(0)=0.$$ So the coefficients $a_{n}$ satisfy and ODE $$a_{n}''(t)+\lambda_{n}a_{n}(t) = (F,e_{n}),\;\; a_{n}(0)=a_{n}'(0)=0.$$ The function $(F,e_{n})$ is an inner-product in $L^{2}(\Omega)$, and $(F,e_{n})$ is a continuous function of $t$ because $t\mapsto F(t)\in L^{2}(\Omega)$ is continuous. The solution I find by variation of parameters is the same as yours: \begin{align} a_{n} & =-\frac{1}{\sqrt{\lambda_{n}}}\cos(\sqrt{\lambda_{n}}t)\int_{0}^{t}\sin(\sqrt{\lambda_{n}}s)(F,e_{n})|_{s}\,ds \\ & +\frac{1}{\sqrt{\lambda_{n}}}\sin(\sqrt{\lambda_{n}}t)\int_{0}^{t}\cos(\sqrt{\lambda_{n}}s)(F,e_{n})|_{s}\,ds \\ & = \frac{1}{\sqrt{\lambda_{n}}}\int_{0}^{t}\sin(\sqrt{\lambda_{n}}(t-s))(F,e_{n})|_{s}\,ds. \end{align} The coefficients $a_{n}(t)$ are twice continuously differentiable and satisfy $$|\sqrt{\lambda_{n}}a_{n}|^{2} \le \int_{0}^{t}|\sin(\cdots)|^{2}\,ds\int_{0}^{t}|(F,e_{n})|^{2}\,ds.$$ So, as a crude estimate, $|\sin(\cdots)|\le 1$ and $$\sum_{n}|\sqrt{\lambda_{n}}a_{n}|^{2} \le t\int_{0}^{t}\sum_{n}|(F,e_{n})|^{2}\,ds = t \int_{0}^{t}\|F(s)\|^{2}\,ds$$ Therefore, $u(t)=\sum_{n}a_{n}e_{n}$ is in $H^{1}_{0}$ for each $t \ge 0$. It's not too hard show that $t \mapsto u(t)\in H_{0}^{1}$ is continuous. Taking a derivative with respect to $t$ of $u$ leads to sum where the $1/\sqrt{\lambda_{n}}$ disappears, and $t\mapsto u'(t)\in L^{2}(\Omega)$ is continuous.

Note: The previous arguments used to show that you have a weak solution no longer work because the assumptions you have given are not enough to guarantee that $t\mapsto u(t)$ is a twice continuously differentiable vector function. However, it is still the case that $\Box T_{u} = F$ in the weak sense, which adds an interesting and important twist to the problem. The argument is still simple, just a little different. If $\psi \in C_{c}^{\infty}(I^{0}\times \Omega)$, then $\Box(a_{n}(t)e_{n}(x))=(F,e_{n})_{L^{2}(\Omega)}e_{n}$ gives \begin{align} (\Box T_{u})(\psi) = \int_{I^{0}\times\Omega}u\Box \psi\,dx\,dt & = (u,\Box\psi)_{L^{2}(I\times\Omega)} \\ & =(\sum_{n}a_{n}(t)e_{n}(x),\Box\psi(t,x))_{L^{2}(I\times\Omega)} \\ & = \sum_{n}(a_{n}e_{n},\Box\psi)_{L^{2}(I\times\Omega)} \\ & = \sum_{n}(\Box(a_{n}e_{n}),\psi)_{L^{2}(I\times\Omega)} \\ & = \sum_{n}((F,e_{n})_{L^{2}(\Omega)}e_{n},\psi)_{L^{2}(I\times\Omega)}\\ & = (\sum_{n}(F,e_{n})_{L^{2}(\Omega)}e_{n},\psi)_{L^{2}(I\times\Omega)}\\ & = (F,\psi)_{L^{2}(I\times\Omega)}. \end{align} Of course, this assumes you know that the eigenfunctions $e_{n}(x)$ are smooth (which they are) and vanish at the boundary of the smooth region $\Omega$ (which they do.)

Justification of interchange of sum and inner-product: You have edited the question again, to ask about the justification for going from line (2) to (3) of the multi-line equation above. All you need to do in order to establish that equality is to show that $\sum_{n} a_{n}(t)e_{n}(x)$ converges in $L^{2}(I\times\Omega)$. First notice that $(a_{n}e_{n},a_{k}e_{k})_{L^{2}(I\times\Omega)}=0$ for $n\ne k$, which follows from the orthogonality of $\{ e_{n} \}$ in $L^{2}(\Omega)$. So, look at $$\|a_{n}e_{n}\|^{2}_{L^{2}(I\times\Omega)}=\|a_{n}\|^{2}_{L^{2}(I)}\|e_{n}\|^{2}_{L^{2}(\Omega)}=\|a_{n}\|^{2}_{L^{2}(I)}$$ To get what you want, all you need to show is that $\sum_{n}\|a_{n}\|^{2}_{L^{2}(I)} < \infty$. From the crude estimate given above, $$\sqrt{\lambda_{1}}\sum_{n}|a_{n}(t)|^{2} \le \sum_{n}|\sqrt{\lambda_{n}}a_{n}(t)|^{2} \le t \int_{0}^{t}\|F(s)\|_{L^{2}(\Omega)}^{2}\,ds.$$ The integral over $I$ of the right side is finite; so that gives you all you need to get convergence in $L^{2}(I\times\Omega)$ of the orthogonal series $u = \sum_{n}a_{n}(t)e_{n}(x)$.

• Your proof is exellente. I understand all the steps. Thank you very much. – varphi May 24 '14 at 18:18
• Please, i have a last question – varphi May 24 '14 at 18:19
• Let $\Omega$ an open bounded on $\mathbb{R}^n$, $I$ an interval on $\mathbb{R}$, $t_0 \in I$. Let $F=(F_t)\in C^0(I,L^2(\Omega))$. There exist a unique$u\in C^0(I,H^1_0(\Omega)) \cap C^1(I^0,L^2(\Omega))$ such that $$\Box u= F \quad \mbox{dans} \mathcal{D}'(I^0 \times \Omega), \quad u_{t_0}=0,\quad u^{(1)}_{t_0}=0$$ and this solution is $$u_t=\sum_{n\geq 1}(\displaystyle\int_{t_0}^t\dfrac{\sin(\sqrt{\lambda_n} (t-s))}{\sqrt{\lambda_n}} F_n(s)ds)e_n$$, where $$F_t=\sum_{n\geq 1}F_n(t)e_n$$ – varphi May 24 '14 at 18:24
• My problem is to prouve that this u satisfies $u\in C^0(I,H^1_0(\Omega)) \cap C^1(I^0,L^2(\Omega))$ ans $\Box u= F$ in $\mathcal{D}'(I^0 \times \Omega)$. Help me please. – varphi May 24 '14 at 18:25
• Assume the solution is found by separation of values as $\sum_{n=1}F_{n}(t)e_{n}$ and arrive at $F_{n}''(t)+\lambda_{n}F_{n}=(F,e_{n})$. This gives one ODE for each coefficient--you'll need initial conditions $F_{n}(0)=F_{n}'(0)=0$. The verification should follow the same basic steps that I outlined for you. – DisintegratingByParts May 24 '14 at 18:33