Parameterization for method of characteristics

  • #1
perishingtardi
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Consider the PDE [tex] xu_x + yu_y = 4u. [/tex] Suppose that we want to find the solution that satisfies [itex]u=1[/itex] on the circle [itex]x^2 + y^2 = 1[/itex] using the method of characteristics.

I have read that the boundary condition can be parameterized as
[tex] x=\sigma, \qquad y=(1-\sigma^2)^{1/2}, \qquad u=1.[/tex]

My question is this: couldn't we also have [itex]y=-(1-\sigma^2)^{1/2}[/itex]?? Don't we have to assume that it could be either the positive or negative square root ([itex] y = \pm\sqrt{1-x^2} [/itex])?
 

Answers and Replies

  • #2
vanhees71
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Let's start with the characterics of the equation and then work in the boundary condition.

The general idea is to set [itex]x=x(t), \quad y=y(t)[/itex] into the equation and take the total derivative
[itex]\frac{\mathrm d}{\mathrm{d} t} u[x(t),y(t)]=\dot{x} \partial_x u + \dot{y} \partial_y u[/itex]
such that the PDE implies an ODE for [itex]u[/itex]. This leads to
[tex]\dot{x}=x, \quad \dot{y}=y, \quad \dot{u}=4u.[/tex]
Now you can solve these (in this case even decoupled!) equations with arbitrary initial conditions, i.e.,
[tex]x(t)=x_0 \exp t, \quad y(t)=y_0 \exp t, \quad u(t)=u_0 \exp(4 t).[/tex]
Now we can set
[tex]x_0=\cos \varphi_0, \quad y_0=\sin \varphi_0, \quad u_0=1,[/tex]
and the so determined characteristics obviously lead to the solution of the boundary problem. The trick is to parametrize the boundary curve as the initial values of the characteristics somehow.

The remaining step is to eliminate the parameters [itex]\varphi_0[/itex] and [itex]t[/itex] in [itex]u[/itex] in terms of [itex]x[/itex] and [itex]y[/itex]. In our case that's simple, because [itex]\exp(2t)=x^2+y^2[/itex] and the solution of the boundary problem reads
[tex]u(x,y)=(x^2+y^2)^2.[/tex]
Indeed we have
[tex]\partial_x u=4 x (x^2+y^2), \quad \partial_y u=4y(x^2+y^2) \; \Rightarrow \; x \partial_x u + y \partial_y u=4(x^2+y^2)^2=4 u[/tex]
and
[tex]u(x,y)=1 \quad \text{for} \quad x^2+y^2=1.[/tex]
 

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