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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.

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If $p$ is a function of $x$ only, then its derivative with respect to $y$ is 0. You should clarified whether $p$ depends on both $x$ and $y$ or only on $x$. –  Alex B. Jun 15 '12 at 7:42
    
p is not a function of y. yes dp/dy is 0 –  sani Jun 15 '12 at 12:46

1 Answer 1

up vote 1 down vote accepted

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.

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