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Let $$ x:[0,\infty) \to \mathbb{C}^n$$ such that $$|x(t)|\leq Me^{\alpha t}$$ for some constants $M \geq 0$ and $a\in \mathbb{R}$. Then the Laplace transform $$\mathbb{L(x)(s)}=\int_0^\infty e^{-st}x(t)dt$$ exists and is analytic for $\Re(s)>a$. Show that for $x\in C^1([0,\infty))$ satisfying $$|x(t)|+|\dot x(t)|\leq Me^{\alpha t}$$ we have $$\mathbb L(\dot x)(s)=sL(x)(s)-x(0)$$ for $\Re(s)>a$. Moreover, I have to show that the Initial Value Problem $$\dot x(t)=Ax(t)+g(t)\quad\text{with}\quad x(0)=x_0$$ is transformed into a linear system of equations by the Laplace transform.

As Artem suggested I am integrating by parts: $$\mathbb{L(\dot x)(s)}=\int_0^\infty e^{-st}\dot x(t)dt$$ So using integration by parts we have $$\mathbb{L(\dot x)(s)}=\int_0^\infty e^{-st}\dot x(t)dt=e^{-st}x(t)+s\int_0^\infty e^{-st}x(t)dt=e^{-st}x(t)|_{t=0}^{t=\infty}+s\mathbb{L(x)(s)} =s\mathbb{L(x)(s)}-x_0$$

I'm thinking: Take the Laplace transformation of both sides of $$\dot x(t)=Ax(t)+g(t)$$ $$\mathbb L ({\dot x(t)})=\mathbb L( {Ax(t)+g(t)})$$ $$s\mathbb L ({x(t)})-x_0=A\mathbb L( {x(t)}+sL(g(t)))$$ $$(sI-A)\mathbb L ({x(t)})=x_0+sL(g(t)))$$ which is a a linear system of equations.

Am I done?

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    Hint: integration by parts.2012-11-25
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    @Artem I am working on it up. Thank you!2012-11-25
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    Here is a [reference](http://faculty.atu.edu/mfinan/4243/section75.pdf) to the method.2012-11-25
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    @MhenniBenghorbal Thank you so much, great reference!2012-11-25
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    @Klara: You are very welcome.2012-11-25

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