# How to find a function with given initial values of other function?

I am looking for a continuous function $$\theta(t)$$ to steer a particle on the trajectory $$x(t)$$.

$$x$$ and $$\theta$$ relate through this constraint

$$\ddot x(t) = Tsin(\theta(t))$$ where $$T$$ is a non-zero positive constant.

$$\theta(t)$$ must satisfy these initial conditions

$$\theta(0) = 0$$

$$\theta(t_0) = 0$$

$$x(t)$$ must satisfy these initial values.

$$x(0)=x_a, x(t_0)=x_b$$

$$\dot x(0)=0, \dot x(t_0)=0$$

$$\ddot x(0)=0, \ddot x(t_0)=0$$

The two functions $$x(t)$$ and $$\theta(t)$$ only need to be continuous between $$x_a$$ and $$x_b$$. $$x_b$$ can be greater or lesser than $$x_a$$. Further, the function $$x(t)$$ should be bijective on $$[0, t_0]$$

This is how I got a solution for $$x(t)$$. I just guessed what seemed right until I got it working. Perhaps there also exists a polynomial solution.

$$\ddot x=-rsin(t)$$

Thus $$\ddot x(0)=0$$ and $$\ddot x(t_0) \implies t_0 \in 0,\pi,2\pi , ...$$

$$\dot x(t) = rcos(t) + C_1$$

$$\dot x(0) = 0 = r + C_1$$

$$\dot x(t_0) = 0 = rcos(t_0) + C_1$$

Since $$t_0 \in 0, \pi, 2\pi,...$$. We can isolate $$C_1$$ if we force $$t_0 = 2\pi$$

Thus $$\dot x(t_0) = 0 = r + C_1 \implies C_1 = -r$$

$$x(t)=rsin(t) - rt + C_2$$

$$x(0) = x_a = C_2$$

$$x(t_0) = x_b = rsin(2\pi) - r2\pi + C_2 \implies r = \dfrac{x_a - x_b}{2\pi}$$

Therefore, $$x(t) = rsin(t) - rx + x_a$$, where $$r = \dfrac{x_a - x_b}{2\pi}$$

Now here is where I am stuck. I do not understand how I can find $$\theta(t)$$ from the given equation. I tried doing

$$\ddot x(t) = Tsin(\theta(t)) = -rsin(t)$$

but I'm not sure how to maintain the $$x(t)$$ constraints

Here is a engineering type solution for some range of parameters. There is some flexibility in the choice of a function $$f$$, but I give a hard bound below based only on the information given. At the end, I draw a representative example.

## Step 1: candidate path $$x(t)$$

Set $$x_a=0, x_b=L$$. $$L=0$$ is impossible by the bijection requirement. Set $$L>0$$ for notational simplicity. (If $$x_b, then let the particle go from $$x_b$$ to $$x_a$$ instead, and then go backwards in time.) Let $$f$$ be any function such that $$f:[0,1]\to[0,\infty), \quad f(0)=f'(0)=f''(0)=f''(1)=f'(1)=f(1) = 0,\\ x\in(0,1)\implies f(x)>0$$ A polynomial example exists;you can consider $$-t^3(t-1)^3$$. Or you can use the following $$f:[0,1]\to [0,\infty) , \quad f(t):= \begin{cases}\exp\frac{-1}{1-(2t-1)^2}& x\neq 0 ,1,\\ 0 & x=0,1\end{cases}$$ This is up to scaling, a classical function in the theory of PDEs known as a "bump function" or the "standard mollifier". It is $$C^\infty$$, bounded above, and verifies $$0=f(0)=f(1)=f'(0)=f'(1)=\dots=f^{(k)}(0)=f^{(k)}(1)=\dots$$ It follows that if we integrate it from $$0$$ to $$t$$ to get $$g(t)$$, then $$g$$ and all required derivatives vanish at 0, while all required derivatives of $$g$$ vanish at $$1$$ (but not $$g(1)$$). Now consider $$f_{t_0} : [0,t_0] \to \mathbb [0,\infty)$$, achieved by a rescaling $$f_{t_0}(t) = cf(t/t_0)$$ where $$c$$ is chosen so that $$\int_0^{t_0} f_{t_0} (s)ds = 1$$. The same is true of $$f_{t_0}$$ now, but at $$0$$ and $$t_0$$. Now define $$x:[0,t_0]\to[0,L],\quad x(t)= L\int_0^t f_{t_0}(s) ds.$$ It follows from $$f>0$$ in $$(0,1)$$ that $$x$$ is a bijection, and satisfies all boundary conditions.

## Step 2: satisfying the ODE

One now needs to check if $$|x''(t)|\le T$$ for all $$t$$. In terms of $$f$$ this is $$|Lf_{t_0}'(t)|=\frac{cL}{t_0}|f'(t/t_0)|\le T,$$ so if $$\sup_{s\in[0,1]}|f'(s)| \le \frac{Tt_0}{cL},$$ then this procedure works. (In snappier notation this is $$\|f\|_{L^1}\|f'\|_{L^\infty} \le \frac{Tt_0}{L}$$.) Since $$c$$ depends on $$f$$ you may get better results with other $$f$$.

In fact, the furthest you can go in the + direction is controlled by the solution to $$x'' \equiv T$$, which leads to $$x(t) = Tt^2/2 + At + B$$. $$A=x'(0)=0$$. $$B=x(0)=x_a=0$$. $$x_b = x(t_0)= Tt_0^2/2$$. So this computation gives $$L\le Tt_0^2/2$$. If this isn't satisfied, you can go home.

## Step 3: path to angles

This part is easy. Given the above calculations we can invert the relationship for $$\theta$$. Under the above assumption, we have $$x''/T\in[-1,1]$$. Just define $$\theta(t) = \arcsin(x''(t)/T).$$ Here, arcsin is a function $$[-1,1]\to[-\pi,\pi]$$. There's a unique solution for $$\arcsin(x)=0$$ which is $$x=0$$ so the boundary conditions are met. This finishes the construction.

Here's a Desmos graph using the degree 6 polynomial I gave. I followed my construction exactly (but the variable names in Desmos are different). The blue line is $$x$$. The yellow line is $$x'$$. The dotted yellow line is $$x''$$. The green horizontal is $$y=T$$. The shaded region shows the parts where $$x''>T$$ ; so this is not a solution but would become one if $$T$$ were increased enough. In particular (response to comments) note that $$x''(0)=0=x''(L)$$.

• Thanks for the detailed response. However since $x(t_0) = x_b$ shouldn't $f(1)= L$? (Instead of 0 as you wrote) – Cedric Martens Jan 3 at 6:58
• @CedricMartens no, I defined my $x$ using the integral of $f$. – Calvin Khor Jan 3 at 6:59
• @CedricMartens in particular $L= x(t_0) = cL\int_0^{t_0} f(s/t_0)ds$ and the value of $f(1)$ is not the important number here – Calvin Khor Jan 3 at 7:05
• I am sorry for this, but over an hour ago I edited my question to fix a conflict. It seems that you started before the edit. I changed $\ddot x(t) = Tsin(\theta(t))$ (before it was a cosine). It follows that at time zero the acceleration should be zero and not $T$. I'm sorry for the mistake and I greatly appreciate the answer. – Cedric Martens Jan 3 at 7:13
• @CedricMartens Thank you very much for your apology. In fact I already know this, because right when you changed the question, I had already written an answer to the older version of the question. The acceleration at time 0 for my answer is 100% without a doubt, 0. This is stated at the beginning of step 1. I will make you a graph to prove this. – Calvin Khor Jan 3 at 7:17