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I am trying to solve the following 2nd order PDE analytically, but haven't succeeded so far. I have tried to separate the variables, but it doesn't work here. I would be very grateful for any suggestions that could lead me in the right direction in solving it and for all suggestions about where I could look further. The equation is the following:
$$\frac{\partial u(x,t)}{\partial x}=- \frac{b(x)}{2} (1-u(x,t))^2+\frac{b(x)}{c} \frac{ \partial^2 u(x,t)} {\partial x \partial t},$$ where $u(x,t)$ is the unknown function, $b(x)$ is a known function of $x$ only, and $c$ is a constant.

Any help would be much appreciated. Thanks!

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Do you even know it has a solution? The vast majority of nonlinear PDE do not... – Danny W. Nov 6 '14 at 14:09

$$\frac{\partial u(x,t)}{\partial x}=- \frac{b(x)}{2} (1-u(x,t))^2+\frac{b(x)}{c} \frac{ \partial^2 u(x,t)} {\partial x \partial t},$$ Let $u(x,t)=v(x,t)-1$ $$\frac{\partial v(x,t)}{\partial x}=- \frac{b(x)}{2} v(x,t)^2+\frac{b(x)}{c} \frac{ \partial^2 v(x,t)} {\partial x \partial t},$$ Change of variable : $X(x)=\int{b(x)}{dx}$


Let $V(X(x),t)=v(x,t)$ $$\frac{\partial V(X,t)}{\partial X}=- \frac{1}{2} V(X,t)^2+\frac{1}{c} \frac{ \partial^2 V(X,t)} {\partial X \partial t},$$ Change of variable : $T(t)=ct$

Let $W(X,T(t))=V(X,t)$ $$\frac{\partial W(X,T)}{\partial X}=- \frac{1}{2} W(X,T)^2+ \frac{ \partial^2 W(X,T)} {\partial X \partial T},$$ Now, there is no parameter (constant nor variable) in the PDE, which is a major simplification.

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$1$ is the solution of the partial differential equation.

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$1$ is a solution to the ODE. But it is certainly not the general solution. – Semiclassical Nov 6 '14 at 12:55

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