Max temperature on circle
Consider the heat problem on the unit disk $D = D_1(0,0)$ described by the heat equation with source
$$u_t = \Delta U + 2x+3y,$$
where $u$ also satisfies $u(x,y,0) = 0$ for all $(x,y)\in D$ and $\nabla u\cdot \vec{n} = 0$ at all points on the boundary of $D$ (i.e., the boundary is insulated).
As it turns out, $u(x,y,t)$ has a unique maximum value for all $t>0$; that value increases as $t$ increases but its location does not move. Thus, there is a unique point $(x_m,y_m)$ where the maximum occurs. Where exactly is that point?
Comments
An answer close to the point of maximum temperature can be obtained using the Heat Explorer tool, but it may be difficult to locate the precise point using the graph. Here was the point of greatest temperature that I could locate on the graph, $(0.55508, 0.83179)$:
Here, we will calculate the exact point through calculus, then verify whether our answer is reasonably close to the estimate obtained above.
First, let us determine what we are looking for. We want the point at which the temperature is greatest. There is only one source of heat, $f(x,y) = 2x + 3y$, and it does not change with respect to time. This means the point of greatest temperature does not change location over time, and can be found by finding the point at which $f(x,y)$ is at its maximum. Thus, the problem reduces to a classic max-min problem we are used to from calculus.
If we were to simply take the max and min of $f(x,y)$, we would set $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$, and obtain $2 = 0$ and $3 = 0$. These impossible results indicate that $f(x,y)$ has no extrema, which is also evident from the fact that $f$ is a linear function of $x$ and $y$, and thus increases indefinitely with $x$ and $y$.
However, while we may only have one heat source function, we do have another function we have to consider, namely that of the unit disk itself. Our space is limited to coordinates $(x,y)$ that obey the relation $x^2 + y^2 \leq 1$. Furthermore, since we know that $f(x,y)$ increases with increasing $x$ and $y$, it follows that $f$ is maximized somewhere at the edge of the disk in the first quadrant (where $x,y \geq 0$).
Now we are finally in a position to calculate $(x_m, y_m)$. First, we rewrite the unit disk formula,
$$x^2 + y^2 = 1$$
to obtain y as a function of x:
$$y = \sqrt{1 - x^2}$$
(Recall we are only interested in the first quadrant, so we do not have to worry about allowing for negative $y$ values.)
We then substitute this function for y in the $f(x,y)$ formula to find $f$ in terms of $x$
$$f = 2x + 3\sqrt{1 -x^2}.$$
Now we can set $\frac{df}{dx} = 0$ to find $x_m$:
$$\frac{df}{dx} = 2 - \frac{3x}{\sqrt{1-x^2}} = 0.$$
We then rearrange to solve for x:
$$x_m = \frac{2}{\sqrt{13}} \approx 0.55470$$
Finally we plug $x_m$ into our formula to solve for $y_m$:
$$y_m = \frac{3}{\sqrt{13}}. \approx 0.83205.$$
Thus, we arrive at our final answer
$$(x_m, y_m) = \left(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right) \approx (0.55470, 0.83205).$$
This seems seems very close to the hottest point I could locate on the graph!