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Consider the following PDE $$ \frac{\partial \Phi}{\partial t} - \frac{1}{2}y \frac{\partial \Phi}{\partial x} + \alpha \beta y^{3/2} \frac{\partial^2 \Phi}{\partial x \partial y} + \frac{1}{2} y \frac{\partial^2 \Phi}{\partial x^2} + \frac{1}{2} \alpha^2 y^2 \frac{\partial^2 \Phi}{\partial y^2} = 0 $$ What is a good substitution to solve this PDE ? I once used affine change of variable of type $$ \Phi(t,x,y) = \exp\left\{ A(t)x+B(t)y\right\} $$ and then deal with Ricatti PDE but it does not seem to help here.

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  • $\begingroup$ Are you interested in a closed form solution? $\endgroup$ Sep 25, 2012 at 15:32
  • $\begingroup$ Yes exactly, I do not care here about numerical approaches. $\endgroup$
    – vanna
    Sep 25, 2012 at 15:34
  • $\begingroup$ When $\beta=0$ , solve it by separation of variables. When $\beta\neq0$ , solve it by separation of variables + kernel method. $\endgroup$ Sep 26, 2012 at 3:04
  • $\begingroup$ @doraemonpaul : any reference for the kernel method ? Thanks for your ideas. $\endgroup$
    – vanna
    Sep 26, 2012 at 8:21

2 Answers 2

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Case $1$ : $\alpha=0$

Then $\dfrac{\partial\Phi}{\partial t}-\dfrac{y}{2}\dfrac{\partial\Phi}{\partial x}+\dfrac{y}{2}\dfrac{\partial^2\Phi}{\partial x^2}=0$

Case $1$a : $\text{Re}(yt)\geq0$

Let $\Phi(x,y,t)=X(x)T(y,t)$ ,

Then $X(x)\dfrac{\partial T(y,t)}{\partial t}-\dfrac{yT(y,t)}{2}\dfrac{dX(x)}{dx}+\dfrac{yT(y,t)}{2}\dfrac{d^2X(x)}{dx^2}=0$

$\biggl(\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}\biggr)\dfrac{yT(y,t)}{2}=-X(x)\dfrac{\partial T(y,t)}{\partial t}$

$\dfrac{\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}}{X(x)}=-\dfrac{2\dfrac{\partial T(y,t)}{\partial t}}{yT(y,t)}=\dfrac{4(f(u))^2-1}{4}$

$\begin{cases}\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}-\dfrac{4(f(u))^2-1}{4}X(x)=0\\\dfrac{\partial T(y,t)}{\partial t}=-\dfrac{(4(f(u))^2-1)yT(y,t)}{8}\end{cases}$

$\begin{cases}X(x)=\begin{cases}c_1(u)e^{\frac{x}{2}}\sinh(xf(u))+c_2(u)e^{\frac{x}{2}}\cosh(xf(u))&\text{when}~f(u)\neq0\\c_1xe^{\frac{x}{2}}+c_2e^{\frac{x}{2}}&\text{when}~f(u)=0\end{cases}\\T(y,t)=c_3(u)e^{-\frac{yt(4(f(u))^2-1)}{8}}\end{cases}$

$\therefore\Phi(x,y,t)=c_1xe^{\frac{4x+yt}{8}}+c_2e^{\frac{4x+yt}{8}}+\int_uC_3(u)e^{\frac{4x-yt(4(f(u))^2-1)}{8}}\sinh(xf(u))~du+\int_uC_4(u)e^{\frac{4x-yt(4(f(u))^2-1)}{8}}\cosh(xf(u))~du$

or $c_1xe^{\frac{4x+yt}{8}}+c_2e^{\frac{4x+yt}{8}}+\sum\limits_uC_3(u)e^{\frac{4x-yt(4(f(u))^2-1)}{8}}\sinh(xf(u))+\sum\limits_uC_4(u)e^{\frac{4x-yt(4(f(u))^2-1)}{8}}\cosh(xf(u))$

Case $1$b : $\text{Re}(yt)\leq0$

Let $\Phi(x,y,t)=X(x)T(y,t)$ ,

Then $X(x)\dfrac{\partial T(y,t)}{\partial t}-\dfrac{yT(y,t)}{2}\dfrac{dX(x)}{dx}+\dfrac{yT(y,t)}{2}\dfrac{d^2X(x)}{dx^2}=0$

$\biggl(\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}\biggr)\dfrac{yT(y,t)}{2}=-X(x)\dfrac{\partial T(y,t)}{\partial t}$

$\dfrac{\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}}{X(x)}=-\dfrac{2\dfrac{\partial T(y,t)}{\partial t}}{yT(y,t)}=-\dfrac{4(f(u))^2+1}{4}$

$\begin{cases}\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}+\dfrac{4(f(u))^2+1}{4}X(x)=0\\\dfrac{\partial T(y,t)}{\partial t}=\dfrac{(4(f(u))^2+1)yT(y,t)}{8}\end{cases}$

$\begin{cases}X(x)=\begin{cases}c_1(u)e^{\frac{x}{2}}\sin(xf(u))+c_2(u)e^{\frac{x}{2}}\cos(xf(u))&\text{when}~f(u)\neq0\\c_1xe^{\frac{x}{2}}+c_2e^{\frac{x}{2}}&\text{when}~f(u)=0\end{cases}\\T(y,t)=c_3(u)e^{\frac{yt(4(f(u))^2+1)}{8}}\end{cases}$

$\therefore\Phi(x,y,t)=c_1xe^{\frac{4x+yt}{8}}+c_2e^{\frac{4x+yt}{8}}+\int_uC_3(u)e^{\frac{4x+yt(4(f(u))^2+1)}{8}}\sin(xf(u))~du+\int_uC_4(u)e^{\frac{4x+yt(4(f(u))^2+1)}{8}}\cos(xf(u))~du$

or $c_1xe^{\frac{4x+yt}{8}}+c_2e^{\frac{4x+yt}{8}}+\sum\limits_uC_3(u)e^{\frac{4x+yt(4(f(u))^2+1)}{8}}\sin(xf(u))+\sum\limits_uC_4(u)e^{\frac{4x+yt(4(f(u))^2+1)}{8}}\cos(xf(u))$

Case $2$ : $\alpha\neq0$ and $\beta=0$

Then $\dfrac{\partial\Phi}{\partial t}-\dfrac{y}{2}\dfrac{\partial\Phi}{\partial x}+\dfrac{y}{2}\dfrac{\partial^2\Phi}{\partial x^2}+\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2 \Phi}{\partial y^2}=0$

Let $\Phi(x,y,t)=X(x)Y(y)T(t)$ ,

Then $X(x)Y(y)\dfrac{dT(t)}{dt}-\dfrac{yY(y)T(t)}{2}\dfrac{dX(x)}{dx}+\dfrac{yY(y)T(t)}{2}\dfrac{d^2X(x)}{dx^2}+\dfrac{\alpha^2y^2X(x)T(t)}{2}\dfrac{d^2Y(y)}{dy^2}=0$

$\dfrac{\dfrac{dT(t)}{dt}}{T(t)}+\dfrac{y}{2}\dfrac{\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}}{X(x)}+\dfrac{\alpha^2y^2}{2}\dfrac{\dfrac{d^2Y(y)}{dy^2}}{Y(y)}=0$

$\dfrac{\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}}{X(x)}=-\dfrac{\alpha^2y\dfrac{d^2Y(y)}{dy^2}}{Y(y)}-\dfrac{2\dfrac{dT(t)}{dt}}{yT(t)}=-\dfrac{4(f(u))^2+1}{4}$

$\begin{cases}\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}+\dfrac{4(f(u))^2+1}{4}X(x)=0\\\dfrac{\alpha^2y^2\dfrac{d^2Y(y)}{dy^2}}{2Y(y)}+\dfrac{\dfrac{dT(t)}{dt}}{T(t)}=\dfrac{(4(f(u))^2+1)y}{8}\end{cases}$

$\begin{cases}\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}+\dfrac{4(f(u))^2+1}{4}X(x)=0\\\dfrac{\alpha^2y^2\dfrac{d^2Y(y)}{dy^2}}{2Y(y)}-\dfrac{(4(f(u))^2+1)y}{8}=-\dfrac{\dfrac{dT(t)}{dt}}{T(t)}=\dfrac{\alpha^2((g(v))^2-1)}{8}\end{cases}$

$\begin{cases}\dfrac{d^2X(x)}{dx^2}-\dfrac{dX(x)}{dx}+\dfrac{4(f(u))^2+1}{4}X(x)=0\\\begin{cases}y^2\dfrac{d^2Y(y)}{dy^2}-\biggl(\dfrac{(4(f(u))^2+1)y}{4\alpha^2}+\dfrac{(g(v))^2-1}{4}\biggr)Y(y)=0\\\dfrac{\dfrac{dT(t)}{dt}}{T(t)}=-\dfrac{\alpha^2((g(v))^2-1)}{8}\end{cases}\end{cases}$

According to http://eqworld.ipmnet.ru/en/solutions/ode/ode0215.pdf,

$\begin{cases}X(x)=\begin{cases}c_1(u)e^{\frac{x}{2}}\sin(xf(u))+c_2(u)e^{\frac{x}{2}}\cos(xf(u))&\text{when}~f(u)\neq0\\c_1xe^{\frac{x}{2}}+c_2e^{\frac{x}{2}}&\text{when}~f(u)=0\end{cases}\\\begin{cases}Y(y)=\begin{cases}c_3(u,v)\sqrt{y}I_{g(v)}\biggl(\dfrac{\sqrt{(4(f(u))^2+1)y}}{\alpha}\biggr)+c_4(u,v)\sqrt{y}I_{-g(v)}\biggl(\dfrac{\sqrt{(4(f(u))^2+1)y}}{\alpha}\biggr)&\text{when}~g(v)~\text{is not an integer}\\c_3(u,v)\sqrt{y}I_{g(v)}\biggl(\dfrac{\sqrt{(4(f(u))^2+1)y}}{\alpha}\biggr)+c_4(u,v)\sqrt{y}K_{g(v)}\biggl(\dfrac{\sqrt{(4(f(u))^2+1)y}}{\alpha}\biggr)&\text{when}~g(v)~\text{is an integer}\end{cases}\\T(t)=c_5(v)e^{-\frac{\alpha^2t((g(v))^2-1)}{8}}\end{cases}\end{cases}$

Case $3$ : $\alpha\neq0$ and $\beta\neq0$

Then $\dfrac{\partial\Phi}{\partial t}-\dfrac{y}{2}\dfrac{\partial\Phi}{\partial x}+\alpha\beta y^{\frac{3}{2}}\dfrac{\partial^2\Phi}{\partial x\partial y}+\dfrac{y}{2}\dfrac{\partial^2\Phi}{\partial x^2}+\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2\Phi}{\partial y^2}=0$

Let $\Phi(x,y,t)=F(x,y)T(t)$ ,

Then $F(x,y)\dfrac{\partial T(t)}{\partial t}-\dfrac{yT(t)}{2}\dfrac{\partial F(x,y)}{\partial x}+\alpha\beta y^{\frac{3}{2}}T(t)\dfrac{\partial^2F(x,y)}{\partial x\partial y}+\dfrac{yT(t)}{2}\dfrac{\partial^2F(x,y)}{\partial x^2}+\dfrac{\alpha^2y^2T(t)}{2}\dfrac{\partial^2F(x,y)}{\partial y^2}=0$

$\biggl(\dfrac{y}{2}\dfrac{\partial^2F(x,y)}{\partial x^2}+\alpha\beta y^{\frac{3}{2}}\dfrac{\partial^2F(x,y)}{\partial x\partial y}+\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2F(x,y)}{\partial y^2}-\dfrac{y}{2}\dfrac{\partial F(x,y)}{\partial x}\biggr)T(t)=-F(x,y)\dfrac{\partial T(t)}{\partial t}$

$\dfrac{\dfrac{y}{2}\dfrac{\partial^2F(x,y)}{\partial x^2}+\alpha\beta y^{\frac{3}{2}}\dfrac{\partial^2F(x,y)}{\partial x\partial y}+\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2F(x,y)}{\partial y^2}-\dfrac{y}{2}\dfrac{\partial F(x,y)}{\partial x}}{F(x,y)}=-\dfrac{\dfrac{\partial T(t)}{\partial t}}{T(t)}=-f(u)$

$\begin{cases}\dfrac{y}{2}\dfrac{\partial^2F(x,y)}{\partial x^2}+\alpha\beta y^{\frac{3}{2}}\dfrac{\partial^2F(x,y)}{\partial x\partial y}+\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2F(x,y)}{\partial y^2}-\dfrac{y}{2}\dfrac{\partial F(x,y)}{\partial x}+f(u)F(x,y)=0\\\dfrac{\dfrac{\partial T(t)}{\partial t}}{T(t)}=f(u)\end{cases}$

Let $F(x,y)=\int_ve^{xg(v)}K(v,y)~dv$ or $\sum\limits_ve^{xg(v)}K(v,y)$ ,

Then $\begin{cases}\dfrac{y}{2}\int_v(g(v))^2e^{xg(v)}K(v,y)~dv+\alpha\beta y^{\frac{3}{2}}\int_vg(v)e^{xg(v)}\dfrac{\partial K(v,y)}{\partial y}dv+\dfrac{\alpha^2y^2}{2}\int_ve^{xg(v)}\dfrac{\partial^2K(v,y)}{\partial y^2}dv-\dfrac{y}{2}\int_vg(v)e^{xg(v)}K(v,y)~dv+f(u)\int_ve^{xg(v)}K(v,y)~dv=0~\text{or}\dfrac{y}{2}\sum\limits_v(g(v))^2e^{xg(v)}K(v,y)+\alpha\beta y^{\frac{3}{2}}\sum\limits_vg(v)e^{xg(v)}\dfrac{\partial K(v,y)}{\partial y}+\dfrac{\alpha^2y^2}{2}\sum\limits_ve^{xg(v)}\dfrac{\partial^2K(v,y)}{\partial y^2}-\dfrac{y}{2}\sum\limits_vg(v)e^{xg(v)}K(v,y)+f(u)\sum\limits_ve^{xg(v)}K(v,y)=0\\T(t)=c(u)e^{tf(u)}\end{cases}$

$\begin{cases}\int_ve^{xg(v)}\biggl(\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2K(v,y)}{\partial y^2}+\alpha\beta g(v)y^{\frac{3}{2}}\dfrac{\partial K(v,y)}{\partial y}+\dfrac{((g(v))^2-g(v))y+2f(u)}{2}K(v,y)\biggr)dv=0~\text{or}\sum\limits_ve^{xg(v)}\biggl(\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2K(v,y)}{\partial y^2}+\alpha\beta g(v)y^{\frac{3}{2}}\dfrac{\partial K(v,y)}{\partial y}+\dfrac{((g(v))^2-g(v))y+2f(u)}{2}K(v,y)\biggr)=0\\T(t)=c(u)e^{tf(u)}\end{cases}$

$\therefore\begin{cases}F(x,y)=\int_ve^{xg(v)}K(v,y)~dv~\text{or}~\sum\limits_ve^{xg(v)}K(v,y)\\\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2K(v,y)}{\partial y^2}+\alpha\beta g(v)y^{\frac{3}{2}}\dfrac{\partial K(v,y)}{\partial y}+\dfrac{((g(v))^2-g(v))y+2f(u)}{2}K(v,y)=0\\T(t)=c(u)e^{tf(u)}\end{cases}$

But $\dfrac{\alpha^2y^2}{2}\dfrac{\partial^2K(v,y)}{\partial y^2}+\alpha\beta g(v)y^{\frac{3}{2}}\dfrac{\partial K(v,y)}{\partial y}+\dfrac{((g(v))^2-g(v))y+2f(u)}{2}K(v,y)=0$ is very difficult to solve.

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Here is a solution to the pde

$$ \Phi \left( x,y,t \right) = A\left( 2\,{\frac {\ln \left( y \right) }{ {\alpha}^{2}}} + t \right) + B\left( {\frac {y\ln \left( y \right) }{{\alpha}^{2}}} - {\frac {y}{{\alpha}^{2}}}+x \right)+{\it C_1}+C_2{{\rm e}^{x}}+{\it C_3}\,y\,,$$

where $A,B,C_i\,,i=1,2,3 \, $ are constants.

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  • $\begingroup$ The initial conditions being $\Phi(x,y,T)=f(x,y)$ and $\Phi(0,0,t)=1$ this leads to a degenerated solution because of the linearity in $t$... Any chance to get a non t-linear solution ? $\endgroup$
    – vanna
    Sep 25, 2012 at 21:30
  • $\begingroup$ @vanna:I'll give it a try. $\endgroup$ Sep 25, 2012 at 22:40
  • $\begingroup$ Where is the case of $\beta\neq0$ ? $\endgroup$ Sep 26, 2012 at 3:06
  • $\begingroup$ @doraemonpaul : Mhenni Benghorbal solution is such that the cross derivative is zero so the solution does not depend on $\beta$. $\endgroup$
    – vanna
    Sep 26, 2012 at 8:20

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