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I just entering new world called Partial Differential Equations , now i just start with Classification PDE , in my Stanley J. Farlow's Text book there are six classification of PDE . But now I little bit struggle with linear or non-linear PDE

Second Order Linear Equation in two variables is PDE can be written in the form $$Au_{xx}+Bu_{xy}+Cu_{yy}+Du_{x}+Eu_{y}+Fu=G$$

So , when i face with PDE like $u_{t}=\alpha^{2}u_{xx}$ i can identify its linear by saying it can be written as general form above with $A=\alpha^{2},B=0,C=0,D=0,E=-1,F=0$

but what if i face with some PDE with higher order let say The Vibrating Beam Equation $u_{tt}+u_{xxxx}=0$ ?

If i must identify this equation like the previous one , i really need a massive variable in general form , is there other good way to show that PDE is linear ?

Znik Dzulqarnain
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  • You can think of a linear PDE as one in which derivatives of the independent variable u are not multiplied together or raised to powers other than $1$ like say $u u_x$ or $(u_{xx})^2$. So your beam equation has no non-linear terms and has a highest order derivative of $4$, so it is a linear fourth order PDE – Triatticus Jul 05 '17 at 15:55
  • @Triatticus $u$ is the dependent variable, not independent. – Robert Israel Jul 05 '17 at 15:58
  • Yes that's right I was definitely abusing the terminology there – Triatticus Jul 05 '17 at 16:01

2 Answers2

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To add to Batman's answer: If we consider the vector space of smooth functions $C^\infty(\mathbb{R}^n)$ recall that the partial derivatives $$ \frac{\partial}{\partial x_i} : C^\infty(\mathbb{R}^n) \to C^\infty(\mathbb{R}^n) $$ are linear transformations since given $f,g\in C^\infty(\mathbb{R}^n)$ and $\alpha,\beta\in \mathbb{R}$ we have that $$ \frac{\partial}{\partial x_i} (\alpha f + \beta g) = \alpha \frac{\partial f}{\partial x_i} + \beta \frac{\partial g}{\partial x_i} $$ Also, note that multiplication by a smooth function $$ \cdot f: C^\infty(\mathbb{R}^n) \to C^\infty(\mathbb{R}^n) $$ is a linear transformation. Since we can compose linear transformations to get a new linear transformation, we should call PDE's described via linear transformations linear PDE's. So, for your example, you are considering solutions to the kernel of the differential operator (another name for linear transformation) $$ D = \frac{\partial^4}{\partial x^4} + \frac{\partial}{\partial t^2} $$ You can take this further. For example, solutions to the pde $$ cos(x)u_{ttt} - u_y $$ corresponds to the operator $$ D = \cos(x)\frac{\partial^3}{\partial t^3} - \frac{\partial}{\partial y} $$

54321user
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Assume you have two arbitrary solutions of the PDE, $u$ and $v$. Then, if you can show that for any scalar $\alpha$ that $u+\alpha v$ is also a solution, the PDE is linear.

In your vibrating beam equation, you have $u_{tt} + u_{xxxx} =0$ and $v_{tt} + v_{xxxx} =0$. Then, $(u+\alpha v)_{tt} + (u+\alpha v)_{xxxx} = u_{tt} + \alpha v_{tt} + u_{xxxx} + \alpha v_{xxxx} = (u_{tt} + u_{xxxx}) + \alpha (v_{tt} + v_{xxxx}) = 0 + \alpha (0) = 0$, so the PDE is indeed linear. The first step is simply linearity of (partial) derivatives, then some rearrangement and using the assumption that $u,v$ satisfy the PDE.

Batman
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