Let us consider $u(x,t)=\sum_{i=1}^2a_i(t)e^{-|x-b_i(t)|}$ where $a_i,b_i$ are differentiable. Now consider the equation $$u_t+uu_x+\left(\frac{1}{2}e^{-|x|}\ast (u^2+\frac{1}{2}u_x^2)\right)_x+a_1\delta(x-b_1)+a_2\delta(x-b_2)=0.$$ I have calculate convolution for $\frac{1}{2}e^{-|x|}\ast (u^2+\frac{1}{2}u_x^2)$ for different $x'$s i.e $x<b_1, x>b_2, b_1<x<b_2, x=b_1, x=b_2$. Let us consider the case $b_1<b_2$ and for $x<b_1$, we have $$\frac{1}{2}e^{-|x|}\ast (u^2+\frac{1}{2}u_x^2)=-\frac{1}{2}a_1^2e^{2x-2b_1}+a_1^2e^{x-b_1}-\frac{1}{2}a_2^2e^{2x-2b_2}+a_2^2e^{x-b_2}-a_1a_2e^{-b_1-b_2+2x}+4a_1a_2e^{x-b_2}=0.$$ Note that all derivative are meaningful in weak sense. Now if we substitute $u$ in the above equation, can we obtain some equation in terms of $\dot{a}_i$ and $\dot{b}_i$. I have the following try $$u_t=\sum_{i}^2\dot{a}_ie^{(x-b_i)}-\sum_{i=1}^2a_i\dot{b}_ie^{(x-b_i)},$$$$u_x=\sum_{i=1}^2a_ie^{(x-b_i)}.$$ Hence $$\sum_{i}^2\dot{a}_ie^{(x-b_i)}-\sum_{i=1}^2a_i\dot{b}_ie^{(x-b_i)}+a_1^2e^{(x-a_1)}+a_2^2e^{(x-b_2)}+2p_1p_2e^{(x-b_2)}+a_1\delta(x-b_1)+a_2(x-b_2)=0.$$ Can we proceed further? Is it possible to calculate convolution in a general form of $x$ rather than different intervals of $x$?
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