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How can I solve the following pde: $$y\dfrac {\partial{f(x,y)}}{\partial{x}}+x\dfrac {\partial{f(x,y)}}{\partial{y}}=1$$

With given initial condition $f(x_0,y_0)=k$.

My attempt with separation of variables:

$$f(x,y)=X(x)Y(y)$$ $$yY(y)X'(x)+xX(x)Y'(y)=1$$

And got stuck here.

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    $\begingroup$ Try the method of characteristics $\endgroup$ – Winther Jan 10 '17 at 7:29
  • $\begingroup$ @medved : The PDE is inhomogeneous, so the separation of variables isn't convenient. First, apply the separation of variables to the associated homogeneous PDE. Second, find a particular solution of the inhomogeneous PDE and add them. For a direct approach, you can apply the method of characteristics. $\endgroup$ – JJacquelin Jan 10 '17 at 8:29
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From this equation, $a·f_x + b·f_y = c$ we can get this set of coupled ODE's:

$$\frac{dx}{a}=\frac{dy}{b}=\frac{df}{c}$$

In your case:

$$\frac{dx}{y}=\frac{dy}{x}=\frac{df}{1}$$

Considering that whenever $\frac{m}{n}=\frac{o}{p}$ then $\frac{m+o}{n+p}=\frac{m}{n}$, we get:

$$\frac{dx+dy}{x+y}=\frac{df}{1};\;u=x+y;\;du=dx+dy$$ $$\frac{du}{u}=\frac{df}{1};\;\ln{u} = f + c_1;\;\ln{(x+y)}=f+c_1$$ Now, with $\frac{dx}{y}=\frac{dy}{x}$, $xdx=ydy$, so: $$x^2=y^2+c_2\;or\;x^2-y^2=c_2$$ We have now the equations of two surfaces, the intersection of which form the characteristics. It must exist a relation beetween them. The equality $c_1=g(c_2)$ is such a relation, with g an arbitrary single variable differentiable function. Then, $c_1=g(x^2-y^2)$ At last, we get the general solution: $$\ln{(x+y)}=f+g(x^2-y^2)\;or$$ $$f(x,y)=\ln{(x+y)}+g(x^2-y^2)$$ We are ready to impose the boundary conditions. We need at least the values of f along some curve in order to get an unique solution. Unfortunately yours only offers the value k at a single point $(x_0,y_0)$, so we have a whole set of surfaces solution of the PDE.

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  • $\begingroup$ Thanks a lot. I wish I could vote this up! $\endgroup$ – medved Jan 11 '17 at 20:30
  • $\begingroup$ You are welcome! $\endgroup$ – Rafa Budría Jan 12 '17 at 6:19

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