35 miscellaneous PDE’s

 35.1 A second order PDE
 35.2 second order PDE in Polar coordinates
 35.3 Laplace like PDE with polynomial solution
 35.4 Third oder PDE
 35.5 PDE solved by Laplace transform
 35.6 Linear PDE, initial conditions at \(t=1\)
 35.7 Linear PDE, initial conditions at \(t=t_0\)
 35.8 second order in time, Linear PDE, initial conditions at \(t=t_0\)
 35.9 Einstein-Weiner PDE

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35.1 A second order PDE

problem number 186

Taken from Maple pdsolve help pages, problem 4.

Solve for \(S \left ( x,y \right ) \) \begin{align*} S(x,y) \left ( \frac{\partial ^2 S}{\partial x \partial y} \right ) + \frac{\partial S}{\partial x} \frac{\partial S}{\partial y} &=1 \end{align*}

Mathematica

\[ \text{Failed} \]

Maple

\[ S \left ( x,y \right ) ={\frac{\sqrt{2\,{\it \_c}_{{1}}x+{\it \_C1}}\sqrt{{\it \_C2}\,{{\it \_c}_{{1}}}^{2}+{\it \_c}_{{1}}y}}{{\it \_c}_{{1}}}} \]

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35.2 second order PDE in Polar coordinates

problem number 187

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(u(r,\theta )\)

\[ \frac{\partial ^2 u}{\partial r^2} + \frac{\partial ^2 u}{\partial \theta ^2} = 0 \]

With boundary conditions

\begin{align*} u(2,\theta )&=3 \sin (2 \theta )+1 \end{align*}

Mathematica

\[ \text{Failed} \]

Maple

\[ u \left ( r,\theta \right ) =-3/2\,i{{\rm e}^{-2\,r+4+2\,i\theta }}+3/2\,i{{\rm e}^{2\,r-4-2\,i\theta }}+1 \]

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35.3 Laplace like PDE with polynomial solution

problem number 188

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(u(x,y)\)

\[ \frac{\partial ^2 u}{\partial x^2} + y \frac{\partial ^2 u}{\partial y^2} = 0 \]

With boundary conditions

\begin{align*} u(x,0)&=0 \\ \frac{\partial u}{\partial y}(x,0) &=x^2 \end{align*}

Mathematica

\[ \left \{\left \{u(x,y)\to -y \left (y-x^2\right )\right \}\right \} \]

Maple

\[ u \left ( x,y \right ) =y \left ({x}^{2}-y \right ) \]

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35.4 Third oder PDE

problem number 189

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(u(x,y)\)

\[ \frac{\partial u}{\partial t} = - \frac{\partial ^3 u}{\partial x^2} \]

With initial conditions

\begin{align*} u(x,0)&=f(x) \end{align*}

Mathematica

\[ \text{Failed} \]

Maple

\[ u \left ( x,t \right ) =1/4\,{\frac{1}{{\pi }^{2}}\int _{-\infty }^{\infty }\!4/3\,{\frac{\pi \,f \left ( -\zeta \right ) }{\sqrt [3]{-t}}\sqrt{-{\frac{x+\zeta }{\sqrt [3]{-t}}}}\BesselK \left ( 1/3,-2/9\,{\frac{\sqrt{3} \left ( x+\zeta \right ) }{\sqrt [3]{-t}}\sqrt{-{\frac{x+\zeta }{\sqrt [3]{-t}}}}} \right ) }\,{\rm d}\zeta } \]

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35.5 PDE solved by Laplace transform

problem number 190

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(u(x,y)\)

\[ \frac{\partial u^2}{\partial x y} = \sin (x) \sin (y) \]

With boundary conditions

\begin{align*} u(x,0)&=1+\cos (x) \\ \frac{\partial u}{\partial y}(0,y) &= -2 \sin y \end{align*}

Mathematica

\[ \text{Failed} \]

Maple

\[ u \left ( x,y \right ) =1/2\,\cos \left ( x-y \right ) +1/2\,\cos \left ( x+y \right ) +\cos \left ( y \right ) \]

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35.6 Linear PDE, initial conditions at \(t=1\)

problem number 191

Added December 20, 2018.

Example 25, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(w(x_1,x_2,x_3,t)\)

\[ \frac{\partial w}{\partial t} = \frac{\partial w^2}{\partial x_1^2} + \frac{\partial w^2}{\partial x_2^2} + \frac{\partial w^2}{\partial x_3^2} \]

With initial condition \(w(x_1,x_2,x_3,1) = e^a x_1^2 +x_2 x_3\)

Mathematica

\[ \text{Failed} \]

Maple

\[ w \left ({\it x1},{\it x2},{\it x3},t \right ) = \left ({{\it x1}}^{2}+2\,t-2 \right ){{\rm e}^{a}}+{\it x2}\,{\it x3} \]

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35.7 Linear PDE, initial conditions at \(t=t_0\)

problem number 192

Added December 20, 2018.

Example 26, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(w(x_1,x_2,x_3,t)\)

\[ \frac{\partial w}{\partial t} = \frac{\partial w^2}{\partial x_1 x_2} + \frac{\partial w^2}{\partial x_1 x_3} + \frac{\partial w^2}{\partial x_3^2} + \frac{\partial w^2}{\partial x_2 x_3} \]

With initial condition \(w(x_1,x_2,x_3,t_0) = e^{x_1} +x_2 -3 x_3\)

Mathematica

\[ \text{Failed} \]

Maple

\[ w \left ({\it x1},{\it x2},{\it x3},t \right ) ={{\rm e}^{{\it x1}}}+{\it x2}-3\,{\it x3} \]

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35.8 second order in time, Linear PDE, initial conditions at \(t=t_0\)

problem number 193

Added December 20, 2018.

Example 27, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(w(x_1,x_2,x_3,t)\)

\[ \frac{\partial w^2}{\partial t^2} = \frac{\partial w^2}{\partial x_1 x_2} + \frac{\partial w^2}{\partial x_1 x_3} + \frac{\partial w^2}{\partial x_3^2} - \frac{\partial w^2}{\partial x_2 x_3} \]

With initial condition \begin{align*} w(x_1,x_2,x_3,t_0) &= x_1^3 x_2^2 + x_3 \\ \frac{\partial w}{\partial t}(x_1,x_2,x_3,t_0) &= -x_2 x_3 + x_1 \end{align*}

Mathematica

\[ \text{Failed} \]

Maple

\[ w \left ({\it x1},{\it x2},{\it x3},t \right ) =1/2\,{{\it t0}}^{4}{\it x1}+1/6\, \left ( -12\,{\it x1}\,t-1 \right ){{\it t0}}^{3}+1/6\, \left ( 18\,{\it x1}\,{t}^{2}+18\,{{\it x1}}^{2}{\it x2}+3\,t \right ){{\it t0}}^{2}+1/6\, \left ( -36\,t{{\it x1}}^{2}{\it x2}+ \left ( -12\,{t}^{3}-6 \right ){\it x1}-3\,{t}^{2}+6\,{\it x2}\,{\it x3} \right ){\it t0}+{{\it x1}}^{3}{{\it x2}}^{2}+3\,{t}^{2}{{\it x1}}^{2}{\it x2}+1/6\, \left ( 3\,{t}^{4}+6\,t \right ){\it x1}+1/6\,{t}^{3}-{\it x2}\,{\it x3}\,t+{\it x3} \]

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35.9 Einstein-Weiner PDE

problem number 194

Added January 2, 2018.

Solve for \(u(x,t)\) with \(x>0,t>0\)

\[ u_t = -\beta u_x + D u_{xx} \]

Assuming \(\beta >0,D>0\)

Mathematica

\[ \text{Failed} \]

Maple

\[ u \left ( x,t \right ) ={\frac{{\it \_C3}\,{\it \_C1}}{{{\rm e}^{{\it \_c}_{{1}}t}}}\sqrt{{{\rm e}^{{\frac{x\beta }{d}}}}}{{\rm e}^{1/2\,{\frac{x\sqrt{{\beta }^{2}-4\,d{\it \_c}_{{1}}}}{d}}}}}+{\frac{{\it \_C3}\,{\it \_C2}}{{{\rm e}^{{\it \_c}_{{1}}t}}}\sqrt{{{\rm e}^{{\frac{x\beta }{d}}}}}{{\rm e}^{-1/2\,{\frac{x\sqrt{{\beta }^{2}-4\,d{\it \_c}_{{1}}}}{d}}}}} \]