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First order PDEs

Markov

Member
Feb 1, 2012
149
1) $u_x+u_y=0,\,x\in\mathbb R,\,y>0$ and $u(x,0)=\cos x,\,x\in\mathbb R.$

2) $xu_x+u_y+uy=0,\,x\in\mathbb R,\,y>0$ and $u(x,0)=F(x),\,x\in\mathbb R.$

3) Solve the following equation $2xu_y-u_x=4xy,$ where the initial curve is given by $x=0,\,y=s,\,z=s.$

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1) Laplace transform or Fourier transform? Can I try separation of variables?

2) Same as 1).

3) I don't get the part of the initial curve, what does it mean?
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
For 1 and 2, I would use separation of variables. For 3, I believe it should read

$x = 0, y = s, u = s$ which really says that $u(0,y) = y$ so separation of variables would work.

The form of the IC in 3 suggests a different way. Supoose we introduce a change of variables $(r,s)$ such that $(x,y) \to (r,s)$. Now the chain rule would give us

$u_r = u_x x_r + u_y y_r$

If we choose $x_r = -1, y_r = 2x$ then

$u_r = -1 u_x + 2x u_y = 4xy$ (from the actual PDE).

Now we create a boundary in the $(r,s)$ plane let's say this is $r=0$ on which $x = 0, y = s, u = s$.

Thus, we are required to solve

$
\begin{alignat}{3}
x_r &= -1, &\;\;\;\;x(0,s) &= 0\\
y_r &= 2x, &\;\;\;\;y(0,s) & = s\\
u_r &= 4xy, &\;\;\;\; u(0,s) &= s.
\end{alignat}$

Once you have the solution, eliminate $r$ and $s$.
 
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