5
$\begingroup$

From section 7.5 in this source, I see that, for $\vec{x} \in \mathbb{R}^3$, if $$ \frac{\partial^2 u\left(\vec{x},t\right)}{\partial t^2} - \nabla^2u\left(\vec{x},t\right) =f\left(\vec{x},t\right),\qquad u\left(\vec{x},0\right) = \phi\left(\vec{x}\right), \qquad \left.\frac{\partial u\left(\vec{x},0\right)}{\partial t}\right|_{t=0} = \psi\left(\vec{x}\right) $$ ... then ... $$ u\left(\vec{x},t\right) = \frac{1}{4\pi t^2} \oint_{\partial B \left(\vec{x},t\right)} \left[ \phi\left(\vec{y}\right) + t\psi\left(\vec{y}\right) + \nabla \phi\left(\vec{y}\right)\cdot\left(\vec{y}-\vec{x}\right) \right]\;dS\left(\vec{y}\right) + \int_{0}^{t} \frac{1}{4\pi\left(t-s\right)} \left[ \oint_{\partial B\left(\vec{x},t-s\right)} f\left(\vec{y},s\right)\; dS\left(\vec{y}\right)\right]\;ds $$ I have two questions.

My First Question is, if I have: $$ \color{blue}{ \color{red}{\frac{1}{c^2}}\frac{\partial^2 u\left(\vec{x},t\right)}{\partial t^2} - \nabla^2u\left(\vec{x},t\right) =f\left(\vec{x},t\right),\qquad u\left(\vec{x},0\right) = \phi\left(\vec{x}\right), \qquad \left.\frac{\partial u\left(\vec{x},0\right)}{\partial t}\right|_{t=0} = \psi\left(\vec{x}\right) } $$ ...is it true that ... $$ \large{ \color{blue}{ u\left(\vec{x},t\right) = \frac{1}{4\pi \color{red}{c^2}t^2} \oint_{\partial B \left(\vec{x},\color{red}{c}t\right)} \left[ \phi\left(\vec{y}\right) + t\psi\left(\vec{y}\right) + \nabla \phi\left(\vec{y}\right)\cdot\left(\vec{y}-\vec{x}\right) \right]\;dS\left(\vec{y}\right) + \int_{0}^{t} \frac{1}{4\pi\left(t-s\right)} \left[ \oint_{\partial B\left(\vec{x},\color{red}{c}t-\color{red}{c}s\right)} f\left(\vec{y},s\right)\; dS\left(\vec{y}\right)\right]\;ds } } $$ (I have highlighted in red the difference from the earlier expression)

For $c \neq 1$, the first term is worked out in section 7.2 of the same source which details the case where $f\left(\vec{x},t\right)\equiv0$ (Kirchhoff's Formula). I have tried to apply a change of variables the same way on the second term, but I'm looking for some confirmation (or correction!) that in this term $\partial B\left(\vec{x}, \color{red}{c}t - \color{red}{c}s\right)$ is the only place where $c$ will appear.

My Second Question is, is the expression above actually identical to the following (focus on the second term)? $$ \large{ \color{blue}{ u\left(\vec{x},t\right) = \frac{1}{4\pi} \left[\oint_{\partial B \left(\vec{x},{c}t\right)} \frac{ \phi\left(\vec{y}\right) + t\psi\left(\vec{y}\right) + \nabla \phi\left(\vec{y}\right)\cdot\left(\vec{y}-\vec{x}\right) }{c^2t^2}\;dS\left(\vec{y}\right)\right] + \frac {1}{4\pi} \left[ \iiint_{B\left(\vec{x},ct\right)} \frac{f\left(\vec{y},t-\dfrac{\left|\vec{y}-\vec{x}\right|}{c}\right)}{\left|\vec{y}-\vec{x}\right|} \; dV\left(\vec{y}\right) \right] } } $$

Thanks...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.