Every day, you walk from point A to point B which are exactly 2 miles apart straight line distance, however, each day, there is a 50% chance of there being an obstructing wall perpendicular to the direct AB segment. The wall cannot be seen so you wont know it is there until you actually bump it. It is like an invisible force field that forces you to walk around it when you bump it and you will know immediately when you have cleared it , thus you can change your path once cleared. The wall extends 1 mile in both directions perpendicular to the direct AB path so if that wall is at the midpoint of AB, it would create a symmetrical + shape with the direct AB path. Additionally, there is a 2nd obstructing wall that we have to deal with 25% of the time (on average) but is only half of the length of the larger wall so it extends only .5 miles in both directions perpendicular to the direct AB path. The 2 walls can be present independently of each other.

You can assume all ground is flat and that neither wall will ever be within the first or last half mile of the direct path line segment between A and B. That is, the 2 walls can only be in the middle mile between A and B if at all. For any given walk, there could be 0, 1, or 2 walls present. Also, any walls will be uniformly distributed in the middle mile. If the 2 walls are at the same exact spot, you can just treat that as if only the large wall is present since the obstruction would be identical.

What walk strategy will minimize the walk distance on average going from A to B? That is, if you were to connect the dots of all the optimal (x,y) coordinates to be at during an average walk, what would the shape of the path look like (on a non-wall day)?

What is that minimum average distance to walk?

**Answer**

This answer has been through so many revisions, but now I’ve gotten all the pieces completed and I am trying to make something readable out of it all. There are really 3 main cases to deal with:

1. The path before any wall has been encountered

2. The path after the small wall but before the big wall

3. The path after the big wall but before the small wall.

Then the results of the 3 parts can be combined to find an average distance walked.

**Part 1**. The path before the first wall.

The big wall is present 50% of the time, and it’s uniformly distributed in [12,32] when present. Thus the probability that the big wall has not been seen for 12≤x≤32 is Pb=12+12(32−x)=54−12x, and the small wall is only present 25% of the time, so the same probability for the it is Ps=34+14(32−x)=98−14x. So the probability that no walls have been encountered is Pbs=PbPs=18x2−78x+4532. Then the probability of the first wall being between x and x+dx is Pbs(x)−Pbs(x+dx)=−dPbsdxdx=(−14+78x)dx. We now compute the mean path length from the starting point to the first wall encountered, subtracting the distance to its center, or to the goal at B if no wall is ever encountered. This is

ˉs=∫3212[√14+(y(x1))2+∫x212√1+(y′(x))2dx−y(x2)](−14x2+78)dx2+38[√14+(y(x1))2+∫3212√1(+y′(x))2dx+√14+(y(x3))2]

Where x1=12 and x3=32. Now we change order of integration as usual to get

∫3212(−14x2+78)∫x212√1+(y′(x))2dxdx2=∫3212√1+(y′(x))2∫32x(−14x2+78)dx2dx=∫3212√1+(y′(x))2(18x2−78x+3332)dx

Combining this with that −y in the original integral and a similar integral from the part that took into account the contribution to the mean path length if no walls were present and the straight line parts, the path simplifies to

ˉs=∫3212[(18x2−78x+4532)√1+(y′(x))2−(78−14x)y(x)]dx+√14+(y(x1))2+38√14+(y(x3))2

The effect of a small deviation from the optimal path, δy(x) is

δˉs=∫3212[(18x2−78x+4532)y′(x)√1+(y′(x))2δy′(x)−(78−14x)δy(x)]dx+y(x1)√14+(y(x1))2δy(x1)+38y(x3)√14+(y(x3))2δy(x3)=−∫3212[ddx{(18x2−78x+4532)y′(x)√1+(y′(x))2}+(78−14x)]δy(x)dx+[y(x1)√14+(y(x1))2−y′(x1)√1+(y′(x1))2]δy(x1)+38[y(x3)√14+(y(x3))2+y′(x3)√1+(y′(x3))2]δy(x3)

Since the variation in ˉs must be invariant under small but arbitrary variations in path, the contents of each of the square brackets must be zero. The first line will yield the differential equation for the curved part of the path, and the second line has the solution

y′(x1)=2y(x1)=y(x1)−012−0

Which shows that the first derivative is continuous across x=12. The third line similarly has the solution

y′(x3)=−2y(x3)=0−y(x3)2−32

Which is the statement of continuity of the first derivative across x=32.

Now back to that differential equation. We have an immediate integral which we can write as

(18x2−78x+4532)y′(x)√1+(y′(x))2=(18x2−78x+4532)+1−c214

We can solve for y′(x) to get

y′=(72−x)2−c21−(c21−1)2√c21−1√(72−x)2−c21

Then we can substitute 72−x=c1coshθ so this translates to

y′=c21sinh2θ−(c21−1)2c1√c21−1sinhθ=c21cosh2θ+1−2c212c1√c21−1sinhθ

Recalling that dx=−c1sinhθdθ we can integrate to get

y=−∫c21sinh2θ−(c21−1)2c1√c21−1sinhθc1sinhθdθ=14√c21−1(−c21sinhθcoshθ+(3c21−2)θ)+c2

We can insert these expressions into the first derivative continuity conditions to get

10−2c212√c21−1√9−c21=12√c21−1(−3√9−c21+(3c21−2)ln(3+√9−c21c1))+2c2

5−2c212√c21−1√4−c21=−12√c21−1(−2√4−c21+(3c21−2)ln(2+√4−c21c1))−2c2

We can eliminate c2 between these two equations so we only have one difficult equation to solve for c1 with the result c1=1.814022405933661 and c2=0.004495637179259.

We have a relatively nice expression for

√1+(y′(x))2=c21cosh2θ−12c1√c21−1sinhθ

In terms of θ.

With full knowledge of the curve in hand, from the continuity equations we have a simplified expression for the straight line path from (0,0) to (12,y(12))

L1=√14+(y(12))2=12√1+(y′(12))2=2√c21−1√9−c21=0.553038982525363

L3=√14+(y(32))2=12√1+(y′(32))2=34√c21−1√4−c21=0.588379287212860

We can integrate to find the curved part of the undisturbed path

L13=∫3212√1+(y′(x))2dx=∫θ3θ11−c21cosh2θ2√c21−1dθ=14√c21−1[3√9−c21−2√4−c21+(c21−2)ln(3+√9−c212+√4−c21)]

Where c1coshθ1=3 and c1coshθ3=2. Since this works out to L13=1.042233793405411, the total path length on a non-wall day will be L1+L13+L3=2.183652063143634 miles.

The maximum excursion from the straight line path happens when y′=0. This happens at θmax=0.759253746502349, x_{\max}=1.137510935801247, and has a value of y_{\max}=0.401134051668599.

To get the average path length we have one more integral to evaluate

\begin{align}\bar s&=L_1+\frac38L_3+\int_{\frac12}^{\frac32}\left[\left(\frac18x^2-\frac78x+\frac{45}{32}\right)\sqrt{1+\left(y^{\prime}(x)\right)^2}-\left(\frac78-\frac14x\right)y(x)\right]dx\\

&=\frac1{32\sqrt{c_1^2-1}}\left\{(-4c_1^2-2)\sqrt{4-c_1^2}+(-2c_1^4+15c_1^2-10)\ln\left(\frac{2+\sqrt{4-c_1^2}}{c_1}\right)\right.\\

&\left.+(6c_1^2+18)\sqrt{9-c_1^2}+(2c_1^4-30c_1^2+20)\ln\left(\frac{3+\sqrt{9-c_1^2}}{c_1}\right)\right\}\tag{17}\\

&-\frac58c_2+L_1+\frac38L_3=1.250510547155483=\text{Cost}_1\end{align}

This is the average length up to the first wall encountered, with the distance to the wall center subtracted, or to point B if no wall was present.

I have included a graph of the path and a table for those who like discrete solutions.

\begin{array}{cc}x&y\\

0.000000&0.000000\\

0.500000&0.236331\\

0.540000&0.254769\\

0.580000&0.272258\\

0.620000&0.288778\\

0.660000&0.304305\\

0.700000&0.318815\\

0.740000&0.332282\\

0.780000&0.344676\\

0.820000&0.355965\\

0.860000&0.366113\\

0.900000&0.375080\\

0.940000&0.382821\\

0.980000&0.389287\\

1.020000&0.394421\\

1.060000&0.398157\\

1.100000&0.400423\\

1.140000&0.401131\\

1.180000&0.400180\\

1.220000&0.397452\\

1.260000&0.392800\\

1.300000&0.386048\\

1.340000&0.376976\\

1.380000&0.365301\\

1.420000&0.350650\\

1.460000&0.332511\\

1.500000&0.310145\\

2.000000&0.000000\\

\end{array}

**Part 2**. The path after the small wall but before the big wall.

The probability of meeting the big wall between x_2 and x_2+dx_2 was P(x_2)dx_2=\frac12dx_2 at the outset, but the probability that we would get to x_1 without seeing the big wall was

P(x_1)=\frac12+\int_{x_1}^{\frac32}P(x_2)dx_2=\frac54-\frac12x_1\tag{18}

So the probability of meeting the big wall between x_2 and x_2+dx_2 given that we made it to x_1 without seeing it is

P(x_2|x_1)dx_2=\frac{P(x_2)}{P(x_1)}dx_2=\frac{2dx_2}{5-2x_1}\tag{19}

With this probability density function, we can assess the average cost of a given path y(x) starting from (x_1,0). It is

\begin{align}\bar s(x_1)&=\int_{x_1}^{\frac32}\left[\frac12+\int_{x_1}^{x_2}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx+1-y(x_2)+\sqrt{(2-x_2)^2+1}\right]\frac{2dx_2}{5-2x_1}\tag{20}\\

&+\left(1-\int_{x_1}^{\frac32}\frac{2dx_2}{5-2x_1}\right)\left[\frac12+\int_{x_1}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx+\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}\right]\end{align}

The integral on the first line is to average the path length over all possible remaining positions of the big wall and the second line multiplies the probability of reaching x_2=\frac32 unscathed by the undisturbed (after hitting the small wall at x_1) path length to get the cost of the undisturbed path.

The first item in each of the square brackets is the \frac12 mile to get around the small wall because our accounting scheme subtracted our distance from the straight line path when the first wall was met.

The integral that comes next is the length of the curved part of our path up to x_2 where the big wall is or to \frac32 if the big wall leaves us alone.

If we meet the big wall, we will have to detour 1-y(x_2) but then we can go in a straight line from (x_2,1) to (2,0).

If no big wall, then after making it through the danger zone we can go straight from (\frac32,y\left(\frac32\right)) to (2,0). The probability of no big wall can be seen to be \frac2{5-2x_1} and the cost contributed by the constants is

K_1=\frac12\left(\frac{3-2x_1}{5-2x_1}\right)+\left(\frac{3-2x_1}{5-2x_1}\right)+\frac12\left(\frac2{5-2x_1}\right)=\frac12+\frac{3-2x_1}{5-2x_1}\tag{21}

The average cost of all the straight line paths from the edge of the big wall to point B is

\begin{align}K_2&=\frac2{5-2x_1}\int_{x_1}^{\frac32}\sqrt{(2-x_2)^2+1}dx_2\\

&=\frac2{5-2x_1}\left(-\frac12\right)\left[(2-x_2)\sqrt{(2-x_2)^2+1}+\ln\left((2-x_2)+\sqrt{(2-x_2)^2+1}\right)\right]_{x_1}^{\frac32}\tag{22}\\

&=\frac1{5-2x_1}\left[(2-x_1)\sqrt{(2-x_1)^2+1}+\ln\left((2-x_1)+\sqrt{(2-x_1)^2+1}\right)-\frac{\sqrt5}4-\ln\left(\frac{\sqrt5+1}2\right)\right]\end{align}

The average cost of the curved path is

\begin{align}V_1&=\frac2{5-2x_1}\int_{x_1}^{\frac32}\int_{x_1}^{x_2}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\,dx_2+\frac2{5-2x_1}\int_{x_1}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\\

&=\frac2{5-2x_1}\int_{x_1}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}\int_{x}^{\frac32}dx_2\,dx+\frac2{5-2x_1}\int_{x_1}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\\

&=\frac2{5-2x_1}\int_{x_1}^{\frac32}\left(\frac32-x\right)\sqrt{1+\left(y^{\prime}(x)\right)^2}dx+\frac2{5-2x_1}\int_{x_1}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\\

&=\frac1{5-2x_1}\int_{x_1}^{\frac32}\left(5-2x\right)\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\tag{23}\end{align}

The average benefit of our path’s deviation from the straight line path is

V_2=\frac2{5-2x_1}\int_{x_1}^{\frac32}-y(x)dx\tag{24}

And the average cost of the straight line part of the undisturbed path is

V_3=\frac2{5-2x_1}\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}\tag{25}

The part of we can do something about by varying our path is

V_1+V_2+V_3=\frac1{5-2x_1}\left\{\int_{x_1}^{\frac32}\left[\left(5-2x\right)\sqrt{1+\left(y^{\prime}(x)\right)^2}-2y(x)\right]dx+2\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}\right\}\tag{26}

We want our path to be invariant to first order to small changes to the path \delta y. The change in the contents of the curly braces above is

\begin{align}\delta V&=\int_{x_1}^{\frac32}\left[\left(5-2x\right)\frac{\partial}{\partial y^{\prime}}\sqrt{1+\left(y^{\prime}(x)\right)^2}\delta y^{\prime}-2\frac{\partial}{\partial y}y(x)\delta y\right]dx+\left[2\frac{\partial}{\partial y}\sqrt{\frac14+\left(y\left(x\right)\right)^2}\delta y\right]_{x=\frac32}\\

&=\left[\left(5-2x\right)\frac{y^{\prime}(x)}{\sqrt{1+\left(y^{\prime}(x)\right)^2}}\delta y\right]_{x_i}^{\frac32}-\int_{x_1}^{\frac32}\left[\frac d{dx}\left(\left(5-2x\right)\frac{y^{\prime}(x)}{\sqrt{1+\left(y^{\prime}(x)\right)^2}}\right)+2\right]\delta ydx\tag{27}\\

&+\left[2\frac{y(x)}{\sqrt{\frac14+\left(y\left(x\right)\right)^2}}\delta y\right]_{x=\frac32}\end{align}

Now, we know the value of y(x_1), so \delta y(x_1)=0, but y\left(\frac32\right) is free to wander so \delta y\left(\frac32\right) can take on any value. Thus

\begin{align}\delta V&=\left[\frac{2y^{\prime}\left(\frac32\right)}{\sqrt{1+\left(y^{\prime}\left(\frac32\right)\right)^2}}+\frac{2y\left(\frac32\right)}{\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}}\right]\delta y\left(\frac32\right)\tag{28}\\

&-\int_{x_1}^{\frac32}\left[\frac d{dx}\left(\left(5-2x\right)\frac{y^{\prime}(x)}{\sqrt{1+\left(y^{\prime}(x)\right)^2}}\right)+2\right]\delta y(x)dx\end{align}

Since this has to be invariant to first order in \delta y, the contents of both sets of square brackets must be zero. The solution for the algebraic brackets is y^{\prime}\left(\frac32\right)=-2y\left(\frac32\right). This has the physical significance that the slope of the path from \left(\frac32,y\left(\frac32\right)\right) to (2,0) is

m=\frac{0-y\left(\frac32\right)}{2-\frac32}=-2y\left(\frac32\right)=y^{\prime}\left(\frac32\right)\tag{29}

This proves the continuity of the first derivative of the optimal path at the kink point x=\frac32. This was considered likely because there wasn’t an obvious kink in the paths there, but it wasn’t a given because light follows an optimal, too, but has kinks in its path where the refractive index of the medium changes discontinuously.

Now we have to solve the differential equation implied by the contents of the second set of square brackets.

\frac d{dx}\left(\left(5-2x\right)\frac{y^{\prime}(x)}{\sqrt{1+\left(y^{\prime}(x)\right)^2}}\right)=-2\tag{30}

This integrates to

\left(5-2x\right)\frac{y^{\prime}(x)}{\sqrt{1+\left(y^{\prime}(x)\right)^2}}=5-2x-2c_1^2\tag{31}

Solving for y^{\prime}(x) and letting u=5-2x-c_1^2,

y^{\prime}(x)=\frac{5-2x-2c_1^2}{2c_1\sqrt{5-2x-c_1^2}}=\frac{u-c_1^2}{2c_1u^{1/2}}=\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=-2\frac{dy}{du}\tag{32}

Integrating,

y=\frac1{2c_1}\left(-\frac13u^{3/2}+c_1^2u^{1/2}\right)+c_2\tag{33}

At the start of the curved path,

y(x_1)=\frac12=\frac1{2c_1}\left(-\frac13u_1^{3/2}+c_1^2u_1^{1/2}\right)+c_2\tag{34}

Where u_1=u(x_1)=5-2x_1-c_1^2. Then

y=\frac1{2c_1}\left(-\frac13u^{3/2}+c_1^2u^{1/2}+\frac13u_1^{3/2}-c_1^2u_1^{1/2}\right)+\frac12\tag{35}

Our condition that determines c_1 is then

-2\left(\frac1{2c_1}\right)\left(u_3^{1/2}-c_1^2u_3^{-1/2}\right)=-2\left(\frac1{2c_1}\right)\left(-\frac13u_3^{3/2}+c_1^2u_3^{1/2}+\frac13u_1^{3/2}-c_1^2u_1^{1/2}\right)-1\tag{36}

Where u_3=u\left(\frac32\right)=2-c_1^2. Now, this equation is much easier to solve than was the case before I had proved continuity of the first derivative! With this value of c_1 in hand, we do the integral implied by V_1+V_2 to get

\begin{align}V_1+V_2&=-\frac12\left(\frac1{5-2x_1}\right)\left[\frac1{3c_1}u_3^{5/2}+c_1^3u_3^{1/2}\right.\tag{37}\\

&\left.-\frac1{3c_1}u_1^{3/2}u_3+c_1u_1^{1/2}u_3-c_1^3u_1^{1/2}-c_1u_1^{3/2}-u_3+u_1\right]\end{align}

Then we can compute c_1(x_1) and \text{Cost}_2(x_1)=K_1+K_2+V_1+V_2+V_3 and tabulate some results and plot some paths.

\begin{array}{ccc}x_1&c_1&\text{Cost}_2(x_1)\\

0.5&1.2653039739&2.5451782877\\

0.6&1.2610954013&2.4328788766\\

0.7&1.2576330600&2.3174265702\\

0.8&1.2551457826&2.1983182953\\

0.9&1.2539248955&2.0749566112\\

1.0&1.2543450503&1.9466409733\\

1.1&1.2568937316&1.8125729376\\

1.2&1.2622157699&1.6719005163\\

1.3&1.2711914424&1.5238692094\\

1.4&1.2851258355&1.3682848227\\

1.5&1.3065629649&1.2071067812\\

\end{array}

Now we want the contribution to the total cost of this part of the path.

In part 1 that we separated the probability of not having encoutered a wall by x_1 into P_{bs}(x_1)=P_b(x_1)P_s(x_1).

Then the probability of encountering any wall between x_1 and x_1+dx_1 is

-\frac{dP_{bs}}{dx_1}dx_1=-P_b\frac{dP_s}{dx_1}dx_1-P_s\frac{dP_b}{dx_1}dx_1\tag{38}

This separates the probability into probabilities that the small wall or big wall will be encountered, so the one we want is that for the small wall

-P_b\frac{dP_s}{dx_1}dx_1=\left(\frac54-\frac12x_1\right)\left(\frac14\right)dx_1=\frac1{16}(5-2x_1)dx_1\tag{39}

Then we multiply by \text{Cost}_2(x_1) and integrate over the domain of the small wall to get

\text{Cost}_2=\int_{\frac12}^{\frac32}\text{Cost}_2(x_1)\frac{5-2x_1}{16}dx_1=0.374353894107649\tag{40}

**Part 3**. The path after the big wall but before the small wall.

When starting out on our journey, the probability of finding the small wall between x_3 and x_3+dx_3 was P(x_3)dx_3=\frac14dx_3 because of its uniform distribution.

However, there was only a probability of P(x_1)=\frac34+\frac14\left(\frac32-x_1\right)=\frac18(9-2x_1) of making it to x=x_1 without encountering the small wall between x=\frac12 and x=x_1,

so the probability of finding the small wall between x_3 and x_3+dx_3 given that we have made it to x_1 unscathed is

P(x_3|x_1)dx_3=\frac{P(x_3)dx_3}{P(x_1)}=\frac{2dx_3}{9-2x_1}\tag{41}

Now that we have the right probability distribution function, we can use it to find the mean distance to point B given a path y(x):

\begin{align}\bar s&=\int_{x_2}^{\frac32}\left[1+\sqrt{(x_2-x_1)^2+\frac14}+\int_{x_2}^{x_3}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\right.\tag{42}\\

&\left.+\frac12-y(x_3)+\sqrt{(2-x_3)^2+\frac14}\right]\frac{2dx_3}{9-2x_1}\\

&+\left(1-\int_{x_2}^{\frac32}\frac{2dx_3}{9-2x_1}\right)\left[1+\sqrt{(x_2-x_1)^2+\frac14}\right.\\

&\left.+\int_{x_2}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx+\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}\right]\end{align}

The integral on the first line above computes the contribution to the path length from days when the small wall was present.

The first 2 terms are the width of the big wall, 1 mile, and the straight-line path taken from the far edge of the big wall at (x_1,1) down to where we are in danger of bumping the small wall at (x_2,\frac12).

Then there is the length of the curved path up to where the small wall is at (x_3,y(x_3)), then the detour we must make around the small wall to (x_3,\frac12) and then the straight line from there to point B at (2,0).

The stuff in the parentheses at the beginning of the second line is the probability that we would make it all the way through to (\frac32,y\left(\frac32\right)) without the small wall being there,

P_{\text{undisturbed}}=1-\frac{3-2x_2}{9-2x_1}=\frac{6-2x_1+2x_2}{9-2x_1}\tag{43}

And then we have the 1-mile detour around the big wall (recall that we subtracted y(x_1) in part 1), the straight line from (x_1,1) to (x_2,\frac12), the length of the curved path from (x_2,\frac12) to (\frac32,y\left(\frac32\right)), and the straight line path from there to point B.

We can change the order of integration to simplify the arc length integral.

\begin{align}&\frac2{9-2x_1}\int_{x_2}^{\frac32}\int_{x_2}^{x_3}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\,dx_3+\frac{6-2x_1+2x_2}{9-2x_1}\int_{x_2}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\tag{44}\\

&=\frac2{9-2x_1}\int_{x_2}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}\int_{x}^{\frac32}dx_3\,dx+\frac{6-2x_1+2x_2}{9-2x_1}\int_{x_2}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\\

&=\frac1{9-2x_1}\int_{x_2}^{\frac32}(9-2x_1+2x_2-2x)\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\end{align}

And that integral that represents the average straight line path from the small wall to point B may be evaluated as

\begin{align}&\frac2{9-2x_1}\int_{x_2}^{\frac32}\sqrt{(2-x_3)^2+\frac14}dx_3=\frac2{9-2x_1}\int_{x_3=x_2}^{x_3=\frac32}-\frac14\cosh^2\theta\,d\theta\tag{45}\\

&=\frac1{9-2x_1}\left[(2-x_2)\sqrt{(2-x_2)^2+\frac14}\right.\\

&\left.+\frac14\ln\left(2-x_2+\sqrt{(2-x_2)^2+\frac14}\right)-\frac1{2\sqrt2}-\frac14\ln\left(\frac{1+\sqrt2}2\right)\right]\end{align}

The rest of the integral doens’t vary in the interval of integration so we can add things up to

\begin{align}\bar s&=1+\sqrt{(x_2-x_1)^2+\frac14}+\frac12\left(\frac{3-2x_2}{9-2x_1}\right)+\frac{6-2x_1+2x_2}{9-2x_1}\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}\tag{46}\\

&+\frac1{9-2x_1}\left[(2-x_2)\sqrt{(2-x_2)^2+\frac14}\right.\\

&\left.+\frac14\ln\left(2-x_2+\sqrt{(2-x_2)^2+\frac14}\right)-\frac1{2\sqrt2}-\frac14\ln\left(\frac{1+\sqrt2}2\right)\right]\\

&+\frac1{9-2x_1}\int_{x_2}^{\frac32}\left[(9-2x_1+2x_2-2x)\sqrt{1+\left(y^{\prime}(x)\right)^2}-2y(x)\right]dx\end{align}

Now that we have the length expressed so simply in term of the path y(x) we will apply a variation y(x)+\delta y(x) and make the dependence on \delta y(x) vanish to first order.

\begin{align}&\delta\frac1{9-2x_1}\int_{x_2}^{\frac32}\left[(9-2x_1+2x_2-2x)\sqrt{1+\left(y^{\prime}(x)\right)^2}-2y(x)\right]dx\tag{47}\\

&=\frac1{9-2x_1}\left[(6-2x_1+2x_2)\frac{y^{\prime}\left(\frac32\right)}{\sqrt{1+\left(y^{\prime}\left(\frac32\right)\right)^2}}\delta y\left(\frac32\right)-(9-2x_1)\frac{y^{\prime}(x_2)}{\sqrt{1+\left(y^{\prime}(x_2)\right)^2}}\delta y(x_2)\right]\\

&-\frac1{9-2x_1}\int_{x_2}^{\frac32}\left[\frac d{dx}\left((9-2x_1+2x_2-2x)\frac{y^{\prime}(x)}{\sqrt{1+\left(y^{\prime}(x)\right)^2}}\right)\delta y(x)+2\delta y(x)\right]dx\end{align}

The integrand above must vanish for any small variation \delta y(x) so in due course that will give us a differential equation for the path but we don’t know the values of the endpoints (x_2,\frac12) and (\frac32,y\left(\frac32\right)) so we have to take into account how the rest of the terms in \bar s vary with the path.

The right endpoint is relatively simple

\delta\frac{6-2x_1+2x_2}{9-2x_1}\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}=\frac{6-2x_1+2x_2}{9-2x_1}\frac{y\left(\frac32\right)}{\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}}\delta y\left(\frac32\right)\tag{48}

(Neglecting for the moment the dependence on x_2) Combining with the \delta y\left(\frac32\right) term from integration by parts and considering that the variation must be invariant to first order in \delta y,

\frac{6-2x_1+2x_2}{9-2x_1}\frac{y\left(\frac32\right)}{\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}}+\frac{6-2x_1+2x_2}{9-2x_1}\frac{y^{\prime}\left(\frac32\right)}{\sqrt{1+\left(y^{\prime}\left(\frac32\right)\right)^2}}=0\tag{49}

The solution is y^{\prime}\left(\frac32\right)=-2y\left(\frac32\right)\tag{50}

Since that is the slope of the line from \left(\frac32,y\left(\frac32\right)\right) to (2,0) this demonstrates the continuity of the first derivative at the right endpoint. The left endpoint is messier. As the path y(x) varies it moves the left endpoint around because y(x_2)=y(x_2+\delta x_2)+\delta y(x_2)=y(x_2)+y^{\prime}(x_2)\delta x_2+\delta y(x_2)=\frac12 is fixed. Thus

\delta x_2=-\frac{\delta y(x_2)}{y^{\prime}(x_2)}\tag{51}

So we have to differentiate \bar s with respect to x_2, including the lower bound of the integral via the fundamental theorem of calculus, divide by y^{\prime}(x_2), and subtract from the term we got from the lower limit from integration by parts to get

\begin{align}0&=-\frac{y^{\prime}(x_2)}{\sqrt{1+\left(y^{\prime}(x_2)\right)^2}}-\frac1{y^{\prime}(x_2)}\left[\frac{x_2-x_1}{\sqrt{(x_2-x_1)^2+\frac14}}-\frac1{9-2x_1}+\frac2{9-2x_1}\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}\right.\\

&\left.-\frac2{9-2x_1}\sqrt{(2-x_2)^2+\frac14}-\frac1{9-2x_1}\left\{(9-2x_1)\sqrt{1+\left(y^{\prime}(x_2)\right)^2}-2y(x_2)\right\}\right.\tag{52}\\

&\left.+\frac2{9-2x_1}\int_{x_2}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\right]\end{align}

Not pretty but it cleans up to

\begin{align}&\frac{x_2-x_1}{\sqrt{(x_2-x_1)^2+\frac14}}-\frac1{\sqrt{1+\left(y^{\prime}(x_2)\right)^2}}\tag{53}\\

&=\frac2{9-2x_1}\left[\sqrt{(2-x_2)^2+\frac14}-\sqrt{\frac14+\left(y\left(\frac32\right)\right)^2}-\int_{x_2}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx\right]\end{align}

The left hand side can be recognized as the difference between \sin\theta_i-\sin\theta_r, so there is refraction at this boundary!

Now we can get back to the differential equation for the path y(x). We had

\frac d{dx}\left((9-2x_1+2x_2-2x)\frac{y^{\prime}(x)}{\sqrt{1+\left(y^{\prime}(x)\right)^2}}\right)=-2\tag{54}

We can integrate to

(9-2x_1+2x_2-2x)\frac{y^{\prime}(x)}{\sqrt{1+\left(y^{\prime}(x)\right)^2}}=9-2x_1+2x_2-2x-2c_1^2\tag{55}

Solving for y^{\prime}(x),

y^{\prime}(x)=\frac{9-2x_1+2x_2-2x-2c_1^2}{2c_1\sqrt{9-2x_1+2x_2-2x-c_1^2}}=\frac{dy}{dx}=-2\frac{dy}{du}=\frac{u-c_1^2}{2c_1u^{1/2}}\tag{56}

Where we have made the substitution u=9-2x_1+2x_2-2x-c_1^2, u_2=u(x_2)=9-2x_1-c_1^2, and u_4=u\left(\frac32\right)=6-2x_1+2x_2-c_1^2. We can integrate this to get

y=\frac1{2c_1}\left(-\frac13u^{3/2}+c_1^2u^{1/2}\right)+c_2\tag{57}

We know that

y(x_2)=\frac12=\frac1{2c_1}\left(-\frac13u_2^{3/2}+c_1^2u_2^{1/2}\right)+c_2\tag{58}

So we have our path

y=\frac1{2c_1}\left(-\frac13u^{3/2}+c_1^2u^{1/2}+\frac13u_2^{3/2}-c_1^2u_2^{1/2}\right)+\frac12\tag{59}

To solve for c_1 and x_2 first we have to evaluate that integral

\begin{align}\int_{x_2}^{\frac32}\sqrt{1+\left(y^{\prime}(x)\right)^2}dx&=\int_{x_2}^{\frac32}\frac1{2c_1}\left[u^{1/2}+c_1^2u^{-1/2}\right]\left(\frac{-du}{2}\right)\tag{60}\\

&=\frac1{2c_1}\left[\frac13u_2^{3/2}+c_1^2u_2^{1/2}-\frac13u_4^{3/2}-c_1^2u_4^{1/2}\right]\end{align}

So now we can put everything in the equations for continuity of the first derivative at the right endpoint and our version of Snell’s law at the left endpoint and solve for c_1 and x_2. Then we have paths plotted in the large and in detail.

Even in the closeup it’s pretty much impossible to see the refraction at y=\frac12.

We can then evaluate the integral we optimized with such difficulty and find the cost of each path. Here is a table.

\begin{array}{ccccc}x_1&c_1&x_2&y\left(\frac32\right)&\text{Cost}_3(x_1)\\

0.5&2.4687532134&1.3131625215&0.3733857985&2.8090608793\\

0.6&2.4622847219&1.3501274970&0.3917097193&2.7249375529\\

0.7&2.4566369485&1.3874433837&0.4127942541&2.6429339266\\

0.8&2.4518060881&1.4249923978&0.4372691891&2.5634550764\\

0.9&2.4477477457&1.4625992592&0.4659645229&2.4870044450\\

1.0&2.4443554844&1.5000000000&0.5000000000&2.4142135624\\

1.1&\unicode{x2014}&\unicode{x2014}&\unicode{x2014}&2.3453624047\\

1.2&\unicode{x2014}&\unicode{x2014}&\unicode{x2014}&2.2806248475\\

1.3&\unicode{x2014}&\unicode{x2014}&\unicode{x2014}&2.2206555616\\

1.4&\unicode{x2014}&\unicode{x2014}&\unicode{x2014}&2.1661903790\\

1.5&\unicode{x2014}&\unicode{x2014}&\unicode{x2014}&2.1180339887\\

\end{array}

If 1\le x_1\le\frac32, there was no interference from the small wall because the straight line path to B went around \left(\frac32,\frac12\right).

To get the cost we multiply by the probability of encountering the big wall first at x_1, which from our analysis of parts 1 and 2 is

-P_s\frac{dP_b}{dx_1}dx_1=\left(\frac98-\frac14x_1\right)\left(\frac12\right)dx_1=\frac1{16}(9-2x_1)dx_1\tag{61}

And integrate over the domain of the big wall to get

\begin{align}\text{Cost}_3&=\int_{\frac12}^1\text{Cost}_3(x_1)\frac{9-2x_1}{16}dx_1+\int_1^{\frac32}\left[1+\sqrt{(2-x_1)^2+1}\right]\frac{9-2x_1}{16}dx_1\tag{62}\\

&=0.611754747093525+0.458894932339652=1.070649679433177\end{align}

And we can add up the contributions to the total cost from each of the parts to get

\begin{align}\text{Cost}&=\text{Cost}_1+\text{Cost}_2+\text{Cost}_3\\ \tag{63}

&=1.250510547155483+0.374353894107649+1.070649679433177\\

&=2.695514120696309\end{align}

**Attribution***Source : Link , Question Author : David , Answer Author : user5713492*