I have been given two equations $ \frac{d^4 \varphi}{dy^4} - M^2 \frac{d^2\varphi}{dy^2} =0 $ and $ \frac{d^3 \varphi}{dy^3} - M^2 \left( \frac{d \varphi}{dy} +1 \right) = \frac{dp}{dx} $ and asked to find the $dp/dx$ with respect to certain boundary conditions. $\varphi$ is a function of $y$ and $p$ is a function of $x$.
What I did was to find the derivative of second equation w.r.t. $y$ $\frac{d^4 \varphi}{dy^4}-M^2 \frac{d^2 \varphi}{dy^2}=\frac{d^2 p}{dxdy}$ and I have two equations $\frac{d^2 p}{dxdy}=0$ and $\frac{d^4 \varphi}{dy^4}-M^2 \frac{d^2 \varphi}{dy^2}=0$ integrated both w.r.t $y$ and substituted the value of $y$ and subjected to certain boundary conditions I believe there should be another way to do it, because it doesnt feel right. Any suggestions? thanks.
When two ordinary differential equations are given
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0p is not a function of y. yes dp/dy is 0 – 2012-06-15
1 Answers
Note that if $p=p(x)$ and $\varphi=\varphi(y)$, then the second equation forces the left hand side and the right hand side to be equal to a constant $\lambda$, since the only way for two functions that depend on different variables to be identically equal is that they are both constant. Therefore $p=ax+b$.
Now, if you have two boundary conditions for $p$, you're done. If you have only one boundary condition for $p$ (you need at least one to solve the problem, otherwise you won't be able to determine $b$!) and some bondary conditions for $\varphi$, then do this. Let $u=\varphi''$, from the first equation you get
$u''-M^2u=0$ which gives
$u(x)=c_1e^{Mx}+c_2e^{-Mx}$ Integrating twice you get
$\varphi(x)=\frac{c_1}{M^2}e^{Mx}+\frac{c_2}{M^2}e^{-Mx}+c_3x+c_4$
Now use the boundary conditions on $\varphi$ to determine the $c_i$'s, put $\varphi$ in the second equation and find $a$.
This is how I would tackle the problem.