Can't get rid of integrals solving this differential equation.

Question

Assuming $\mathbf{A}\equiv \vec A$ , $\dot q\equiv \frac{d}{dt}q$ ,and $\ddot q\equiv \frac{d^2}{dt^2}q$ ,

And Using the Right-hand Cartesian coordinate system with base vectors $\mathbf{\hat i\, \hat j \, \hat k}$ corresponding to the axes $x \, y\, z$,

This is a derived given equation:

$$\mathbf{\ddot r}(t)=\cos(t)\mathbf{\hat i}-\dot r_z(t) \sin(t) \mathbf{\hat i}+\dot r_x(t) \sin(t)\mathbf{\hat k}$$

Having the initial conditions such that $\mathbf{r}(0)=\mathbf0$ and $\mathbf{\dot r}(0)=\mathbf{0}$.

The method is to integrate both sides twice. But after the first integration, due to having product of two t-dependant functions in the second term, an integral will remain unsolved. (using integration by parts).

If you couldn't get what I mean, try integrating the second term of the right side of the equation using this method:

Then you'll find out what I mean by "how to get rid of the integral".

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The question is that: is there any alternative to solving this? (Other than integration by parts, I mean).

If not, then what is my mistake? or how should my approach be revised to get the solution?

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EDIT: The approach to the equation. This isn't a problem or homework.

I just wondered how the motion of a charged particle in a combination of electric and magnetic field (with the specified function) would be. picture

Answer 1

The $y$ component is not coupled to the other two components and has the trivial solution $r_y(t)=0$.

For the other two components, since $\mathbf r(t)$ itself doesn't appear in the equation, we can solve for $\mathbf v(t)=\dot{\mathbf r}(t)$ and then integrate. So we want to solve

$$ \pmatrix{\dot v_x(t)\\\dot v_z(t)}=\pmatrix{0&-\sin t\\\sin t&0}\pmatrix{v_x(t)\\v_z(t)}+\pmatrix{\cos t\\0}\;. $$

These equations are decoupled if we introduce $v_\pm(t)=v_x(t)\pm\mathrm iv_z(t)$ to obtain

$$ \dot v_+(t)=\hphantom -\mathrm i\sin tv_+(t)+\cos t\;, \\ \dot v_-(t)=-\mathrm i\sin tv_-(t)+\cos t\;. $$

To solve the homogeneous versions of these equations, divide by $v_\pm(t)$, yielding

\begin{align} \frac{\dot v_\pm(t)}{v_\pm(t)}=\pm\mathrm i\sin t\;, \end{align}

and integrate to obtain

$$ \log v_\pm(t)=\mp\mathrm i\cos t+C_\pm\;,\\ v_\pm(t)=c_\pm\mathrm e^{\mp\mathrm i\cos t}\;, $$

and thus

$$ v_x(t)=\hphantom -c\cos\cos t+d\sin\cos t\;,\\ v_z(t)=-c\sin\cos t+d\cos\cos t\;. $$

A particular solution of the inhomogeneous equations can in principle be obtained by varying the constants, but in the present case the resulting integral,

$$ c_\pm(t)=\int_0^t\mathrm e^{\pm\mathrm i\cos\tau}\cos\tau\,\mathrm d\tau\;, $$

can't be expressed in terms of elementary functions. The same is then also true for $r_x(t)$ and $r_z(t)$, which result from integrating $v_x(t)$ and $v_z(t)$.

You may also be interested in this treatment of the problem.