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I want to know the sufficient condition to change the order of integral and differential. following is the problem:

$$f \in C([0,T];L^2(R^N))$$ $$\frac{d}{dt}f=g(t,x)$$ Then we multiply $f$ on both sides and take $L^2$-innerproduct, so we get $$\frac{1}{2}\frac{d}{dt}\|f\|^2_{L^2}=<g,f>.$$

In this process, I have a question in LHS. We implicitly used the change of differential and integral operator, and I want to know why it holds. I tried to understand it by Dominate Convergence Theorem, but I don't know how to apply it in this problem because we don't have any condition of g. Anybody would help me please?

Lalit Tolani
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yunjoo
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1 Answers1

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Starting from the left-hand side: $$\langle g, f\rangle = \int_{\mathbb R^n} f(t, x)g(t, x) dx$$ Substituting in $g = \frac{\partial f}{\partial t}$ we get $$\int_{\mathbb R^n} f(t,x) f_t(t,x) dx = \int_{\mathbb R^n} \partial_t \left[\frac12 f^2(t,x)\right] dx$$ Now consider the general statement of the Leibniz integral rule seen here. (apologies for referencing Wikipedia)

Restrictions $1$ and $2$ are met pretty trivially, and I believe we can say that restriction $3$ is met if the inner product between $f$ and $g$ converges, but I'm not very sure of that.

Assuming the restrictions are met, we can continue with

$$\int_{\mathbb R^n} \partial_t \left[\frac12 f^2(t,x)\right] dx = \frac12 \partial_t \int_{\mathbb R^n} f^2(t,x) dx = \frac12 \partial_t ||f||^2_{L^2}$$

Stephen Donovan
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