# Non-trivial solutions in PDE

In the following question I am trying to find a non trivial solution to the following partial differential equations by inspection and integrating, where many solutions may be possible

1) $$\frac{\partial u}{\partial y} + 2yu = 0$$

So here my thinking is just to move $2yu$ to the right side of the equation and solve this DE which is just linear,

$$\frac{\partial u}{\partial y} = -2yu$$ $$P(x) = -2x$$ $$Q(x) = 0$$ $$e^{\int -2ydy} = e^{-y^2}$$ $$\frac{\partial u}{\partial y} e^{-y^2} = -2yu\cdot e^{-y^2}$$ $$u = \int-2yu\cdot e^{-y^2}dy$$

Then how would I solve this integral? and I am not sure if this is correct since it says to solve by inspection and integration

2) $$\frac{\partial^2 u}{\partial x\partial y}=0$$

I am not sure what to do with this equation, if you integrated wouldn't you just get a constant C?

• For $(2)$, no, integration gives you an arbitrary function of the other variable. i.e $$u_{xy} = 0 \implies u_{x} = f(x)$$ – Mattos Oct 12 '17 at 2:37
• @Mattos then how do I find a non trivial solution? – fr14 Oct 12 '17 at 2:38
• Integrate now with respect to $x$.. – Mattos Oct 12 '17 at 2:40
• so then $u_x=x+c$? – fr14 Oct 12 '17 at 2:42
• sorry your workings just appeared now with the text – fr14 Oct 12 '17 at 2:43

This reduces to $$\frac{1}{2y} \frac{\partial u}{\partial y} + u = 0$$

So we can get away with some old school tricks

$$\frac{1}{2y} \frac{\partial u}{\partial y} + u = 0 \rightarrow \frac{1}{2y} \frac{\partial u}{\partial y} = -u \rightarrow \frac{1}{u} \partial u = -2y \partial y$$

Yielding

$$\ln(u) = -2y^2 + F(x)$$ $$u = e^{F(x) - 2y^2}$$

For any $F(x)$ of your choice. Example

$$u = e^{e^x - 2y^2}$$

Is one solution

• Isn't $\ln(u)=-y^2+F(x)$ ? – JJacquelin Oct 12 '17 at 15:34

$$\frac{\partial u}{\partial y} + 2yu = 0 \tag 1$$

Since there is no differential wrt $x$ in the equation, $x$ can be considered as a parameter. For each value of $x$, Eq.$(1)$ is no longer a PDE but is an ODE : $$\frac{du}{dy} + 2yu = 0$$ You know how to solve this separable ODE : $$u=C\:e^{-y^2}$$ $C$ can be different for each value of $x$. Thus, in fact, $C$ is an arbitrary function of $x$.

The general solution of Eq.$(1)$ is $$u(x,y)=F(x)\:e^{-y^2}$$ where $F(x)$ is an arbitrary function.  

$$\frac{\partial^2 u}{\partial x\partial y}=0 \tag 2$$ Let $\quad v=\frac{\partial u}{\partial x} \quad\to\quad \frac{\partial^2 u}{\partial x\partial y}=\frac{\partial v}{\partial y}$ $$\frac{\partial v}{\partial y}=0 \tag 3$$ Since there is no differential wrt $x$ in the equation, $x$ can be considered as a parameter. For each value of $x$, Eq.$(3)$ is no longer a PDE but is an ODE : $$\frac{dv}{dy} = 0$$ $$v=C$$ $C$ can be different for each value of $x$. Thus, in fact, $C$ is an arbitrary function of $x$. $$v(x,y)=f(x)$$ $$\frac{\partial u}{\partial x}=f(x) \tag 4$$ Since there is no differential wrt $y$ in the equation, $y$ can be considered as a parameter. For each value of $y$, Eq.$(4)$ is no longer a PDE but is an ODE : $$\frac{du}{dx}=f(x)$$ Integrating wrt $x$ gives : $$u=\int f(x)dx+c$$ $f(x)$ is an arbitrary function, thus $F(x)=\int f(x)dx$ is an arbitrary function.

$c$ can be different for each value of $y$. Thus, in fact, $c$ is an arbitrary function of $y$, say $G(y)$. $$u(x,y)=F(x)+G(y)$$ where $F$ and $G$ are arbitrary functions.

• thanks for your submission! since there are many possibilities of solutions, is the other answer posted correct also for question 1)? – fr14 Oct 12 '17 at 13:28
• I don't agree with $u = e^{F(x) - 2y^2}$. I agree with $u = e^{F(x) - y^2}$. Since $F$ is arbitrary, this is the same as $u = f(x)e^{ - y^2}$ with $f(x)=e^{F(x)}$. – JJacquelin Oct 12 '17 at 15:32
• okay I will look at your answer then, thanks! – fr14 Oct 12 '17 at 16:09