Solving a partial linear differential equation (wave equation) with higher order terms.

Question

I know that the solutions of linear differential equations can be added with any scalar coefficients and it is quite easy to prove too. What I am not able to understand is if we have an equation as follows:-

$$\frac{du}{dt} + a\frac{du}{dx} = \mu_2\frac{d^2u}{dx^2} + \mu_3\frac{d^3u}{dx^3} + \mu_4\frac{d^4u}{dx^4} + ...$$

Can we individually solve the equation for$$\frac{du}{dt} + a\frac{du}{dx} = \mu_2\frac{d^2u}{dx^2}$$ and $$\frac{du}{dt} + a\frac{du}{dx} = \mu_3\frac{d^3u}{dx^3}$$ and so on and simply add the solutions due to each of these.

I have encountered this situation while solving discretised wave equation but I feel that splitting the RHS might change the solution completely. Kindly tell me whether the approach is correct or not? If yes, then why and if no, then also why?

Answer 1

No, you can't. If $u$ solves the first equation and $v$ the second, then $$ \frac{\partial}{\partial t}(u+v)+a\,\frac{\partial}{\partial x}(u+v)=\mu_2\,\frac{\partial^2u}{\partial x^2}+\mu_3\,\frac{\partial^3v}{\partial x^3}, $$ and there is no reason to expect that the right hand side equals $$ \mu_2\,\frac{\partial^2(u+v)}{\partial x^2}+\mu_3\,\frac{\partial^3(u+v)}{\partial x^3}. $$

Answer 2

You can use the separation of variables: $u(x,t)=f(x)g(t)$

$$a\frac{f'}{f}-\mu_2\frac{f''}{f}-\mu_3\frac{f'''}{f}-\mu_4\frac{f''''}{f} = -\frac{g'}{g} = \lambda$$

$g(t)=e^{-\lambda t}$

The roots of the equation $\quad \mu_4r^4+\mu_3r^3+\mu_2r^2-ar+\lambda =0\quad$ which are functions of $\lambda$ are: $\quad r_1(\lambda)\:,\: r_2(\lambda)\:,\: r_3(\lambda)\:,\: r_4(\lambda)$

$$f(x)g(t)=e^{-\lambda t} \left( c_1(\lambda)x^{r_1(\lambda)x} + c_2(\lambda)x^{r_2(\lambda)x} + c_3(\lambda)x^{r_3(\lambda)x} + c_4(\lambda)x^{r_4(\lambda)x} \right)$$

where $\quad c_1(\lambda)\:,\:c_2(\lambda)\:,\:c_3(\lambda)\:,\:c_4(\lambda)\quad$ are any functions of $\lambda$, at least four times derivable.

Solution on discret form : $$u(x,t)=\sum_{\text{any }\lambda}e^{-\lambda t} \left( c_1(\lambda)x^{r_1(\lambda)x} + c_2(\lambda)x^{r_2(\lambda)x} + c_3(\lambda)x^{r_3(\lambda)x} + c_4(\lambda)x^{r_4(\lambda)x} \right)$$

Solution on integral form : $$u(x,t)=\int e^{-\lambda t} \left( c_1(\lambda)x^{r_1(\lambda)x} + c_2(\lambda)x^{r_2(\lambda)x} + c_3(\lambda)x^{r_3(\lambda)x} + c_4(\lambda)x^{r_4(\lambda)x} \right)d\lambda$$

For complex $\lambda$ and/or if they are complex roots among $\quad r_1(\lambda)\:,\: r_2(\lambda)\:,\: r_3(\lambda)\:,\: r_4(\lambda)$ there will be sinusoidal terms among the terms constituing the general solution.