Change of coordinates for partial differential equation

Question

I'm struggling with a solution to:

$ \frac{∂u}{∂x} + 3(\frac{∂u}{∂y}) + 10u = 20e^{10x}$.

I've been told to use the change of coordinates method, of the form $ξ = ax + by, η = bx - ay$ for a suitable choice of $a$ and $b$. I'm not sure what to do with the $20e^{10x}$ component. I'm not also sure on what the "for a suitable choice of $a$ and $b$" part means.

Answer 1

Let $ru_x+su_y+tu=v$ a PDE with real functions $r,s,t$ and $v$. Note the following: If you have such a linear first-order-PDE, in general a variable transformation will help to gain the solutions, whenever $\xi$ is a solution of the PDE $ru_x+su_y=0$. So we first need $\xi$.

Note further the following: Concerning the upper PDE, we call

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{s}{r}$$ the characteristic equation of the PDE. In the theory of the method of characteristics, you can do the following (with our example here, we have $\frac{s}{r}=3$): Write $\frac{\mathrm{d}y}{3}=\mathrm{d}x,$ then integration yields $x+C=\frac{1}{3}y$ with a real number $C$. Equivalently, one gets $C=\frac{1}{3}y-x$ and you can choose

$$\xi:=C=\frac{1}{3}y-x$$

because this is an invariant of the characteristic equation. From your definitions, we immediately have

$$(a,b)=(-1,\frac{1}{3})$$

and so we get

$$\eta=\frac{1}{3}x+y.$$

So we now put our unknown function $u(x,y)$ as a new function $U$, i.e. $u(x,y)=U(\xi, \eta)$ and in more detail, it is $U=U(\xi(x,y), \eta(x,y))$. We plug this into the PDE using the multivariable chain rule:

\begin{align}u_x+3u_y+10u&=(U_\xi \xi_x+U_\eta \eta_x)+3(U_\xi \xi_y + U_\eta \eta_y)+10U\\ &=-U_\xi + \frac{1}{3}U_\eta + 3(\frac{1}{3}U_\xi + U_\eta)+10U \\ &=\frac{10}{3}U_\eta +10U \\ &=20\mathrm{e}^{10x}. \end{align}

We still have to express $x$ in terms of $\xi$ and $\eta$. From the definition of $\xi$ and $\eta$ one immediately becomes $3\xi-\eta=-\frac{10}{3}x$, that means $x=-\frac{3}{10}(3\xi-\eta)$. This leads to

$$\frac{10}{3}U_\eta+U=20\mathrm{e}^{10(-\frac{3}{10}(3\xi-\eta))}=20\mathrm{e}^{-9 \xi}\mathrm{e}^{3\eta}.$$

As you can see, this is an ODE with respect to $\eta$. The ODE is linear, so we can use the structur theorem for lineary ODE to solve it: The general solution is the sum of the homogenous ODE and a particular solution of the inhomogenous ODE. Looking at the homogenous ODE $\frac{10}{3}U_\eta+10U=0$ one gets quick

$$U_{\text{hom}}=f(\xi)\mathrm{e}^{-3\eta},$$

where $f$ denotes a differentiable, real-valued function. Based on the right hand side, for the particular solution we use the approach $U_\mathrm{p}=A\mathrm{e}^{3\eta}$, whereby $A \in \mathbb{R}$. Plugging this into the inhomogeneous ODE gives us

$$\frac{10}{3}3A\mathrm{e}^{3\eta}+10A\mathrm{e}^{3\eta}=20\mathrm{e}^{-9\xi}\mathrm{e}^{3\eta},$$ so $A=\mathrm{e}^{-9\xi}$ and we have

$$U_\mathrm{p}=\mathrm{e}^{-9\xi}\mathrm{e}^{3\eta}$$ as a particular solution. For the general solution we obtain

$$U=U_\mathrm{hom}+U_\mathrm{p}=f(\xi)\mathrm{e}^{-3\eta}+\mathrm{e}^{-9\xi}\mathrm{e}^{3\eta}.$$

Transforming it back into the original variables we get the general solution of your PDE:

$$u(x,y)=f(-x+\frac{1}{3}y)\cdot \mathrm{e}^{-x-3y}+\mathrm{e}^{10x},$$

whereby $f$ is a differentiable function. For the sake of completeness, we check this now. We calculate the partial derivatives step by step:

$$u_x=\partial_x \left(f(-x+\frac{1}{3}y)\cdot \mathrm{e}^{-x-3y}+\mathrm{e}^{10x}\right)=-f'\cdot \mathrm{e}^{-x-3y} + f\cdot (- \mathrm{e}^{-x-3y})+10 \mathrm{e}^{10x}, $$

and with respect to $y$ we have

$$u_y=\partial_y \left(f(-x+\frac{1}{3}y)\cdot \mathrm{e}^{-x-3y}+\mathrm{e}^{10x}\right)=\frac{1}{3}\cdot f'\cdot \mathrm{e}^{-x-3y}+ f \cdot (-3\mathrm{e}^{-x-3y}).$$

Now we put this together into the PDE:

\begin{align}u_x+3u_y+10u&=-f'\cdot \mathrm{e}^{-3x-3y}+f\cdot (-\mathrm{e}^{-3x-3y})+10\mathrm{e}^{10x} +3(f'\cdot \frac{1}{3}\cdot \mathrm{e}^{-x-3y}+f\cdot(-3)\mathrm{e}^{-x-3y})\\ &+10(f\cdot \mathrm{e}^{-3x-3y} + \mathrm{e}^{10x})\\ &=20\mathrm{e}^{10x}. \end{align}

Answer 2

$$ \frac{∂u}{∂x} + 3(\frac{∂u}{∂y}) + 10u = 20e^{10x}$$

$u(x,y)=v(\xi,\eta)\quad$ with the change of variables $\quad\begin{cases} ξ = x \\ η = x - \frac13 y \end{cases}$

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial \xi}+\frac{\partial v}{\partial \eta}$$ $$\frac{\partial u}{\partial y}= -\frac13 \frac{\partial v}{\partial \eta}$$

$$ \frac{∂u}{∂x} + 3(\frac{∂u}{∂y}) + 10u =\frac{\partial v}{\partial \xi}+\frac{\partial v}{\partial \eta}+3(-\frac13 \frac{\partial v}{\partial \eta})+10v= 20e^{10\xi}$$ $$ \frac{\partial v}{\partial \xi}+10v= 20e^{10\xi}$$ There is no $\eta$ in the equation which then is reduced to an ODE, linear, first order. Solving it leads to :

$v=e^{10\xi}+Ce^{-10\xi}$

where $C$ is a constant with respect to $\xi$, but of course not with respect to $\eta$. $$v(\xi,\eta)=e^{10\xi}+e^{-10\xi}f(\eta)$$ Back to $(x,y)$ : $$u(x,y)=e^{10x}+e^{-10x}f(x-\frac13 y)$$

Answer 3

Equation is linear. Then solution is $$u=u_h+u_p$$

$u_p=e^{10x}$ is particular solution.

$u_h$ is solution of homogenous equation $$u_x+3u_y+10u=0.\qquad(1)$$ Solution of $(1)$ we search in form $$u=ve^{-10x}$$ For $v$ we get equation $$v_x+3v_y=0.$$ It solution is $$v=f(y-3x)$$ You can easily check this. Then general solution is $$u=f(y-3x)e^{-10x}+e^{10x}$$