Find the solution to the nonlinear PDE: $U - XUx - (1/2)*(Uy)^2 + X^2 = 0$ with $U(X,0) = X^2 - (1/6)*(x^4), 0<X<1$

187 Views Asked by At

Step 1: Rewrite the PDE $P= Ux$ and $Q= Uy$

$$F = U - XP - (1/2)Q^2 + x^2$$

Step 2: Charpits Equations

$dx/dτ = Fp = -x$

$dy/dτ = Fq = -2Q$

$dp/dτ = -Fx-P*Fu = -2x$

$dq/dτ = -Fy-Q*Fu = -Q$

$du/dτ = PFp+QFq = -xp - 2Q^2$

Step 3: Integrate

At this point I am confused as to where to start?

Next Steps: Parameterise Initial data

$Xo = S, Uo = S^2 - (1/6)*S^4, Yo = 0$

$dUo/dS = Po*dxo/ds + Qo*dyo/ds$

$Po = 2S -(2/3)*S^3$

2

There are 2 best solutions below

0
On

Hint:

Let $U=X^2-V$ ,

Then $U_X=2X-V_X$

$V_Y=-U_Y$

$\therefore 2X^2-V-X(2X-V_X)-\dfrac{(-V_Y)^2}{2}=0$ with $V(X,0)=\dfrac{X^4}{6}$

$XV_X-V-\dfrac{(V_Y)^2}{2}=0$ with $V(X,0)=\dfrac{X^4}{6}$

Let $V=XW$ ,

Then $V_X=XW_X+W$

$V_Y=XW_Y$

$\therefore X(XW_X+W)-XW-\dfrac{(XW_Y)^2}{2}=0$ with $W(X,0)=\dfrac{X^3}{6}$

$X^2W_X=\dfrac{X^2(W_Y)^2}{2}$ with $W(X,0)=\dfrac{X^3}{6}$

$(W_Y)^2=2W_X$ with $W(X,0)=\dfrac{X^3}{6}$

$W_Y=\pm\sqrt{2W_X}$ with $W(X,0)=\dfrac{X^3}{6}$

$W_{XY}=\pm\dfrac{W_{XX}}{\sqrt{2W_X}}$ with $W(X,0)=\dfrac{X^3}{6}$

Let $Z=W_X$ ,

Then $Z_Y=\pm\dfrac{Z_X}{\sqrt{2Z}}$ with $Z(X,0)=\dfrac{X^2}{2}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dY}{dT}=1$ , letting $Y(0)=0$ , we have $Y=T$

$\dfrac{dZ}{dT}=0$ , letting $Z(0)=Z_0$ , we have $Z=Z_0$

$\dfrac{dX}{dT}=\mp\dfrac{1}{\sqrt{2Z}}=\mp\dfrac{1}{\sqrt{2Z_0}}$ , letting $X(0)=f(Z_0)$ , we have $X=f(Z_0)\mp\dfrac{T}{\sqrt{2Z_0}}=f(Z)\mp\dfrac{Y}{\sqrt{2Z}}$ i.e. $Z=F\left(X\pm\dfrac{Y}{\sqrt{2Z}}\right)$

$Z(X,0)=\dfrac{X^2}{2}$ :

$F(X)=\dfrac{X^2}{2}$

$\therefore Z=\dfrac{\left(X\pm\dfrac{Y}{\sqrt{2Z}}\right)^2}{2}$

$2Z=X^2\pm\dfrac{\sqrt2XY}{\sqrt{Z}}+\dfrac{Y^2}{2Z}$

0
On

Solving with reduction of to a first order PDE and method of characteristics :

enter image description here