I am trying to find functions $p_0(x)$ and $p_1(x)$ that will minimise my Loss function: $$\mathcal{L} = \frac{1}{2}\left( \int_0^1p_0(x)(x^2\alpha + (1-x)^2\beta)\,dx + \int_0^1p_1(x)(x^2\beta + (1-x)^2\alpha) \, dx \right) + \\ + \frac{1}{2}\int_0^1 p_0(x) \log \frac{2p_0(x)}{p_0(x) + p_1(x)} \, dx + \frac{1}{2}\int_0^1 p_1(x)\log \frac{2p_1(x)}{p_0(x) + p_1(x)} \, dx $$ $\alpha$ and $\beta$ are parameters, $\alpha+\beta = 1$, $0<\alpha<\beta$. The meaning of $x$ is a probability, so $x$ is from 0 to 1.
To find $p_0(x)$ and $p_1(x)$, I take functional derivatives of $\mathcal{L}$ w.r.t. them, and set the derivatives to 0. I get this system of equations: $$\log{\frac{2p_0(x)}{p_0(x) + p_1(x)}} + x^2\alpha + (1-x)^2\beta = 0 \\ \log{\frac{2p_1(x)}{p_0(x) + p_1(x)}} + x^2\beta + (1-x)^2\alpha = 0$$
Now the problem is that when I try to solve this system, I do not get expressions for $p_0(x)$ and $p_1(x)$, but instead an expression for their ratio $\frac{p_1(x)}{p_0(x)}$, and another equation on $x$ and parameters $\alpha$ and $\beta$: $$\frac{p_1(x)}{p_0(x)} = \exp^{(\beta-\alpha)(1-2x)} \\ \log\frac{4\exp^{(\beta-\alpha)(1-2x)}}{(1 + \exp^{(\beta-\alpha)(1-2x)})^2} = -(1-x)^2-x^2 $$ The second equation does not have solutions for possible values of $\alpha$, $\beta$, and $x$. Graphically solving the second equation: coloured lines are lefthand side for different values of $\alpha$, the black line is the righthand side. No intersections => no solutions.
What could this mean for the problem? Is there no such functions that minimise the Loss?
Update (27.02):
Here I show how I calculated functional derivatives of $\mathcal{L}$. I am going to write $p_0(x)$ and $p_1(x)$ as $p_0$ and $p_1$ for simplicity.
At first let's look at the f.d. of $\mathcal{L}$ w.r.t. $p_0(x)$.
Suppose $p_0$ minimises $I_1 = \int_0^1 p_0(x) \log \frac{2p_0(x)}{p_0(x) + p_1(x)} \, dx$, and $\tilde{p_0} = p_0 + \varepsilon \eta(x)$, where $\eta(x)$ is an arbitrary function. Then $$\left.\frac{dI_1(\varepsilon)}{d\varepsilon}\right|_{\varepsilon = 0} = 0\\ I_1(\varepsilon) = \int_0^1 \tilde{p_0} \log \frac{2\tilde{p_0}}{\tilde{p_0} + p_1} \, dx$$ So now I can find out what the functional derivative of $I_1$ is: $$\left.\frac{dI_1(\varepsilon)}{d\varepsilon}\right|_{\varepsilon = 0} = \left.\int_0^1 \left( \frac{\partial \tilde{p_0}}{\partial \varepsilon} \log{\frac{2\tilde{p_0}}{\tilde{p_0} + p_1}} + \tilde{p_0}\frac{\partial}{\partial \varepsilon} \left(\log{2\tilde{p_0}} - \log{(\tilde{p_0} +p_1)} \right)\right)dx\right|_{\varepsilon=0} = \\ = \int_0^1 \left.\left(\eta(x)\log{\frac{2\tilde{p_0}}{\tilde{p_0} + p_1}} + \tilde{p_0}\frac{1}{2\tilde{p_0}}2\frac{\partial \tilde{p_0}}{\partial \varepsilon} - \tilde{p_0}\frac{1}{\tilde{p_0}+p_1}\frac{\partial \tilde{p_0}}{\partial \varepsilon} \right)dx\right|_{\varepsilon = 0} = \\ = \int_0^1 \eta(x)\left(\log{\frac{2p_0}{p_0+p_1}} + 1 - \frac{p_0}{p_0+p_1}\right) dx$$ So $$\frac{\delta I_1}{\delta p_0(x)} = \log{\frac{2p_0}{p_0+p_1}} + 1 - \frac{p_0}{p_0+p_1}$$
Now let's look at the integral $I_2 = \int_0^1 p_1(x)\log \frac{2p_1(x)}{p_0(x) + p_1(x)} \, dx $. Similarly,
$$\left.\frac{dI_2(\varepsilon)}{d\varepsilon}\right|_{\varepsilon = 0} = \int_0^1 \left. p_1\frac{\partial}{\partial \varepsilon} \left( \log{2p_1} - \log{(\tilde{p_0} + p_1)}\right)dx \right|_{\varepsilon = 0} = \\ = \int_0^1 \left. p_1\left(-\frac{1}{\tilde{p_0} + p_1}\frac{\partial \tilde{p_0}}{\partial \varepsilon}\right) dx \right|_{\varepsilon = 0} = \int_0^1\eta(x)\left(-\frac{p_1}{p_0+p_1}\right)dx$$ $$\frac{\delta I_2}{\delta p_0(x)} = - \frac{p_1}{p_0+p_1}$$
The sum of these two functional derivatives is: $$\frac{\delta I_1}{\delta p_0(x)} + \frac{\delta I_2}{\delta p_0(x)} = \log{\frac{2p_0}{p_0+p_1}} + 1 - \frac{p_0}{p_0+p_1} - \frac{p_1}{p_0+p_1} = \log{\frac{2p_0}{p_0+p_1}}$$
For the first two terms in $\mathcal{L}$ the functional derivatives are simpler: $$\frac{\delta \int_0^1p_0(x)(x^2\alpha + (1-x)^2\beta)\,dx }{\delta p_0(x)} = x^2\alpha + (1-x)^2\beta$$
$$\frac{\delta \int_0^1p_1(x)(x^2\beta + (1-x)^2\alpha) \, dx}{\delta p_0(x)} = 0$$
All in all, $$ \frac{\delta \mathcal{L}}{\delta p_0(x)} = x^2\alpha + (1-x)^2\beta + \frac{\delta I_1}{\delta p_0(x)} + \frac{\delta I_2}{\delta p_0(x)} = x^2\alpha + (1-x)^2\beta + \log{\frac{2p_0}{p_0+p_1}}$$
Since $I_1$ and $I_2$ are symmetric, functional derivative w.r.t. $p_1(x)$ is going to look similarly.