2.15.14 Fisher’s \(u_t = u(1-u)+u_{xx}\)

problem number 123

Added December 27, 2018.

Taken from https://en.wikipedia.org/wiki/List_of_nonlinear_partial_differential_equations

Fisher’s equation. Solve for \(u(x,t)\) \[ u_t = u(1-u)+u_{xx} \]

Mathematica


\begin {align*} & \left \{u(x,t)\to \frac {1}{4} \left (1+\tanh \left (\frac {1}{12} \left (5 t-\sqrt {6} x-12 c_3\right )\right )\right ){}^2\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {1}{12} \left (5 t-i \sqrt {6} x-12 c_3\right )\right )\right ) \left (1+\tanh \left (\frac {1}{12} \left (5 t-i \sqrt {6} x-12 c_3\right )\right )\right )\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {1}{12} \left (5 t+i \sqrt {6} x-12 c_3\right )\right )\right ) \left (1+\tanh \left (\frac {1}{12} \left (5 t+i \sqrt {6} x-12 c_3\right )\right )\right )\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (1+\tanh \left (\frac {1}{12} \left (5 t+\sqrt {6} x-12 c_3\right )\right )\right ){}^2\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (1+\tanh \left (\frac {5 t}{12}-\frac {x}{2 \sqrt {6}}+c_3\right )\right ){}^2\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {5 t}{12}-\frac {i x}{2 \sqrt {6}}+c_3\right )\right ) \left (1+\tanh \left (\frac {5 t}{12}-\frac {i x}{2 \sqrt {6}}+c_3\right )\right )\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {5 t}{12}+\frac {i x}{2 \sqrt {6}}+c_3\right )\right ) \left (1+\tanh \left (\frac {5 t}{12}+\frac {i x}{2 \sqrt {6}}+c_3\right )\right )\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (1+\tanh \left (\frac {5 t}{12}+\frac {x}{2 \sqrt {6}}+c_3\right )\right ){}^2\right \}\\ \end {align*}

Maple


\begin {align*} & \{u \left (x , t\right ) = 1\}\\& \left \{u \left (x , t\right ) = \frac {\left (\tanh ^{2}\left (c_{1}-\frac {5 t}{12}+\frac {\sqrt {6}\, x}{12}\right )\right )}{4}-\frac {\tanh \left (c_{1}-\frac {5 t}{12}+\frac {\sqrt {6}\, x}{12}\right )}{2}+\frac {1}{4}\right \}\\& \left \{u \left (x , t\right ) = \frac {\left (\tanh ^{2}\left (c_{1}+\frac {5 t}{12}-\frac {\sqrt {6}\, x}{12}\right )\right )}{4}+\frac {\tanh \left (c_{1}+\frac {5 t}{12}-\frac {\sqrt {6}\, x}{12}\right )}{2}+\frac {1}{4}\right \}\\& \left \{u \left (x , t\right ) = \frac {\left (\tanh ^{2}\left (c_{1}-\frac {5 t}{12}-\frac {\sqrt {6}\, x}{12}\right )\right )}{4}-\frac {\tanh \left (c_{1}-\frac {5 t}{12}-\frac {\sqrt {6}\, x}{12}\right )}{2}+\frac {1}{4}\right \}\\& \left \{u \left (x , t\right ) = \frac {\left (\tanh ^{2}\left (c_{1}+\frac {5 t}{12}+\frac {\sqrt {6}\, x}{12}\right )\right )}{4}+\frac {\tanh \left (c_{1}+\frac {5 t}{12}+\frac {\sqrt {6}\, x}{12}\right )}{2}+\frac {1}{4}\right \}\\& \left \{u \left (x , t\right ) = -\frac {\left (\tanh ^{2}\left (c_{1}-\frac {5 t}{12}+\frac {i \sqrt {6}\, x}{12}\right )\right )}{4}-\frac {\tanh \left (c_{1}-\frac {5 t}{12}+\frac {i \sqrt {6}\, x}{12}\right )}{2}+\frac {3}{4}\right \}\\& \left \{u \left (x , t\right ) = -\frac {\left (\tanh ^{2}\left (c_{1}+\frac {5 t}{12}-\frac {i \sqrt {6}\, x}{12}\right )\right )}{4}+\frac {\tanh \left (c_{1}+\frac {5 t}{12}-\frac {i \sqrt {6}\, x}{12}\right )}{2}+\frac {3}{4}\right \}\\& \left \{u \left (x , t\right ) = -\frac {\left (\tanh ^{2}\left (c_{1}-\frac {5 t}{12}-\frac {i \sqrt {6}\, x}{12}\right )\right )}{4}-\frac {\tanh \left (c_{1}-\frac {5 t}{12}-\frac {i \sqrt {6}\, x}{12}\right )}{2}+\frac {3}{4}\right \}\\& \left \{u \left (x , t\right ) = -\frac {\left (\tanh ^{2}\left (c_{1}+\frac {5 t}{12}+\frac {i \sqrt {6}\, x}{12}\right )\right )}{4}+\frac {\tanh \left (c_{1}+\frac {5 t}{12}+\frac {i \sqrt {6}\, x}{12}\right )}{2}+\frac {3}{4}\right \}\\ \end {align*}

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