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I've just started learning about PDE's and am struggling with this question.

Consider the equation $$\cos^2(x) \frac{\partial{u}}{\partial x} + y \frac{\partial{u}}{\partial y} =0$$ Find the characteristic curves in the explicit form. Hence obtain the general solution of the PDE. For each of the following boundary conditions, find the particular solution or explain why it does not exist:

  • (a) $u(0, y) = y^2$.
  • (b) $u(x, 0) = x^2$.

So far I have done this but am unsure of how to complete the question:

$\dfrac{dy}{dx} = \dfrac{y}{\cos^2(x)}$

$\int_y\dfrac{dy}{y} = \int_ x\dfrac{dx}{\cos^2(x)}$

$\ln(y)= \tan(x) +C$

$y= C e^{\tan(x)} $

$C=y e^{-\tan(x)} $

$u=f(e^{-\tan(x)}) $

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    Just as an aside about the TeX, most special functions in math can be turned into their normal font by putting a \ before them $sin$ becomes $\sin$, $ln$ becomes $\ln$ etc. You can also make fractions by putting in \frac{}{} where the braces contain the numerator, followed by the denominator. Then $dy/dx$ becomes $\frac{dy}{dx}$. If you use \dfrac in an answer you get a bigger, more prominent fraction - the d stands for 'display mode.' $\dfrac{dy}{dx}$ – 2017-01-22
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    Note that the integral $\int \frac{{\rm d}x}{\cos^2(x)} = \tan(x) + C$ not $2\tan(x)\sec^2(x)$ so the characteristic curves should be on the form $y = y_0e^{\tan(x) - \tan(x_0)}$. – 2017-01-22
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    Thanks, I tried to clean it up a bit – 2017-01-22
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    Sorry that was a silly error. Am more struggling with the second part of substituting in. I think that $f(y)=u= y^2$? – 2017-01-22
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    There is a missing $y$ in the last equation. The solution is $u(x,y) = f(\color{red}{y}e^{-\tan(x)})$. To solve for $f$ simply substitute $x=0$ and use $u(x,y) = y^2$ to solve for $f$. – 2017-01-22
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    Ok so the first one cannot be solved as $u=f(ye^{-2tan(x)})$ so when y = 0, we have u = f(0) which is not enough information to obtain the whole function f(s) as y=0 is a characteristic curve? – 2017-01-22
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    That sounds good. – 2017-01-22

1 Answers 1

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I claim that (a) is solvable and (b) is a characteristic problem. Okay, we have the PDE

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

with initial conditions

$$\begin{cases} \text{(a) } ~ u(0, y) = y^2 \\ \text{(b) } ~ u(x, 0) = x^2 \end{cases}$$

We do this with the method of characteristics. There is a nice determinant criterion to check if our Cauchy problem is non-characteristic. For an initial curve $\gamma$ and a PDE a $u_x+bu_y=c$ it is given by

$$\begin{vmatrix}a|_\gamma & x_0' \\ b|_\gamma & y_0'\end{vmatrix}\neq0 \Longrightarrow \text{non-characteristic Cauchy problem} $$

(a) $~\begin{vmatrix} 1 & 0 \\ y_0 & 1\end{vmatrix}=1 \neq 0$

$$\begin{cases} x_t&=\cos(x)^2 \\ y_t&=y \\ u_t&=0 \end{cases} \Rightarrow \begin{cases} x&=\arctan(t) \\ y&=y_0 e^{t}\\ u&=u_0=y_0^2 \end{cases} \Rightarrow t=\tan(x), y=y_0 \exp(\tan(x))$$ $$\Longrightarrow u(x,y)=\frac{y^2}{\exp(2\tan(x))}$$ You can easily check that this is indeed the solution to this Cauchy problem.

(b) $~\begin{vmatrix} \cos^2(x_0) & 1 \\ 0 & 0\end{vmatrix}=0$

$$\begin{cases} x_t&=\cos(x)^2 \\ y_t&=y \\ u_t&=0 \end{cases} \Rightarrow \begin{cases} x&=\arctan(t+\tan(x_0)) \\ y&=0 \\ u&=x_0^2 \end{cases}$$

As you noticed correctly this system is underdetermined.