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$u_{xx}+4u_{xy}+3u_{yy} + 3u_x-u_y+2u=0$

I found that $\xi(x,y) = y-3x$ and $\eta(x,y)=y-x$, then $0= u - 5u_{\xi} - 2u_{\eta} - 2u_{\xi \eta}$. I try to use manipulation like SFFT: $$6u = \left (\frac{\partial}{\partial \xi}+1 \right )\left (2\frac{\partial}{\partial \eta} + 5\right )u.$$ Let $v=2\frac{\partial u}{\partial \eta}+5u$ and hence $6u=\frac{\partial v}{\partial \xi}+v$. From $v=2\frac{\partial u}{\partial \eta} + 5u$ give us $$u = e^{-5\eta/2}\int ve^{5\eta/2}\;d\eta,$$ but I don't know how to solve $u$ using this form to $6u=\frac{\partial v}{\partial \xi}+v$.

Wildan B. W.
  • 171

1 Answers1

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We start with the following PDE (see above):

$$u-5 u_{\xi}-2 u_{\eta}- 2 u_{\xi \eta} = 0$$

We try to separate variables by multiplication

$$u(\xi,\eta)=f(\xi)\cdot g(\eta)$$

This leads to the following equation

$$1-2 \frac{f_{\xi}\cdot g_{\eta}}{f\cdot g}- 5 \frac{f_{\xi}}{f}- 2 \frac{g_{\eta}}{g}=0$$

We can get 2 separate ODEs by defining

$$\frac{f_{\xi}}{f}=k_1 \implies f(\xi)=c_1\cdot e^{k_1 \xi}$$ $$\frac{g_{\eta}}{g}=k_2 \implies g(\eta)=c_2\cdot e^{k_2 \eta}$$

and

$$1-2 k_1 k_2-5 k_1-2 k_2=0$$

This equation is satisfied if

$$k_2=-\frac{5k_1-1}{2(k_1+1)}$$

The solution then is

$$u(\xi,\eta)=c(k_1)\cdot exp(k_1 \xi)\cdot exp(k_2 \eta)=c(k_1)\cdot exp(k_1 \xi)\cdot exp(-\frac{5k_1-1}{2(k_1+1)} \eta)=c(k1)\cdot exp(k_1\xi-\frac{5k_1-1}{2(k_1+1)} \eta)$$

Finally we sum over all possible $k_1$ (may be discrete $\sum_{k_1}$ or continuous $\int_{k_1}$)

$$u(\xi,\eta)=\sum_{k_1} c(k_1)\cdot exp(k_1\xi-\frac{5k_1-1}{2(k_1+1)} \eta)$$

General solution in original $x, y$ coordinates

$$\hat{u}(x,y)=u(y-3x,y-x)=\sum_{k_1} c(k_1)\cdot exp(k_1(y-3x)-\frac{5k_1-1}{2(k_1+1)}(y-x))$$

gpmath
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