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I am working on this problem on PDE's :

$f(x, y)\frac{{\partial}f}{{\partial}x} + \frac{{\partial}f}{{\partial}y} = 1$ with $f(u,u) = \frac{u}{2} , 0< u < 1$.

I tried to solve this PDE with the method of characteristics but since $f(x, y)$ is not given, I did not know how to continue.

Any ideas?

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  • $\begingroup$ $f(x,y)$ is what you're solving for, if you were given it there wouldn't be a need for a PDE. $\endgroup$ – ultrainstinct Dec 4 '17 at 18:19
  • $\begingroup$ @Huffman_Coding But how can I apply the method of characteristics here? $\endgroup$ – user507313 Dec 4 '17 at 18:21
  • $\begingroup$ List out your three odes and parametrize, i'll give you a hint. They are $\frac{dx}{dt} = f(x,y)$, $\frac{dy}{dt} = 1$, and $\frac{df}{dt} = 1$. You should be able to get what $f(x,y)$ is from the third ode, then plug it back into the first. $\endgroup$ – ultrainstinct Dec 4 '17 at 18:24
  • $\begingroup$ @Huffman_Coding Thank you for this. However I have already done that. By the third ODE, I got that $f = t + c_1$. After that is the part that I am struggling. $\endgroup$ – user507313 Dec 4 '17 at 18:31
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Remark: You can use existence theorems to conclude that a solution exists around the initial curve. Compare Theorem 1.1. in "Partial Differential Equations" by Prasad & Ravindran:

Let $x_0,y_0,u_0$ be $C^1$ in a closed interval and $a,b,c \in C^1$ in some domain of $(x,y,u)$-space containing the initial curve $\Gamma:\tau \mapsto (x_0(\tau),y_0(\tau),u_0(\tau))$. Further, let $$\begin{vmatrix} x_0' & y_0' \\ a & b \end{vmatrix}_\Gamma \neq 0.$$ Then there exists a unique solution $u=u(x,y)$ of the quasilinear equation $$a(x,y,u) u_x + b(x,y,u) u_y = c(x,y,u)$$ in the neighborhood of the curve $\gamma : \tau \mapsto (x_0(\tau),y_0(\tau))$ and satisfying $u_0(\tau)=u(x_0(\tau),y_0(\tau))$.

In your case we have $\Gamma :\tau \mapsto (\tau,\tau,\tfrac{\tau}{2})$ hence $$\begin{vmatrix} x_0' & y_0' \\ a & b\end{vmatrix}_\Gamma=\begin{vmatrix} 1 & 1 \\ \tfrac{\tau}{2} & 1 \end{vmatrix}=1-\tfrac{\tau}{2}$$ and for $\tau\neq 2$ we have unique solution of the PDE. By assumption we have $0<\tau<1$ so this is fine.


Computation: We have the characteristic ODEs $$x_s=f, y_s=1, f_s=1$$ with the initial curve $\gamma(\tau)=(\tau,\tau)=(x_0,y_0)$ and $f|_\gamma=\tau/2$.

We solve them and get $y=s+\tau$, $f=s+\frac{\tau}{2}$, $x'(s)=s+\frac{\tau}{2}$ i.e.

\begin{align} x(s) &=\int s+ \frac{\tau}{2} ds+\tau=\frac{s^2}{2}+ \tau(\frac{s}{2}+1) \\ &=\frac{s^2}{2}+(y-s)(s/2+1)=s^2/2+ys/2+y-s^2/2-s \end{align}

hence $x-y=s(y/2-1)$ which yields

\begin{align} s&=2\frac{x-y}{y-2} \\ \tau&=y-2\frac{x-y}{y-2}=\frac{y^2-2x}{y-2} \end{align}

thus

$$f=2\frac{x-y}{y-2}+\frac{1}{2} \frac{y^2-2x}{y-2}=\frac{x-2y+y^2/2}{y-2}$$

Checking the initial condition $$f(u,u)=\frac{-u+u^2/2}{u-2}=\frac{u(-1+u/2)}{u-2}=u/2$$ and $f \cdot f_x+f_y=1$ as it can easily be computed. The 'solution space' seems to be $\mathbb{R} \times \mathbb{R}_{y>2}$ or $\mathbb{R} \times \mathbb{R}_{y<2}$.

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  • $\begingroup$ Thank you so much. That helped me a lot to understand how this method works. $\endgroup$ – user507313 Dec 4 '17 at 20:36

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