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Theorem: Let $X$ be a Banach space, $\{T(t)\}_{t\geq 0}$ a $C_0$-semigroup on $X$ and $U_0\in X$. If $A:D(A)\subset X\to X$ is the infinitesimal generator of $\{T(t)\}_{t\geq0}$, then the function $U:[0,\infty)\to X$ given by $U(t)=T(t)U_0$ is a solution of $(1)$. $$\left\{\begin{align*} U_t(t)=AU(t);&~~~~t\in[0,\infty)\\ U(0)=U_0& \end{align*}\right.\tag{1}$$

Now consider the problem

$$\left\{\begin{align*} y_t(x,t)=y_{xx}(x,t);&~~~~&&x\in\mathbb{R};\;t\in[0,\infty)\\ ~y(x,0)=f(x);&&&x\in\mathbb{R} \end{align*}\right.\tag{2}$$

where $y_{xx}$ is the weak derivative of second order of $y$. By theorem above, it's possible to show that $(2)$ has a solution.

So, could someone explain me (with some details) how can we rewrite $(2)$ in order to get a equivalent system, analogous to $(1)$?

The solution that I saw just says that it's enough to show that the operator $A:H^2(\mathbb{R})\to L^2(\mathbb{R})$ defined by $A(y)=y_{xx}$ is a infinitesimal generator of a $C_0$-semigroup on $L^2(\mathbb{R})$ (for this, the Hille-Yosida theorem is used however my question is not about the application of the Hille-Yosida Theorem. I need help to understand how to transform the original system in a system like $(1)$ and why the existence of a solution for $(1)$ implies the exitece of a solution for the original system).

Thanks.

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I think if you see the comments on the last 6 lines on page 206 of Pazy's book, he mentioned this. I understand that the solution of the abstract equation (using semigroup theory) does not satisfy necessarily the partial differential equation. To be a solution of the original partial differential equations, it should satisfy some regularity property which is outside the field of the abstract theory (semigroup). So you can call the solution given by semigroup theory a weak solution, because the semigroup method doesn't care about the spatial variable, it only cares about the time variable. –  user144542 May 30 at 20:31
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2 Answers 2

It's pretty straightforward. Some of it is already in the solution you cited. Here $X=L_2(\mathbb R)$, $A=\partial^2_x$, $D(A)=H^2(\mathbb R)$, $U_0$=$f$. Applying the theorem one gets the result for (2).

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Strictly speaking, applying the theorem one gets a function $U:[0,\infty)\to L^2(\mathbb{R})$ that satisfies $$(3)\left\{\begin{align*} U_t(t)=\partial_x^2U(t);&~~~~t\in[0,\infty)\\ U(0)=f& \end{align*}\right.$$ that is, $$(4)\left\{\begin{align*} &\lim_{h\to 0}\left\|\frac{U(t+h)-U(t)}{h}-\partial_x^2U(t)\right\|_{L^2(\mathbb{R})}=0;~ t \in [0,\infty)\\ &U(0)(x)=f(x);\;x\in\mathbb{R} \end{align*}\right.$$ Why does $(4)$ imply $(2)$? Is it pretty straightforward? –  Pedro Jan 7 at 12:46
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In order to get the solution of the Cauchy problem (2), you need some boundary conditions, otherwise you cannot write che domain for the operator $A$. Moreover, you have to consider the abstract Cauchy problem (1), which is much more easy to treat. Over suitable functional spaces (1) is equivalent to (2), then if you prove that (1) is well posed, i.e. its solution exists, is unique and depend continuously from data then (2) is well posed as well.

The theory says that for a closed operator $A$ the associated abstract Cauchy problem is well posed if and only if $A$ generates a strongly continuous semigroup

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Could you give me a concrete example of a system of PDE that is equivalent to $(1)$? The theory that I have seen gives theorems that ensures existence of solutions for an ACP (abstract Cauchy problem). I can understand how to treat an ACP, but I need help to understand the applications to pde. For me, the word "equivalence" is the problem. If you see my first comment in the Andrew's answer, probably you will better understand what I say. –  Pedro Jan 7 at 13:04
    
You can see here en.wikipedia.org/wiki/Semigroup_theory or for more detailed studies, you can see on the books of Engel, Nagel "One parameter semigroups for linear evolution equations", Pazy "Semigroups of linear operators and applications to PDE", Angenent's "one parameter semigroups". My advice: the matter is difficult so in order to tackle the problems you have to study everything step by step. –  twin prime Jan 7 at 14:06
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