Find the solution of PDE $\frac{\partial u}{\partial x}=4\frac{\partial u}{\partial y}$

Question

The solution of partial differential equation: $\frac{\partial u}{\partial x}=4\frac{\partial u}{\partial y}$ given $u(0, y)=8e^{-3y}$ is

a) $u(x, y)=8e^{-12x-3y}\quad $
b) $u(x, y)=8e^{3x-4y}\quad$

c) $u(x, y)=8e^{-3y-4x}\quad$
d) $u(x, y)=8e^{-12x+3y}\quad$

My answer: I checked the options

The option a) $\ u(x, y)=8e^{-12x-3y}\ $ satisfies both the conditions $$u(0, y)=8e^{0-3y}=8e^{-3y}$$

$$\frac{\partial u}{\partial x}=8e^{-12x-3y}(-12)=-96e^{-12x-3y}$$

$$\frac{\partial u}{\partial y}=8e^{-12x-3y}(-3)=-24e^{-12x-3y}$$

$$\implies\frac{\partial u}{\partial x}=4\frac{\partial u}{\partial y}$$

So I guessed option (a) that is correct answer, but I am looking for analytic solution of this P.D.E. please help somebody solve this PDE. thanks

Answer 1

HINT: Note that $\nabla u$ is a scalar multiple of the vector $(4,1)$, so level curves of $u$ are lines orthogonal to $(4,1)$. Thus $u(x,y) = f(4x+y)$ for some function $f$.

Answer 2

We try to find a solution of the type $u(x,y)=f(x)g(y)$. If we plug this into the original equation, we get $$\frac{\partial f(x)}{\partial x}g(y)=4f(x)\frac{\partial g(y)}{\partial y}$$ Rearranging the terms (the functions are not identically $0$) you get $$\frac{1}{f(x)}\frac{\partial f(x)}{\partial x}=4\frac{1}{g(y)}\frac{\partial g(y)}{\partial y}$$ The left hand side is just a function of $x$, the right hand side is just a function of $y$. They can be equal only if they are equal to a constant, say $C$. Solving those, you get $f(x)=Ae^{Cx}$ and $g(y)=Be^{Cy/4}$. Putting it all together, you get $$u(x,y)=Ke^{Cx+Cy/4}$$ Plugging in your initial condition, you get $K=8$ and $C/4=-3$, which yields $$u(x,y)=8e^{-12x-3y}$$

Note: If it's at an exam, note that you can immediately get rid of answers b and d, since they don't obey the condition at $x=0$, so you have fewer solutions to check.

Answer 3

To solve this differential equation you should consider the method of characteristics.

Step 1: To begin re write the differential equation in the following form

$$ \frac{\partial u}{\partial x} - 4\frac{\partial u}{\partial y} = 0.$$

Step 2: Look at the coefficients of both derivatives namely $\mathbf{v} = (1,-4)$.

Step 3: Notice then that you can write the PDE as

$$ (1,-4) \cdot \nabla u = 0, $$

which is equivalent to

$$ \mathbf{v} \cdot \nabla u =0, $$

where I have used $\nabla$ as the gradient operator. We say that $\mathbf{v} \cdot \nabla u $ is the directional derivative for $u$ in the direction of $\mathbf{v}$. It must always be zero. We can define a new vector $\mathbf{w}=(-4,-1)$ which is orthogonal to $\mathbf{v}$. You should check this. Lines parallel to $\mathbf{v}$ are given by $-4x - y = C$ for some constant $C$ which are called the characteristic lines. We say that $u(x,y)$ depends on $-4x - y $ such that

$$ u(x,y) = f(-4x-y) = f(C),$$

and this formula holds for all values of $x,y$.

Step 4: Consider your initial boundary condition given by $u(0,y) = 8e^{-3y}$ which gives

$$ u(0,y) = 8e^{-3y} = f(-y).$$

Finally our solution is then given by

$$ u(x,y) = 8 e^{3C} = 8e^{-12x-3y}.$$